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

(N/A) In the given figure,$LM \parallel CB$.
By using the Basic Proportionality Theorem (Thales Theorem) in $\triangle ABC$,we obtain:
$\frac{AM}{AB} = \frac{AL}{AC} \quad ...(i)$
Similarly,in $\triangle ADC$,since $LN \parallel CD$,by using the Basic Proportionality Theorem,we obtain:
$\frac{AN}{AD} = \frac{AL}{AC} \quad ...(ii)$
From equation $(i)$ and equation $(ii)$,we observe that the right-hand sides are equal.
Therefore,we obtain:
$\frac{AM}{AB} = \frac{AN}{AD}$