Question

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Then: 


a) BE>CF

b) BE<CF

c) BE=CF

d) None of the above

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Answer 1

Answer:

I think the b option is the right answer

I am not sure about the answer

Answer 2

Answer:

C : BE = CF

given: a ∆ABC is an isosceles

With AB and AC are equal....(1)

BE and CF are the altitudes of ∆ABC

so, angleAEB=90° and angle AFC=90°..(2)

to prove- BE =CF

proof= In ∆ABE and ACF,

= angle AEB =AFC (from 2 )

= angle A= angle A. (common)

= AB=AC (by 1)

hence , ∆ABE and ACF are congruent ( by AAS rule )

and, BE =CE (by cpct) ...hence proved.