Question

ABC is an isosceles triangle in which altitudes BF and CF are drawn to equal sides AC and AB respectively.. show that this altitudes are equal​

Answer 1

Step-by-step explanation:

Angle AEB = angle AFC... (given)

AB = AC... ( Sides of isosceles triangle)

Proof: In ABE and ACF

Angle AEF = Angle AFC... (given)

Angle A = Angle A ....(common angle)

AB = AC ...(given)

Therefore ABE = ACF ...(AAS rule)

BE = CF..... Hence proved

Answer 2

Hi there!

Your Answer :-

$\tt\underline\red{Given \: :- }$

AC = AB

BE and CF are altitudes.

$\tt\underline\red{To \: \: Prove\::-}$

BE = CF

$\tt\underline\red{Proof\::-}$

Angle A = Angle A (Common)

Angle AEB = Angle AFC (Given)

AB = AC (Given)

∆AEB ≅ ∆AFC (By AAS).

$<font color=blue>So, BE = CF (By C.P.C.T)$