AMC 12 A

Place the figure in the coordinate plane with $D = (0, 0)$ and $A = (5, 0)$. Because $\angle CDA = 120^\circ$ and $CD = 3$, it follows that $C = \left(-\frac{3}{2}, \frac{3}{2}\sqrt{3}\right)$. The perpendicular bisectors of $\overline{CD}$ and $\overline{AD}$ intersect at the center of the circumscribing circle. The midpoint of $\overline{CD}$ is $\left(-\frac{3}{4}, \frac{3}{4}\sqrt{3}\right)$. The line through $\overline{CD}$ has slope $-\sqrt{3}$, so a line perpendicular to that has slope $\frac{\sqrt{3}}{3}$. The perpendicular bisector of $\overline{CD}$ therefore has equation
$$y - \frac{3}{4}\sqrt{3} = \frac{\sqrt{3}}{3}\left(x + \frac{3}{4}\right).$$
The perpendicular bisector of $\overline{AD}$ has equation $x = \frac{5}{2}$. Solving this system of equations locates the center of the circumscribing circle at $O\left(\frac{5}{2}, \frac{11}{6}\sqrt{3}\right)$.

By the Distance Formula, the radius of the circle is
$$r = OD = \sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{11}{6}\sqrt{3}\right)^2} = \frac{7\sqrt{3}}{3}.$$
Let $\theta$ be the measure of $\angle BCO$. By the Law of Cosines applied to $\triangle BCO$,
$$\frac{49}{3} = BO^2 = BC^2 + CO^2 - 2BC \cdot CO \cos \theta = 9 + \frac{49}{3} - 2 \cdot 3 \cdot \frac{7\sqrt{3}}{3} \cos \theta,$$
which gives $\cos \theta = \frac{3}{14}\sqrt{3}$. A Double Angle Formula then gives $\cos 2\theta = 2 \cdot \frac{27}{196} - 1 = -\frac{71}{98}$. Triangles $\triangle BOC$ and $\triangle COD$ are congruent isosceles triangles, so $\angle BCD$ has measure $2\theta$. By the Law of Cosines applied to $\triangle BCD$,
$$BD^2 = BC^2 + CD^2 - 2BC \cdot CD \cos 2\theta = 9 + 9 - 2 \cdot 3 \cdot 3 \cdot \left(-\frac{71}{98}\right) = \frac{1521}{49} = \left(\frac{3 \cdot 13}{7}\right)^2,$$
so $BD = \frac{39}{7}$. The Law of Cosines applied to $\triangle ADC$ gives $AC = 7$, so the shorter diagonal has length $\frac{39}{7}$.

The Law of Cosines applied to $\triangle ADC$ gives
$$CA^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^\circ.$$
Because $\cos 120^\circ = -\frac{1}{2}$, diagonal $\overline{AC}$ has length $\sqrt{9+25+15} = 7$. Because $ABCD$ is cyclic, $\angle ABC$ is supplementary to $\angle ADC$, so $\angle ABC = 60^\circ$. The Law of Cosines applied to $\triangle ABC$ gives
$$7^2 = 3^2 + BA^2 - 2 \cdot 3 \cdot BA \cdot \cos 60^\circ.$$
Simplifying yields $BA^2 - 3 \cdot BA - 40 = 0$, so $BA = 8$. By Ptolemy's Theorem
$$CA \cdot BD = BA \cdot CD + BC \cdot AD.$$
Substituting gives $7 \cdot BD = 8 \cdot 3 + 3 \cdot 5$, so $BD = \frac{39}{7}$, which is less than the length of diagonal $\overline{AC}$.