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. ${\displaystyle \angle QPR={90}^{\circ }-\frac{1}{2}\angle BAC}$.

Answer 1

Join PQ and PR

i. BQ is the bisector of ∠ABC

${\displaystyle \Rightarrow \angle ABQ=\frac{1}{2}\angle ABC}$

Also, ∠APQ = ∠ABQ

(Angle in the same segment)

∴ ∠ABC = 2∠APQ

ii. CR is the bisector of ∠ACB

${\displaystyle \Rightarrow \angle ACR=\frac{1}{2}\angle ACB}$

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

${\displaystyle \Rightarrow }$ 180° – ∠BAC = 2∠QPR

${\displaystyle \Rightarrow \angle QPR={90}^{\circ }-\frac{1}{2}\angle BAC}$