Question

In a Δ ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q, Prove that ∠BPC + ∠BQC = 180°.

Answer 1

In the given problem, BP and CP are the internal bisectors of respectively. Also, BQ and CQ are the external bisectors of respectively. Here, we need to prove:

We know that if the bisectors of anglesand of ΔABC meet at a point O then .

Thus, in ΔABC

……(1)

Also, using the theorem, “if the sides AB and AC of a ΔABC are produced, and the external bisectors of and meet at O, then”.

Thus, ΔABC

$\angle BQC=90°-\frac{1}{2}\angle A......\left(2\right)$

Adding (1) and (2), we get

Thus,

Hence proved.