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

Solution:

Here, ABCE is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of the opposite angles is 180°.

So, ∠AEC+∠CBA = 180°

As ∠AEC and ∠AED are linear pair,

∠AEC+∠AED = 180°

Or, ∠AED = ∠CBA … (1)

We know in a parallelogram; opposite angles are equal.

So, ∠ADE = ∠CBA … (2)

Now, from equations (1) and (2) we get,

∠AED = ∠ADE

Now, AD and AE are angles opposite to equal sides of a triangle,

∴ AD = AE (proved).