Problem Statement
Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Verified Solution & Marking Scheme
Given, To Prove, and Construction
**Given:** In $\triangle ABC$, line $DE \parallel BC$ intersecting $AB$ at $D$ and $AC$ at $E$.
**To Prove:** $\frac{AD}{DB} = \frac{AE}{EC}$.
**Construction:** Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.
Express ratios of areas with base on AB
$\text{Area}(\triangle ADE) = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times AD \times EN$
$\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EN$
$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} AD \times EN}{\frac{1}{2} DB \times EN} = \frac{AD}{DB} \quad \text{--- (1)}$
Express ratios of areas with base on AC
$\text{Area}(\triangle ADE) = \frac{1}{2} \times AE \times DM$
$\text{Area}(\triangle CDE) = \frac{1}{2} \times EC \times DM$
$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} AE \times DM}{\frac{1}{2} EC \times DM} = \frac{AE}{EC} \quad \text{--- (2)}$
Equate areas on common parallel base DE
$\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallel lines $DE$ and $BC$.
$\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE) \quad \text{--- (3)}$
From (1), (2), and (3):
$\frac{AD}{DB} = \frac{AE}{EC}$
Hence proved.