For the parallelogram, Prove $XY=CD$

Question

$ABCD$ is a parallelogram. The bisectors of $\angle A$ and $\angle B$ meet BC and AD at X and Y respectively. Prove that $XY=CD$?

Please give me some hint to prove it. I can't initiate the problem so unable to show any work.

Answer 1

Hint:

• A line that crosses some other two parallel lines forms the same angle with both.
• Assume that $\angle A$ has some value $\alpha$ and then calculate all the other angles with respect to that.
• Vertices $ABXY$ form a special kind of polygon.

I hope this helps $\ddot\smile$

Answer 2

HINT.

As angles $\angle A$ and $\angle B$ are supplementary, it follows that their bisectors are perpendicular.