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Given a generic parallelogram $ABCD$ as in the figure, we can extend the angle bisectors to get a central quadrilateral $EFGH$. Now it is easy to show that $EFGH$ is a rectangle using $\alpha + \beta = 90^\circ$. My question is, how do we show it is not always a square and precisely when is it a square? Diagram of parallelogram

I tried showing that $EG$ is parallel to $AD$, but could not figure any way to prove that. Directly showing $EF \neq FG$ if $AB \neq BC$ did not work either. Any hints (without using trigonometry, coordinate geometry, vectors etc. but only geometry)?

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  • $\begingroup$ $$\dfrac{EF}{FG}=\dfrac{\sin \beta}{\sin \alpha},$$ so $EFGH$ can be square only when $\alpha=\beta=45^\circ$, other words: when $ABCD$ is a rectangle. $\endgroup$ – Oleg567 Sep 10 '14 at 16:25
  • $\begingroup$ @Oleg567 I don't know how you got that, but do you have a solution without trigonometry, as mentioned, if possible. $\endgroup$ – Nemo Sep 10 '14 at 16:27
  • $\begingroup$ You can use ratios of cathetus/hypotenuse instead of $\sin$ or $\cos$. $\endgroup$ – Oleg567 Sep 10 '14 at 16:35
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enter image description here

Angles in gray triangles are $\alpha, \beta$ and $90^\circ$. Catheti (green and red) are equal, when $\alpha=\beta$.

So, central rectangle can be square when $\alpha=\beta ~~$ too.

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There is a solution but impliying trigo :

$HE = HB - EB = BC*cos(\beta) - AB*cos(\beta) = (BC-AB)*cos(\beta)$ $HG = HC - GC = BC*cos(\alpha) - CD*cos(\alpha) = (BC-CD)*cos(\alpha)$
But $CD = AB$
So $HE = HG \Leftrightarrow cos(\alpha) = cos(\beta) \Leftrightarrow \alpha = \beta$ ($45°$ since $\alpha + \beta = 90°$)

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  • $\begingroup$ Thanks, will wait for any answer without trigonometry for a while though. $\endgroup$ – Nemo Sep 10 '14 at 17:18
  • $\begingroup$ No problem. I think Oleg567 got it without trigo. $\endgroup$ – Deuteu Sep 10 '14 at 20:29

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