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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 728 details.
Note that centers O and O’ will be on AC.Let NG cut EF at L and HM cut EF at L’NG Perpen. To AB & HM perpen. To BCQuadrilaterals ANQL and CDL’M are cyclicPower from point P to circle ANQL= BN.BA=BL.BDPower from point P to circle CDL’M=BM.BC=BL’.BDPower of point P to circles O and O’ =BN.BA=BE.BF=BM.BCFrom above we have BL.BD=BL’.BD => BL=BL’ => L coincide to L’So BF, GN and MH are concurrent
A beautiful application of Ceva's Theorem (Trigonometric Version)Sketch:Let EAC = x, ECA = yAE, BE, CE are concurrent Cevians satisfying sin ABD.sin ECB.sin EAC = sin EBC.sin ECA.sin EABwhich impliescos A.sin(C-y).sin x = cos C.sin y.sin (A-x)...(#)EF,GN,HM are Cevians thro'the vertices E,G,H of triangle EGH.sin HED=sin HCE=sin y ;sin EGN=sin EAN=sin(A-x);sin CHM=sin(90-C)=cos Cwhilesin DEG=sin GAE=sin x;sin NGA=sin(90-A)=cos A;sin MHE=sin MCE= sin(C-y)Clearly in view of (#)The product of the above first three ratios = the product of the above next three ratios.Hence by Ceva, EF,GN,HM are concurrent.
Problem 728The ΒF is the radical axis of centers of circles O and O ',then BN.BA=BE.BF=BM.BC. So A,N,M,C are concyclic.Then <MNB=<BCA.But If I bring the altitudes ck and AL (T = orthocenter).Is <AKC=<ALC=90 then A,K,L,C are concyclic. So <LKB=<BCA.ThereforeKL//NM then BK/BN=BL/BM ,but if NG and HM intersect the BF to P, P ' respectively(KT//PN,LT//MP’). Is BK/BN=BT/BP, BL/BM=BT/BP’. So BT/BP=BT/BP’ then BP=BP’.Therefore NG,MH,BD are concyrrent.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE