‘ GEOMETRICAL TREATMENT OF CURVES

W H IC H A R E

ISOGONAL CONJUGATE TO A STRAIGHT LINE WITH RESPEC T TO A TRIANGLE

IN T WO PA R TS

PA R T FIR S T

S A T T P n . D I . H J. C W , .

UNIV E RS IT Y O F P E NNS Y L V A NIA

II SH EWE LL AND SANBORN L EAC , ,

BO STO N NE W YO RK C HIC AGO

CONTENTS O F PART FIRST .

Tm: HYPE R B O LA

Tm: EL LIPS E

Second Pa rt cont a in

i T a n E LLIPS E ( cont nu ed ) .

Tm: PA R A B O LA .

H IGH E R PL A NE CUR V E S .

G OM TRICAL ATMEN F E E TRE T O CURVES.

Two a are said be a con u ate h 1 . str ight lines to isogon l j g wit respect to an angle when they form equal angles with the Le n bisector of the angle . t the poi t of intersection of three a ac a an a a a an str ight lines, e h dr wn from ngul r point of tri gle, a a a be 0 , then their isogon l conjug tes will lso intersect in one sa a a a o point, y which is c lled the isogon l conjug te p int to

a a . the point 0 , with respect to the three ngles of the tri ngle The isogona l conjugate points to all points on a straight line with respect to a triangle will form a curve which is said to be isogonal conjugate to the straight line with respect to the tri ' If l a a L Fi . r a . al s L 1 a e ngle the point on str ight line , g , ’ n a u a B and C a AB joi ed with the ng l r points , , of the tri ngle C, then we get two proj ective ! at pencils which are also perspec a tive to one nother, since the intersection of the corresponding The na i all rays are collinear . isogo l conjugate po nts to points ' ’ LL if B and C will ! at on , joined to , form two projective The a are pencils . intersections of two corresponding r ys in a a and are general not on the s me str ight line, therefore not and h u s a as w perspective ; t eir loc is conic section, is kno n l a r from the fundamenta l ws of projective geomet y .

a u a a a 2. The conic isogon l conj g te to str ight line with respect to a triangle will pass through the angular points of ’ a B and C Fi . 1 the triangle . It will p ss through , g , since they

in ou r !at . For are case the vertices of the pencils , let the point of intersection of the given line LL ' with the side B C be 2 I S OGONAL CUR VE S .

P BA a u a B P and CA na a , then is isogon l conj g te to , ; is isogo l i u a CP . n B A and CA s conj g te to , The point of intersectio of a a the o BP the isogon l conjug te point to p int of intersection of , and CP A a a and , ; or, is the point isogon l conjug te to P,” the n i e r Ta and A as A . A C co ic w ll, ther fore, p s th ough king , or and B as !at ha B a , the vertices of the pencils, we ve the isogon l u a i P and a u a i conj g te po nt to , , C the isogon l conj g te po nt to P and n a a h u B and C. , ; the co ic p sses lso t ro gh

The a ac a a 3. ch r ter of the conic isogon l conjug te to the line ' LL with respect to the triangle AB C depends u pon the posi tion of the line with respect to the circumcircle of the triangle

AB C. We know that the point isogonal conjugate to a point In on the circu mcircle of the triangle is a point at infinity . order to find the isogonal conj u gate point to P on the circum of AB a e s circle C, we h ve to find the inters ction of the line ' Fi i re AA a u a e PA PB P . a isogon l conj g t to , , C ( g wh ch , ' ' B B , and CC . ' ' AA a a AP a BAA u a If is isogon l conjug te to , then ngle eq ls ' e B P BB angle PA C w) . L t the isogonal conj u gate to be ' P = x an le . B then g PBA equals angle CBB Since angle C , ' an le PBA al a BB as therefore g equ s ngle C B , denoting the I B 'BA a B . t ha a whole ngle by B follows, therefore, t t ngle ' ' I w ma u a as and BB i a a to AA . n am a eq ls , s p r llel the s e y it y ' ' ' ' to AA BB and n a a BB . be show that CC is p r llel Thus, , , ’ CC are ll ra ac and a u a a pa llel to e h other, the isogon l conj g te

n P at i at n . poi t to , their intersect on, is therefore infi ity

' 1 e u m L L Fi . 4. If the line , g , do s not cut the circ circle of a i a a the tri ngle, then the con c isogon l conjug te to this line with e an h at i and resp ct to the tri gle as no point nfinity, the conic n s a will be a ellipse . If the line intersect the circle bout the a u a w a triangle, the conic isogon l conj g te to the line ill h ve two

at t and an a . If points infini y, will, therefore, be hyperbol the m an h line touches the circu circle of the tri gle, t en the conic will

a at fin and w e be a a a la . h ve one point in ity, ill , ther fore, p r bo 3 THE H YP E RB OLA .

1 . THE HYPE RBOLA .

The n na a a t the 5. co ic isogo l conjug te to line which cu s

a r a . to r a n circumcircle of t i ngle is, with respect this t i ngle, a ’ Kie ert s H a a a . hyperbol , which is c lled p yperbol m a u P and Fi . 3 If the str ight line cuts the circ circle in Q, g , ' will AP a a e AP m then the line , isogon l conjug t to , deter ine

at i. e. di the direction of the point infinity , the rection of one of ' as o e a. The i A a con the ympt t s of the hyperbol l ne Q , isogon l a A di e m jug te to Q , will determine the rection of the s cond asy p The as a a tote . ( ymptotes will be p r llel to these directions . ) P and A a a as A Q form the s me ngle their isogonal conj u gates . ' S an B AP a a P A C and an A ince gle equ ls ngle , gle Q C equals ' ' a BA BAP CAB A C a P A ngle Q , therefore 4 + X + 4Q equ ls X C ' ' E BA an AF a a P ’ OA A . S K X Q , or gle Q equ ls ngle Q ince the asymptotes of the hyperbola isogonal conjugate to P Q with ' ' to a AB C are a a AP and A respect the tri ngle p r llel to Q , there fore the angle between the asympto tes of the hyperbola will be equ al to the angle between AP and A Q . If the line P Q a r u i n p sses th o gh the centre of the circle, then w ll the a gle formed by A P and A Q be a right angle ; the asymptotes of

a a a to er en the hyperbol will be p r llel these lines, hence p p dicu lar ac and a i u i to e h other ; the hyperbol w ll be eq lateral .

We ha am o a a n u 6. s ll now ex ine the hyperb l isogon l co j gate ’ ’ to Brocard s Diameter (produced) . Brocard s diameter is the ’ line joining the centre of the circu mcircle and Grebe s point of ’ has the triangle . Grebe s point the property that its distances from the sides of the triangle are to each other as the respec and a tive sides ; the sum of the squ res of these is , for the tri ’ A B a am e a a . s a ngle, minimum roc rd s di et r p sses through the r rc a n u cent e of the ci umcircle, the conic isogon l co j gate to it is an equ ilateral hyperbola . We shall locate points with respect ’ to the triangle isogonal conjugate to points on Brocard s di ame

ter. All be a a these points will on the equil ter l hype rbola . ’ Let the points of intersection of Broc ard s diameter and the

e s a be P P and as w thre ides of the tri ngle , , , , R ; then, as ’ IS OGONA L CUR VE S .

w A B and C are a a sho n , , points isogon l conjug te to P“, d P P an . A B and C are , , C, respectively . , , , therefore, on the a equ il teral hyperbola . The median point E is isogonal conj u ’ ’ a G II B a g te to rebe s point , which is one extremity of roc rd s diameter ; E is therefore also a point on the equilateral hyper ’ bola isogonal conjugate to Brocard s diameter with respect to the triangle . F 4 a a AB H and A UC Fi . rom the simil r tri ngles , , g , it follows a AH a m n a A U. S a ca th t , is isogon l conjug te to i il rly it be shown th at the altitude from B is isogonal conjugate to the diameter through B ; and the altitu de from C is isogonal con

a am h C. The o H e jug te to the di eter t rough orth centre, , ther a a hi u fore, is isogon l conjug te to , the centre of the circ mcircle ’ a . And M B a diam of the tri ngle since is on roc rd s eter, there H u a fore is on the eq ilater l hyperbola .

If A B nd ’ . a C are th e a u B a 7 ,, 1 , , ng lar points of roc rd s tri h a AA B B and are . Let ngle, t en the lines I, , , CC, concurrent Fi the point of concurrency be I) ( g . This may be proved as follows

Let Fi . 5 P a 2 31 M ( g , l te ) 0 , , , M be the middle points of the B C A C AB a L n B . a sides , , of the tri ngle A C et the t gent drawn at C to the circumcircle of the triangle meet the symme a a a a A di n line (isogon l conjug te to the medi n line) through . ' at K — we w a the point , kno th t the tangent to the circum circle at B will meet the symmedian line throu gh A also in K ' — K 'JI a B C. Bu t A 31 then , is perpendicul r to since 1 , is ’ d a B A also perpen icul r to C ( l is an ang ular point of Brocard s A E a ll and K are a . It k w tri ngle) , therefore ,, a , colline r is no n ' ’ at C E H AB an a m and K K AB an th i , } is h r onic pencil , I , } is

a a . If 31 a as a a h rmonic r nge , be t ken vertex of the h rmonic c Ai AB al an a and as pen il , then d § is so h rmonic pencil ; ' A a a M B a a A is p r llel to G , therefore its h rmonic conjug te will

’ ' b A I AM s A K. In a wa ma isect J , or , bisect , the s me y it y be

’ a E JI e B i and C3I C K. L et proved th t , bis cts J , , bisects , the c n point of interse tion of AA , a d B C be A “ ; then since Arm s A K a A K A A lli a bi ects I in the tri ngle a h , therefore , will lso TH E H YP E RB OLA . 5

'

c I A A O u a I1 B . T AA and AK bise t L , ” or ,, eq ls , herefore “ , If are isotomic conjugate lines . the intersection of BB , and and AB be B B and B K A C be and of CC, then ,, , a re isotomic a e as are a CC and CK Now s conjug t s, lso ,c , ince AK B K and CK are at ” , ” , concurrent ( K) , therefore their isotomic c u a AA BB CC are a u onj g tes ,” “, ,c lso conc rrent, the point of concu rrency being D.

