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

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#### Solution

It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.

∠AEC + ∠CBA = 180°

∠AEC + ∠AED = 180° (Linear pair)

∠AED = ∠CBA ... (1)

For a parallelogram, opposite angles are equal.

∠ADE = ∠CBA ... (2)

From (1) and (2),

∠AED = ∠ADE

AD = AE (Angles opposite to equal sides of a triangle)

Concept: Cyclic Quadrilateral

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