## Saturday, August 25, 2012

### Problem 799: Parallelogram, Angle Bisector, Parallel, Triangle, Circumcircle, Circumcenter

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

Click the figure below to see the complete problem 799.

1. http://img19.imageshack.us/img19/9383/problem799.png

Draw angles and lines per attached sketch
We have ∆ABE and ∆ECF are isosceles triangles
QE=QC (Q is circumcenter of ∆CEF)
AB=CD=BE
∠EQH=∠CQH=∠EFC= α
∠ECD=2 α ( external angle of ∆ ECF)
∠QCD= α +90 and ∠QEB=external angle of EQH= α +90
So ∆BQE congruence to ∆DQC ( Case SAS) =>∠ BQD=2 α = ∠BCD
So quadrilateral BDCQ is cyclic

2. Our target is to prove that B,Q,C,D concyclic.

Let x be the angle FAD (=angle BAE = angle BEA = angle FEC = EFC) [from previous problem797]

Since EFC is isos., QC is the perpendicular bisector of EF and angle QCE = 90-x. Then angle QCD = 90 + x.
Now observe that
QC = QE
BE = BA = CD
Angle QCD = 90 + x = 180 - (90+x) = 180 - (angle QEC) = angle QEB

Hence triangle QEB is congruent to triangle QCD.

Therefore angle QBC = angle QBE = angle QDC.
By the converse of angle in the same segment, it is proved.

3. ∆'s ABE, CEF are isosceles and CQ ⊥ bisector of EF
imply ∠QEB = Π - ∠QEC = Π - ∠QFC = Π - ∠QCF = ∠QCD
So ∆QBE ≡ ∆QDC, ∠QBC = ∠QBE = ∠QDC
Hence Q, B, D, C are concyclic

4. Also BD is tangential to circle BEQ