Fi . 1 M Ill 8. Let 6 P a o s ( g , l te ) ., M, , be the middle p int of h B C A C AB r a AB C and M the t ree sides , , of the t i ngle ; y, M M' the d B C A C A B » , be mid le points of the sides , ,, , ,, , , of ’ B r a i S the i c roca d s tri ngle ; then w ll , po nt of interse tion of M and M a a e to S M ” M , ”be isogon l conjug t the point ' ’ e cf AM BM and CM of int rsection ” , In order to prove this ll e ar to a i it wi be n cess y give few propert es of the triangle .

. A i Fi . r ula to B 9 If d . ( g which is pe pendic r C B C so ha A u a to A H be produced below , t t M is eq l , ., then r A C,A, B, is a pa allelogram.

We a e C BA C B C CB A h v Z , , Z , Z , ; ’ CBA A B C 8 B ar a but since I , Z , ( roc d s ngle) ,

C BA C C 1 A B C. therefore X , , Z , B 3 , Bu t A B C C BA 8 I , Z , , = C BA C C C BA . hence I , , Z ,B X , 43

° a A BM is m a to r a C E M The tri ngle , . si il r t i ngle , ,, (since they ’ are A u a B right triangles having X , B C eq l to C, BA rocard s angle) .

A B : BM Therefore , ,

Bu t whence

d u the and since the included angle between A, R an eq als ' 6 I S OGONAL C UR VE S .

n le a. and 0 an B O A a g included between , therefore tri gle , , is In am wa ma similar to the triangle AB C. the s e y it y be a a B A al a a AB C proved th t tri ngle , C , is so simil r to the tri ngle , n B A Bu t or triangle C,BA, is similar to tria gle , C , since

A B a A O as - an are , equ ls , , therefore the l t mentioned tri gles not

i a a and u B A a B . only s mil r, but equ l, conseq ently , , equ ls C, Bu t al C B u a C A B A u al C A and C A so , eq ls , , hence , 2 eq s , , , , a B C u a B A u A C A B a a al o equ ls , , eq ls , ; or the fig re , , , is p r lel

gram.

1 The a AB C and its B ca t a av 0 . tri ngle ro rd ri ngle h e the

A A B Fi . 6 a a am dia E . S C a o s e me n point, ince , , , ( g ) is p r llel a a a AA and B a at the gr m, the di gon ls , , C, bisect e ch other ' M A M a m a an as point C, , ¢ is edi n line in the tri gle ll A a in we as in the triangle AA,A, . second medi n line the a AA A AM A I a e tri ngle , , is , (since M, we h ve, ther ’ ' Z a A E a 2 E M and AE al E M . Bu t fore, th t , equ ls a , equ s , ARI al a a a AB C and , is so medi n line in the tri ngle , therefore ll E is the median point in as we as in AB C. By means as ma a of these l t two properties, we y proceed to prove th t ' ' M M AP and JI JI are n . a a , M ” L c co current Since E is the common median point of the triangles AB C and therefore,

' S M a a AA as a M E M and AE A or is p r llel to ,, the tri ngles G , , ' M are a and a M E a an A AM . simil r, therefore ngle , , equ ls gle , , ’ am an ma M a a B B and In the s e m ner we y show M , p r llel to ,, ' M a al CC and a AB C a to M , p r lel to , ; since tri ngle is simil r r a a M M M a a t i ngle therefore tri ngle , , c is simil r to tri n ' ' M M M T a A B C has a a gle , , ri ngle , , , evidently the s me rel tive position with respect to AB C as triangle M has w M M T AA B B CC in ith respect to JL , herefore, since ,, ,, ,, ' ’ ter ect a a JI H AP M s in one point, D, their p r llels, L M , ” M T will intersect in one point . his point, which is the centre of ' ' perspective of the triangles and AI JI QM may be denoted by S . TH E H YP E RB OLA . 7

a AB C and M I I a re e i the The tri ngles J J , p rspect ve, The centre of perspective being E . line joining the two per S ective D and S e r an AB p points, , with resp ct to the t i gles C

and M U I u a u e E . J M, m st p ss thro gh the centre of p rspective,

S E and D are a and DE : E S 2 : 1 . Therefore , , colline r,

AM CM are r 1 2. The lines 2. concu rent, the point of ' u S a at S 1 conc rrency being , which is isogon l conjug e to ( 0) B nd re h Fi . O a C a ( g , ,, , concyclic, t erefore the lines ' 0 0 and B, C, are antipa rallel with respect to A . Triangles ’ i S AS a d a A and AB are a . 0 0 , C, , therefore, s mil r ince is me i n ' the a A O and AM a a line in tri ngle O , is medi n line in the a AB AS and AM are i a tri ngle , C,, therefore the lines sogon l d BM are a e . S a B S an a conjugat lines imil rly, isogon l conjug te, as are a S and CM Bu t AS B S and CS are n lso C , , co cur a a AM BM and CM are rent, hence their isogon l conjug tes ' T ' n u and i u S . S co c rrent, the po nt of conc rrency is herefore ’ a a i to S Bro a s is the isogon l conjug te po nt , which is on c rd ' i diam and S a u ate a a . eter, is point on the eq l r l hyperbol

S dl 1 3. is the mid e point of In order to prove this ' m r a a nd i po t nt property , we will first find the dist nces of O a 0 r i and and D m th from the ve t ces, of O, fro the sides of e triangle AB C.

Fi K nd KK KK . I a 1 4. a s If , ( g Q , , represent the dist nce K a of from the three sides of the tri ngle, then

2 a, A 2 b A 2 c - A

r A a a a . We a C AI whe e represents the re of the tri ngle h ve , . e u al to E R C a a AB and C M and KB q } , since K , is p r llel to , , , ; are each perpendicular to AB We eas ily obtain

’ ’ a -f a o e t w . 8 IS OGONAL C UR VE S .

2 A al Whence, so, z ’ ’ ’ ” 01A V u b a c No

L n et C (Fig . 7) be the foot of the perpe dicular from B to ’ A O n the m a a A C and AB C , the in si il r right tri ngles , M, C Alhf BA G 8 (X , ,2 K ) 2 A

» s BA 0 1A ea a a

2 whence 0 A

A B OA u a 1 80 and B G a ngle eq ls B, O B, where Bis ngle H AB C a AB C. a B OG and BA of the tri ngle The tri ngles , are m la and si i r,

h = altitu d e A1 , ( , 1 ) , £8 If 51

c“ “8 B O = B G 2 A vm e m

A simila r consideration gives

' A n B and A O C are a a gai , 0 0 simil r tri ngles,

’ therefore A O b B 0 a

A O’ B 0 9

Let C C CC 0 0 d ta 1 5. , , , , , be the is nces ' of 0 and 0 from the sides of the triangle AB C. B O m a a A C M c Triangle C, is si il r to tri ngle , , , hen e

9 2 ac - A f r there o e ’ ’ 2 2 ’ ’ a b “ 0 b c THE H YP E RB OLA . 9

’ 2 a o A

’ 2 b c A

’ ’ i z ’ ’ a b c. e b c

’ ' a C O O m a to a AC M e tri ngle , is si il r tri ngle , , , whenc

' C ' C I 2 A C . J , 0 0 ' 0 A ’ . vm .

2 b’ A h a w ence ’ ’ ’ ’ a o b c

’ ' ' 2 bc A S a C C “ imil rly , , ’ ’ ’ ” We a c b c

’ 2 a o A

’ ’ ’ ’ ’ 2 a o a c b c

Let X the dl h ta ce 1 6. be mid e point of t en will its dis n XX B C . from be ’ ’ 2 A c 2 A a b XX a ’ ’ ’ ’ ’ ’ 2[a b a o b c ]

g ” ’ ’ ’ ’ a o a o b c

Let A A and A A be a A A C , ,, , ,o the dist nces of , from the sides and AB a A BC A A C and COO are of the tri ngle ; then , ,, . T r a a . simil r tri ngles herefo e,

A 1 44“ A1 0 .

0 0 , C C

a A M O and are a Tri ngles , , simil r, 0 0 whence . A M 1 .

Mu a and ta a n ltiplying ( ) (B) , we ob in, fter reductio ,

’ 0 0 C . , 4A OT) A M it? , , ,

° A m 556 , I S OGONAL CUR VE S .

n ma av In a similar ma ner we y h e,

A A I IO

whence

Let DD DD be i a D r m res ec , , , , DD” the d st nces of f o the p tive sides of the triangle ; then from the similar triangles n A M and DAD and 1 1 a DADb a d. I “ , ” l we h ve

and by combining these equ alities we get

’ A 0 DD, ,A ,, A A bf DD, , ,,

S m a « i il rly, DD, 2 by DD.

3 at DD,

DD A Now a . DD o. c. DD 2 . , , , ,

3 a u and DD If we find from ( ) v l es for DD, c in B B ma terms of , , we y put

2 w A ? 3 ’ ’ a.[a b a o W ]

’ ’ L w 2 a o A ike ise, ’ ’ b c ]

’ ’ 2 a b A

’ ’ 2 2 ’ ’ c[a b (1 0 b c I

distances of E from the sides of the triangles are

2 A

3 0

1 2 IS OGONAL CUR VE S .

ac n al e h other, the pe cil formed by these isogon conjugates will

- d a a a an a a] . h ve the s me cross r tio, will therefore be h rmonic Ta AM AE and AH a n a king, then, for , , their isogon l co jug tes AS AK AM and a at , , respectively, letting the isogon l conjug e ' ' ' AD AD A A S KD an a to be , then i , , } is h rmonic pencil, cut ’ B a am a a IS KX . ting roc rd s di eter in the h rmonic r nge H , } In am wa ma n a B MS KX and the s e y it y be show th t { , } I C AIS KX are a i . t at a , } h rmonic penc ls follows once th t ' ’ X D and B a am te . coincides with , is on roc rd s di e r ' Fu MK a and 0 0 at rthermore, is perpendicul r to bisects the ' S and D a n a S point , is the h rmonic co jug te to with respect to AI and a and a K; hence, from the rel tions of poles pol rs, it is ' ' ’ evident that D is the pole of 0 0 with respect to Brocard s

circle .

