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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 perpendiculars to the side AC and AB respectively.

Such that,

BE = CF

We have to prove that, ΔABC is isosceles triangle.

Now, consider ΔBCF and ΔCBE

We have,

∠BFC = ∠CEB = 90°(Given)

BC = CB (Given)

CF = BE (Given)

So, by RHS congruence rule, we have

ΔBFC ≅ ΔCEB

Now,

∠FBC = ∠ECB (By c.p.c.t)

∠ABC = ∠ACB (By c.p.c.t)

AC = AB (Opposite sides of equal angles are equal in a triangle)

Therefore, ΔABC is isosceles.

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