S T P E C N & S O T C S TE G E O M E T R Y GEOMETRY TRIANGLE CENTRES Rajasthan AIR-24 SSC SSC (CGL)-2011 CAT Raja Sir (A K Arya) Income Tax Inspector CDS : 9587067007 (WhatsApp) Chapter 4 Triangle Centres fdlh Hkh triangle ds fy, yxHkx 6100 centres Intensive gSA defined Q. Alice the princess is standing on a side AB of buesa ls 5 Classical centres important gSa ftUgs ge ABC with sides 4, 5 and 6 and jumps on side bl chapter esa detail ls discuss djsaxsaA BC and again jumps on side CA and finally 1. Orthocentre (yEcdsUnz] H) comes back to his original position. The 2. Incentre (vUr% dsUnz] I) smallest distance Alice could have jumped is? jktdqekjh ,fyl] ,d f=Hkqt ftldh Hkqtk,sa vkSj 3. Centroid (dsUnzd] G) ABC 4, 5 lseh- gS fd ,d Hkqtk ij [kM+h gS] ;gk¡ ls og Hkqtk 4. Circumcentre (ifjdsUnz] O) 6 AB BC ij rFkk fQj Hkqtk CA ij lh/kh Nykax yxkrs gq, okfil vius 5. Excentre (ckº; dsUnz] J) izkjafHkd fcUnq ij vk tkrh gSA ,fyl }kjk r; dh xbZ U;wure nwjh Kkr djsaA 1. Orthocentre ¼yEcdsUnz] H½ : Sol. A fdlh triangle ds rhuksa altitudes (ÅapkbZ;ksa) dk Alice intersection point orthocentre dgykrk gSA Stands fdlh vertex ('kh"kZ fcUnw) ls lkeus okyh Hkqtk ij [khapk x;k F E perpendicular (yEc) altitude dgykrk gSA A B D C F E Alice smallest distance cover djrs gq, okfil viuh H original position ij vkrh gS vFkkZr~ og orthic triangle dh perimeter (ifjeki) ds cjkcj distance cover djrh gSA Orthic triangle dh perimeter B D C acosA + bcosB + c cosC BAC + BHC = 1800 a = BC = 4 laiwjd dks.k (Supplementary angles- ) b = AC = 5 BAC = side BC ds opposite vertex dk angle c = AB = 6 BHC = side BC }kjk orthocenter (H) ij cuk;k 2 2 2 b +c -a 3 x;k angle. cosA = = 2bc 4 Proof : a2+c2-b2 9 cosB = = AEHF ,d cyclic quadrilateral ¼ pØh; prqHkqZt ½ gSA 2ac 16 0 0 F = 90 , E = 90 a2+b2-c2 cosC = = 1 FAE + EHF = 1800 2ab 8 FAE = BAC 3 9 105 Perimeter = 4 + 5 + 6 1 = cm. EHF = BHC 4 16 8 16 (vertically opposite angles - f'k"kkZfHkeq[k dks.k) BAC + BHC = 1800 Download our app : NEON CLASSES 1 geometry NEON CLASSES, JAIPUR • 9828728833/34 Orthic triangle 4. Orthic triangle dsoy acute angled triangles esa gh gS rFkk o Triangle ds rhuks altitudes ds feet dks feykus ls cuus possible right angled obtuse angled okyk dgykrk gSA ;g (vf/kd dks.k f=Hkqt) ds fy, orthic triangle ugh cuk;k triangle orthic triangle tk ldrk gSA triangle orthocentre ds fy, pedal triangle Hkh gksrk gSA A Orthic triangle esa cuus okys cyclic quadrilaterals (pdzh; prqHkqZt) AFHE, BFHD, CEHD BCEF, ACDF, ABDE F E Orthic triangle DEF dh inradius and circumradius H Inradius = 2RcosAcosBcosC Circumradius = R/2 Triangle ABC ds vertices ('kh"kZ fcUnw) pedal B C D triangle DEF ds excentres gksrs gSaA vFkkZr~ ABC dh Triangle DEF orthic triangle gSA Hkqtk;sa orthic triangle DEF ds fy, exterior angle cká dks.k lef}Hkktd gksrh gSA 1. Triangle ABC dk orthocentre H, orthic triangle bisectors ( ) DEF dk incentre (vUr% dsUnz) gksrk gSA vFkkZr~ original Vertex A, B rFkk C ls orthic triangle DEF dh Hkqtkvksa triangle ABC ds altitudes orthic triangle DEF ds ij Mkys