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 .
Answers
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
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]