' DH is parallel to MD .

F r E z E o , D S

’ and E ul by er s proposition,

H E E lli: 2 1 .

Therefore,

and a are ca , since the ngles included by these lines verti l, it u follows that H D is parallel to S M. B t SM and D 2} ! are on ' the same straight line ; therefore H D is parallel to D M

’ 1 Tarr s N Fi . 5 P a a a e 1 8. y point, ( g , l te is isogon l conjug t ’ a to the point at infinity on Broc rd s diameter. N is the point of intersection of the perpendicu lars let fall from the angular the a AB C B C A O and A B points of tri ngle to the sides , ,, , ,, , , of ’ I N Broc ard s triangle . t is easy to see that is on the circum er ference of the circumcircle of AB C. Suppose the p pendicu

a A B C and B A O e at N. l rs from to , ,, from to , ,, to inters ct Since the sides of the angle A NB are perpendicular to the sides of the angle therefore is ANB equal to 1 80

B a l A B . T Bu t the angle A , C, , equals ng e C herefore is ANB a 1 0 A CB N u m r a equ l to 8 , or is on the circ circle of t i ngle

AB C. 1 3 TH E H YP E RB OLA .

1 N has am r o A BC as M 9 . the s e position with espect t , has wi a a NAB al th respect to th t is, ngle equ s a B A M S A e u a to ngle , , ince N is perp ndic l r MC,, therefore N angle AB equals angle B , C, M ; but angle B , C, M equ als angle

B A M the are c . T e o n , , , since points con yclic her fore the p i ts are similarly situ ated .

S AX d aw na u a AN ha 20 . uppose r n isogo l conj g te to ; t t is , angle NAB angle X A C. We have seen (1 9) that NAB is al MA B a B KM a NAB u a equ to , ,, which is equ l to , , or ngle eq ls w A M B E M X A . No as B K a al to K , C , is p r lel C, therefore a a X A a a must be p r llel to , or the line which is isogon l conjug te ’ to AN runs parallel to Broca rd s diameter . a ma a B Y a con u In like m nner we y show th t , the isogon l j a BN and Z al u a CN are a g te to , C , the isogon conj g te to , lso ’ a con u parallel to Brocard s diameter . Since the lines isogon l j ’ a AN B N N are all aral B a am g te to , , C , p lel to roc rd s di eter, it easily follows that N is isogonal conjugate to a point at infinity ’ in the direction of Brocard s diameter Thus N will also ’ be on the hyperbola isogonal conjugate to Brocard s diameter

with respect to the triangle .

' Let Z nal c u a e Z c be the point isogo onj g t to , the entre ’ ' of Brocard s circle (Fig . It can be proved that Z is on ’ i H S . A B C E H S Z N the l ne joining with The points , , , , , , , are al n a P P P AI S Z and isogon co jug te to the points , , , , , , K, , , , n B ’ Th the point at infinity o rocard s diameter respectively . e ’ A B C E H S Z and N are points , , , , , , , therefore on the hyper ’ bola isogonal conjugate to Brocard s diameter with respect to B a . to a a the tri ngle efore proceeding ex mine this hyperbol , a a e a we shall give few properties of the equil t r l hyperbola .

1 An an at a a a a 22. ( ) y chord of equil er l hyperbol ppe rs the extremities of a diameter of the hyperbola either u nder the same (when the diameter without being produced meets the a a am chord) , or under supplement ry ngles (when the di eter

must be prod uced to meet the chord) . IS OGONA L CUR VE S .

Fi 4 1 am nd f Let AA . 1 P a a a B C a o , ( g , l te ) be di eter chord an equilateral hyperbola whose asymptotes are the lines a and a a BA he a , ; then it is to be proved th t the ngle C equals t

A Let A B a a and a at the angle B , C. meet the symptotes , and D and A C at E and E and A B points D ,, let meet them ,, , F D a A C a a A B an AMC at F and , . r w p r llel to , , then tri gle is nd equ al to triangle A ,MF (since AM equals A,M a the an gles are a MP u a MC and A C a A F equ l) , therefore is eq l to , equ ls ,

u a B F A and B are n a. and eq ls , (since , points o the hyperbol , the intercepts of any chord between the cu rve and its asymp But A C a a totes are equ al) . runs p r llel to therefore

ABF a a al l a and AB a CF . Bu t ME , C is p r le ogr m, equ ls , a MC and ME ula F C FF ual equ ls , , is perpendic r to , hence , eq s

F C and a FF M a an M G. , , ngle , equ ls gle E

B a e ula a a If K be dr wn perp ndic r to the symptote , then

and since

n B a F D a d K is perpendicul r to , ,, therefore,

BF, B D, A D A C .

a wa AH a a a A O can be In the s me y, if be dr wn p r llel to , , it w Bu t AD as ua A . shown that AE equ als AH . eq l to C T DAE HA C herefore, X X ; DA E BA C but X X , d H A BA an X C X , C ; BA C BA C whence, X X , , and the proposition is proved .

The a and ad a u p e 23. bisectors of the ngle its j cent s pl mentary angle formed by the lines joining any point of an equilateral hyperbola with the extremities of a diameter of a re r the hyperbola pa allel to the as ymptotes . Fi 1 h and A nd A Let B . 4 a a a ( g ) be point of the yperbol , , the extremities of a diameter ; it is to be proved tha t the bisecf E 1 TH H YP E RB OLA . 5

a ABA a a to a and o tor of the ngle , is p r llel , , the bisect r of the

ad ac a r . j ent supplement y angle A ,BD, is parallel to a The a FF and FBA are a al and sides of the ngles , C p r lel, therefore the angles are equa l . The bisec tor of the angle

FF C a m e a h a FB , is the sy ptot ,, t erefore the bisector of ngle A

a al a and o will be p r lel to ,, the bisect r of its supplementary a dj ac ent angle F, BD, will be parallel to the perpendicular to

a l a a to the a this line ; th t is, it wi l be p r llel asymptote .

’ a B a t a m 24. The ngles which the sides of roc rd s ri ngle for with the respective homologous sides of the given triangle ’ are equ al to the angles which Brocard s diameter forms with the rad ii of the circumcircle drawn to the vertices of the gi ven triangle. We are to prove that B C B C AMR ( , , ,) X , A C A C B MK ( , , ,) X ,

’ Let the point of intersection of AM and Brocard s circle be " Fi denoted by A ( g . then it is to be proved that B "M B C C A K. ( , , ,) X

A AMM u a and a A M]II u a c ngle , eq ls X B, ngle , , eq ls X 7 (sin e A M u a AC t , is perpendic l r to ) ; herefore, A M l{ A A = A " A = AM l M M . X M X , , X , X , X B X y B A "MA = u t X , X

and . therefore, X X B X 7 " A a A B C A B A g in, X X , , , X , , = l B (X B ¥ v 4AxClB u X X A "A C X X , ,, A ' 'A 0 4 Z 1 1 " A . whence the line A , is parallel to B C The angle formed " " A A and A K A A H h u al by , , (or X , ) is t erefore eq to the 1 6 I S OGONA L CUR VE S . a n e m B C and B u t A and M a re on gl for ed by , C, . B since , B oc a ’ r rd s circle, " " A A K : A ME X , X , A "MK B C X X ( , A similar cou rse of reasoning will establish the tru th a s re a r s a g d the other ngles .

’ 2 a t r s o n t 5. The line joining the orthocentre II with T y p i N is a diamete r of the equ ilateral hyperbola isogonal c on j u g a t e to B ’ roc a rd s d iameter with respect to the triangle A B C . It h as been proved that any chord of an equila teral h y p e r bola a ppea rs the extremities of a diameter of the hy pe rb o la u n der equal or su pplementary angles Suppo se NH is ’ a dia meter of the hyperbola isogonal conj u gate to B roc a r d s h e n dia meter with respect to the triangle AB C (Fig . t H s A B A C B C DA etc . a a as e N a s mu t the chords , , , , , ppe r w ll

u n der the same or under supplementary angles . prove in the two following sta tements 1 AH C = 1 80 ANC ( ) X X , 2 A H D 1 80 AN ( ) X X D .

From Fig . 1 5 we see

AH : 1 — X C 80 § X HA C + Z H CA } ;

HA C : 90 L X e. C a e A B C X 7 ( X of tri ngl ) , — X II CA 90 X a ; AH C = 1 80 90 — — X 2 KY+ 9O X a }

a = 1 80 — X 7 + X X B.

A H C = 1 80 X X A NO. N w H D ' o , , , and N are on a a the s me str ight lin e . F or , d ra w ' h u D a a a H A and t ro gh line p r llel to , let this pa ra llel int e rs e c t " " ' A N at H A H D and M the point , then , , , a re on c c lic c y . ' " T ma v w ’ his y be pro ed as follo s : D H is pa rallel to A H , a n d i AH a B C ' " s nce is perpendicul r to , therefore D E i s a ls o p e r i l " end cu ar B C. S NH p to ince is perpendicula r to E C , , by THE H YP E RB OLA . 1 7

" ' n re a NH D an an e constructio , it follows, therefo , th t is gl a a B C and B C i w e equ l to the ngle formed by , ,, wh ch as qu al to M the angle A K.

