On the sides of parallelogram ABCD erect squares  all either on the outside or the inside of the parallelogram.
Prove that their centers form a square.
Since the opposite sides of a parallelogram are equal the opposite squares are congruent.
Call the centers of the externally erected squares starting at side AB by E, F, G, H.
Form the four triangles AEH, BEF, CGF and DGH.
Suppose angle A is acute with measure x then angle D = 180x.
angle EAH = x+90
angle GDH=180x+90=270x which is more than 180 so subtract from 360 to get
angle GDH = x+90
By SAS the four triangles are congruent.
So EF=FG=GH=HE (the centers form a rhombus)
Consider angle GHE = DHA +DHG  AHE.
Since DHA is right and DHG=AHE, GHE is a right angle so the rhombus is a square.
If the erected squares were internal, the angles become A = x90 = 90x
and
D = 180x  90 = 90x
and the rest of the proof is the same.

Posted by Jer
on 20140102 14:37:05 