Geometry Problem

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

Click the figure below to see the problem 726 details.

## Wednesday, February 8, 2012

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

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http://img860.imageshack.us/img860/4517/problem726.png

ReplyDeleteDraw 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

By considering triangle FBG and collinear points A-H-C, apply menelaus theorem

ReplyDelete(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.

Extend AC to K such that AH = HK

ReplyDeleteAGKF 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°

For figure of Problem 726:

ReplyDeletehttp://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

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