Bisectors of adjacent angles of a parallelogram meet on midline?

Question

Suppose $KLMN$ is a parallelogram, and that the bisectors of angle $K$ and angle $L$ meet at point $A$. Prove that $A$ is equidistant from $\overline{LM}$ and $\overline{KN}$, without using trigonometry.

Answer 1

Drop perpendicularly from $A$ to $KN, KL, LM$. You should find two pairs of congruent triangles.

Answer 2

Drop perpendiculars from A to the sides and let them be AB and AC as shown in the figure. The actual angle values are not important except those shown as 90 degrees.

First show that angle at A is 90 degrees and hence conclude that AKB, AKL, ALC are all similar and use similarity to prove your result.