The bisectors of angles A and D of parallelogram ABCD meet at point M

The bisectors of angles A and D of parallelogram ABCD meet at point M lying on side BC. Find the sides of the parallelogram if its perimeter is 36 cm.

Consider all the angles obtained as a result of the intersection of the bisectors AM and DM at point M.

<BAM = <MAD, since the bisector AM divides <BAC into two equal angles, and <MAD = <BMA, as they lie in a cross with parallel AD and BC.

And since the angles in the triangle ABM at the base of AM are equal, then the lateral sides AB = BM.

Also consider the angles of the triangle MCD: <ADM = <MDC = <CMD, and the triangle MCD has equal sides MC = CD.

We got that BC = AD = BM + MC = AB + AB = 2 * AB.

Perimeter p = AB + BC + CD + AD = 6 * AB = 36 cm.

AB = CD = 36/6 = 6cm, BC = CD = 2 * 6 = 12cm.