Question

In ΔABC, P is any point inside it. Prove that ∠BPC > ∠A.

[Hint: Extend BP to meet AC in Q. Ext. ∠BPC > Int. opp. ∠PQC and Ext. ∠PQC > ∠A]

Answer 1

Given:

• Triangle ABC and a point P lies inside triangle ABC.
• We are to prove that ∠BPC > ∠A.

Hint provided:

• Extend BP to meet AC at point Q.
• Use exterior angle properties.

Proof (Using Exterior Angle Properties):

1. Extend BP to meet AC at point Q, forming triangle PQC. 

2. In triangle PQC, angle BPC becomes an exterior angle at vertex Q.

Hence, ∠BPC > ∠PQC    ... (1) (Exterior Angle Theorem)

3. Now consider triangle ABQ. Since P lies inside triangle ABC, point Q lies beyond P on AC, so triangle ABQ exists.

In triangle ABQ, angle PQC is an exterior angle.

So, ∠PQC > ∠A    ... (2) (Exterior Angle Theorem again)

4. From (1) and (2), we combine:

∠BPC > ∠PQC > ∠A ⇒ ∠BPC > ∠A