Given that ABC is a triangle in which BE and CF are perpendicular to the sides AC and AB respectively such that BE = CF

We have to prove that ΔABC is isosceles.

Now, consider ΔBCF and ΔCBE,

We have

So, by RHS congruence criterion, we have ΔBFC ≅ ΔCEB

Now,

∠FBC = ∠EBC [∵ Incongruent triangles corresponding parts are equal]

⇒ ∠ABC = ∠ACB

⇒ AC = AB [∵ Opposite sides to equal angles are equal in a triangle]

∴ ΔABC is isosceles