Question

In a ΔABC, 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

Given that ABC is a triangle.

BP and CP are internal bisector of ∠B and ∠C respectively

BQ and CQ are external bisector of ∠B and ∠C respectively.

External ∠B = 180° - ∠B

External ∠C = 180° - ∠C

In ΔBPC

∠BPC + \(\frac{1}{2}\)∠B + \(\frac{1}{2}\)∠C = 180°

∠BPC = 180° - \(\frac{1}{2}\)(∠B + ∠C) (i)

In ΔBQC

∠BQC + \(\frac{1}{2}\)(180° - ∠B) + \(\frac{1}{2}\)(180° - ∠C) = 180°

∠BQC + 180° - \(\frac{1}{2}\)(∠B + ∠C) = 180°

∠BPC + ∠BQC = 180°[From (i)]

Hence, proved