Question

ABCD is a parallelogram if the bisector DP and CP of angles D and C respectively meet at P on side AB then show that P is the mid point of side AB

Answer 1

in triangle ADP and triangle BCP,
Angle A = angle B =90 degree(Property of parallelogram)
angle D = angle C(because DP & CP are the bisectors)
AD=CD (Sides of parallelogram)
so, by ASA cong. condition triangle ADP is congruent to triangle BCP
by C. P. C. T. AP =BP

Answer 2

in abd & dcb
ab = dc (opp. side of pallarlogram)
ad = bc (opp. side of pallarlogram)
angle a = c (opp. angle of pallarlogram)abddcb

abd congruent to dcb by S.S.S.rule