30.6. Let $A B C$ be an isosceles triangle with $A B=A C$. Let $D$ and $E$ be the feet of the perpendiculars from $B$ and $C$ to $\overline{A C}$ and $\overline{A B}$, respectively. Suppose that $\overline{C E}$ and $\overline{B D}$ intersect at point $H$. If $E H=1$ and $A D=4$, find $D E$.
30.6. $\frac{8 \sqrt{17}}{17}$

Refer to Figure 6. By Pythagorean Theorem, we get $A H=\sqrt{17}$. Since $\angle A E H+\angle A D H=90^{\circ}+90^{\circ}=180^{\circ}$, the quadrilateral $A D H E$ is cyclic. By Ptolemy's Theorem (in a cyclic quadrilateral, the product of the lengths of the diagonals equals the sum of the products of the lengths of the opposite sides), we have
$$
D E \cdot \sqrt{17}=1 \cdot 4+1 \cdot 4,
$$
or $D E=\frac{8 \sqrt{17}}{17}$.