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ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2AD.

Rectilinear Figures

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Answer

ABCD is a parallelogram in which bisector of ∠A and ∠B meets DC at E.

To prove: AB = 2AD

ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2AD. Rectilinear Figures, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

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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