ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE=CF, prove that ΔABC is isosceles.

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, ∆ABCis isosceles triangle.

Now, consider ∆BCF and ∆CBE,

We have,

∠BFC = ∠CEB = 90^o (Given)

BC = CB (Given)

CF = BE (Given)

So, by RHS congruence rule, we have

∆BFC ≅CEB

∠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, ∆ABC is isosceles.