Triangles: Basic Proportionality Theorem (Thales' Theorem)

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

- **Given:** A triangle $ABC$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively ($DE \parallel BC$). - **To Prove:** $\frac{AD}{DB} = \frac{AE}{EC}$. - **Construction:** Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.

Recall: $\text{Area}(\triangle) = \frac{1}{2} \times \text{base} \times \text{height}$. $\text{ar}(\triangle ADE) = \frac{1}{2} \times AD \times EN$ $\text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN$ $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{...(1)}$ Similarly: $\text{ar}(\triangle ADE) = \frac{1}{2} \times AE \times DM$ $\text{ar}(\triangle CDE) = \frac{1}{2} \times EC \times DM$ $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{...(2)}$

Note that $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels $DE$ and $BC$. Therefore: $\text{ar}(\triangle BDE) = \text{ar}(\triangle CDE) \quad \text{...(3)}$ From (1), (2), and (3): $\frac{AD}{DB} = \frac{AE}{EC}$ Hence proved.