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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 

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)) [1]
  BC^2 = CM^2 + BM^2 - 2*CM*BM*cos(angle(CMB)) [2]

Adding [1] and [2], 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 

Best regards,
- Doctor Floor, The Math Forum
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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