Q. In the given figure, ABCD is a parallelogram and bisector of angle A bisects the side BC at P. Prove that AD = 2AB.

Soln:
i. <DAP = <PAB      i. AP bisects <A
ii. <DAP = <APB    ii. Being alternate angles
iii. <PAB = <APB   iii. From statements (i) and (ii)
iv. AB = BP            iv. From state. (ii), being sides of an isoceles triangle
v. AB = BC/2          v. P is mid-point of BC
    or,2AB = BC        
vi. 2AB =  AD         vi. Being opposite side of ||gm ABCD.(BC = AD)

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