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

Asked by  | 11th Feb, 2012, 10:53: AM

Expert Answer:

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

Answered by  | 13th Feb, 2012, 11:20: AM

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