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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 1266.
https://goo.gl/photos/ywDg9feFQ4TrtjkP6Let ED meet AC at P.Apply Menelaus’s theorem in secant EDP of triangle ABCPC/PA x EA/EB x DB/DC= 1Replace EB=DB in above expression and simplify it . we get PC/PA=DC/EA= FC/FASo (ACFP)= -1 Apollonius circle with diameter FP will pass through GSo GF is an angle bisector of angle AGC
If 2s = a+b+c with the usual notation, AE = AF = s-c and CF = CD = s-c. If FD = m and FE = n and if X is the midpoint of FD then Tr.s GXD and EAF are similar being isoceles and having equal angles A/2 Hence (m/2)/GD = (s-c)/n ....(1)Similarly (n/2)GE = (s-a)/m....(2)From (1) and (2) we have (s-a)/(s-c) = GD/GE and since < BED = < BDE ( each angle = 90-B/2), Tr.s AEG and CDG are similar.So < AGE = < CGD, hence GF bisects < AGCSumith PeirisMoratuwaSri Lanka
Problem 1266Is triangleFEG similar triangleCOD(<CDO=90=<FGE,<FEG=<FOD/2=<COD) so FG/CD=EG/OD.But triangleFGD is similar with triangle AEO then FG/AE=GD/EO. By dividing by members we have AE/CD=EG/GD(OE=OD). Is BE=BD so <AEG=<CDG, therefore triangleAEG is similar with triangle CDG.So<AGE=<CGD.Then <FGA=<FGC (as complementary angles ).APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE