Question

Be and cf are two equal altitudes of a triangle abc.prove that triangle abc is isosceles.

Answer 1

Area of Δ ABC = 1/2 Base * altitude
           = 1/2 AC * BE    =  1/2 AB * CF  = 1/2 AB * BE,    given BE = CF
So AC = AB
Hence it is an isosceles Δ.
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Consider the triangles ABE and ACF.
Angle A is common. angle BEA = 90 = angle CFA. The triangles are similar.

 

 So isosceles triangle.

Answer 2

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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