State and prove Basic Proportionality Theorem

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.

Proof: Consider $\triangle ABC$ with line $DE \parallel BC$ intersecting $AB$ at $D$ and $AC$ at $E$. Join $BE$ and $CD$. Draw $EN \perp AB$ and $DM \perp AC$. Area of $\triangle ADE = \frac{1}{2} \times AD \times EN$ Area of $\triangle BDE = \frac{1}{2} \times DB \times EN$ $\frac{\text{area}(\triangle ADE)}{\text{area}(\triangle BDE)} = \frac{AD}{DB}$ Similarly, $\frac{\text{area}(\triangle ADE)}{\text{area}(\triangle DEC)} = \frac{AE}{EC}$ Since $\triangle BDE$ and $\triangle DEC$ lie on the same base $DE$ and between the same parallel lines $DE$ and $BC$: $\text{area}(\triangle BDE) = \text{area}(\triangle DEC)$ Therefore, $\frac{AD}{DB} = \frac{AE}{EC}$. Hence proved.