" ' " ' X AH D 1 80 X NH D 1 80 X AMK ;

' " ' d K on MD an AH D ua 1 0 an since is , therefore gle eq ls 8 ’ ' AMD A H D and M are co and X or, the points , , ncyclic, ' ' ' X D E M X MAD . ’ M and H and and D are a u Since , D isogon l conj gate ' i a MA D a a H AD . points, therefore w ll the ngle equ l ngle ' " ' Bu t MAD was a MH D HAD u a X equ l to X , hence X eq ls ' ' " ' S D H a al HA MD a a to ME D . X ince is p r lel to , is p r llel " ' DH AH D ua H D M and o N 31 and N , X eq ls X the p ints , , D, , " " ’ H A are a . a l ADH and E MB , , colline r Therefore the tri ng es are a a d i a a m simil r n s mil rly situ ted, the centre of si ilitude being ' The N. line joining the two corresponding vertices D and H of " ' the triangles ADH and H MD must pass through the centre ' m N. T H D and N are of si ilitude, herefore the points , , collinear .

h AHD 1 80 An AH "D" 2 a AND . 6. To prove t t X X gle ua 1 80 AMK is eq l to X , " H " ' therefore X H AM X D M 1 80. In a ANJII NM a A lli e n u a a ii the tri ngle , equ ls , b i g eq l r d , hence X H "AM X NAM X ANM and thus X ANM X AHD 1 80. S N M and D are a ince , , colline r (25) therefore, ANM A ND X X , AND AH D 1 80 X X , A X HD 1 80 X AND .

D a e a iam is therefore point on the hyp rbol , the d eter of which N h W is H . T e centre of the hyperbola is the middle point N ’ of H . The centre of the hyperbola is a point on Feuerbach s r ci cle of the triangle AB C. ’ We a F F a is know th t the centre , , of euerb ch s circle the middle point of the line joining the centre of the circumcircle 1 8 IS O NA R V GO L CU E S .

a and n 0 1 of the tri ngle, the orthocentre . F is the middle poi t H M , and W is m l i N is u a the idd e po nt of H , therefore F W eq l

- one al M - N . to h f of , or one half the radiu s of the circu mcircle ’ W re e F u bac is, the for , on e er h s circle .

2 7. Let as m h the y ptotes of the hyperbola be a and a ,, t ey are s Wm and Wm F ’ m and then chord , , of euerbach s circle, m ’ , being the extremities of a diameter of Feu erbac h s circle a a DH . S e D and a p r llel to inc H are points on the hyperbol , therefore will the line joining the middle point U of H D and th e th e D centre of hyperbola bisect all chords parallel to H , and will a u ’ B u t p ss thro gh the centre of Feuerbach s circle . since the intercepts of a chord between the hyperbola an d its as m are ua If r u h y ptotes eq l, therefore C U, equals UU2. th o g ’ the centre of Feuerbac h s circle is drawn a parallel to H D u a m u al ntil it intersects the sy ptotes in m, and m” then Fm, eq s as a m a M It is e y to see th t , and m, are the extremities of ’ am F a th e di eter of euerb ch s circle . Since W is the centre of a and at am F ’ m m is hyperbol the s e time on euerbach s circle, , ,, a an u F ’ and sec t line thro gh the centre of euerbac h s circle, the an m Wm al a a m gle , , equ s right ngle, therefore m, and , must be ’ a am and m d F ac the extremities of di eter, , an m, are on euerb h s

circle . r un an e In o der to derst d the relations of the asymptot s, it a to am e will be necess ry ex ine some properties of the tria ng l , e a s S ’ esp ci lly tho e of imson s line .

’ SIMSON S LINE .

a n . 28. If from poi t P (Fig on the circumcircle of the e A B P P and dra n u l a r to tri a ngl C, ,, PP” PP, be w perpendic r a o P P P the respective sides of the t i ngle, then the p ints , , ,, , ’ re a a w a S m of a on the s me str ight line, hich is c lled i son s line

P with respect to the tri angle AB C . This may be proved a s P d P P follows : Join P , , a n , , ; then these lines form one stra igh t For P an d PA and a PF nd . C a P F line , join , since ngles , C , C th e are B u e a PP CP . t a r right ngles, points , , concyclic sin c e

’ NA E I S OGO L CUR V S .

Since 2; P UC xB C U therefore X BA T X BA U

A T la A and ce al er or the line is perpendicu r to U, hen so p pen ’ dicu lar to Simson s line

The Simson lines belonging to the extremities of a a are ula di meter of the circumcircle perpendic r to each other . I i f and Fi . 9 are a diam e P Q, g , the extremit es of et r, then ° a P a 90 u al PP B and ar ngle UQ equ ls eq s , , therefore UQ is p l D T d P T a B C. a a B C a al lel to r w Q perpen icul r to , then is p r B and are E T a are P C and u lel to C, is equ l to , , conseq ently,

X TAB X PA C. Bu t by a previous proposition the Simson line for Q is parallel to A T ; and A T is perpendicu lar ’ S e S l to imson s line b longing to P, or the two imson ines be longing to the extremities of the diameter P Q are perpendicu

lar to eac h other .

’ Let and i 32. P Q be the po nts of intersec tion of Brocard s diameter produced to meet the circumcircle of the triangle A BC ; then they will be the extremities of a diameter of the u m circ circle itself. AP and A are a to ac and e Q perpendicul r e h other, therefor their isogonal conjugates will also be perpendicular to eac h As a other . we h ve seen the isogonal conjugates to AP and A at nit Q determine the directions of the points infi y, or the directions of the as ymptotes of the hyperbola isogonal ’ conjugate to Brocard s diameter with respect to the triangle . i lar S S e Fi . u ince the imson l n belonging to P, g 9, is perpendic A T al a AP and S to , the isogon conjug te to , the imson line to n a AP a a A Q is perpe dicul r to , the isogon l conjug te to Q and also the Simson lines belonging to the extremities of a diameter are perpendicular to each other therefore the asymptotes of the hyperbola will be parallel to the Simson a lines for P and Q. Subsequently we shall prove th t the ’ asymptotes of the hyperbola isogonal conjugate to Brocard s diameter a re not only parallel to the Simson lines belonging TH E H YP E R B OLA . 21

’ to the extremities of the prolonged Brocard s diameter on the

a n i m. circu mcircle, but th t they coi cide w th the

The r and e a a. a re 33. o thocentre the m di n point of tri ngle a respectively the external and internal centres of similitu de of ’ u and F u a the circ mcircle of e erb ch s circle of the triangle . In order to find the centres of similitu de of the circu mcircle ’ d F ac r a a mu an of euerb h s ci cle of tri ngle, we st draw parallel

radii in both circles . The line joining the extremities of the radii drawn parallel in the same or in the opposite direction will intersect the line joining the centres of the circles in the a extern l or the internal centre of similitude . ' Let YY Fi . 1 0 a a MX mu X , g , be p r llel to , then st Y pass " ’ n The a H a d I E . F through , X through r diu s of eu erbach s

circle is equal to one- half of the radius of the circ umcircle B a F Y u a MX. ru F Y th t is, eq ls 5 y const ction, is T a all MX and FM al MH . p r el to , equ s herefore,

d H ar i X Y an e a a . or , , on the s me str ight l ne

I a ME u al 2 f E E F. 34. is the medi n point, then eq s This

is easily established. Since

and

therefore and ME z E F = § Ms MH =

ME 2 E F.

' ME X and E Y F Hence in the triangles ,

' MX : F Y ME E F,

' X E and Y are a . or , , colline r

’ The u m and F u ac a a 35. circ circle e erb h s circle of tri ngle n a n in h 1 divide a y line dr w from the orthocentre t e ratio 2 : .

For,

bu t IH H F : 2 1 since J , = therefore H X II Y 2 1 . 22 IS OGONAL CUR VE S .

The same will be the cas e with any line drawn from H to the circumcircle of AB C.

The S um AB C 36. imson line for P, on the circ circle of with e t a AB C ni w resp ct to ri ngle , bisects the line joi ng P ith the orthocentre of AB C.

Let H P Fi . 1 1 P a 3 Sim at D are ( g , l te ) cut the son line ; we ’ to prove that D is a point on Feuerbac h s circle . dl P o . S o D is i H P (Indirect ro f ) upp se the mid e po nt of , or, ’ a a i D a o F ac r wh t is the s me th ng, is p int on euerb h s ci cle ; then we have only to prove that the line joining D with the foot of ’

P . a B C i. e. D S n the perpendicul r from P to , ,, is imson s li e S A C and are ula B C and H D ince PP, perpendic r to , DP, Let the triangle GDP, is isosceles . E be the middle point of

AH a DE . T n DE al AP . The o E C ; dr w he equ s Q p ints , , ’ and D are F ac A and C are the on euerb h s circle ; , P, on i P . S E D u a A and a us circumcircle nce eq ls 5 , the r di of ’ F a al - a a u um rc euerb ch s circle equ s one h lf the r di s of the circ ci le, ’ the angle E C D inscribed in Feuerbach s circle will be equ al to a A P um AB the ngle O inscribed in the circ circle of C.

We a E GD A OP . h ve then, X X E D C DP P . X X , T DP P A OP . herefore, X , X

Let the intersection of DP, and A C be Then DP P A OP C X , X X P,, P, o P P and C are or the p ints ,, ,, P, concyclic ; therefore, PP C PF C X , X , h T e point P, is therefore the foot of the perpendicular from ’ P to A C ; i. e. P, P, is Simson s line .

’ N ra —The oint is ll d o . p D ca e the centre of Simson s line.

