$P$ is any point inside the triangle $A B C$ . Prove that: $∠ B P C > ∠ B A C$ .

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Solution

Consider $△ A B C$
A point $P$ can only be an Interior Angle $⟺ ∠ P B C < ∠ A B C$ and $∠ P C B < ∠ A C B \dots (1)$
By angle sum property
$∠ B A C + ∠ A B C + ∠ A C B = 180^{\circ} \dots (2)$
$∠ B P C + ∠ P B C + ∠ P C B = 180^{\circ} \dots (3)$
From $(1) (2) (3)$
$⟹ ∠ B P C > ∠ B A C$

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