Solution:-

From the given figure, ∠D = ∠E and AD/BD = AE/EC,

We have to prove that, BAC is an isosceles triangle

So, consider the ∆ADE

∠D = ∠E … [from the question]

AD = AE … [sides opposite to equal angles]

Consider the ∆ABC,

Then, AD/DB = AE/EC … [equation (i)]

Therefore, DE parallel to BC

Because AD = AE

DB = EC … [equation (ii)]

By adding equation (i) and equation (ii) we get,

AD + DB = AE + EC

AB = AC

Therefore, ∆ABC is an isosceles triangle.