NIMCET 2019 Question Paper with solutions

Given, AD is median of ΔABC
∴ D is mid-point of BC
and E is mid-point of AD
Draw DM parallel to EFIn ΔADM,
By mid-point theorem,
F mid-point of AM
$\therefore \mathit{A}\mathit{F}=\frac{1}{2}\mathit{A}\mathit{m}=\mathit{F}\mathit{M}$
In ΔBFC,
M is parallel to BF and D is mid-point of BC.
∴ By mid-point theorem,
M is mid-point of FC
$\therefore \mathit{F}\mathit{M}=\frac{1}{2}\mathit{F}\mathit{C}$
$\mathit{A}\mathit{F}=\frac{1}{2}\mathit{F}\mathit{C}$  $\left[\because \mathit{A}\mathit{F}=\mathit{F}\mathit{M}\right]$
$\frac{\mathit{A}\mathit{F}}{\mathit{F}\mathit{C}}=\frac{1}{2}$
$\mathit{A}\mathit{F}:\mathit{F}\mathit{C}=1:2$