10. Given trapezoid $A B C D$ with bases $A D=3, B C=1$, the diagonals intersect at point $O$, two circles intersect base $B C$ at points $K, L$, these two circles are tangent at point $O$, and are tangent to line $A D$ at points $A, D$ respectively. Find $A K^{2}+D L^{2}$.
10. As shown in Figure 4, draw the common tangent of the two circles through point $O$, intersecting the lower base $AD$ at point $P$. By the properties of tangents, we have
$$
A P=O P=D P \text {. }
$$

This indicates that $\triangle A O D$ is a right triangle and is similar to $\triangle C D B$, with the similarity ratio $k=\frac{B C}{A D}=\frac{1}{3}$.
It is easy to see that $\triangle A K C \backsim \triangle A O K$.
Thus, $A K^{2}=A O \cdot A C$
$$
=A O \cdot \frac{A O+O C}{A O} \cdot A O=\frac{4}{3} A O^{2} \text {, }
$$

Similarly, $D L^{2}=\frac{4}{3} O D^{2}$.
Therefore, $A K^{2}+D L^{2}=\frac{4}{3} A O^{2}+\frac{4}{3} O D^{2}$
$$
=\frac{4}{3} A D^{2}=\frac{4}{3} \times 3^{2}=12 \text {. }
$$