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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1267.
https://goo.gl/photos/1anav7EpeN3Uvv8e9Let ED meet AC at P and E1D1 meet AC at P1Note that triangles BED and BE1D1 are isosceles and ED//E1D1CF=AF1= p- AB where p= semi perimeter of triangle ABC1. Per the result of problem 1265 and 1266 we have FC/FA= PC/PA= kWe can calculate FP in term of k and AC FP= ( (k/1+k) + k/(1-k)). AC Similarly F1A/F1C=P1A/P1C= k and F1P1= ( (k/1+k) + k/(1-k)). ACSo FP=F1P1Triangle PGF congruent to P1G1F1 ( case ASA) => FG=F1G1We also have FG//F1G1 => FGF1G1 is a parallelogram2. we have triangle AFG congruent to CF1G1 ( case SAS) and triangle CGF1 congruent to AG1F1 ( case SAS)So AG= CG1 and CG= AG1 And AGCG1 is a parallelogram
Problem 1267Is AF_1=CF(O=incircle, O_1=excircle corresponding to AC) or AE_1=AF_1=CF=CD. But triangleAE_1G_1 is similar with triangle G_1D_1C then E_1G_1/G_1D_1=AE_1/CD_1=DC/CD_1 so than vice versa Thales' theorem we haveE_1D//G_1C. Also triangle AEG is similar with triangle GDC so EG/GD=AE/DC=AE/AE_1then AG//DE_1.Terefore AG//CG_1.Similar CG//AG_1 so AG_1CG is parallelogram.If AC Intersects GG_1 in M then G_1M=MG and F_1M=MF (AF_1=FC) so F_1G_1FG isparallelogram.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
With the usual notation, 2s = a+b+c,AE1= AF1 = CF =CD = s-c andAE = AF = CF1 =CD1 = s-aFrom Problems 1265 and 1266 we have<E1AG = < G1CD1 and < EAG = < DCG, so < GAG1 = GCG1Further AG1/CG1 = AF1/CF1 = (s-c)/(s-a) = CF/AFHence Tr.s GAG1 and GCG1 are similar. However the side GG1 is common tothese 2 triangles. Hence they must also be congruent and therefore AGCG1 is aparallelogram.Hence Tr.s AF1G1 ≡ CGF (SAS), so AG1 = GCSimilarly Tr.s AF1G ≡ CFG1 (SAS), so AG = CG1So in quadrilateral FGF1G1, the opposite sides are equal, hence the sameis a parallelogram.Sumith PeirisMoratuwaSri Lanka