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

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Solution

Given : In $Δ A B C,$

$B E ⊥ A C a n d C F ⊥ A B$

$B E = C F$

To prove : $Δ A B C$ is an isosceles triangle

Proof : In right $Δ B C E a n d Δ B C F$

Side BE = CF (Given)

Hyp. BC = BC (Common)

$∴ Δ B C E ≅ Δ B C F (RHS axiom)$

$∴ ∠ B C E = ∠ C B F (c . p . c . t .)$

$∴ A B = A C (Sides opposite to equal angles)$

$∴ Δ A B C is an isosceles triangle$

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