Option : (B)
Given,
In ∆ABC
E is mid point of median AD
AC = 10.5 cm
Draw DM || EF
∵ E is mid point of AD so F is mid point of AM
AF = FM ……………………(i)
In ∆BFC
EF || DM
So,
FM = MC ………………….(ii)
From(i) &(ii)
AF = MC……………..(iii)
AC = AF + MC + FM
= AC = AF + AF + AF
From (i) (ii) &(iii)
AC = 3AF
AF = \(\frac{1}{3}\)AC
AF = \(\frac{1}{3}\) x 10.5 = 3.5 cm