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Question
In triangle ABC, altitudes BE and CF are equal. Prove that the triangle is isosceles.
Sum
Solution
In ΔABE and ΔACF,
∠A = ∠A .........[Common]
∠AEB = ∠AFC = 90° ......[Given: BE ⊥ AC; CF ⊥ AB]
BE = CF ..........[Given]
∴ ΔABE ≅ ΔACE .........[AAS Criterion]
⇒ AB = AC
Therefore, ABC is an isosceles triangle.
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