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

Proved below.

Step-by-step explanation:

Given:

Here in a ΔABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q.

To prove:

∠BPC + ∠BQC = 180°.

Proof:

∠ABC + ext.∠ABC = 180°       [Angles on a straight line]

Dividing both the sides by 2, we get

$\frac{1}{2} (\angle ABC + ext. \angle ABC) = \frac{180}{2}$

$\frac{1}{2} (\angle ABC + ext. \angle ABC) = 90$

∠PBC + ∠QBC = 90°           [PB bisect Interior ∠B, QB bisects ext.∠B]

∠PBQ = 90°             [shown in the figure]               [1]

Similarly ∠PCQ = 90°                                                  [2]

Sum of angles of quadrilateral PBCQ =360°

∠BPC + ∠PBQ + ∠PCQ + ∠BQC = 360°          

∠BPC + 90° + 90° + ∠BQC = 360°                 [From Eq (1) and (2)]

∠BPC + 180° + ∠BQC = 360°

∠BPC + ∠BQC = 360° - 180°

∠BPC + ∠BQC = 180°

Hence proved.