The i S m ’ i 37. point of intersect on of i son s lines belong ng to the extremities of a diameter of the circumcircle of a triangle ’ is on the circu mference of Feuerbach s circle for that triangle . i 2 P H Let W F . 1 a 3 e ( g , l te ) be the point of int rsection of Q ’ and S m on s ne and D the n i i s li for Q, let be point of i tersect on 23 THE H YP E RB OLA . of HP and the Simson line for P ; then W and D are middle o s H and PH and W and D are p int of Q respectively, on ’ to and a one- a Feuerbach s circle . WD is parallel equ l to h lf P and il all a H P and of Q , w l bisect lines dr wn from to Q , ’ therefore WD will bisect MH at the centre of Feuerbach s ’ H WD m a iam e F ac circle, F. ence beco es d et r of euerb h s i and an Z D w S u a c rcle, since gle W bet een the imson lines is eq l to a right angle therefore Z is on the circu mference of ’ r a WD as a dia a F ac . the ci cle bout meter ; th t is, euerb h s circle

— We Norm This point Z is called the vertex of either Simson line. ’ see then that Feu erbach s circle intersects a Simson line in its centre and vertex

We see then that the asymptotes of the equilateral hyperbola ’ isogonal conjugate to Broca rd s diameter mu st coincide with ’ ’ Simson s lines belonging to the points in which Brocard s diam eter intersects the circumcircle of triangle AB C

’ as. In connection with the properties of Simson s line a r a a a i r a a n l e dy given, we sh ll ex mine some mpo t nt rel tio s, which shall hereafter be made use of in the chapter on H igher a r Pl ne Cu ves .

If 1 wi H and H Fi . 3 H 39 . P, g , be joined th the points of intersection of the altit udes upon the sides produced h and to meet t e circumcircle, if the points of intersection of of PH PH and PH with the three sides be respectively U W o W o r , V, ; then the p ints U, V, , t gether with the o tho e H are n a an d al to S ms centr , colli e r, UV W is par lel the i on line to P.

' " ’ m WH H 2 WE H F a H H K . ; , (since euerb ch s circle bisects

B ' " " ' u t WH H PH C PB C a . X , X X (being in s me segment) A a UH H UH 'H PH 'A PBA g in, X , X , X X ; hence, by a di n d tio , = WHH + UH H PB C + PBA AB O. X , X , X X X 24 IS N R OGO AL CU VE S .

A d H HH a d ing X , , to both sides, we h ve, WE H HH H HH AB H HH , U , , , C , X X X X X , t H BH 2 H H H a . . + ; . . X W H and are n a and a n Therefore, , , U colli e r ; in like m n er m a n a a W H and a . y be shown th t , , V re colli e r

. PP W at X 40 If , is produced to meet the line UV , then r PX is pa allel to E H since both are perpendicular to A C. 1 ' a F The portion H H , equ ls the portion E , H since euer ’ ac an b h s circle bisects y line from the orthocentre ; therefore,

" A P X V = P P V = H H V = P CP and the lso, X , X , X , X , ; ’ ' u H P CB an a ria a its fig re being inscribed qu d l ter l, the po n P P C and are and ,, ,, , P concyclic, P X = H W PP P V P P . X Q , or X 2 C , X E V

’ r n W is a a S l Wh UV m . erefo e, the li e p r llel to i son s ine for P

All straight lines drawn from P to WUV are bisected

S n . H as a al a PH by the imso line for P ence, speci c se, is T a the a e bisected. his is simpler proof of s me theor m given some pages back e al a S e W sh l c ll D, the point in which the imson line is int r e d and H s cte by the line joining P the orthocentre, , the centre of Simson’ s line

’ The a S e 42. ngle between two imson s lines b longing to two ’ and AB C al points P P on the circumcircle of , is equ to the ' inscribed angle upon the are between P and P on the circle ’ u AB and a a F a r a bo t C, lso to the ngle in euerb ch s ci cle upon the are between the centres of the Simson lines . ' ’ ' P to H Fi 1 3 B t Let the line joining ( g ) intersect C a U , ' ' then H U is parallel to the Simson line belonging to P . Since ' ' ' ’ ’ H = H a H E U and n U U, lso U , UU is commo , hence ' and X UH U = X Bu t ' since 11 U and H U are pa ra llel to the Simson lines for P and 2 TH E H YP E RB OL A . 5

' a e S al to a P , the ngle between th se imson lines is equ the ngle T ' ’ ' ' between PH and P H or X PH P We have already shown that H is the centre of similitude e AB and F a of the circumcircl of C, the euerb ch circle ’ ' ’ and D and H and H a re n therefore P , P D, , , correspondi g ' n F n i . a H D IF H D a d points the circles It follows th t , , , , , ' ' ' ’ ’ H P are and DH D PH P corresponding lines, X , X , or the ’ inscribed angle in Feuerbach s circle upon the are between the S al a centres of the imson lines, is equ to the ngle between the Simson lines . We ha ve seen tha t the Simson lines corresponding to the a a are a ac and extremities of di meter perpendicul r to e h other, ’ they intersect on the circu mference of Feuerbac h s circle T and (31 , his point of intersection of both lines the cir

a al vertex S 37 NC . cle we h ve c led the of either imson line ( , ) ’ F a a a nd euerb ch s circle, therefore, p sses through both vertex c the S entre of imson line (37,

A B C and a u a a are Sim 43. side, , its ltit de in tri ngle the son lines corresponding to the extremities of the diameter ' i AA (F g . For the i la A AB and , since feet of the perpend cu rs from to A A and t ula A C coincide with , the foo of the perpendic r from

B C H S A AH . A ai to is , , therefore the imson line for is , g n, ' the u a A AB A C the feet of perpendic l rs from to the sides , , B a are B C and c B C C of the tri ngle , , respe tively ; hence is ' the Simson line for A .

If h an a u a A a an AB Fi . 44. t rough ng l r point of tri gle C, g 1 6 a be a to , the di meter of the circumcircle dr wn meet the cir ' ' at A n A H cle , then the line joini g with the orthocentre will bisect the side B C .

F r T a n . o Si his follows e sily from know properties , the mson ' ' i A B C e B C A H and a s l ne to is ther fore must bisect , , ’ a F ac i we h ve shown, this bisection is on euerb h s c rcle ’ Bu t F ac u B C at a a at euerb h s circle c ts two pl ces only , th t is ' H and a A H as M , M ; hence it is obvious th t p ses through , ,

a nd the proposition is established . 26 I S OGONAL CUR VE S .

’ Si d a 45. mson s lines correspon ing to the extremities of diameter are said to be conjugate .

The E and F Fi . 1 6 a S 46. points of intersection, , g , of imson a a a Si are u a dita line, with p ir of conjug te mson lines, eq lly s nt ' ’ ' ' T T S and E F 2 T S . from the centre on , ' ' L et T S be the Simson line of a point on the circumcircle of AB and a Sim DS and TS at E C, cutting the conjug te son lines ' ' and F T E D T S D i a , then X X , s nce the ngle between two ’ Simson lines equals the inscribed angle in Feuerbach s circle ' T a on the are between the centres . herefore the tri ngle E T S ' ' " ' and T E S T S E E 1 u al T S . is isosceles, X X , whence eq s ' ' ' ' S E S E T S F T FS and T S T F ince X X X , ; or ' ' = ' E T T E T S .

The are the Sh son 47. between vertices of two lines (not conjugate) is twice as large as the a rc between their centres . ' For E T S an a and is isosceles tri ngle, F T’S ' T’ 2 T'S D X X S S , or the arcs meas u ring these angles are in the ratio of 1 2.

' ’ ' T E T E = T S a am F u 48. If is less th n the di eter of e er ’ ac a a b h s circle, or less th n the r dius of the circumcircle of ' ’ a AB a T as and T E tri ngle C, then circle from centre with as ’ a u F ac a at S and r di s will cut euerb h s circle second time, T 'S " T'E T 'E S since , will be the vertex of another pair ’ at S as of conjug e imson s lines, which will p s through E and F. It is evident that there are always two pa irs of conjugate Sim as E and son lines p sing through F .

If T'S T 'S " d 49 . diminishes, the points S an S will ' a ac T ppro h the position of , but from opposite directions . If ' ' " T S T S z S S E and F becomes ero, then , , coincide with ' ' T . T a S T S hen the ngle between the vertices, or X ° a 90 th e two whence the ngle between the centres is or, ' Simson lines passing through T will be perpendicular to eac h ' ' T are T and a ' . S ula other hese lines perpendic r to it at T , ' T being the common centre .

S A 28 I OGON L CUR VE S .

’ a of a E KF and b ch s circle the tri ngle , therefore the second ’ intersection of Feuerbach s circle with the side E F must be ’ ’ a e E F KS . H the foot of the ltitud to , or ence S is the foot Bu a tu KN. t as a an of the lti de , we h ve seen before y an d its a u a a at S a nd side ltit de is p ir of conjug e imson lines , since E F is a Simson line of a point on the circumcircle of AB C a KN S m n , with respect to the tri ngle, is the i son li e con jugate to E F .

An a E KF Si n 52. y tri ngle like formed by three mson li es, a re S u a to the altitudes of which imson lines conj g te the sides, ’ and having Feuerbach s circle in common with the triangle ’

AB a call S imson s Tria n le. C, we sh ll g

’ S F ac to an 53. ince euerb h s circle is common both tri gles ’ d E KF a F a - al AB C an , the r dius of euerb ch s circle is one h f the radius of the circu mcircle of either triangle ; therefore the radius of the circu mcircle of any Simson triangle is equ al to the radiu s of the circumcircle of the original triangle .

’" h m S a S 54. T e co mon vertex of the p ir of limiting imson ’ ’ a as N lines belonging to T S is on the same str ight line K, , and S ' ’" ’ For T S a am F ac , since is di eter of euerb h s circle , X or S 'S is perpendicular to E F ; therefore S is on the altitude to E F in the triangle E KE .

