Thales' intercept theorem

Assume first that $DE \parallel BC$. Compare $S_{\triangle BDE}$ and $S_{\triangle CDE}$: both triangles have the same base $DE$, and since $DE \parallel BC$ means $B$ and $C$ lie on a line parallel to $DE$, the perpendicular distance from $B$ to line $DE$ equals the perpendicular distance from $C$ to line $DE$. Equal base and equal height give $S_{\triangle BDE}$ $=$ $S_{\triangle CDE}$.

Now compare $S_{\triangle ADE}$ to $S_{\triangle BDE}$: viewed with apex $E$, triangle $ADE$ has base $AD$ along line $AB$ and triangle $BDE$ has base $DB$ along the same line, sharing the same height from $E$ down to line $AB$. Two triangles with equal height have areas in the ratio of their bases, so $S_{\triangle ADE}$$/$$S_{\triangle BDE}$ $=$ $\frac{AD}{DB}$. By the same argument with apex $D$ and base line $AC$, $S_{\triangle ADE}$$/$$S_{\triangle CDE}$ $=$ $\frac{AE}{EC}$.

Since $S_{\triangle BDE}$ $=$ $S_{\triangle CDE}$ from the first step, the two ratios above share the same denominator once rewritten, so $\frac{AD}{DB}$ $=$ $S_{\triangle ADE}$$/$$S_{\triangle BDE}$ $=$ $S_{\triangle ADE}$$/$$S_{\triangle CDE}$ $=$ $\frac{AE}{EC}$. This proves $\frac{AD}{DB} = \frac{AE}{EC}$.

For the converse, suppose $\frac{AD}{DB} = \frac{AE}{EC}$ holds. Let $E'$ be the point on $AC$ such that $DE' \parallel BC$; the forward direction just proved gives $AD/DB = AE'/E'C$. Combining with the hypothesis $AE/EC = AD/DB$ gives $AE'/E'C = AE/EC$, and since a point dividing segment $AC$ in a fixed ratio is unique, $E' = E$. Therefore $DE = DE' \parallel BC$.