In the adjoining figure, AB = AC and D is the midpoint of BC. Use SSS rule of congruency to show that

(i) △ABD ≅ △ACD

(ii) AD is bisector of ∠A

(iii) AD is perpendicular to BC.

(i) In △ABD and △ACD,

Given,

AB = AC (Given)

BD = CD (As D is the midpoint of BC)

AD = AD (Common)

Hence, by SSS axiom △ABD ≅ △ACD.

(ii) Since, △ABD ≅ △ACD

We know that corresponding angles of congruent triangle are equal.

∴ ∠BAD = ∠CAD.

Hence, proved that AD is bisector of ∠A.

(iii) △ABD ≅ △ACD

We know that corresponding angles of congruent triangles are equal.

∠ADB = ∠ADC.

Let ∠ADB = ∠ADC = x.

We know that,

⇒ ∠ADB + ∠ADC = 180°
⇒ x + x = 180°
⇒ 2x = 180°
⇒ x = 90°.

Hence, ∠ADB = ∠ADC = 90°.

Hence, proved that AD is perpendicular to BC.