What's the length of the chord?

Since $ACBD$ is a cyclic quadrilateral , then $\angle BCD = \angle BAD = 30^\circ$ and $\angle CDB = \angle CAB = 30^\circ$ . And $\triangle BCD$ is isosceles. Let $BC = DB = b$ . Then $DC = 2b \cos 30^\circ = \sqrt 3b$ . By Ptolemy's theorem

$\begin{aligned} AC \cdot BD + BC \cdot AD & = AB \cdot CD \\ ab + b(a+2) & = 6\sqrt 3 \cdot \sqrt 3 b & \small \blue{\text{Divide both sides by }b} \\ a + a + 2 & = 18 \\ \implies a & = \boxed 8 \end{aligned}$

It is straight forward to find that $\angle ACB + \angle ADB = 180^{\circ}$ , and that $\overline{BC} = \overline{BD}$ . Hence, we can rotate the lower triangle $\triangle ABD$ about point $B$ by $120^{\circ}$ clockwise to obtain an isosceles triangle having the two equal sides $= 6\sqrt{3}$ and the base angle $30^{\circ}$ . The base is $a + (a + 2) = 2 (a + 1)$ , and therefore, $a + 1 = 6 \sqrt{3} \sin 60^{\circ} = 9$ , hence $a = 8$ .

In right $\triangle{ACE}: \:\ \:\ \overline{AE} = a\cos(30^{\circ}) = \dfrac{\sqrt{3}}{2}$ and $h = \overline{CE} = \dfrac{a}{2}$

Using law of cosines on $\triangle{ACD}$ with included $\angle{CAD} \implies \overline{CD}^2 = a^2 + 2a + 4$

Using law of cosines on $\triangle{CBD}$ with included $\angle{CBD} \implies$

$a^2 + 2a + 4 = 2x^2(1 + \cos(60^{\circ})) = 3x^2 \implies x = \sqrt{\dfrac{a^2 + 2a + 4}{3}}$

and $h = \dfrac{a}{2} = \sqrt{\dfrac{a^2 + 2a + 4}{3}}\sin(\theta) \implies \sin(\theta) = \dfrac{a}{2}\sqrt{\dfrac{a^2 + 2a + 4}{3}} \implies$

$\cos(\theta) = \dfrac{\sqrt{a^2 + 8a + 16}}{2\sqrt{a^2 + 2a + 4}} \implies \overline{EB} = \sqrt{\dfrac{a^2 + 2a + 4}{3}}\cos(\theta) =$

$\dfrac{\sqrt{a^2 + 8a + 16}}{2\sqrt{3}} \implies \overline{AB} = \overline{EB} + \overline{AE} =$ $\dfrac{1}{2\sqrt{3}}(\sqrt{a^2 + 8a + 16} + 3a) = 6\sqrt{3}$

$\implies \sqrt{a^2 + 8a + 16} = 3(12 - a) \implies a^2 + 8a + 16 = 1296 - 216a + 9a^2 \implies$

$8(a^2 - 28a + 160) = 0 \implies (a - 20)(a - 8) = 0$ and $a < 12 \implies \boxed{a = 8}$ .