The bisector of angles A and B of parallelogram ABCD intersects at point M

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

In a right-angled triangle AMB, the sides of which are the height of the rhombus AM and the diagonal of the rhombus AB, <MCA = 180 ° – <AMC – <MAC = 180 ° – 90 ° – 20 ° = 70 °.

<BAC = <MCA = 70 °.

<ABC = 180 ° – <BAC – <MCA = 180 ° – 70 ° – 70 ° = 40 °;

<ADC = <ABC = 40 °;

<BCD = <BAD = <BCA * 2 = 70 ° * 2 = 140 °.

Answer: 2 angles of a rhombus are equal to 40 ° and two angles are equal to 140 °.