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?
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)?
Best Answer
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.