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Geometry Problem 1108: Right Triangle, External Squares, Catheti, Cathetus, Congruence, Harmonic Mean. Level: High School, SAT Prep, College Geometry

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The figure shows a right triangle ABC and the external squares ABDE and BCFG. AF meets BC at M and CE meets AB at N. Prove that BM = BN = one-half the harmonic mean of AB and BC that is Problem 1108 Formula to prove.
 
 

Geometry Problem 1108 Right Triangle, External Squares, Catheti, Congruence, Harmonic Mean
 
 
 

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Last updated: Apr 10, 2015 by Antonio Gutierrez