Statement: 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. Given: $\triangle ABC$ in which $DE \parallel BC$, meeting AB at D and AC at E. To Prove: $\frac{AD}{DB}=\frac{AE}{EC}$. Construction: Join BE and CD. Draw $EM \perp AB$ and $DN \perp AC$. Proof: Area of $\triangle ADE = \frac{1}{2}\times AD \times EM$ and Area of $\triangle DBE = \frac{1}{2}\times DB \times EM$. So $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DBE)}=\frac{AD}{DB}$ ...(i). Similarly, $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)}=\frac{AE}{EC}$ ...(ii). Since $\triangle DBE$ and $\triangle DEC$ lie between the same parallels DE and BC and on the same base DE, $\text{ar}(\triangle DBE)=\text{ar}(\triangle DEC)$ ...(iii). From (i), (ii) and (iii): $\frac{AD}{DB}=\frac{AE}{EC}$ (Proved)