Geometry Problem

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

Click the figure below to see the complete problem 702.

## Sunday, December 11, 2011

### Problem 702: Triangle, Circumcircle, Parallel, Concyclic Points

Labels:
circumcircle,
concyclic,
parallel,
triangle

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Join G to F. /_EDF=/_EGF & /_DGE=/_DFE so /_DGF=/_DGE+/_EGF=/_EDF+/_DFE=/_DEB=/_ACB.

ReplyDeleteHence etc.

http://img809.imageshack.us/img809/2501/problem702.png

ReplyDeleteConnect GF ( see picture)

Since quadrilateral GDEF cyclic so m(GDE)=m(GFC)

Note that m(GDE) supplement to angle(GAC) ( DE//AC)

So angle(GFC) supplement to angle (GAC)

And A,G,E,F concyclic

Peter Tran

DE//AC ---> angle (BED)=(BCA)

ReplyDeletecyclic quadrilateral DEFG ---> (BED)=(BGF)

Therefore: angle(BGF)=(BCA).

Then the quadrilateral AGFC is cyclic.

Join GF

ReplyDeleteAngle DGF = angle BED = angle BCA

So A,G,F,C are concyclic

Triangle BGF ~ Triangle BED ~ Triangle BCA

ReplyDeleteBG:BC = BF:BA

BG.BA = BF.BC

A, G, F, C are concyclic