Triangles: Basic Proportionality Theorem (Thales)

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.

In $\triangle ABC$, $DE \parallel BC$. Draw $DM \perp AC$ and $EN \perp AB$. Join $BE$ and $CD$. $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2}AD \cdot EN}{\frac{1}{2}DB \cdot EN} = \frac{AD}{DB}$ $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2}AE \cdot DM}{\frac{1}{2}EC \cdot DM} = \frac{AE}{EC}$

Since $\triangle BDE$ and $\triangle CDE$ lie on same base $DE$ between parallels $DE$ and $BC$, $\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE)$. Therefore, $\frac{AD}{DB} = \frac{AE}{EC}$.