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In Fig., AE bisects ∠CAD and ∠B =∠C. Prove that AE || BC.

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Best answer

Given,

AE bisects ∠CAD

∠B = ∠C

In ΔABC

∠CAD = ∠B + ∠C

∠CAD = ∠C + ∠C

∠CAD = 2∠C

∠1 + ∠2 = 2∠C (Therefore, ∠CAD = ∠1 + ∠2)

∠2 + ∠2 = 2∠C (Therefore, AE bisects ∠CAD)

2∠2 = 2∠C

∠2 = ∠C (Alternate angles)

Therefore, AE ‖ BC

Hence, proved

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