 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Mutually Perpendicular Medians (Posted on 2008-06-14) 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) Comments: ( Back to comment list | You must be logged in to post comments.) 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 Please log in:

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