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).

Answer 1

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.