Proposed Problem

Click the figure below to see the complete problem 389 about Triangle, Parallel lines, and Harmonic Mean.

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Complete Problem 389

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Sunday, November 15, 2009

### Problem 389. Triangle, Parallel lines, Harmonic Mean

Labels:
harmonic mean,
parallel,
triangle

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ReplyDeleteA' such that GA'//AC and C' such that HC'//AC. EC/BC=n/(n+m) and FA/BA=m/(n+m). AC'/AF=n/(n+m) and CA'/EC=m/(n+m). Multiplying proportions we get that AC'/BA=nm/(n+m)=CA'/BC which proves GH parallel to AC and C'GHA' to be a line. Then GH/m=n/(n+m).

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