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Question
In the given figure, AE bisects ∠CAD and ∠B= ∠C. Prove that AE || BC.

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Solution
In the given problem, AE bisects ∠CAD and ∠B = ∠C
We need to prove AE || BC

As, ∠CAD is bisected by AE
∠CAD = 2∠CAE - 2∠DAE ..........(1)
Now, using the property, “an exterior angle of a triangle in equal to the sum of the two opposite interior angles”, we get,
∠CAD ∠B +∠C
∠CAD = 2∠C (∠B = ∠C)
2∠CAE = 2∠C (using 1)
∠CAE = ∠C
∠CAE = ∠ACB
Hence, using the property, if alternate interior angles are equal, then the two lines are parallel, we get,
∠CAE = ∠ACB
Thus,AE || BC
Hence proved.
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