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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 .

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Home | Search | Geometry | Problems | All Problems | Open Problems | Visual Index | 1101-1110 | Right Triangle | Square | Triangle & Squares | Congruence | Similarity | Harmonic Mean | Email | Post a comment or solutionLast updated: Apr 10, 2015 by Antonio Gutierrez