Question

In a triangle, ABC, the internal bisectors of angles B and C meet at P and the external bisectors of the angles B and C meet at Q. Such that: $\angle BPC$ + $\angle BQC$ = ${120}^o$.

State true or false:

• A. True
• B. False

Answer 1

The correct option is B False

$\angle ABC+ext.\angle \angle ABC=180$ (Angles on a straight line)
$\frac{1}{2}(\angle ABC+ext.\angle ABC)=90$
$\angle PBC+\angle QBC=90$ (PB bisect Interior $\angle B$, QB bisects $ext.\angle B$)
$\angle PBQ=90$
Similarly, $\angle PCQ=90$
Sum of angles of quadrilateral PBCQ = 360
$\angle BPC+\angle PBQ+\angle PCQ+\angle BQC=360$
$\angle BPC+\angle BQC=180$
$\angle BPQ+\angle BQC$ = 2 rt. angles