Question

In the given Figure. P is any point on the chord BC of a circle such that AB = AP. Prove that CP = CQ.

Answer 1

We have to prove that CP = CQ i.e., Δ CPQ is an isosceles triangle. for this it is sufficient to prove that ∠ CPQ = ∠ CQP.
In Δ ABP, 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      ....(ii)( ∵ ∠PQC = ∠CQP )

From (i) and (ii), we get
∠ CPQ = ∠CQP
⇒ CQ = CP
Hence proved.