Q. 34.6( 28 Votes )
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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