Question

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

• A. 5 cm
• B. 2.5 cm
• C. 3 cm
• D. 3.5 cm

Answer 1

The correct option is D

3.5 cm

In $\Delta ABC$, E is the mid -point of median AD such that BE produced meets AC at F. Also, AC = 10.5 cm.

Draw DG || AF

In $\Delta ADG$

E is mid -point of AD and EF || DG

$\therefore$ F is mid -point of AG

$\Rightarrow AF=FG$ ....(i)

In $\Delta BCF$

D is mid -point of BC and DG || BF

$\therefore$ G is mid -point of FC

$\therefore FG=GC$ ....(ii)

From (i) and (ii)

AF = FG = GC = $\frac{1}{3}AC$

But AC = 10.5 cm

$\therefore AF=\frac{1}{3}AC=\frac{1}{3}\times 10.5=3.5cm$