Question

in triangle abc , be and cf are altitudes on the sides ac and ab respectively such that be = cf . using rhs congruence rule prove that ab = ac . ​

Answer 1

Step-by-step explanation:

ACCORDING TO QUESTION YOUR FIGURE IS WRONG.WHICH I SHOW YOU THAT IS THE CORRECT FIGURE AS PER THE QUESTION

In right-angle triangles BCE and CBF, we have,

BC = BC (common hypotenuse);

BE = CF (given).

Hence BCF and CBF are congruent, by RHS theorem. Comparing the triangles, we get ∠B=∠C.

This implies that

AC = AB (sides opposite to equal angles).

Similarly,

AD=BE⇒∠B=∠A

⇒AC=BC

Together, we get AB=BC=ACor △ABC is equilateral. [henceproved]

MARK IT AS BRAINLIESTS ANSWER

Answer 2

In right-angle triangles BCE and CBF, we have,

BC = BC (common hypotenuse);

BE = CF (given).

Hence BCF and CBF are congruent, by RHS theorem. Comparing the triangles, we get ∠B=∠C.

This implies that

AC = AB (sides opposite to equal angles).

Similarly,

AD=BE⇒∠B=∠A

⇒AC=BC

Together, we get AB=BC=ACor △ABC is equilateral. [henceproved]