Question

P is any point inside the triangle ABC therefore $\angle BPC<\angle BAC$.

If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

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