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# ABCD is a parallelogram. The circle through $ \mathrm{A}, \mathrm{B} $ and $ \mathrm{C} $ intersect $ \mathrm{CD} $ (produced if necessary) at $ \mathrm{E} $. Prove that $ \mathrm{AE}=\mathrm{AD} $.

Given:

ABCD is a parallelogram. The circle through \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) intersect \( \mathrm{CD} \) (produced if necessary) at \( \mathrm{E} \).

To do:

We have to prove that \( \mathrm{AE}=\mathrm{AD} \).

Solution:

$ABCE$ is a cyclic quadrilateral.

We know that,

In a cyclic quadrilateral, the sum of the opposite angles is $180^o$.

This implies,

$\angle AEC+\angle CBA = 180^o$.........(i)

$\angle AEC+\angle AED = 180^o$.........(ii) (Linear pair)

From (i) and (ii), we get,

$\angle AED = ∠CBA$….....(iii)

$\angle ADE = \angle CBA$….........(iv) (Opposite sides of a parallelogram are equal)

From (iii) and (iv) we get,

$\angle AED = \angle ADE$

We know that,

Sides opposite to equal angles of a triangle are equal.

$AD$ and $AE$ are angles opposite to equal sides of a triangle.

Therefore,

$AD = AE$

Hence proved.

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