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

Home > Shapes > Geometry
Proof requested (Posted on 2016-04-14) Difficulty: 3 of 5
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

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution 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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information