Online Geometry theorems, problems, solutions, and related topics.
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 1265.
https://goo.gl/photos/mniwJNrj8t6jnXvG7Let 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
Problem 1265Is BE=BD then <BED=<BDE or <AEG=<CDG.But <FOC=<FOD/2=<FEG and <CFO=90=<FGE soTriangle EGF is similar the triangleOFC so EG/OF=GF/FC similar GD/OF=GF/AF with division by states we have EG/GD=AF/FC.So triangleAEG is similar triangleGDC then <EGA=<DGC.Therefore <AGF=<FGC (as equal angles supplements).APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
Really excellent work Apostolis
Thank you for your nice words Sumith.
Trigonometry solutionEG = 2.AE.sin(A/2).sin(C/2)GD = 2.CD.sin(C/2).sin(A/2)So EG/GD = AE/CDHence Tr.s AEG and CDG are similar and the result followsSumith PeirisMoratuwaSri Lanka
Geometry SolutionLet M be the midpoint of EF and let EF =m, FD =n, EG = p and GD = qTr.s EGM and DCM are similar (isosceles and both having <C), so(m/2)/p = (s-c)/n ….(1) where 2s = a+b+c with the usual notation.Similarly we can show that(n/2)/q = (s-a)/m…(2)(2)/(1) gives, p/q = (s-a)/(s-c)Hence Tr.s AEG and CDG are similar and the result follows.Sumith PeirisMoratuwaSri Lanka
To prove GF is angle bisector of AngleAGC, it would suffice to prove that the triangles AEG and CDG are similar. Let A,B,C be the angles of the triangle ABCJoin EF and form the isosceles triangle AEF. Similarly join FD and form the isosceles triangle CDF.In triangle AEF the angles are A, 90-A/2, 90-A/2=> Angle GDF = 90-A/2 ( per Ext Segment theorm) => AngleGFD = A/2 Similarly AngleGFE = C/2Join OE and consider the triangle OAE. The Angles in the triangle are 90,A/2 & 90-A/2 which is similar to the triangle GFDHence AE.GD = OE.GF -------------(1)Similarly join OD and form the triangle ADC whose angles are 90,C/2 & 90-C/2, which is similar to the triangle GFEHence EG.DC = OD.GF => EG.DC = OE.GF (Since OE=OD) ---------(2)FROM (1) & (2) AE.GD = EG.DC => Triangles AEG and CDG are similar Therefore AG/AE = CG/CD => AG/AF = CG/CF ----------- (3)Hence (3) is the result of Angle bisector theorem and GF is the bisector of Angle AGC