Question

In figure 3.7, bisectors of $\angle B$ and $\angle C$ of $△ABC$ intersect at point P. Prove that $\angle BPC=90+\frac{1}{2}\angle BAC$.
Complete the proof filling in the blanks.

Answer 1

$\angle ABC+\angle BAC+\angle ACD={180}^o$- [Sum of all angles of a triangles]

$\angle ABC+\angle ACB=180-\angle BAC$

Dividing throughout by $2$

$\frac{\angle ABC}{2}+\frac{\angle ACB}{2}=90-\frac{\angle BAC}{2}$ ....... $\left(1\right)$

Now, $\angle PBC=\frac{1}{2}\angle ABC....\left(2\right)$ [$PB$ is bisector of angle $B$]

$\angle PCB=\frac{1}{2}\angle ACB.....\left(3\right)$ [$PC$ is the bisector of angle $C$]

Substituting equations $\left(2\right)$ and $\left(3\right)$ in equation

$\angle PBC+\angle PCB=90-\frac{\angle BAC}{2}........\left(4\right)$

In $△BPC$

$\angle PBC+\angle PCB+\angle BPC={180}^o$ [Sum of all angles of triangle]

$\angle PBC+\angle PCB=180-\angle BPC$ .......... $\left(5\right)$

From equations $\left(4\right)$ and $\left(5\right)$ we get

$180-\angle BPC=90-\angle BAC$

$\angle BPC=90+\frac{1}{2}\angle BAC$