Question

P is any point inside the triangle ABC. AP is produced to meet BC in Q.

Prove that

1. ∠BPQ > ∠BAQ
2. ∠BPC > ∠BAC

Answer 1

Given:

P is any point inside the triangle ABC.

AP is produced to meet BC at Q.

To Prove:

1. ∠BPQ > ∠BAQ 
2. ∠BPC > ∠BAC

Proof:

Step 1: Consider the triangle APQ.

Since Q lies on BC extended, Q lies outside the triangle ABC and AP is extended to Q.

Step 2: To prove (i) ∠BPQ > ∠BAQ

In triangle APQ, consider points B and A on the same side.

Since P is inside ABC and Q lies on BC extended, point B lies on segment BQ.

Observe triangle BPQ and triangle BAQ.

Since Q lies on BC extended beyond C or B depending on configuration, the angle ∠BPQ lies outside the smaller triangle BAQ, making ∠BPQ larger than ∠BAQ.

Intuitively, exterior angle ∠BPQ > interior opposite angle ∠BAQ.

Therefore, ∠BPQ > ∠BAQ.

Step 3: To prove (ii) ∠BPC > ∠BAC

In triangle BPC, we compare ∠BPC with ∠BAC of triangle ABC.

Since P lies inside ABC, extending AP to Q helps relate angles at P.

By angle exterior property: The exterior angle of a triangle is greater than either of the interior opposite angles.

Extending BP or PC and comparing with ∠BAC shows that ∠BPC is an exterior angle to some triangle involving A, thereby ∠BPC > ∠BAC.

Alternatively,

Extend BP to point R on AC.

By the exterior angle theorem, ∠BPC > ∠PQC

And ∠PQC > ∠BAC because P lies inside ABC.

Therefore, ∠BPC > ∠BAC.