60. As shown in the figure, given that $B D$ is the bisector of $\angle A B C$, $D E \perp B C$ at $E, S_{A B C}=24, A B=8, B C=12$, then $\mathrm{DE}=$ . $\qquad$
Answer: 2.4
Solution: Draw $D F \perp A B$ at $F$.
Since $B D$ is the bisector of $\angle A B C$, and $D E \perp B C$,
we have
$$
D E=D F,
$$

By
$$
\begin{aligned}
S_{A B C} & =S_{A B D}+S_{B C D} \\
& =\frac{1}{2} A B \cdot D F+\frac{1}{2} B C \cdot D E, \\
& =24,
\end{aligned}
$$

which means
$$
\frac{1}{2} \cdot 8 \cdot D E+\frac{1}{2} \cdot 12 \cdot D E=24,
$$

Solving for $D E$ gives
$$
D E=2.4 \text {. }
$$