3. As shown in Figure 2, in $\odot O$, $MN$ is a chord, and the sides $AB$ and $GH$ of the squares $ABCD$ and $EFGH$ are also chords. Points $D, E, F, C$ are all on $MN$. If the distance from $O$ to $MN$ is $OS=h$, then $AB-HG$ equals ( ).
(A) $\frac{4}{3} h$
(B) $\frac{7}{4} h$
(C) $\frac{8}{5} h$
(D) $\frac{11}{6} h$
3.C.

As shown in Figure 7, it is easy to see that $S$ is the midpoint of $MN$, and also the midpoint of $DC$ and $EF$.

Obviously, points $H$, $S$, and $B$ are collinear.
Draw $OK \perp HB$ at $K$, then $BK=HK$.

Also, $\frac{SK}{OK}=\frac{HE}{ES}=2$, thus, $SK=\frac{2}{\sqrt{5}} h$.
From $2 SK=BS-HS$
$$
=\frac{\sqrt{5}}{2} BC-\frac{\sqrt{5}}{2} HE=\frac{\sqrt{5}}{2}(AB-HG) \text {, }
$$

Therefore, $AB-HG=\frac{2}{\sqrt{5}} \times 2 \times \frac{2}{\sqrt{5}} h=\frac{8}{5} h$.