Geometry Problem

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

Click the figure below to see the complete problem 798.

## Thursday, August 23, 2012

### Problem 798: Parallelogram, Secant Line, Tangent Circles, Circumcircles, Triangle

Labels:
circle,
circumcircle,
parallelogram,
secant,
tangent

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Since ABCD is a parrallelogram, GC//EB and therefore angle HGC = angle HEB.

ReplyDeleteSo the tangent lines for the 2 circles must be exactly the same at H.

q.e.d.

http://img577.imageshack.us/img577/6738/problem798.png

ReplyDeleteDraw lines per attached sketch

Note that ∆ HGC and ∆ HED are similar

Consider dilation transformation centered H , scale factor= HC/HD=HG/HE

This transformation , circumcircle of ∆ HGC will become circumcircle of ∆ HED and O will become O’

So H, O and O’ are collinear => circumcircles of 2 triangles will tangent at H