Question

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

1. ∠ABC = 2∠APQ,
2. ∠ACB = 2∠APR,
3. `∠QPR = 90^circ - 1/2 ∠BAC`.

Answer 1

Join PQ and PR

i. BQ is the bisector of ∠ABC

`=> ∠ABQ = 1/2 ∠ABC`

Also, ∠APQ = ∠ABQ

(Angle in the same segment)

∴ ∠ABC = 2∠APQ

ii. CR is the bisector of ∠ACB

`=> ∠ACR = 1/2 ∠ACB`

Also, ∠ACR = ∠APR

(Angle in the same segment)

∴ ∠ACB = 2∠APR

iii. Adding (i) and (ii)

We get

∠ABC + ∠ACB = 2(∠APR + ∠APQ) = 2∠QPR

`=>` 180° – ∠BAC = 2∠QPR

`=> ∠QPR = 90^circ - 1/2 ∠BAC`