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.
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Posted by Jer
on 2016-04-15 13:59:05 |
simplified geometric proof
