P is any point inside the triangle ABC. Prove that : ∠BPC > ∠BAC.

Let ∠PBC = x and ∠PCB = y.

In △ BPC,

By angle sum property of triangle,

⇒ ∠BPC + ∠PBC + ∠PCB = 180°

⇒ ∠BPC + x + y = 180°

⇒ ∠BPC = 180° - x - y

⇒ ∠BPC = 180° - (x + y).

Let ∠ABP = a and ∠ACP = b

In △ BAC,

By angle sum property of triangle,

⇒ ∠ABC + ∠BCA + ∠BAC = 180°

⇒ (∠ABP + ∠PBC) + (∠BCP + ∠ACP) + ∠BAC = 180°

⇒ (a + x) + (y + b) + ∠BAC = 180°

⇒ (a + b) + (x + y) + ∠BAC = 180°

⇒ ∠BAC = 180° - (x + y) - (a + b)

⇒ ∠BAC = ∠BPC - (a + b)

⇒ ∠BPC = ∠BAC + (a + b)

⇒ ∠BPC > ∠BAC.

Hence, proved that ∠BPC > ∠BAC.