​ In ΔABC, AD, BE are medians BE ∥ DF To prove CF = 1/4 AC .

Question

In ΔABC,\(\overline{AD}\) , \(\overline{BE}\) are medians  \(\overline{BE}\) ∥ \(\overline{DF}\)

To prove CF = 1/4 AC it was written as In ΔABC ‘D’ is the mid-point of \(\overline{BC}\) and BE || DF. As per triangle mid-point of  \(\overline{CE}\).

CF = 1/2 CE

Which of the following is the next step ? 

A) CF = 1/2 AC 

B) 2CF = CE 

C) CF = 1/2(1/2 AC) 

D) 4CF = AC

Answer 1

Correct option is  B) 2CF = CE