If in a triangle ABC, medians from A and B are perpendicular to each other, then show that a²+b²=5c²

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