# Area of square created by intersection of segments from a square vertexes to their opposite sides

There will be an square created when we draw segments from a square vertexes to their opposite sides' middle.

What is the relation between smaller square's area and the side length of the bigger one?

• Are the sides divided into equal lengths? – Varun Iyer Oct 20 '14 at 12:08
• yes, I'm sorry I forgot that! – MTP1376 Oct 20 '14 at 12:09
• The area of the square is one fifth the area of the larger square – Varun Iyer Oct 20 '14 at 12:17
• is there a clear reason for "line segment from D to DE∩CH is also a"? – MTP1376 Oct 20 '14 at 12:48
• that is not correct. I will post my solution late but the length is not equal to a – Varun Iyer Oct 20 '14 at 12:51

Let $a$ be the side length of the grey square (and $1$ the side lengthg of the original square). By similarity, the length of the line segment from $D$ to $DE\cap CH$ is also $a$. Then the triangle with base $AE$ complete the quadrilateral with top edge $DE$ to a square of area $a^2$. We can do the same with the other triangles and conclude that $1^2= 5a^2$.