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
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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