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