Prove that the medians corresponding to equal sides of an isosceles triangle are equal.

In ∆ ABC,

⇒ AB = AC (Since, ABC is an isosceles triangle)

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

From figure,

BD and CE are medians of triangle.

As,

⇒ AB = AC

Dividing both sides of the equation by 2, we get :

⇒ $\dfrac{AB}{2} = \dfrac{AC}{2}$

⇒ BE = CD

In Δ EBC and Δ DCB,

⇒ ∠B = ∠C (Proved above)

⇒ BC = BC (Common side)

⇒ BE = CD (Proved above)

∴ Δ EBC ≅ Δ DCB (By S.A.S. axiom)

We know that,

Corresponding sides of congruent triangles are equal.

∴ BD = CE

Hence, proved that medians bisecting the equal sides of an isosceles triangle are also equal.