[Math] Diagonals of the rectangle formed by the angle bisectors of a parallelogram

Question

Let $ABCD$ be a parallelogram. I proved that the angle bisectors of $A$, $B$, $C$, $D$ form a rectangle. How can I prove that the diagonals of this rectangle are parallel to the sides of $ABCD$? And is there a relation between the length of these diagonals and $AB$ or $BC$?

I'm looking for an elementary solution only using parallelograms, congruent triangles.

Answer 1

Let the angle bisectors at $A$ and $B$ meet at $P$. Drop perpendiculars from $P$ to points $A^\prime$, $B^\prime$, $Q$ on the sides of the parallelogram as shown:

Clearly, we have constructed a few similar right triangles, and, in particular, two pairs of congruent right triangles. We see that $P$ must be halfway between opposite sides of the parallelogram; likewise for $P^\prime$. This guarantees the desired parallelism property. $\square$

For a relation about the lengths, lop-off the trapezoid on one side and paste it to the other, getting a rectangle whose width is equal to the original base of the parallelogram, $\overline{AD}$.

Then, for the configuration shown (where $|AD|>|AB|$):

$$|AD| = a + b + d = |AB| + d \qquad\to\qquad d = |AD| - |AB|$$