A triangle ABC is inscribed in a circle. The bisectors of angle BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that :

(i) ∠ABC = 2∠APQ

(ii) ∠ACB = 2∠APR

(iii) ∠QPR = 90° - $\dfrac{1}{2}$∠BAC.

Join PQ and PR.

(i) BQ is the bisector of ∠ABC.

∴ ∠ABQ = $\dfrac{1}{2}$∠ABC ………(1)

We know that,

Angles in same segment are equal.

∴ ∠APQ = ∠ABQ …………(2)

From (1) and (2) we get,

⇒ ∠APQ = $\dfrac{1}{2}$∠ABC ………..(3)

⇒ ∠ABC = 2∠APQ.

Hence, proved that ∠ABC = 2∠APQ.

(ii) CR is the bisector of ∠ACB.

∴ ∠ACR = $\dfrac{1}{2}$∠ACB ………(4)

We know that,

Angles in same segment are equal.

∴ ∠ACR = ∠APR …………(5)

From (3) and (4) we get,

⇒ ∠APR = $\dfrac{1}{2}$∠ACB ………..(6)

⇒ ∠ACB = 2∠APR.

Hence, proved that ∠ACB = 2∠APR.

(iii) Adding equations (3) and (6), we get :

⇒ ∠APQ + ∠APR = $\dfrac{1}{2}$∠ABC + $\dfrac{1}{2}$∠ACB

⇒ ∠QPR = $\dfrac{1}{2}$(∠ABC + ∠ACB)

⇒ ∠QPR = $\dfrac{1}{2}$(180° - ∠BAC)

⇒ ∠QPR = 90° - $\dfrac{1}{2}$∠BAC.

Hence, proved that ∠QPR = 90° - $\dfrac{1}{2}$∠BAC.