x, yEc (perpendiculars) ABC ds fy, interior angle bisectors gksrs gSa] vFkkZr~ circumcentre (ifjdsUnz) O ij feyrs gSa vFkkZr~ ABC dh ifjo`r dh f=T;k f=Hkqt dh Hkqtkvksa ∠BEF = ∠BED, ∠ADE = ∠ADF, ∠CFE = ∠CFD circumradius ( ) DEF ij yEcor~ gksrh gSA 2. Triangle ABC esa inscribed fd, tk ldus okys lHkh T A triangles esaa lcls de perimeter (ifjeki) orthic triangle dh gksrh gSA Orthic triangle dh sides = F E (i) R Sin2A = a Cos A O Circumcentre (ii) R Sin2B = b Cos B ¼ifjdsUnz½ (iii) R Sin 2C = c Cos C B D C vr% orthic triangle dh perimeter = Orthocentre and circumcircle ds e/; laca/k & acosA + bcosB + ccosC. A abc cosAcosBcosC Area of orthic triangle DEF = 2R K R = ABC dh circumradius F E H 3. Orthic triangle DEF dk circumcircle original triangle ABC dh Hkqtkvksa ds mid points ls xqtjrk gS rFkk ;g circumcircle, orthocentre ls vertices dh B C nwjh ds e/; fcanqvksa ls Hkh xqtjrk gSA ;g bl triangle dk ij dk gSA nine point circle dgykrk gS] tks fd orthic triangle FK, line AB orthocentre H reflection vr% ,d gS rFkk dk DEF ds fy, circumcircle ¼ifjo`r½ gksrk gSA HBKA kite line segment HK midpoint F gSA Nine point circle dh radius, ABC dh circum radius dh vk/kh gksrh gSA ∠ABH = ∠ABK Like Our FB Page : @neon.classes 2 Web.: www.neonclasses.com rFkk nksuks gS rFkk FE 1 A Orthic DEF JKL similar KJ = 2 A F E K J F E H B D C bl case esa D, E rFkk F altitudes gSaA B D C BFC ~ BDA L (B common gS rFkk ∠F = ∠D = 900, vr% AA property ls) AEB ~ AFC ⇨ CDA ~ CEB Concurrent Lines BF BC AC AB CD CA vr% = ; = ; = 3 ;k 3 ls vf/kd lines tks fdlh ,d point ij intersect BD BA AF AC CE BC djrh gS vFkkZr~ fdlh ,d gh fcanw ij vkdj feyrh gS] bu rhuksa equations dks multiply djus ij] concurrent lines dgykrh gSA rFkk og fcanw ftl ij ;s BF . AE . CD = BC . AB . CA = 1 vkil esa dkVrh gSa BD AF CE BA AC BC lines point of concurrency dks djus ij dgykrk gSA Left Hand Side rearrange AF BD CE Line a . = 1 Eg.: FB DC EA vr% ABC esa rhuksa altitudes ,d single point ij feysaxsa Line b m ftls orthocentre (yEcdsUnz) dgrs gSaA Incentre ds fy,& Line c fdlh ABC esa incentre exist djrk gS bldk proof dk gS ;gka fcanw m, point of concurrency dgykrk gSA Ceva's theorem trigonometric version ftlds vuqlkj AD, BE rFkk CF rHkh concur djsaxh tc& fdlh triangle ds fy, fofHkUu lines (vyx&vyx ds vuqlkj fdlh ,d fcanw ij vkil esa dkVrh gSA Sin ∠ BAD . Sin ∠CBE . Sin ∠A CF = 1 concepts ) Sin ∠ABE Sin ∠BCF Sin ∠CAD bl fcanw dks mu lines ds vuqlkj vyx&vyx centres A (Eg. orthocentre, incentre, centroid, circumcentre, θ θ excentre etc.) ds :Ik esa define fd;k x;k gSA 2 2 F E Ceva'a theorem fdUgh Hkh rhu lines ds fy, concurrence crkrh gS vr% ;g orthocentre, incentre rFkk circumcentre ds fy, Hkh true gS rFkk Ceva's theorem dh lgk;rk ls ge bu centres dk B D C existence prove dj ldrs gSaA Incentre ds case esa D, E rFkk F angle bisectors ds feet gSa vr% Proof of existence ∠BAD = ∠CAD, ∠ABE = ∠CBE rFkk ∠ACF = ∠BCF fdlh Hkh esa yEcdsUnz triangle orthocentre ( ) exist vr% Sin ∠ BAD . Sin ∠ABE . Sin ∠A CF = 1 : 1.1 = 1 djrk gS bldk lcls simple proof Ceva's theorem