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Mutually Perpendicular Medians (Posted on 2008-06-14) Difficulty: 3 of 5
If in a triangle ABC, medians from A and B are perpendicular to each other, then show that a2+b2=5c2

See The Solution Submitted by Praneeth    
Rating: 2.0000 (3 votes)

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Solution Solution | Comment 2 of 4 |

Let G be the centroid and A' and B' the
midpoints of sides BC and CA respectively.
a^2 + b^2 = |BC|^2 + |CA|^2
          = (|BA'| + |A'C|)^2 + (|CB'| + |B'A|)^2
          = (2|BA'|)^2 + (2|B'A|)^2
          = 4|BA'|^2 + 4|B'A|^2
          = 4(|BG|^2 + |GA'|^2) + 4(|B'G|^2 + |GA|^2)
          = 4[(2|B'G|)^2 + |GA'|^2] + 4[|B'G|^2 + (2|GA'|)^2]
          = 4(4|B'G|^2 + |GA'|^2) + 4(|B'G|^2 + 4|GA'|^2)
          = 5(4|GA'|^2 + 4|B'G|^2)
          = 5[(2|GA'|)^2 + (2|B'G|)^2]
          = 5(|AG|^2 + |GB|^2)
          = 5|AB|^2
          = 5c^2
 

  Posted by Bractals on 2008-06-15 03:30:38
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