Question

In a triangle ABC, the internal bisectors of angle B and C meet at P and the external bisector of the angle B and C meet at Q.
Prove that : $\angle$BPC + $\angle$BQC = 2 rt. angles.

Answer 1

$\angle ABC+ext.\angle \angle ABC={180}^o$ (Angles on a straight line)

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

$\angle PBC+\angle QBC={90}^o$ (PB bisect Interior $\angle B$, QB bisects $ext.\angle B$)

$\angle PBQ={90}^o$

Similarly, $\angle PCQ={90}^o$
Sum of angles of quadrilateral PBCQ $={360}^o$

$\angle BPC+\angle PBQ+\angle PCQ+\angle BQC={360}^o$

$\angle BPC+\angle BQC={180}^o$

$\therefore \angle BPQ+\angle BQC$ = 2 rt. angles