In fig., ABC is a triangle in which AB = AC. D and E are points on the sides AB and AC respectively such that AD = AE. Show that the points B, C, E and D are concyclic.
                                                     

Given: In ΔABC, AB = AC. D and E are points on the sides AB and AC respectively such that AD = AE.
To Prove: Points B, C, E and D are concyclic.
Proof: In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that ∠ABC + ∠CED = 180° and
∠ACB + ∠BDE = 180°.
In ΔABC, we have
AB = AC and AD = AE
AB - AD = AC - AE
DB = EC
Thus, we have
AD = AE and DB = EC.

DE || BC                                               [By the converse of Thale's Theorem]
∠ABC = ∠ADE                                         [Corresponding angles]
∠ABC + ∠BDE = ∠ADE + ∠BDE                   [adding ∠BDE on both sides]
∠ABC + ∠BDE = 180° 
∠ACB + ∠BDE = 180°                                  [Since AB = AC Therefore ∠ABC = ∠ACB]
Again DE||BC
∠ACB = ∠AED
∠ACB + ∠CED = ∠AED + ∠CED                   [Adding ∠CED on both sides]
∠ACB + ∠CED = 180° 
∠ABC + ∠CED = 180°                                        [Since ∠ABC = ∠ACB]
Therefore BDEC is a cyclic quadrilateral.
Hence, B, C, E and D are concyclic points.