Question

(i) In the figure, O is the centre of the circle and AP is thetangent to the circle at point A. Ray AF is the bisectorof angle BACProve : seg AP seg PE​

Answer 1

In Fig. AP is a tangent to the circle at P, ABC is a secant and PD is the bisector.

In Fig. AP is a tangent to the circle at P, ABC is a secant and PD is the bisector of ∠BPC. Prove that ∠BPD = ½(∠ABP = ∠APB

Since, ∠APB and ∠BCP are angles in alternate segments of chord PB.

∴ ∠APB = ∠BCP ……..(i)

Since, PD is bisector of ∠BPC

∴ ∠CPB = 2∠BPD ……..(ii)

In △PCB, 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.

Answer 2

Step-by-step explanation:

this diagram will help you