4.10. From point $A$, two rays are drawn intersecting a given circle: one - at points $B$ and $C$, the other - at points $D$ and $E$. It is known that $A B=7, B C=7, A D=10$. Determine $D E$.
4.10. Since $\triangle A D C \sim \triangle A B E$ (Fig. 4.9), we have $\frac{A E}{A B}=\frac{A C}{A D}$, where $A B=B C=7, A D=10$. Let $D E=x$; then $\frac{10+x}{7}=\frac{14}{10}$, from which $100+10 x=98$, i.e., $x=-0.2$. Such a value of $x$ indicates that point $E$ is located between $A$ and $D$, i.e., $\frac{A D}{A B}=\frac{A C}{A E}$ and, 
thus, $\frac{10}{7}=\frac{14}{10-x}$. From this we find

$x=0.2$.

Answer: 0.2.