Prove that the bisectors of the base angles of an isosceles triangle are equal.

In isosceles triangle △ ABC,

Let AB = AC,

∴ ∠C = ∠B = x (let) [Angles opposite to equal sides are equal]

From figure,

BD and CE are bisectors of angle B and C.

∴ ∠CBD = $\dfrac{∠B}{2} = \dfrac{x}{2}$ and ∠BCE = $\dfrac{∠C}{2} = \dfrac{x}{2}$.

∴ ∠CBD = ∠BCE.

In △ CBD and △ BCE,

⇒ ∠CBD = ∠BCE (Proved above)

⇒ ∠C = ∠B (Proved above)

⇒ BC = BC (Common side)

∴ △CBD ≅ △BCE (By A.S.A. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ BD = CE.

Hence, proved that the bisectors of the base angles of an isosceles triangle are equal.