in a parallelogram ABCD ,the bisector of angle A also bisects BC at X. find AB:AD.
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Given,
In parallelogram ABCD ,
Bisector of ∠A bisects BC at X
∵ AD││BC and AX cuts them so
∠DAX = ∠AXB (alternate angles)
∠DAX = ∠XAB (AX is bisector of ∠A)
∴∠AXB = ∠XAB
AB= BX (sides opposite of equal angles)
Now,AB/AD= BX/BC
=AB/AD=BX/2BX
(since,X is mid point of BC)
=AB/AD=1/2
Hence,
AB:AD=1:2
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