Diagonals of the rectangle formed by the angle bisectors of a parallelogram 
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.

 A: 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 $|AD|>|AB|$):

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

A: 
Without loss of generality, we can assume that $AB>BC$. Let's start with proving that $AW=ZB=DU=VC=CD-AD$.In parallelogram, adjacent angles are supplementary so $m\angle{A}+m\angle{D}=180\circ$. From $\triangle{ADV}$ we have: $0.5 \cdot m\angle{A}+m\angle{D}+m\angle{DVA}=180\circ$ which means $m\angle{DVA}=0.5 \cdot m\angle{A}$. Thus, $\triangle{ADV}$ is isosceles and $DV=AD$, $VC=CD-AD$. Similarly, $CU=BC$ and $DU=CD-BC=CD-AD=VC$.
It's easy to show now that $\triangle{AZM} \cong \triangle{DMV}$: $AZ=AD=DV$ and alternative interior angles are congruent. Thus, $MZ=MD$. Similarly, $\triangle{AMD} \cong \triangle{BCS}$ so $BS=MD=MZ$. Because $DZ||BU$, $MSBZ$ is a parallelogram and $MS||AB||DC$. We can now see that $MS=CD-AD=TN$ (diagonals of rectangle are congruent).
