Question

In a $\Delta ABC$, the internal bisectors of $\angle B$ and $\angle C$ meet at $P$ and the external bisectors of $\angle B$ and $\angle C$ meets at Q. Prove that $\angle BPC+\angle BQC={180}^0$.

Answer 1

$\angle ABC+\angle XBC=180°\because ABX$ is a straight line

$\Rightarrow 2\angle PBC+2\angle QBC={180}^o$

$\Rightarrow \angle PBC+\angle QBC={90}^o$

$\Rightarrow \angle PBQ={90}^o$

$\angle ACB+\angle YCB=180°\because ACY$ is a straight line

$\Rightarrow 2\angle PCB+2\angle QCB={180}^o$

$\Rightarrow \angle PCB+\angle QCB={90}^o$

$\Rightarrow \angle PCQ={90}^o$

Now, in quadrilateral BPCQ, sum of all angles$={360}^o$

So,$\angle BPC+\angle BQC+\angle PBQ+\angle PCQ=360°$

$\Rightarrow \angle BPC+\angle BQC+90°+90°=360°\Rightarrow \angle BPC+\angle BQC=180°$