Proof of the Parallelogram Law
Date: 08/07/99 at 18:49:14 From: Kenneth Subject: Proof of the Parallelogram Law How do you prove the parallelogram law geometrically without using vectors? Given: triangle ABC with midpoint M on side AC Prove: (AB)^2 + (AC)^2 = (AM)^2 + (BM)^2 I was reading a book on vector calculus that gave the vector interpretation and proof of the parallelogram law, which states |vector A + vector B|^2 + |vector A - vector B|^2 = 2|vector A|^2 + 2 |vector B|^2. I tried interpreting the vector form geometrically and came up with the above question. I have tried drawing a line parallel to side AC from B, intersecting the extension of AB at P. Then I tried using ratios of line segments when intersected by parallel lines but I am still getting nowhere.
Date: 08/08/99 at 03:41:14 From: Doctor Floor Subject: Re: Proof of the Parallelogram Law I am afraid that you mistyped the parallelogram law. It should be, in your setting: AB^2 + BC^2 = 2*AM^2 + 2*BM^2 I do not know a purely geometric proof, but I know a trigonometric proof that is fairly easy: Note that angle(AMB) + angle(CMB) = 180 degrees, so cos(angle(AMB)) = -cos(angle(CMB)) Now we can twice use the law of cosines: AB^2 = AM^2 + BM^2 - 2*AM*BM*cos(angle(AMB))  BC^2 = CM^2 + BM^2 - 2*CM*BM*cos(angle(CMB))  Adding  and , and realizing that AM = CM, we find the parallelogram law. By the way, to find the reason for the name, draw a line through C parallel to BM, a line through B parallel to AC, and a line through M parallel to AB. The three lines intersect in a point D, and MBDC is a parallelogram. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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