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. (By the converse of Thale’s theorem)
⇒ AD/DB = AE/EC ⇒ DE||BC
∠ABC = ∠ADE (corresponding angles)
⇒ ∠ABC + ∠BDE = ∠ADE + ∠BDE (Adding ∠BDE on both sides)
⇒ ∠ABC + ∠BDE = 180°
⇒ ∠ACB + ∠BDE = 180° (∵ AB = AC ∴ ∠ABC = ∠ACB)
Again DE || BC
⇒ ∠ACB = ∠AED
⇒ ∠ACB + ∠CED = ∠AED + ∠CED (Adding ∠CED on both sides).
⇒ ∠ACB + ∠CED = 180° and
⇒ ∠ABC + ∠CED = 180° (∵ ∠ABC = ∠ACB)
Thus BDEC is a quadrilateral such that
⇒ ∠ACB + ∠BDE = 180° and
⇒ ∠ABC + ∠CED = 180°
∴ BDEC is a cyclic quadrilateral. Hence B, C, E, and D are concyclic points