Question

BE and CF are two equal altitudes of triangle ABC . prove that the triangle is isosceles

Answer 1

In triangle ABC two equal altitude BE and Cf
taking triangle ABE and ABF
Angle a=angle a. (common)
BE =CF. (given)
angle BEA =BFA. (each 90)
thus ABE congruent ABF by ASA
thus Ab=BC by cpct
thus triangle ABC is an isosceles triangle

Answer 2

Hello mate ^_^

____________________________/\_

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.

Thank you______❤

_____________________________❤