D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.

Join DE.

In ∆ABC,

AB = AC = x (let) [Given]

So, ∠B = ∠C [Angles opposite to equal sides are equal]

Similarly,

In ∆ADE,

AD = AE = y (let) [Given]

So, ∠ADE = ∠AED [Angles opposite to equal sides are equal]

Now, in ∆ABC we have

$\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{y}{x}$.

Hence, DE || BC [By converse of BPT]

So,

⇒ ∠ADE = ∠B [Corresponding angles are equal]

⇒ (180° - ∠EDB) = ∠B

⇒ ∠B + ∠EDB = 180°

∠B = ∠C [Proved above]

So,

⇒ ∠C + ∠EDB = 180°

Thus, opposite angles are supplementary.

Similarly,

⇒ ∠B + ∠CED = 180°

Since, sum of opposite angles of cyclic quadrilateral = 180°.

Hence, proved that B, C, E and D are concyclic.