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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 726 details.
http://img860.imageshack.us/img860/4517/problem726.pngDraw FM //BC , M on AC ( see picture)Triangle AEC and AFM are isosceles => AF=FMTriangle FMH congruence to triangle GCH (case ASA)So CG=FM=AF . Note that EC perpen. To BCTriangle FAE congruence to triangle GCE ( Case SAS)So FE=FG => Triangle FEG is isosceles and EH perpendicular to FG
By considering triangle FBG and collinear points A-H-C, apply menelaus theorem(BC/CG)*(GH/HF)*(FA/AB) = 1As BC = AB & GH = HF, we have FA = CG.Together with AE = EC, and angle FAE = 90 = angle BCE = angle ECG, we can conclude that triangle AEF is congruent to triangle ECG.So EF = EG, so triangle FEG is isosceles.q.e.d.
Extend AC to K such that AH = HKAGKF is a parallelogram (since diagonals AK, GF bisect each other)So AF = GK∆CGK is isosceles (∠GKC = alternate ∠BAC = ∠BCA = vert opp ∠GCK)So AF = GK = GC (note AF∥GK)Join EF, ECRt ∆AEF ≅ Rt ∆CEG (note AE = CE since E lies on ⊥ bisector of AC)So EF = EG. Already FH = HG.Hence ∆EHF ≅ ∆EHG, ∠EHF = ∠EHG = 90°
For figure of Problem 726:http://img546.imageshack.us/img546/7070/gogeometryproblem726.pdf
By symmetry, <ECG=90 and AE=EC. PF is parallel to AC such that P is on BC. AF=PC because AFPC is an isosceles trapezoid, but PC=CG because triangles GCH and GPF are similar, so AF=CG. Last condition makes triangles EAF and ECG congruent. It follows that FE=EG and EH is perpendicular bisector of FG