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Question
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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Solution
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
∠BFC=CEB=90° [Given]
BC=CB [Common side]
And CF=BE [Given]
So, by RHS congruence criterion, we have ΔBFC≅CEB
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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