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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 711 details.
Let E' be the midpoint of AD. Join EE'BA ∥ EE' ∥ CD∆s ABE and DCE are congruent. So ∠CDE = ∠BAE∠AED = ∠AEE' + ∠DEE' = ∠BAE + ∠CDE = 2∠BAE.Similarly ∠AFB = 2∠DAF is proved.For convenience denote ∠AED = 2x, ∠AFB = 2ySo ∠BAG = x, ∠ABG = 90° - y∴ ∠EGN = ∠BGA = 90° + y - x∠GNH = ∠GEH + ∠EGN = 2x + 90° + y - x = 90° + x + yNext ∠EAF = 90° - ∠BAE - ∠DAF = 90° - x - yHence ∠GNH + ∠EAF + 180°and A, G, N, H are concyclic
http://img249.imageshack.us/img249/3880/problem711.pngDraw EX//BA and FX//BC ( see picture)1, 2. We have ( AEX)=(BAE)= a and (AFX)=(FAD)= b ---- (properties of // lines)EX and EY are angle bisectors of (AED) and (AFB)So (AED)=2(BAE)= 2a and (AFB)=2(DAF)=2b3. From the picture we have (EAF)= 90-a-b -------(1)(FNY)=b+(YXF)(EXY)=a+(YXE)(FNY)+(EXY)= a+b+(EXF)=90+a+b= ( GNH) ---- (2)Add (1)+(2) side by side : (EAF)+(GNH)=1804. and A,G,N,H are concyclic
1 and 2 are easily proved considering congruent triangles. Let FIH be parallel to AD such that H is on AE and I is on DE< EAF + < AFH = < EHI = HIE = < DNF + NFI. Now FH bisects < AFB, hence < EAF = DNF from which results 3 and 4 follow easily. Sumith PeirisMoratuwaSri Lanka
Problem 711The AF intersects BC at P,then <FPB=<FBP and <HEP=<AEB. Is <NHA=<DHP=<DEP+<HPB=<AEB+<FBE=<GEB+<GBE=<NGE. Therefore A,G,N,H are concyclic. MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE