D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD=AE.Prove that B,C,D,E are cocyclic
Asked by
| 9th Feb, 2014,
11:43: AM
Expert Answer:

In ΔABC,
∠ABC = ∠ACB (Angles opposite to equal side are equal)
⇒ ∠DBC = ∠ ECB .....(1)
Now,
AB = AC and AD = AE
⇒ AB – AD = AC – AE
⇒ DB = EC
⇒ DE || BC (B.P.T.)
Now DE || BC and BD is the transversal
⇒ ∠DBC + ∠BDE = 180° ....(2) (Sum of adjacent angles is supplementary)
From (1) and (2), we get
∠ECB + ∠BDE = 180°
⇒ BCED are concylic (Opposide angles of a cyclic quadrilateral are supplementary)
In ΔABC,
∠ABC = ∠ACB (Angles opposite to equal side are equal)
⇒ ∠DBC = ∠ ECB .....(1)
Now,
AB = AC and AD = AE
⇒ AB – AD = AC – AE
⇒ DB = EC
⇒ DE || BC (B.P.T.)
Now DE || BC and BD is the transversal
⇒ ∠DBC + ∠BDE = 180° ....(2) (Sum of adjacent angles is supplementary)
From (1) and (2), we get
∠ECB + ∠BDE = 180°
⇒ BCED are concylic (Opposide angles of a cyclic quadrilateral are supplementary)
Answered by
| 10th Feb, 2014,
11:43: AM
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