Question

13. In the given figure, P is any point on the chord BC of a circle such that AB = AP. Prove
that CP =CO

Answer 1

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We have to prove that  CP = CQ  i.e.,

ΔCPQ  is an isosceles triangle. For this it is sufficient to prove that ∠CPQ = ∠CQP.

In ΔABP,  we have

AB = AP

⇒ ∠APB  = ∠ABP

⇒   ∠CPQ = ∠ABP      ……(i) [ ∵ ∠APB and ∠CPQ are vertically opposite angles  ∴  ∠APB = ∠CPQ]

Now, consider arc AC. Clearly , it subtends ∠ABC and ∠AQC at points B and Q.

∴  ∠ABC  = ∠AQC  [∵ Angles in the same segment]

⇒ ∠ABP = ∠PQC      [∵  ∠ABC = ∠ABP and ∠AQC = ∠PQC]

⇒ ∠ABP = ∠CQP       [∵   ∠PQC = ∠CQP]

From (i) and (ii), we get

∠CPQ = ∠CQP

⇒  CQ = CP

Hence proved.