Question

$P$ is any point inside the triangle $ABC$ prove that $\angle BPC>\angle BAC$

Answer 1

Consider the given $\Delta ABC$

Construct:-draw a line AD through point P.

We know that,

$Exteriorangle=sumofopposite\mathrm{int}eriorangle$

$In\Delta ADC$, $\angle DPC$ is exterior angle then,

$\angle DPC=\angle PAC+\angle PCA$

$\therefore \angle DPC>\angle PAC.......\left(1\right)$

$In\Delta ADB$, $\angle DPB$ is exterior angle then,

$\angle BPD=\angle PAB+\angle PBA$

$\therefore \angle BPD>\angle PAB.......\left(2\right)$

Adding equation $\left(1\right)$ and $\left(2\right)$ to, we get

$\angle DPC+\angle BPD=\angle PAB+\angle PAC$

$\therefore \angle BPC>\angle BAC$

Hence proved.