Prove the following theorem:
Given any parallelogram, construct on its sides four squares
external to the parallelogram  the quadrilateral formed by joining the centers
of those four squares is a square.
Simple details left out for brevity.
Let the parallelogram be ABCD and centers of the four squares be E on AB, F on BC, G on CD and H on DA. The quadrilateral in question is EFGH.
It's not hard to show the triangles HAE, FBE, FCG, HDG, are congruent by SAS (the included angle is A+90)
All the sides of the quadrilateral are therefore equal and
Angle HEF = AEB  AEH + BEF = 90 etc.
So it is a square.

Posted by Jer
on 20160415 13:59:05 