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

I mean the gray area

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

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  • $\begingroup$ Are the sides divided into equal lengths? $\endgroup$ – Varun Iyer Oct 20 '14 at 12:08
  • $\begingroup$ yes, I'm sorry I forgot that! $\endgroup$ – MTP1376 Oct 20 '14 at 12:09
  • $\begingroup$ The area of the square is one fifth the area of the larger square $\endgroup$ – Varun Iyer Oct 20 '14 at 12:17
  • $\begingroup$ is there a clear reason for "line segment from D to DE∩CH is also a"? $\endgroup$ – MTP1376 Oct 20 '14 at 12:48
  • $\begingroup$ that is not correct. I will post my solution late but the length is not equal to a $\endgroup$ – Varun Iyer Oct 20 '14 at 12:51
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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 $1^2= 5a^2$.

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  • $\begingroup$ That's right, but please explain why sum of areas of little rectangles and quadrilaterals is equal to area of the smaller square? (or why is line segment from D to DE∩CH is also a) $\endgroup$ – MTP1376 Oct 20 '14 at 12:22
  • $\begingroup$ could you please? :D $\endgroup$ – MTP1376 Oct 20 '14 at 12:43
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    $\begingroup$ Let intersection of DE and CH be X, DE and AG be Y. Since CH and AG are parallel, and H midpoint of AD, then by similar triangles X is midpoint of DY. $\endgroup$ – Sean Lo Oct 20 '14 at 13:32

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