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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 1258.
Let GE = p, EF = q and FG = rLet AG = AF = BD = yTr.s GEF and BCD are similar so, p/4 = q/x = r/y .........(1)Applying Ptolemy to cyclic quad AGEF, py + qy = 5r and hence from (1). 4r + xr = 5r which yields x = 1Sumith PeirisMoratuwaSri Lanka
https://goo.gl/photos/ffCv5yAwZMzaTRv7AConnect FG, EFExtend BC to H such that CH=CD=x ( see sketch)We have AG=AF => ∠ (AGF)= ∠ (AFG)= uLet ∠ (GAE)= w => ∠ (BDC)= ∠ (BAC)= ∠ (GFE)= wLet ∠ (CAD)= ∠ (DBC)=vSince ∠ (DCH)= ∠ (BAD) => Isoceles triangles GAF simillar to triangle HCD ( case AAA)=> ∠ (CDH)= uTriangle AFE congruent to BDH ( case ASA)So AE= BH=> 5=4+x => x=1
Problem 1258 Let the point K on the side AE (no extension) such that AK=BC=4.So KE=1,triangle AFK=triangle BDK(AF=BD,AK=BC,<FAK=<DBC).So CD=KF then <AKF=<BCD=180-<FAG=180-<FKE. So <FAG=<FKE .But <FGA=<GFA=<FEA=<FEK,so <KFE=<KEF(<KFE=180-<FKE-<FEK=180-<FAG-<FGA=<AFG).So KF=KE=CD=1.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE