In the parallelogram abcd, the bisector of the angle bad is drawn, which intersects the side bc at point p.

In the parallelogram abcd, the bisector of the angle bad is drawn, which intersects the side bc at point p. a) prove that the triangle abp is isosceles. b) find the side ad if no = 10 cm, and the perimeter of the parallelogram is 52 cm

Since AP is the bisector of the angle, the angle VAR = DAP.

The angle DAP = BPA as the intersecting angles at the intersection are parallel to the lines AD and BC of the secant AM, then the angle BAP = BPA, and therefore the triangle ABP is isosceles with the base AP, which was required to be proved.

Since the ABP triangle is isosceles, then AB = BP = 10 cm, then SD = 10 cm.

Ravsd = 2 * (AD + AB) = 52 cm.

AD + AB = 52/2 = 26 cm.

AD = 26 – AB = 26 – 10 = 16 cm.

Answer: The length of the side AD is 16 cm.