3. In an acute triangle $\triangle ABC$, $BE$ and $CF$ are two altitudes. Let the area of $\triangle AEF$ be $S_{1}$, and the area of quadrilateral $BCEF$ be $S_{2}$. If the diameter of the circumcircle of $\triangle AEF$ is exactly equal to the length of $BC$, then the relationship between $S_{1}$ and $S_{2}$ is ( ).
(A) $S_{1}>S_{2}$
(B) $S_{1}=S_{2}$
(C) $S_{1}<S_{2}$
(D) cannot be determined
3. B.

As shown in Figure 2, draw the altitude $A D$, and let $H$ be the orthocenter. It is easy to see that points $A, F, H, E$ are concyclic, with $A H$ as the diameter of the circle. Then
$$
\begin{array}{l}
A H=B C . \\
\text { Hence } \triangle A H E \cong \triangle B C E \\
\Rightarrow A E=B E \Rightarrow \angle B A C=45^{\circ} .
\end{array}
$$

It is easy to prove that points $B, C, E, F$ are concyclic.

Therefore, $\triangle A E F \backsim$
$\triangle A B C$.
$$
\begin{array}{l}
\text { Hence } \frac{S_{1}}{S_{1}+S_{2}}=\frac{S_{\triangle A E F}}{S_{\triangle A B C}} \\
=\left(\frac{A E}{A B}\right)^{2}=\left(\frac{1}{\sqrt{2}}\right)^{2} \\
=\frac{1}{2} .
\end{array}
$$

Therefore, $S_{1}=S_{2}$.