Let ABCD be a convex quadrilateral.

Prove that ABCD is a parallelogram
if and only if
|AC|² + |BD|² = |AB|² + |BC|² + |CD|² + |DA|².

Start with the "if":

Call the four vertices of the parallelogram:
A=(0,0) B=(a,0) C=(a+b,c) D=(b,c)
|AC|² + |BD|²
=((a+b)²+c²)+((b-a)²+c²
=2a²+2b²+2c²

|AB|² + |BC|² + |CD|² + |DA|²
=2(b²+c²)+2(a²)
=2a²+2b²+2c²

For the "only if" I'll show if the equation works for a trapezoid, the trapezoid must be a parallelogram.
A=(0,0) B=(a,0) C=((b+x),c) D=(b,c)
|AC|² + |BD|² = |AB|² + |BC|² + |CD|² + |DA|²
((b+x)²+c²)+((b-a)²+c²) = (b²+c²)+x²+((b+x-a)²+c²)+a²
x²-2ax+a²=0
(x-a)²=0
x=a

What I'd like to see is a visual proof such as the three on the wikipedia page
http://en.wikipedia.org/wiki/Pythagorean_theorem

Posted by Jer on 2013-07-09 11:00:04