In the adjoining figure, ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.

In △ABC, ∠A = 90° and AB = AC.

⇒ ∠B = ∠C (As angles opposite to equal sides are equal)

We know,

∠A + ∠B + ∠C = 180°

90° + ∠B + ∠B = 180°

2∠B = 90°

∠B = 45° .

As AD is bisector of ∠A, ∠BAD = ∠CAD = 45°.

AD = AD (Common)

∠ACD = ∠ABD = 45°

Hence, △ABD ≅ △ACD by AAS axiom.

We know that corresponding parts of congruent triangles are equal.

∴ BD = CD ………(i)

In △ABD, ∠BAD = ∠ABD (each = 45°)

⇒ AD = BD (As sides opposite to equal angles are equal) ……..(ii)

∴ BC = BD + CD

= BD + BD (Using (i))

∴ BC = 2BD

Using (ii),

BC = 2AD.

Hence, proved that BC = 2AD.