Q3 of 49 Page 58

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.

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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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