See complete Problem 95

Similar Triangles, Inradii, Parallel. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Geometry Problem 95

Labels:
incenter,
incircle,
inradius,
parallelogram,
similarity,
triangle

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Just saw the problem on the Mathematica Blog and followed the Mathematica Demo at

ReplyDeletehttp://demonstrations.wolfram.com/TheRadiiOfFourIncircles/

So I wondered why it was true. Placing the radii of the incircles to the tangent points on BC and using similar triangles makes it clear:

r/BC=r1/BE =r2/EG =r3/GC=(r1+r2+r3)/(BE+EG+GC).

Since (BE+EG+GC=BC), r = r1+r2+r3.