Geometry Problem

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

Click the figure below to see the complete problem 799.

## Saturday, August 25, 2012

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

Labels:
angle bisector,
circumcenter,
circumcircle,
parallel,
parallelogram,
triangle

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

ReplyDeleteDraw 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

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

ReplyDeleteLet 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.

∆'s ABE, CEF are isosceles and CQ ⊥ bisector of EF

ReplyDeleteimply ∠QEB = Π - ∠QEC = Π - ∠QFC = Π - ∠QCF = ∠QCD

So ∆QBE ≡ ∆QDC, ∠QBC = ∠QBE = ∠QDC

Hence Q, B, D, C are concyclic

Also BD is tangential to circle BEQ

ReplyDelete