Cevians, Symmedians, and Excircles
MA 341 – Topics in Geometry Lecture 16 Cevian A cevian is a line segment which joins a vertex of a triangle with a point on the opposite side (or its extension). B cevian
C A D
05-Oct-2011 MA 341 001 2 Cevian Triangle & Circle • Pick P in the interior of ∆ABC • Draw cevians from each vertex thhhrough P to the opposite side • Gives set of three intersecting cevians AA’, BB’, and CC’ with respect to that point. • The triangle ∆AABC’B’C’ is known as the cevian triangle of ∆ABC with respect to P • Circumcircle of ∆A’B’C’ is known as the evian circle with respect to P.
05-Oct-2011 MA 341 001 3 Cevian circle
CiCevian triangle
05-Oct-2011 MA 341 001 4 Cevians In ∆ABC examples of cevians are: medians –cevian point = G perpendicular bisectors – cevian point = O angle bisectors – cevian point = I (incenter) altitudes – cevian point = H Ceva’s Theorem ddleals with concurrence of any set of cevians.
05-Oct-2011 MA 341 001 5 Gergonne Point In ∆ABC find the incircle and points of tangency of iilincirclewihith sides of ∆ABC. Known as contact triangle
05-Oct-2011 MA 341 001 6 Gergonne Point These cevians are concurrent! Why? Recall that AE=AF, BD=BF, and CD=CE
Ge
05-Oct-2011 MA 341 001 7 Gergonne Point The point is called the Gergonne point, Ge.
Ge
05-Oct-2011 MA 341 001 8 Gergonne Point Draw lines parallel to sides of contact trianggugle through Ge.
05-Oct-2011 MA 341 001 9 Gergonne Point Six points are concyclic!! Called the Adams Circle
05-Oct-2011 MA 341 001 10 Gergonne Point Center of Adams circle = incenter of ∆ABC
05-Oct-2011 MA 341 001 11 Isogonal Conjugates Two lines AB and AC through vertex A are said to be isogonal if one is the reflection of the other through the angle bisector.
05-Oct-2011 MA 341 001 12 Isogonal Conjugates If lines through A, B, and C are concurrent at P, then the isogonal lines are concurrent at Q.
Points P and Q are isogonal conjugates.
05-Oct-2011 MA 341 001 13 Symmedians
In ∆ABC, the symmedian ASa is a cevian through vertex A (Sa BC) isoggyonally conjugate to the median AMa, Ma being the midpoint of BC. The other two symmedians BSb and CSc are defined similarly.
05-Oct-2011 MA 341 001 14 Symmedians
The three symmedians ASa, BSb and CSc concur in a ppyoint commonly denoted K and variably known as either • the symmedian point or • the Lemoine point
05-Oct-2011 MA 341 001 15 Symmedian of Right Triangle
The symmedian point K of a right triangle is the midpoint of the altitude to the hypotenuse. A
M K b
B C D
05-Oct-2011 MA 341 001 16 Proportions of the Symmedian
Draw the cevian from vertex A, through the symmedian point, to the opposite side of the triangle, meeting BC at Sa. Then
BS c2 c a b 2 CSa b
a
05-Oct-2011 MA 341 001 17 Length of the Symmedian
Draw the cevian from vertex C, through the symmedian point, to the opposite side of the triangle. Then this segment has length 222 ab 2a 2b c CS c ab22 Likewise bc 2b222 2c a AS a bc22 ac 2a222 2c b BS b 22 05-Oct-2011 MA 341 001ac 18 Excircles In several versions of geometry triangles are defined in terms of lines not segments.
