Question
$P$ is any point inside the $\triangle ABC.$Prove that$: \angle BPC > \angle BAC.$
✓
Answer
Let $\angle PBC = x$ and $\angle PCB = y$
then,
$\angle BPC =180^{\circ}-( x + y )\ldots (i)$
Let $\angle ABP=a$ and $\angle ACP=b$
then,
$\angle BAC =180^{\circ}-( x + a )-( y + b )$
$ \Rightarrow \angle BAC =180^{\circ}-( x + y )-( a + b )$
$ \Rightarrow \angle BAC =\angle BPC -( a + b )$
$ \Rightarrow \angle BPC =\angle BAC +( a + b )$
$ \Rightarrow \angle BPC >\angle BAC$
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