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Question
In Fig. AP is a tangent to the circle at P, ABC is secant and PD is the bisector of ∠BPC. Prove that ∠BPD = `1/2` (∠ABP - ∠APB).
Solution
Since ∠APB and ∠BCP are angles in the alternate segment of chord PB.
∴ ∠APB = ∠BCP ...(i)
Since PD is the bisector of ∠BPC.
∴ ∠CPB = 2 ∠BPD ...(ii)
In ΔPCB, side CB has been produced to A, forming exterior angle ∠ABP.
∴ ∠ABP = ∠BCP + ∠CPB
⇒ ∠ABP = ∠APB + 2 ∠BPD ....(Using (i) and (ii))
⇒ 2 ∠BPD = ∠ABP - ∠APB
⇒ ∠BPD = `1/2`(∠ABP - ∠APB)
Hence proved.
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