Given that ABC is a triangle in which BE and CF are perpendiculars to the side AC and AB respectively.
Such that,
BE = CF
We have to prove that, ΔABC is isosceles triangle.
Now, consider ΔBCF and ΔCBE
We have,
∠BFC = ∠CEB = 90°(Given)
BC = CB (Given)
CF = BE (Given)
So, by RHS congruence rule, we have
ΔBFC ≅ ΔCEB
Now,
∠FBC = ∠ECB (By c.p.c.t)
∠ABC = ∠ACB (By c.p.c.t)
AC = AB (Opposite sides of equal angles are equal in a triangle)
Therefore, ΔABC is isosceles.