Sunday, November 15, 2009

Problem 389. Triangle, Parallel lines, Harmonic Mean

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

Problem 389. Triangle, Parallel lines, Harmonic Mean.
See also:
Complete Problem 389
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 5:11 PM
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2 comments:

  1. AnonymousNovember 17, 2009 at 8:13 AM

    http://s13.radikal.ru/i186/0911/ea/50312a5a2246.jpg

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  2. Ivan BazarovFebruary 22, 2014 at 6:41 PM

    A' 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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