Question

In Fig., if $LM||CB$ and $LN||CD$, prove that $\frac{AM}{AB}=\frac{AN}{AD}$.

Answer 1

In $△$ABC,

$LM\parallel BC$
$\therefore$ By proportionality theorem,

$\frac{AM}{AB}=\frac{AL}{AC}............\left(1\right)$
Similarly,
In $△$ADC,

$LN\parallel CD$
$\therefore$ By proportionality theorem,

$\frac{AN}{AD}=\frac{AL}{AC}............\left(2\right)$
$\therefore$ from $\left(1\right)$ and $\left(2\right)$,
$\frac{AM}{AB}=\frac{AN}{AD}$