55 The point of contact L and the vertex A of a limiting ' ’ to T S a re a ta Simson line . with respect , equ lly dis nt from the centre of the line ; or R L " ' ’ ' Let L F ac T RS S intersect euerb h s circle in R, then X ’ " ' ’ Bu t L S L hence T E is pa rallel to and X ” ' ’ ' I a ma E S R L . n since L T T L , like m nner it y be ’ ’ proved that R is the middle point between L and 2 THE E LLIP SE . 9

2. THE E LLIPSE .

’ ni i na at to Lemoine s wi 56. The co c sogo l conjug e line, th ’ e the a an a l St r spect to tri ngle, is ellipse, c l ed einer s ellipse, ’ Lem ine whose centre is the median point of the tria ngle . o s ’ line is the polar to Grebe s point with respect to the circum f circle o the triangle . i 1 m A F . 7 as a e to To the point , g , pole with resp ct the circu ” of ABC a AK a to circle belongs the pol r , which is the t ngent P i at A . is r the circle To , , wh ch the point of inte section of a ts a n at B nd C and at am im a the t ngen dr w a , the s e t e o S m a h u A o a p int on the y medi n line t ro gh , belongs the p l r The B C. intersection of the two given polars is the pole of h e e n t line joining the two resp ctive poles of the give polars. ” T o to A S m an i herefore K , is the p le or to the ym edi l ne I ’ m A. f o e S through K , is the p int of int rsection of the y dia AK and a u AB a a me n the circle bo t C, then is t ngent at Sim la to the circle the point i rly, is the pole to " S m a and S m dia the ym edi n K , is the pole to the y me n ” ’ The K r OH . o a e o h , p ints K the p les to the t ree ra AK B K 0K and are st ight lines , , , , , since these concurrent, ’ u G the point of conc rrency being rebe s point K, therefore the " ’ t n i be T poin s K a d K w ll collinear . his line is called ’ Lemoin e AB AK n s a C. a d line with respect to the tri ngle , ’ d ’ B K i . A B an are , ntersect in K , , K , K , concyclic ; but since ’ " ' ’ AK and B11 in AK and B K , , intersect K, therefore , , will L ’ at a i a i. e. to emoine s intersect po nt Q, on the pol r to K,

and CK a . line ; p sses through Q ,

" ' If a CAI and M E 7. 5 the medi n line , be produced, , be ’ a a ME AK A al u a m de equ l to . , , then , or Q , is isogon conj g te " to and B A B a a , or Q , is isogon l conjug te to H A and B a ence the intersection of Q , Q , or Q, is isogon l con ' " E ’ a. jugate to the intersection of A E and B i. the point This may be proved as follows : S M AB and M ince , is the middle point of , " ’ ' " AB BE a a a a and BAE AB C. therefore is p r llelogr m, X X I S OGONAL CUR VE S .

Bu t E and B K are al a since M , isogon conjug tes, therefore ’ ’ ’ ABE KB C K B C. Bu t K B C K A C X X X , X , X , , ' " ’ B AE and A K A a therefore X X , or Q , is isogon l conj u gate to In the same way it can be proved that ’ AB and B K B a at the X , or Q , is isogon l conjug e to B L k w na a to E ' and i e ise, Q, is isogo l conjug te , Q, ” to E .

W a AB a 58. ith respect to the tri ngle C, the ellipse isogon l ’ ' " u a e Lemoine s and a u E E conj g t to line, p ssing thro gh , , i a as an ula o A B and C of w ll lso p s through the g r p ints , , the " a The a u at B K BA and tri ngle isogon l conj g e to , is , the ” " al a CK CA a isogon conjug te to , is ; K , is , therefore , point In w m al a A . am a a isogon conjug te to the s e y, we y show " " K be a a B and K C. , to isogon l conjug te to , , to The ellipse ’ ” wil A B C E E and S l therefore pass through , , , , , ince ' ” ’" E E = AE E E BE and E E E B will , , C , the point be the centre of the ellipse .

One and 59 . of the points of intersection of the ellipse the ’ um a h al S circ circle of the tri ngle is R, w ich is c led teiner s i m e . point . (S e F g Since R is on the circu circle of AB a a at and C, its isogon l conjug te point is infinity ; it is on ’ A aw Lem ine . o s line, as will now be proved line dr n through ’ A a a to Lem ine s a ate to , p r llel o line, will be isogon l conjug

AR .

We shall first examine some properties of the point R. The lines d rawn throu gh the angu lar points of the triangle AB C ’ B a a a parallel to the respective sides of roc rd s tri ngle, or, wh t a ula A N BN and CN 1 is the s me thing, perpendic r to , , ( 8) ’ N Tarr s ( being y point) , intersect in one point R, which is ’ S and a AB C. called teiner s point, is on the circle bout

Let i a n at A la AN and at 60 . the l nes dr w perpendicu r to , N B a BN at . S AR perpendicul r to , intersect R ince X and NB R = A N B R are c n and X the points , , , o cyclic , um i A N and A N and B are R must be on the circ c rcle of B ; , , ’ 31 THE E LL IP S E .

AB u i on the circumcircle of C, whence R is on circ mc rcle of

A S NA : NR a a e i B C. ince X R is di met r of th s circle N and are a and NCR h er en ( , M R colline r) , X t e p p We are dicular to CN at C passes through R. now to prove that R is isogonal conjugate to the point at infinity in the ’ ’ L m in S Lemoine s a direction of e o e s line . ince line is the pol r ’ MK e ula Lemoine s . h ra to K, will be p rpendic r to line T e y ’ drawn through A to the point at infin ity on Lemoine s line is parallel to this line and is perpendicular to MK. ' L et this perpendicular cu t the circumcircle of AB C at R ' Fi n = A nal con u . R AB CAR i. e. R ( g the X X , is isogo j ’ ’ B ca gate to AR . Since AR is parallel to the side B , C, of ro rd s a a and B am as tri ngle, the ngle between AR C is the s e the a B n Bu t an B C and ngle between , C, a d B C. the gle between , , B C al h a AMK ma is equ to t e ngle , which we y denote by ' t The a at M h arc AR 2 q . ngle the centre, , over t e ' hence the inscribed angle over the arc A R equ als Let the AR u B C a ABC R point where c ts the side of the tri ngle be , , AR B We a then X , h ve then ' CAR CAR A E ACB : X , X X R, X

h C a AB . in w ich 7 represents X of tri ngle C ’ — ’ = Arc R B = arc AB arc AR y

' ’ R AB CAR AR al u a AR hence X X , or is isogon conj g te to , ’ i In m wa or to the line parallel to Lemoine s l ne . the sa e y it ” ma a B al a B y be shown th t R is isogon conjug te to R , which is ’ parallel to L emoine s line through B ; a nd that CR is isogonal ’ conj u gate to C which is parallel to L emoine s line through

C. a u a The point R is, therefore, the point isogon l conj g te to L i ’ the point on emo ne s line at infinity .

a and an n If circle ellipse intersect, the directio s of the axes of the ellipse may be found by bisecting the angles mm an between the co on chords of the ellipse d the circle . u as A B C and are to nd Th s, , , , R points common both circle a a ellipse, the directions of the xes of the ellipse will be given by the two bisectors of the angles between the common chords 82 rsoa os L CUR VE S .

and AR B C. S A R ara B C u ince is p llel to , ,, the line thro gh the median point E and pa rallel to the bisector of the angle d B w between B C an , C, ill be one of the a xes ; and a line also E and e u a hi ll ax u . thro gh , p rpendic l r to t s, wi be the other is

’ a l al a e Lemoine s 62. The xes of the e lipse isogon conjug t to line with respect to the triangle are parallel to the asymptotes ’ a s a of the hyperbol i ogon l conjugate to Brocard s diameter. This may be proved as follows i 1 re a AB C F . 8 a If u pon the sides of the tri ngle , g , con mia a BA C CB A and A C stru cted si l r isosceles tri ngles, , , , , ,B, then the triangles AB C and have the same median point The ar t a al as e the E . proof is simil to h t in the speci c when vertices of the similar isosceles triangles are the angu lar points ’ a Let A B and C the of Brocard s tri ngle , , , , be vertices of these similar isosceles triangles constru cted upon the sides of AB nd KA KB and C m B C A C C, a let , , ,” K , eet the sides , , and AB respectively at Aw and then it can be proved that triangle is similar to triangle the centre

of similitude being K. If we erect a perpendicular at Ah to ’ B C m B a dia at A M to eet roc rd s meter Q, then, putting for , ” B u al KK KK e a ,JII, , their eq s, , , , resp ctively , we h ve

A M A -4 M , . sa s Q a A H , ,

h r A B C and B C are a have Since t e t iangles , ,,A simil r, we

A M M C a A M , , , _ , , , Bt MC 6 B rill » M 3 Az . t A M E M B K x . . . . ,, 0. A A B a Therefore e . e . A K E H , ,, we et Similarly, g

r a and are m a and K h or, t i ngles si il r, is t e centre of similitu de . THE E LLIP SE. 33

M Q, Q z F a l a B rom the equ tion it fol ows th t ,, Q, is

a a E M and e iula A p r llel to , , since is perp nd c r to C, there B a di a A C ula at fore ,, Q, is lso perpen cul r to , or the perpendic r

B A C a r u . S m a e ula at ,, to p sses th o gh Q , i il rly, the perp ndic r

C to AB as . a to n ,v p ses through Q , If, now, Q, is c used coi h e ec Bro cide with eit er Q, or Q , the points of int rs tion of ’ a a and u ia A B c rd s di meter the circ mcircle of the tr ngle C, then a 0 l a a n the tri ngle 27 wi l degener te into the str ight li es and which are the Simson lines belonging

and r r . to Q, Q, with respect to the ci cumcircle of the t iangle B The triangle A, , C, will degenerate into the straight lines and which will be parallel to the Simson lines and and a dian belonging to Q, Q 4 ; they will p ss through the me n E and B C a a poi t , for the lines A, , , still h ve the medi n A E ABC. A B C and A B C point in common with lso, , , , , , , , , , are on the perpendiculars at the middle points of the respec

v of a ABC. S S ti e sides the tri ngle ince the imson lines to Q, di and. a a are Q, correspond to the extremities of meter, they a a and a perpendicul r to e ch other, therefore their p rallels

re al u a . and A, B, C, a so perpendic l r to each other

Fur h t ermore,

M : M K M z M E Q 8¢ ¢ c a Q u a . .,

M K A K A M z A A , ; , , . 3 1, = Q « A4 =A4K “ J A I L,

A M 31 4 51 A M 3A A 3 ¢ 3 1 i a 4 1,

A A an a a and E A A whence , , § is h rmonic r nge, , , § i E E A . A A s an harmonic pencil Since 4 , 3 , will bisect the angle

