In parallelogram ABCD, the bisectors of angles A and B meet at a point on the side CD. Find the perimeter of ABCD if AB = 10.

Decision.
Let the bisector of angle A – AF, the bisector of angle B – BF. Then the point F is the intersection point of the bisectors lying on the side DC.
ABCD is a parallelogram hence: AB = CD = 10, AD = BC and opposite sides are parallel.
AB is parallel to CD, so the angle FAB is equal to the angle DFA and the angle ABD is equal to the angle CFB as criss-crossing angles for parallel lines.
Triangle ADF is isosceles in two corners, AD = DF. Triangle FCB is isosceles in two corners, FC = BC.
P = AB + BC + CD + DA = 10 + FC + DF + 10 = 20 + DC = 20 + 10 = 30.
Answer: 30.