A
B C
05-Oct-2011 MA 341 001 19 Excircles Do these sets of three lines define circles? Known as tritangent circles
A
B C
05-Oct-2011 MA 341 001 20 Excircles
I A C IB r c I rb B C
IA ra
05-Oct-2011 MA 341 001 21 Construction of Excircles
05-Oct-2011 MA 341 001 22 Extend the sides
05-Oct-2011 MA 341 001 23 Bisect exterior angle at A
05-Oct-2011 MA 341 001 24 Bisect exterior angle at B
05-Oct-2011 MA 341 001 25 Find intersection
Ic
05-Oct-2011 MA 341 001 26 Drop perpendicular to AB
Ic
05-Oct-2011 MA 341 001 27 Find point of intersection with AB
Ic
05-Oct-2011 MA 341 001 28 Construct circle centered at Ic
Ic rc
05-Oct-2011 MA 341 001 29 05-Oct-2011 MA 341 001 30 Excircles
The Ia, Ib, and Ic are called excenters. ra, rb, rc are called exradii
05-Oct-2011 MA 341 001 31 Excircles Theorem: The length of the tangent from a vertex to the opposite exscribed circle equals the semiperimeter, s.
CP = s
05-Oct-2011 MA 341 001 32 Excircles 1. CQ = CP 2. AP = AY 3. CP = CA+AP = CA+AY 4. CQ= BC+BY
5. CP + CQ = AC + AY + BY + BC 6. 2CP = AB + BC + AC = 2s 7. CP = s
05-Oct-2011 MA 341 001 33 Exradii
1. CPICP
2. t(C/2)tan(C/2)=rC/s I 3. Use Law of c Tangents
C (s a)(s b) s(s a)(s b) rstanc s 2()2 s(sc ) sc
05-Oct-2011 MA 341 001 34 Exradii Likewise
s(s b)(s c) r a sa s(s a)(s c) r b sb s(s a)(s b) r c sc
05-Oct-2011 MA 341 001 35 Excircles Theorem: For any triangle ∆ABC 11 1 1 rrabc r r
05-Oct-2011 MA 341 001 36 Excircles
111 sa sb sc rrrrrr()()a b c s(sb)(sc) s (a)()(s a)(sc) s (a)()(s a)(sb) sa sb sc s(s a)(s b)(s c) s(s a)(s b)(s c) s(s a)(s b)(s c)
3s (a b c) s(()()()sa)(sb)(sc)
ss s(s a)(s b)(s c) K
1 r
05-Oct-2011 MA 341 001 37 Nagel Point In ∆ABC find the excircles and points of tangency of the excilircles wihith sides of ∆ABC.
05-Oct-2011 MA 341 001 38 Nagel Point These cevians are concurrent!
05-Oct-2011 MA 341 001 39 Nagel Point Point is known as the Nagel point
05-Oct-2011 MA 341 001 40 Mittenpunkt Point The mittenpunkt of ∆ABC is the symmedian ppfoint of the excentral triangle
(∆IaIbIc formed from centers of excircles)
05-Oct-2011 MA 341 001 41 Mittenpunkt Point The mittenpunkt of ∆ABC is the point of intersection offfm the lines from the excenters through midpoints of corresppgonding sides
05-Oct-2011 MA 341 001 42 Spieker Point The Spieker center is center of Spieker circle,,.., i.e., the incenter of the m edial triangle of the original triangle .
05-Oct-2011 MA 341 001 43 Special Segments Gergonne point, centroid and mittenpunkt are collinear
GGe =2 GM
05-Oct-2011 MA 341 001 44 Special Segments Mittenpunkt, Spieker center and orthocenter are collinear
05-Oct-2011 MA 341 001 45 Special Segments Mittenpunkt, incenter and symmedian point KKw are collinear with distance ratio
IM 2(a222 +b + c ) = MK (a +b +c)2
05-Oct-2011 MA 341 001 46 Nagel Line The Nagel line is the line on which the incenter ,,g triangle centroid , ,p Spieker center Sp, and Nagel point Na lie.
GNa =2 IG
05-Oct-2011 MA 341 001 47 Various Centers
05-Oct-2011 MA 341 001 48