3. As shown in Figure 3, in isosceles right $\triangle ABC$, $D$ is a trisection point on the hypotenuse, $AD=2DB$, $E$ is the midpoint of $BC$, and $CD$ and $AE$ intersect at point $F$. Then, an equation satisfied by $FA$, $FC$, and $FE$ is
3. $F C^{2}=F A \cdot F E$.

It is only necessary to prove: In the right triangle $\triangle A C E$, $C F$ is the altitude to the hypotenuse.
As shown in Figure 5, connect
$D E$, given $\angle D B E=$
$\angle D A C=45^{\circ}$ and $B E$
$=E C$, we know
$$
\frac{D B}{D A}=\frac{B E}{A C}=\frac{1}{2} \text {. }
$$

Thus, $\triangle D B E \backsim \triangle D A C$.
Figure 5

Therefore, $\angle B E D=\angle A C D$.
Draw $D H \perp B E$, then
$$
H E=B E-B H=\frac{1}{2} B C-\frac{1}{3} B C=\frac{1}{6} B C=\frac{1}{3} C E \text {. }
$$

Thus, $\frac{H E}{C E}=\frac{1}{3}=\frac{D H}{A C}$.
Hence, Rt $\triangle D H E \backsim \mathrm{Rt} \triangle A C E$.
Therefore, $\angle A E C=\angle D E H=\angle A C F$.
Then $\angle A F C=180^{\circ}-\angle A C F-\angle C A F$
$$
=180^{\circ}-\angle A E C-\angle C A E=\angle A C E=90^{\circ} \text {. }
$$

Thus, $C F$ is the altitude to the hypotenuse of the right triangle $\triangle A C E$, so we have
$$
F C^{2}=F A \cdot F E \text { or } \frac{F C}{F A}=\frac{F E}{F C} \text {. }
$$