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

Solution:

In $\triangle \mathrm{ACB}$ (see figure), $\mathrm{LM} \| \mathrm{CB}$ (Given)

$\Rightarrow \frac{A M}{M B}=\frac{A L}{L C}$ ...(1)

(Basic Proportionality Theorem)

In $\triangle \mathrm{ACD}$ (see figure), $\mathrm{LN} \| \mathrm{CD}$ (Given)

$\Rightarrow \frac{A N}{N D}=\frac{A L}{L C}$ ...(2)

(Basic Proportionality Theorem)

From (1) and (2), we get

$\frac{A M}{M B}=\frac{A N}{N D}$

$\Rightarrow \frac{\mathrm{AM}}{\mathrm{AM}+\mathrm{MB}}=\frac{\mathrm{AN}}{\mathrm{AN}+\mathrm{ND}} \Rightarrow \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$