Question

In the adjoining figure, AB = AC and AP is the bisector of ∠A. Prove that:

1. P is the mid-point of BC 
2. AP is perpendicular to BC

Answer 1

Given:

• In triangle ABC, AB = AC (so triangle ABC is isosceles with apex A).
• P lies on BC and AP bisects ∠A (i.e. ∠BAP = ∠PAC).

To Prove:

1. P is the midpoint of BC (i.e. BP = PC). 
2. AP is perpendicular to BC (i.e. AP ⟂ BC).

Proof [Step-wise]:

1. Consider triangles ΔABP and ΔAPC.

2. AB = AC.   ...(Given)

3. AP = AP.   ...(Common side)

4. ∠BAP = ∠PAC.   ...(AP is the bisector of ∠A)

5. From (2), (3) and (4),

ΔABP ≅ ΔPAC by SAS congruence of two sides and the included angle.

6. From congruence, corresponding sides BP and PC are equal.

Hence, BP = PC, so P is the midpoint of BC.

This proves (i).

7. Also from congruence, corresponding angles ∠APB and ∠CPA are equal.

8. Since B, P, C are collinear, ∠APB and ∠APC are a linear pair and therefore ∠APB + ∠APC = 180°.

9. With ∠APB = ∠APC (step 7), each must be 90°.

Thus ∠APB = 90°, so AP ⟂ BC.

This proves (ii).