Given: KLMN is a parallelogram, KA − angle bisector of ∠K LA − angle bisector of ∠L Prove: m∠KAL = 90°

A parallelogram is a four sided closed figure whose opposite sides and both congruent and parallel.

Parallelogram KLMN (see diagram) has KL parallel to NM. We are asked about angle KAL so le's call its measure z.

KA bisents angle K which means that it cuts it into two equal parts. As a result angle LKA and angle NKA are equal in measure. I call the measure of each of these x.

LA bisects angle L so it cuts angle L into two equal parts. That is, angle KLM and angle ALM are equal in measure. I call the measure of each y.

Triangle KLA contains angles measuring x, y and z. Since it is a triangle the sum of its interior angles is 180 degrees. That means x + y + z = 180.

In a parallelogram the consecutive angles sum to 180 degrees.

If you want to know why this is true it is because KN and LM are parallel (opposite sides of a parallelogram) and KL is a transversal (cuts through the parallel lines - it might help to extend KN and LM to see this better and also to turn your paper so that LM is at top). Angles K and L are interior angles on the same side of the transversal and so sum to 180 degrees.

The above means that angles K and L sum to 180 degrees. So, x + x + y + y = 180. That is 2x + 2y = 180 and 2(x+y) = 180 so x+y = 90 degrees.

from before we know that x + y + z = 180 degrees. Since we know that x + y = 90 degrees that means 90 + z = 180 so z = 90 degrees.

As such we have just shown that angle KAL measures 90 degrees as asked.