## Wednesday, February 8, 2012

### Problem 726: Isosceles Triangle, Perpendicular, Midpoint, 90 Degrees

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 726 details.

1. http://img860.imageshack.us/img860/4517/problem726.png
Draw FM //BC , M on AC ( see picture)
Triangle AEC and AFM are isosceles => AF=FM
Triangle FMH congruence to triangle GCH (case ASA)
So CG=FM=AF . Note that EC perpen. To BC
Triangle FAE congruence to triangle GCE ( Case SAS)
So FE=FG => Triangle FEG is isosceles and EH perpendicular to FG

2. By considering triangle FBG and collinear points A-H-C, apply menelaus theorem
(BC/CG)*(GH/HF)*(FA/AB) = 1
As 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.

3. Extend AC to K such that AH = HK
AGKF 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, EC
Rt ∆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°

4. For figure of Problem 726:
http://img546.imageshack.us/img546/7070/gogeometryproblem726.pdf

5. 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