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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.

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Given that ABC is a triangle in which BE and CF are perpendicular to the sides AC and AB respectively such that BE = CF

We have to prove that ΔABC is isosceles.

Now, consider ΔBCF and ΔCBE,

We have

So, by RHS congruence criterion, we have ΔBFC ≅ ΔCEB

Now,

∠FBC = ∠EBC [∵ Incongruent triangles corresponding parts are equal]

⇒ ∠ABC = ACB

⇒ AC = AB  [∵ Opposite sides to equal angles are equal in a triangle]

∴ ΔABC is isosceles

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