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

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Step 1: **Given:** In $	riangle ABC$, $DE \parallel BC$ where $D$ is on $AB$ and $E$ is on $AC$.
**To prove:** $\frac{AD}{DB} = \frac{AE}{EC}$
**Construction:** Draw $EM \perp AB$ and $DN \perp AC$. Join $BE$ and $CD$.
Step 2: In $	riangle ADE$: $	ext{ar}(	riangle ADE) = \frac{1}{2} 	imes AD 	imes EM$ and $	ext{ar}(	riangle BDE) = \frac{1}{2} 	imes DB 	imes EM$
$	herefore \frac{	ext{ar}(	riangle ADE)}{	ext{ar}(	riangle BDE)} = \frac{AD}{DB}$ ... (1)
Step 3: Similarly, $\frac{	ext{ar}(	riangle ADE)}{	ext{ar}(	riangle CDE)} = \frac{AE}{EC}$ ... (2)
Step 4: $	riangle BDE$ and $	riangle CDE$ are on the same base $DE$ and between the same parallels $DE$ and $BC$.
$	herefore 	ext{ar}(	riangle BDE) = 	ext{ar}(	riangle CDE)$
From (1) and (2): $\frac{AD}{DB} = \frac{AE}{EC}$

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