Mathematics

$$P$$ is any point inside the triangle $$ABC$$. Hence $$\angle BPC> \angle BAC$$


ANSWER

True


SOLUTION
Let
$$ \angle PBC = x $$ and $$ \angle PCB = y $$
$$ => \angle BPC = 180 - (x + y) $$

Let
$$ \angle ABP = z $$ and $$ \angle ACP = w $$
$$ => \angle BAC = 180 - (x + z) - (y + w ) $$
$$ => \angle BAC = 180 - (x + y) - (w + z ) $$
$$ \angle BAC = \angle BPC - ( w+z ) $$
$$ => \angle BPC = \angle BAC + (w + z ) $$
$$ => \angle BPC > \angle BAC  $$
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