**Given:** ΔABC in which DE || BC, D on AB, E on AC.
**To Prove:** $\frac{AD}{DB}=\frac{AE}{EC}$
**Construction:** Join BE, CD. Draw $DM\perp AC$ and $EN\perp AB$.
**Proof:**
$$ar(\triangle ADE)=\frac12\times AD\times EN,\quad ar(\triangle DBE)=\frac12\times DB\times EN$$
$$\Rightarrow \frac{ar(\triangle ADE)}{ar(\triangle DBE)}=\frac{AD}{DB}$$
Similarly, $$\frac{ar(\triangle ADE)}{ar(\triangle DEC)}=\frac{AE}{EC}$$
Since ΔDBE and ΔDEC lie on the same base DE and between the same parallels DE and BC, $ar(\triangle DBE)=ar(\triangle DEC)$.
Therefore, $$\frac{AD}{DB}=\frac{AE}{EC}$$ Hence proved.