A quadrangle $ABCD$ which is inscribed in a circle $O$ satisfies $$\mathrm { AB } = \mathrm { BC } = \sqrt { 2 } , \quad \mathrm { BD } = \frac { 3 \sqrt { 3 } } { 2 } , \quad \angle \mathrm { ABC } = 120 ^ { \circ } ,$$ where $$AD > CD .$$ (1) Then $\mathrm { AC } = \sqrt { \mathbf { A } }$, and the radius of circle O is $\sqrt { \mathbf { B } }$. (2) Set $x = \mathrm { AD }$. Since $\angle \mathrm { ADB } = \mathbf { CD }^\circ$, $x$ satisfies $$4 x ^ { 2 } - \mathbf { E } \mathbf { F } x + \mathbf { G H } = 0 .$$ Also, set $y = \mathrm { CD }$. In the same way, it follows that $y$ satisfies $$4 y ^ { 2 } - \mathbf { I J } y + \mathbf { K } \mathbf { L } = 0 .$$ Thus, noting (1), we obtain $AD$ and $CD$.
A quadrangle $ABCD$ which is inscribed in a circle $O$ satisfies
$$\mathrm { AB } = \mathrm { BC } = \sqrt { 2 } , \quad \mathrm { BD } = \frac { 3 \sqrt { 3 } } { 2 } , \quad \angle \mathrm { ABC } = 120 ^ { \circ } ,$$
where
$$AD > CD .$$
(1) Then $\mathrm { AC } = \sqrt { \mathbf { A } }$, and the radius of circle O is $\sqrt { \mathbf { B } }$.
(2) Set $x = \mathrm { AD }$. Since $\angle \mathrm { ADB } = \mathbf { CD }^\circ$, $x$ satisfies
$$4 x ^ { 2 } - \mathbf { E } \mathbf { F } x + \mathbf { G H } = 0 .$$
Also, set $y = \mathrm { CD }$. In the same way, it follows that $y$ satisfies
$$4 y ^ { 2 } - \mathbf { I J } y + \mathbf { K } \mathbf { L } = 0 .$$
Thus, noting (1), we obtain $AD$ and $CD$.