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°.

Question

In a &#x394; ABC , the internal bisectors of &#x2220; B and &#x2220; C meet at P and the external bisectors of &#x2220; B and &#x2220; C meet at Q . Prove that &#x2220; BPC + &#x2220; BQC = 180&#xB0;.

Philoid · Accepted Answer

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^o - ∠B

External ∠C = 180^o - ∠C

In

∠BPC + ∠B + ∠C = 180^o

∠BPC = 180^o - (∠B + ∠C) (i)

In

∠BQC + (180^o - ∠B) + (180^o - ∠C) = 180^o

∠BQC + 180^o - (∠B + ∠C) = 180^o

∠BPC + ∠BQC = 180^o [From (i)]

Hence, proved

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°.