10.282. In triangle $A B C$, a line $D E$ is drawn parallel to the base $A C$. The area of triangle $A B C$ is 8 sq. units, and the area of triangle $D E C$ is 2 sq. units. Find the ratio of the length of segment $D E$ to the length of the base of triangle $A B C$.
Solution.

We have $S_{\triangle D E C}=\frac{1}{2} D E \cdot H K=2$ (Fig. 10.77) and $H K=\frac{4}{D E} ; S_{\triangle A B C}=$ $=\frac{1}{2} A C \cdot B H=8$, from which $B H=\frac{16}{A C}$. Since $\triangle A B C \sim \triangle D B E$, then $\frac{D E}{A C}=\frac{B K}{B H}$.

Further, $B K=B H-H K$ and, therefore,

$$
\frac{B K}{B H}=\frac{B H-H K}{B H}=1-\frac{H K}{B H}=1-\frac{4}{D E}: \frac{16}{A C}=1-\frac{A C}{4 D E}
$$

Let $\frac{D E}{A C}=x$. Then $x=1-\frac{1}{4 x}$ or $4 x^{2}-4 x+1=0 \Rightarrow x=\frac{1}{2}$. We obtained $\frac{D E}{A C}=\frac{1}{2}$.

Answer: 1:2.