gSA Sin ∠CAD Sin ∠CBE Sin ∠BCF Left Hand Side dks rearrange djus ij] Ceva's theorem ds vuqlkj] Sin ∠BAD Sin ∠CBE Sin ∠ACF . = 1 Rkhu lines AD, BE rFkk CF rHkh concur gksaxh vFkkZr~ fdlh Sin ∠ABE Sin ∠BCF Sin ∠CAD ,d fcanw ij feysaxh tc & vr% rhuksa angle bisectors ,d point ij feysaxsa ftls AF. BD . CE = 1, vUr%dsUnz dgrs gSaA FB DC EA incentre ( ) Download our app : NEON CLASSES 3 geometry NEON CLASSES, JAIPUR • 9828728833/34 Orthocentric System ADB esa Triangle ABC rFkk bldk orthocentre H feydj ,d AB > AD ......... (i) orthocentric system cukrs gSa vFkkZr~ fdUgh Hkh rhu BEC esa points dks feykus ls cuus okys triangle ds fy, pkSFkk BC > BE .........(ii) gksrk gSA point orthocentre CFA esa A AC > CF ..........(iii) adding (i), (ii), (iii) we get F E AB+BC+AC > AD+BE+CF H F fdlh triangle ABC esa altitudes dk vuqikr %& 1 1 1 Area of ABC = 2 ABCF = 2 BC AD = 2 AC BE 1 1 1 AB : BC : AC = : : CF AD BE B D C Equilateral triangle ¼leckgq f=Hkqt½ esa rhuksa Hkqtk,sa leku gksrh gS] vr% altitudes ¼yEc½ Hkh leku gksrs gSaA F ABC dk orthocentre = H ABH dk orthocentre = C Position of Orthocentre- BCH dk orthocentre = A (i) Acute angled triangle (U;wu dks.k f=Hkqt) :- ACH dk orthocentre = B blesa orthocentre f=Hkqt ds vUnj fLFkr gksrk gSA F AH.HD = BH.HE = CH.HF A Orthocentre, altitudes dks ftu nks lengths esa divide djrk gS mudk product ges'kk ,d constant gksrk gSA H Similarly, (i) AD.DH = BD.DC (ii) BE.EH = AE.EC B C 0 (iii) CF.FH = AF.FB A, B, C rhuksa 90 ls de gSA AF BD CE (ii) Right angled triangle (ledks.k f=Hkqt) F = 1 FB DC EA F Pair of similar triangles A DBH ~ EHA DHC ~ FHA FHB ~ EHC F Sum of three altitudes of a triangle is less than sum of three sides of a triangle. BH C fdlh f=Hkqt ds rhuksa altitudes ¼yEcksa½ dk ;ksx rhuksa Hkqtkvksa Orthocentre ds ;ksx ls de gksrk gSA fdlh ledks.k f=Hkqt esa] Right angled triangle esaa orthocentre ledks.k okys ij fLFkr gksrk gSA Hypotenuse ¼d.kZ½ > altitude ¼yEc½ vertex Like Our FB Page : @neon.classes 4 Web.: www.neonclasses.com (iii) Obtuse angled triangle (vf/kd dks.k f=Hkqt) (2) Incentre ¼vUr% dsUnz] I ½ Obtuse angled triangle esa orthocenter vf/kd dks.k okys 'kh"kZ dh rjQ f=Hkqt ls ckgj fudy tkrk gSA f=Hkqt ds rhuksa angles ds angle bisectors (dks.k 0 lef}Hkktd) dk intersection point (dVku fcUnw) B > 90 , B obtuse angle gSA Incentre dgykrk gSA A A 2 / 2 / E I r F C B r D /2 r /2 H /2 /2 B C 0 ABC + AHC = 180 Incentre ls rhuksa sides dh nwjh leku gksrh gS bl nwjh dks A vUr% o`r dh f=T;k dgrs gSa rFkk blls cuus inradius (r ) E okyk o`r incircle (vUr%o`r) dgykrk gSA F Incentre ges'kk f=Hkqt ds vUnj gh fLFkr gksrk gSA rFkk F C B okLro esa ml f=Hkqt ds medial triangle ds vUnj fLFkr gksrk gSA A D H K I BAC + KHC = 1800 Proof /2 /2 EBA ~ DBH B C 0 E = D = 90 fdlh Hkqtk }kjk I ij cuus okyk angle rFkk Hkqtk ds EBA = DBH (vertically opposite angles) opposite vertex ij cuus okys angle dk laca/k& EAB =DHB (similar triangles) 0 = 90 + KHC + DHB = 180 (Straight line ij cus angles) 2 EAB = BAC Proof : 0 KHC + BAC = 180 BIC esa ⇒ + + = 1800 .........