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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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...
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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) read less
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