In figure, if LM || CB and LN || CD, prove that

$\dfrac{AM}{AB} = \dfrac{AN}{AD}$.

We know that,

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

In Δ ABC,

LM || CB [∵ Given]

$\therefore \dfrac{AM}{MB} = \dfrac{AL}{LC}$ ………… (1)

In Δ ACD,

LN || CD [∵ Given]

$\therefore \dfrac{AN}{DN} = \dfrac{AL}{LC}$ ………… (2)

From equations (1) and (2),

$\Rightarrow \dfrac{AM}{MB} = \dfrac{AN}{DN} \\[1em] \Rightarrow \dfrac{MB}{AM} = \dfrac{DN}{AN}$

Adding 1 on both sides, we get :

$\Rightarrow \dfrac{MB}{AM} + 1 = \dfrac{DN}{AN} + 1 \\[1em] \Rightarrow \dfrac{(MB + AM)}{AM} = \dfrac{(DN + AN)}{AN} \\[1em] \Rightarrow \dfrac{AB}{AM} = \dfrac{AD}{AN} \\[1em] \Rightarrow \dfrac{AM}{AB} = \dfrac{AN}{AD}.$

Hence, proved that $\dfrac{AM}{AB} = \dfrac{AN}{AD}.$.