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 Proof requested (Posted on 2016-04-14)
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.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 simplified geometric proof Comment 3 of 3 |
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 2016-04-15 13:59:05

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