BE and CF are two equal altitudes of atriangle ABC . Prove that the triangle ABC is isoceles
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if two angles are equal then it is isoceles triangle
BE=CF
BE=CF
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Hello mate ^_^
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Solution:
In ∆BEC and ∆CFB
BE=CF (Given)
∠BEC=∠CFB (Each given equal to 90°)
BC=CB (Common)
Therefore, by RHS rule, ∆BEC≅∆CFB
It means that ∠C=∠B (Corresponding parts of congruent triangles are equal)
⇒AB=AC (In a triangle, sides opposite to equal angles are equal)
Therefore, ∆ABC is isosceles.
hope, this will help you.
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