21. As shown in the figure, in $\triangle \mathrm{ABC}$, $A B=A C=5, B C=8$, point $\mathrm{E}$ is the midpoint of $\mathrm{BC}$, $E F \perp A B$ at point $\mathrm{F}$, then the length of $\mathrm{EF}$ is ( )
(A) $\frac{12}{5}$
(B) $\frac{24}{5}$
(C) $\frac{6}{5}$
(D) $\frac{11}{5}$
Answer: (A)
Solution: Connect $A E$.
Since $A B=A C$, point $E$ is the midpoint of $B C$, and $B C=8$,
Therefore,
$$
A E \perp B C, B E=4,
$$

In $\mathrm{Rt} A B E$, by the Pythagorean theorem, we have
$$
\begin{array}{l}
A E=\sqrt{A B^{2}-B E^{2}}=\sqrt{5^{2}-4^{2}}=3 . \\
S_{A B E}=\frac{1}{2} A E \cdot B E=\frac{1}{2} A B \cdot E F, \\
E F=\frac{A E \cdot B E}{A B}=\frac{3 \times 4}{5}=\frac{12}{5} .
\end{array}
$$

Therefore, the answer is (A).