Now a a a 63. , in two simil r tri ngles the bisectors of the ngles formed by any line in one triangle with the corresponding a e a al a line in the other tri ngle ar p r lel to e ch other, hence the m E and M the n AE bisector of the angle for ed by A , E ” or li e , 34 I SOGONA L CUR VES .

a. ara e i. the line is p ll l to the bisector of the angle

B . Bu formed by B , C, and C t the bisector of the angle formed d by B , C, an B C is parallel to one axis of the ellipse whose is at E n A B C i centre , therefore the li e , , , coincides w th one i a and A s o a . xis, , B, C, coincide with the sec nd x s of the ellipse We a n a A B C and B C are a r h ve see however, th t , , , A, , 4 p ’ ’ a llel to Simson s lines belonging to the extremities of Broca rd s a are a a a m t di meter, which themselves p r llel to the sy pto es ’ of the equ ilateral hyperbola isogonal conjugate to Brocard s diameter ; hence the axes of the ellipse isogonal conjugate ’ to Lemoine s line are parallel to the asymptote of the equ i B ’ lateral hyperbola isogonal conj u gate to rocard s diameter.

61 Before taking up other properties of the ellipse isogona l ’ e to Lomoine s in r a t a hall conjugat l e with espect to ri ngle, we s examine the properties of the fifth remarkable point and its

relations to the other points in the triangle.

ma a at a 0 and 65. It y be of interest to st te this point th t ’ 1 re an l a 0 (Fig . 5) a the foci of e lipse inscribed in the tri ngle a ed AB C. The points of cont ct of the sides with the inscrib are K K and K ellipse a, B’ 7 (the points of intersection of the t r Symmedian lines with the sides of the iangle) . We know tha t the focal radii to the point of contac t of a tangent to an e ellipse form equ al angles with th tangent.

Since

and B 0

’ B 0 0 BK therefore 4 0 0 , b’ 0K:

' and as 1 OB O = O CB 8 a 0 11 3 and , 3 4 , therefore tri ngles 2 a K E = ' are m a . T O O O si il r his proves th t K , X K ; ' n d are and c ac a d 0 an 0 the foci , K, is the point of ont t of

the side B C with the inscribed ellipse .

’ 36 ISOGONAL C UR was.

h We s all now prove that A Q equals twice 0 M” B Q equ als M and ua l O C 0 1 . S OM a a twice ,» Q eq ls twice 1 ince , is p r lle A and a a B a nd M I ara A B to Q , is p r llel to Q , J , is p llel to , a M I a nd A B are a e therefore the tri ngles O J , Q simil r, whenc

A : M = AB : Q O .

or A Q equ als twice

The ma m th 68. a h a and e fifth re rk ble point, t e edi n point, centre of the inscribed circle of a triangle are collinear ; and the median point divides the distance between the fifth remark able point and the centre of the inscribed circle in the ratio

of 2 1 . Join E with Q and 0 ;

A Q :

a ara A lso is p llel to Q ,

AM E M 0 . therefore X Q , X ,

From this it follows that the triangles A Q E and E OM; are a simil r, therefore

m he s m lar and E O are c a . F t the points Q, , olline r ro i i a s A E E 0M a a tri ngle Q , . it lso follows th t

= 2 : 0E 1 ,

E 0 69 . te ana et It is in resting to notice the logy b ween Q, , , and H E M , ,

S E : E = Z 1 ince Q O ,

and

' E M and 0 are a u a n a a z and , Q, , the ng l r poi ts of tr pe oid, the median point of the triangle is the point of intersection of i the diagonals of th s trapez oid . 37 THE EL LIPSE.

0 the e 70. The middle point P, of Q , is the centre of circl inscribed in the triangle P 0 = 50 Q

— P 0 0E = PE = - 0 } Q ,

a AE O m a r a M E P the tri ngle is si il r to the t i ngle . ,

a a M P . p r llel to ,

We a M I a a A C know th t J , is p r llel to ,

OAM PM M therefore z , Z . , , and

Bu t OAM OAM since 4 ; X , 2 PM I A therefore ; J , X P LM ; t a PM a M M In i h t is, the line , bisects the ngle M , l ke ma ma a P a M I I nner it y be proved th t M bisects ngle J J , H ence P is the centre of the circle inscribed in the triangle

’ The F a r r JII centre of euerb ch s ci cle ci cumscribing MM ,

F . We a P O and F is notice th t is the middle point of Q , is H EW and n ana e n the middle point of ; , i deed, the logy b twee Q n m F on n h a d H ay be noticed throughout. is the li e joining t e middle point of B C with th e middle point of the upper portion of the altitude drawn to B C (36) a nd P is on the line joining the middle point of B C with the middle point of the upper part of A Q

This last may be proved as follows Let Fi 1 P a A t n PM . 9 3 a ; ( g , l te ) meet Q the point R ; the

n P O a P a nd 0 rli a a e A a si ce equ ls Q , , is p r ll l to Q , the tri ngle P OJI a r a P R e a R . B u t , is equ l to the t i ngle Q , or qu ls Q

PAM a a A O and a a e AR , is p r llel to , is p r ll l to the line , I SOGONA L CUR VES .

a the u A GM E is a a al ram therefore it follows th t fig re . p r lelog

and OM AR . T 0 M m u a AR and , hen , beco es eq l to QR ,

R is the middle point of A Q .

' ' '

L i . m 2. et A B and F 20 7 , , C ( g ) be the iddle points of the a B A and AB a rcs subtended by the sides C, C, of the tri ngle ' AB C and A d m i a ; let M, be pro uced to eet the c rcumference m in and a a e second ti e on this line, which is di met r of the " " ' a A a J A a u a M A h ur circle, t ke so th t M sh ll eq l . t en the fig e " " ' ' AH A A a ara a m and AA diu a is p llelogr , is perpen c l r to AA ' " A MA IN A " 'M A N c M ; " ' A llf r MM r e h a u , ,” ( b ing t e r di s) ,

" ' " A A r MM r MM 2MM . , , .

Bu t the u pper part of the altitu de drawn to any side is twice the length of perpendicular from the centre of the circum c h sam a ircle to t e e side of the tri ngle, = " ' " A H 2 M2II and A A AH . therefore , ,

" '" A A i a a al AH are ula to s lso p r lel to , since they perpendic r " " ' am B and u HA A a. ara am. the s e side C, the fig re A is p llelogr " " ' In the same way it may be shown that BHB B is a parallelo " " ' a d H an a C a a a r m. gr m, th t C C is p r llelog a " ' ' " ' The a A AA a a A A ula ngle is right ngle, or is perpendic r ' " ' " " AA . Bu t A A a a H and A H to is p r llel to A , is therefore l ' L " ' u a AA . u a B B perpendic r to ikewise E H is perpendic l r to , " ' a n d O H to CC .

" L a A 73. ma a and h oc te Q, the fifth re rk ble point, t e points , " " B C and H are ma us , , Q, concyclic. This y be proved th A a a and a to 0M A O Q is p r llel to equ l twice , If on O be a al A O and be h prolonged, Q , m de equ to , Q, joined wit s M and a re a and M a Q, then the point Q,” ,” Q colline r, G Q , equ ls " ' JI and A JI a i u a ar G O; since “ equ ls c , fig re is p " ’ ' l A a al . a lelo ram. A a a AA S g Q is p r lel to O, . or p r llel to ince ' " " AA is u a A H A a u a to perpendic l r to , Q is lso perpendic l r 39 THE ELLIP SE.

" " a ann a a u ma ha B HA . In m er n logo s to this we y find t t Q " " er endicq H B and C u a to E C is p p to , Q is perpendic l r E h am t h i as i The length Q is, therefore, t e di e er of t e c rcle p s ng " " " u A B and C . thro gh , ,

' c A OC = ACC ACO Sin e X 4 + 1 ,

Z OAB Z B CC' —¥ t A + 4} 0 ; AOC"= OA C ' X K ,

' n B 'A ' C A 0 0 , a d B O.

' ' B C u a to A O at its m dl is, then, perpendic l r id e point, or ' ' ' ' e B a m A O is perp ndicular to B C . y similar re soning B O ay ' ' ' ' ' w e e u a to A and A B . be sho n to be p rp ndic l r C , C C to The r a point 0 is, therefore, the orthocent e of the tri ngle

' ' H u a A O and B C a er en 74. is perpendic l r to , is lso p p " ' ' " dicular A O HA a a B and H E to ; hence is p r llel to C , is ' ' " " parallel to A O . Then the angle B HA is equ al to the angle ’ Bu t H and C are on the circumference of the same nd circle, a B NHA " B N IIA H x x C ’ xa = xa n Likewise and xA' Z A " " " ' ' ' a A B C and A B are The tri ngles C therefore similar.

M ' 75. If is the centre of the circle abou t the triangle M ' M a. r For then O Q is parallelog am. the triangle

i i” a AB C and M O a a M m , . C is simil r to ; since is p r llel to Q , O a a B and 0 a a M p r llel to Q , M p r llel to CQ , therefore the corresponding point to Q in the triangle AB C is the point 0 in a M H 0 is mar a the tri ngle MM , ence the fifth re k ble a M M point in tri ngle JL , 40 I SOGONA L CUR VES .

