By drawing the angle bisectors of a parallelogram, they intersect in points $E,F,G,H$ and the quadrilateral $EFGH$ is formed. It can be proven that $EFGH$ is a rectangle؛ But I've struggled with proving the length of $EG$ (the rectangle's diameter) in terms of $AB$ and $BC$. I would be thankful if someone helped me.
Find the length of $EG$ in terms of $AB$ and $CD$?
Let's denote the intersection of AB and GH as K. Due to the fact that the intersections of a convex quadrilateral make another quadrilateral the sum of its opposite angles is $180^o$ and that quadrilateral EFGH is rectangle (the sum of its opposite angles is $180^o) so GH is bisector of angle BCD. we have:
$\angle BKG=\angle GCD=\angle BCG$
$\angle HGB=90^o$
Therefore triangles BKG and BCG are equal for a.s.a(two angles equal and one common side which is BG) or we may say BG is the altitude of isosceles triangle KBC. So $BK=BC$ and we have:
$EG=AK=AB-BK=AB-BC$