In triangle ABC, bisector of angle BAC meets opposite side BC at point D. If BD = CD, prove that △ ABC is isosceles.

Produce AD upto E such that AD = DE.

In △ ABD and △ EDC,

⇒ AD = DE (Given)

⇒ BD = CD (Given)

⇒ ∠ADB = ∠EDC (Vertically opposite angles are equal)

∴ △ ABD ≅ △ EDC (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ AB = CE ………(1)

⇒ ∠BAD = ∠CED

⇒ ∠BAD = ∠CAD (As AD is the bisector BAC)

∴ ∠CAD = ∠CED

∴ CE = AC (Sides opposite to equal angles are equal) ……….(2)

From equations (1) and (2), we get :

⇒ AB = AC.

Hence, proved that ABC is an isosceles triangle.