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In the figure, ABC is a triangle in which AB = AC. Points D and E are points on the side AB and AC respectively such that AD = AE. Show that the points B, C, E and D lie on a same circle.

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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

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