Mathematics
ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2AD.
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Answer
ABCD is a parallelogram in which bisector of ∠A and ∠B meets DC at E.
To prove: AB = 2AD
Since, AE and BE are bisector of ∠A and ∠B
∠1 = ∠2 and ∠3 = ∠4
In parallelogram ABCD, we have
AB || DC
∠1 = ∠5 [Alternate angles are equal, AE is transversal]
Thus,
∠2 = ∠5 … (i)
∴ DE = AD [∵ Sides opposite to equal angles in ∆AED]
∠3 = ∠6 [Alternate angles]
∠3 = ∠4 [Since, BE is bisector of ∠B (given)]
Thus, ∠4 = ∠6 … (ii)
∴ BC = EC [∵ Sides opposite to equal angles in ∆BCE]
AD = BC [Opposite sides of || gm are equal]
AD = DE = EC
AB = DC [Opposite sides of a || gm are equal]
⇒ AB = DE + EC
⇒ AB = AD + AD
⇒ AB = 2AD
Hence, proved that AB = 2AD.
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