The two possible positions of $D$ are symmetric with respect to the line $(A B)$, so it suffices to consider the case where the triangle $A B D$ is oriented counterclockwise (see figure). The common center of the two circles is denoted by $M$. Since the radius $MD$ is perpendicular to the tangent $B C$, the triangle $M B D$ is right-angled, so from the Pythagorean theorem
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$|B D|^{2}=|M B|^{2}-|M D|^{2}=r_{1}^{2}-r_{2}^{2}=169-64=105$ follows.
By the Thales' theorem, $\angle A C B=90^{\circ}$. Since $\angle A C B=\angle M D B=90^{\circ}$ and the common angle $\angle D B M=\angle C B A$, the triangles $A B C$ and $M B D$ are similar, and since $|M B|=r_{1}=|M A|$, it also follows that $|D C|=|B D|=\sqrt{105}$ and $|C A|=2 \cdot|D M|=16$. Thus, the lengths of the legs in the right-angled triangle $A D C$ are known, and it follows that $|A D|=\sqrt{105+16^{2}}=\sqrt{361}=19$. Therefore, the side $A D$ has a length of 19.

Hints: Numerous other solution methods are possible. Because the problem is relatively simple, points were also deducted for errors in the final calculation step (examples: $\sqrt{360}, \sqrt{461}$, or even $\sqrt{361} \approx 18.5$).