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Question
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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Solution
∠B = ∠C ...(Given AB = AC)
AD + DB = AE + EC
BD = EC ...(Given AD = AE)
DE parallel BC Since AEC is a straight line.
∠AED + ∠CED = 180°
∠CBD + ∠CED = 180°
Similarly of the opposite angles = 180°
∴ BCED is a cyclic quadrilateral
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