Question

In ΔABC, E is the mid-point of median AD such that BE produced meets AC at F. If AC = 10.5 cm, then AF = 

A. 3 cm 

B. 3.5 cm 

C. 2.5 cm 

D. 5 cm

Answer 1

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