In a parallelogram ABCD, the bisector of angle A also bisect BC at X, prove that AB=2AD

Construction : Draw a line from X parallel to AB meeting AD at Y ,

So , Since X is the mid point of BC , therefore Y becomes the mid point of AD

Now Since XB || YA and XB = YA , So ABXY is a parallegram ,

Also Diagonal AX bisects angle A , Thus a parallegram where diagonals bisect the angles is a rhombus ,

Thus ABXY is a rhombus , which implies AB = AY = AD/2

2AB = AD