Question

In trapezium ABCD, $M\in \overline{AD},n\in \overline{BC}$ are the points such that $\frac{AM}{MD}=\frac{BN}{NC}=\frac{2}{3}$.
The diagonal $\overline{AC}$ intersects $\overline{MN}$ at $O$. Then find the value of $\frac{AO}{AC}$.

Answer 1

In $\square ABCD,\overline{AB}||\overline{CD}$, $M$ and $N$ are the points on transversals $\overleftrightarrow{AD}$ and $\overleftrightarrow{BC}$ respectively
Also, $\frac{AM}{MD}=\frac{BN}{NC}$
$\therefore \overleftrightarrow{MN}||\overleftrightarrow{AB}$ and $\overleftrightarrow{AB}||\overleftrightarrow{CD}$
$\overleftrightarrow{MN}$ intersects $\overleftrightarrow{AC}$ at $O$.
$\therefore$ In $\Delta ADC,\overleftrightarrow{MO}||\overleftrightarrow{DC},M\epsilon \overline{AD},O\epsilon \overline{AC}$
$\therefore \frac{AM}{MD}=\frac{AO}{OC}$
But $\frac{AM}{MD}=\frac{2}{3}$
$\therefore \frac{AO}{OC}=\frac{2}{3}$
$\therefore \frac{AO}{AO+OC}=\frac{2}{2+3}$
$\therefore \frac{AO}{AC}=\frac{2}{5}$