Question

In the adjoining figure, AB = AC and D is the mid-point of BC. Use SSS criteria of congruency and prove that:

1. ΔABD ≅ ΔACD
2. AD bisects ∠A.
3. AD is perpendicular to BC.

Answer 1

Given:

• In triangle ABC, AB = AC (so ABC is isosceles with apex A).
• D is the midpoint of BC (so BD = DC).

To Prove: 

1. ΔABD ≅ ΔACD
2. AD bisects ∠A (i.e. ∠BAD = ∠DAC)
3. AD is perpendicular to BC (AD ⟂ BC).

Proof [Step-wise]:

1. BD = DC because D is the midpoint of BC.   ...(Given)

2. AB = AC.   ...(Given)

3. AD = AD.   ...(Common side to triangles ABD and ACD)

4. From (1), (2), (3) the three pairs of corresponding sides are equal:

AB = AC

BD = DC

AD = AD

Hence by SSS congruence, ΔABD ≅ ΔACD.

5. From the congruence in step 4, corresponding parts of congruent triangles are equal (CPCTC).

In particular ∠BAD = ∠DAC. 

Therefore AD bisects ∠A.

This proves (ii).

6. Also by congruence, the angles at D in the two triangles are equal:

∠ADB = ∠ADC. 

These two angles are adjacent and lie on a straight line BC, so ∠ADB + ∠ADC = 180° they form a linear pair.

7. If two adjacent angles are equal and sum to 180°, each must be 90°.

Thus ∠ADB = ∠ADC = 90°, so AD is perpendicular to BC.

This proves (iii).