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°.

#### Solution

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

∠BPC +∠BQC = 180°

We know that if the bisectors of angles ∠ABC and ∠ACB of Δ*ABC *meet at a point O then` ∠BOC = 90° + 1/2 ∠A`.

Thus, in ΔABC

`∠BPC = 90° + 1/2 ∠A.` ……(1)

Also, using the theorem, “if the sides AB and AC of a ΔABC are produced, and the external bisectors of ∠B and ∠Cmeet at O, then `∠BOC = 90° + 1/2 ∠A.` .

Thus, ΔABC

\[\angle BQC = 90^\circ - \frac{1}{2}\angle A . . . . . . \left( 2 \right)\]

Adding (1) and (2), we get

`∠BOC +∠BQC = 90° + 1/2 ∠A + 90° -1/2 ∠A`

∠BQC + ∠BQC = 180°

Thus, `∠BQC + ∠BQC = 180°

Hence proved.