All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 all squares (Posted on 2014-01-01)
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.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Proof Comment 1 of 1
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 = 180-x.
angle EAH = x+90
angle GDH=180-x+90=270-x 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 = x-90 = 90-x
and
D = 180-x - 90 = 90-x
and the rest of the proof is the same.

 Posted by Jer on 2014-01-02 14:37:05

 Search: Search body:
Forums (0)