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

In Δ ABC BE and CF are the perpendiculars drawn on sides AC and AB respectively

Let O be the intersection of the two perpendiculars

In Δ BFC and Δ BEC

BE = FC(Given)

BFC = BEC = 900 (Perpendiculars)

BC = BC(Common side)

So Δ BFC and Δ BEC are congruent to each other by R.H.S. axiom of congruency

FBC = ECB (Corresponding Part of Congruent Triangle)

Since two angles are the same so Δ ABC is isosceles.

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