The line joining H and Q in triangle AB C will be homolo In gous to the line OH in the triangle MJAM. these tri angles all homologous parts except the angles are in the ratio ’ 2 S n M is therefore H Q equals CM. i ce the middle point l ' H OM a M . Bu t GM a ra of Q , therefore will be equ to Q is p l

M and M CM a a a am. lel to Q, therefore Q is p r llelogr

The la a u a s AB C 76. perpendicu rs from the ng l r point of " " " F 20 i to the respective sides of A B C ( ig . ) ntersect in one point, R . Let R be the intersection of the perpendiculars from A and " " " " he B to C B and A C respectively . T perpendicular from C " " For A a as u . an C to B lso p ses thro gh R , let 7 represent gle

a AB C ARB i of tri ngle , 3, hence x w ll be 7 90 - 5 90 a 90 either 2 or this c se it is

A C "B A G’E 1 80 Z Z y,

" n as A and B . he ce C is centre of the circle p sing through , R, " " " a m C AR B The perpendicul r fro to , or the line C , bisects " " AR and a C B R A " ; the perpendicul r from to , or the line O , H " n B R . A a ta B a d and bisects ence is equ lly dis nt from R, " B u l n a and C. S a a a lso from B imil rly, is eq ly dist t from R and C. CR therefore must pass throu gh R and be perpendicu " " lar to A B .

The s rs 77. point of inte ection of the corresponding sides ' ' ' " " " a A B C and A B are a and the of the tri ngles C colline r, line joining these points is perpendicular to a nd bisects OR

(Fig . ' ' " " ' ' Let A B and A B at A B a intersect W. is perpendicul r ' ' and s CC A B a to bisect , therefore every point on is equ lly " " distant from C and 0 . E very point on A B is equally dis H ' ' a C and . W A B and t nt from R ence , the intersection of " " A B u a a C and . S a , is eq lly dist nt from R imil rly, we find ' ' " ” a B C and B C and it th t U, the intersection of , V, the n er ' ' " " of A O and A C are a a and section , equ lly dist nt from 0 R ; r la and o WVU is a straight line perpendicu r to bisecting OR . 4 TH E ELLIP SE. 1

a a u a 2 r am e 78. If dist nce eq l to , the di et r of the inscribed a AB C a off r circle of the tri ngle , be l id from the ve tices on a a u r a AB C ta e ch of the ltit des of the t i ngle , we ob in three ts A B C are r a r a and poin , ” ,, ,, which the ve tices of new t i ngle ; this triangle is concyclic with triangle and is similar

to triangle AB C. In a M M and AB C an the tri ngles . , M, y two corresponding 2 w are a 1 : . 0 a a lines in the r tio , hich is the fifth rem rk ble point of triangle corresponds to Q in the triangle

AB - a n C. u a A a d r is therefore eq l to one h lf Q , the pe

endicular 0 0 0 B C one- al iula p . from to is h f the perpend c r N a m u A a a Q , dr wn fro Q to the line thro gh p r llel to the side B 2 r am C. A diula N a lso this perpen c r Q , is equ l to , the di eter 2 s e c AB C. S AA a ua r of the in crib d cir le of ince , lso eq ls , A a a to B C AA HA Q , is p r llel , or x 4Q 4 , Q therefore A am u B n 4 is on the circumcircle of H Q . The s e is tr e of , a d

C . A B are h , The points , ,, C, then concyclic wit

79 . S A a a B C and B a a ince Q is p r llel to , Q 4 is p r llel to = A O z A Q B M OB Bu t z w a = nd a A CE . S a AB C and therefore imil rly I Z , B A C. a A i r 4 X Tri ngle , B, C, 18 therefore sim la to i the tr a ngle AB C.

- 80 . 0 in c c is the centre of the ir le of A, B , C. and at the

am m For A : 2 7 and s e ti e the orthocentre of , A , A "A ' 2M A ' . ; hence

' " AA, : A A

' a A 0 0 and M BA are m a OAO The tri ngles , . si il r (4 , ’ ' ' BA : a A G : A B = : = M OO M A . But CC r 4 5 ) , hence , , ' ' and A B A O ; hence

' ’ " A 0 : A O A : 1 44 A A .

F om h as a h r an r t e l t proposition it follows th t t e t i gles AA, 0 ’ " and A A are m a G si il r. 42 I SOGONA L CUR VES .

" and ' A 0 : A O A O A O.

' " L B O B O B O : B O ikewise , ,

" nd a ' = a : C C . lso CO C O C, O

There is the same relation between the triangles A, B , C, " " " (similar to A BC) and A B C (similar to with respect 0 as a AB and to the point , between the tri ngles C that

O r in- d is, is the cent e of the c ircle of an at the same time the orthocentre of

" " " " A B n u a n A r 81 . is perpe dic l r to a d bisects OC, ; C is pe " " endicular and di ul p to bisects OB , ; and B C is perpen c ar to and bisects OA, .

2 a e 8 . The dist nce of the centre of th inscribed circle from R (the intersection of perpendicula rs from the angu lar points of the triangle AB C to the respective sides of is

a - equ l to the diameter of the in circle of triangle AB C. " " For A B ula and C and m is perpendic r to OC, R, bisects the

. C OC an a z and R CC Z r. both R , is isosceles tr pe oid, O ,

L m r nd 83. h n L u a a If be t e iddle point of OR , the O eq ls , WVU e d ula R at L W e m a since is perp n ic r to O , VU b co es

a in- a t ngent to the c ircle of AB C. The line WVU is lso the ad a a i to r ic l xis, or the l ne of equal powers with respect the circle abou t and the centre 0 of the in- circle of ABC m E i on is the li iting point of the coaxal system . ac h po nt " " A B Bu t and al a and C . , hence W, is equ ly dist nt from O , W a u a d a and W is lso eq lly ist nt from R C, therefore is the n T a as u h C and C. r a ce tre of circle p sing thro g R, O, , , i ngles

A B n- , , C, and AB C are similar ; 0 is the centre of the i circle ; ’ and M a is the centre of the circumcircle of A, B, C, The ngle ’ OC M u a 1 a A CC , eq ls } ( ngle O , , which is formed by the a u m C and a C al u a ltit de fro the bisector of the ngle , is so eq l 1 — OC a and a CC an an to 50 3) , is chord, ngle O , is gle and c inscribed in the circle passing through ROC, O ; sin e ’ an M O O u a a OCC M a a at gle , eq ls ngle , , 0 , is t ngent the point C H M C , to this circle . ence 0, is perpendicular to W ,, or the

44 I SO ON L CU S G A R VE .

' ' ' 90 . And n E C A 90 Z a 57 si ce 4 57, is on the s me circu mference as We have now to prove that the ' " " e u ar m C A B as a Z a p rpendic l fro to p ses lso through , or th t n ' " " L t “ the line joi ing Z with C is perpendicu lar to A B . e B ' " " ' o i Z B and A and C in be the p int of intersect on O , , the po t ' " " of intersection of Z C and A B then B 'Z C ' B 'A 'C ' B HA NON 4 X z ,

" " " ' ' a nd X 1 80 4 C A B 1 80 Z E Z O . T " ' e Z A and C are c . h refore , , , con yclic

" ' " ' And A B Z A The e since X , 4 C , Z also lin ' " " Z C u to is, therefore, perpendic lar A B . " ' ' ’ A a A is a B C and C er en g in, Q perpendicul r to , is p p " " ' ' " " dic u lar A B h Z O R B A to , ence 4 Z , or 4 Q Z A 'B ' ' ' ' th A . The angle formed by Z A and A B is equal to e " " " angle formed by Q A and A B ; or in the similar tria ngles ’ ' ' A B C and Z has the same relative position on the ' ' ' circu mference of the circ le about A B C with respect to the angular points of as has Q on the circumference of " " " the circle abou t A B C with respect to the a ng u lar points of

We a . 86. a a and Z a re sh ll now prove th t Q, colline r ' ' ' " " " r a A B and A B are m a 0 The t i ngles C C si il r, the point being the orthocentre in both triangles The lines join ' ' ' ing three points in A B C form the same angles as the lines " " " joining three corresponding points in A B C T t M’ M herefore a O Q X O Z . W Bu t OM = OM X Q Z Q a t OJIZ Z’ OM hence a , Q, and t Ill and Z are a a ht the three poin s, , Q, on the s me str ig line .

Th 87. e u a A B B A perpendic l rs from to , C,, from to , O, , a nd m A B fro C to , , intersect at T on the circu mference of the circle abou t AB C. TH E ELLIP SE. 45

The proof for this is similar to that for the point Z in Now T m B the previous discussion. on the circu circle of A C has the same relative position with respect to the ang ular s AB C as has um A B C point of , Q on the circ circle of , , , with respect to the angular points of HA, is perpendicu lar B and T e u a AB to C, C is perp ndic l r to , therefore

= B T = B B T. 4 ,A, H 4 C x A

In and A a al to B AZ a a to B . since C is p r lel Q ,, is p r llel , C, a BZ a al A O and Z a a like m nner we get p r lel to , ,, C p r llel to

The a a to AZ B Z CZ 88. isogon l conjug tes , , , with respect to triangle AB C are perpendicu lar to the line joining the a o orthocentre H with the fifth remark ble p int Q of AB C. If through Z a pa rallel is drawn to AB to meet the circu mcircle ' ' AB C at Z n arc AZ al arc B Z n a of , the the equ s , bei g rcs ' a a ds . T Z CA Z CB and between p r llel chor herefore A X , ' Z C is isogonal conjugate to Z C.

C H GH = C B = CB Z = CZ ' Z X , Q Q X , , Q X ¥ , = ' Z GH Q z CZ Z ;

' ' and Z Z r a CH Z m s since is pe pendicul r to , therefore C u t be

u a H . The a a e to Z A Z B perpendic l r to Q isogon l conjug t s , , and Z C a re a a ac and are all e i p r llel to e h other, perp nd cular to

H Q . The Simson line to Z is perpendicular to the line na a e Z A a al H n isogo l conjug t to , therefore it is p r lel to Q , a d S T the imson line to mu st be perpendicular to H Q .