4. $A B C D$ is a cyclic quadrilateral such that $|D A|=|B C|=2$, and $|A B|=4$. If $|C D|>|A B|$ and the lines $D A$ and $B C$ intersect at an angle of $60^{\circ}$, find the radius of the circumscribing circle.
Answer: $\frac{2 \sqrt{21}}{3}$

Let $E$ be the intersection of $D A$ and $B C$. Since $|D A|=$ $|B C|$, it follows from the power of a point formula that $|A E|=|B E|$, which in turn implies that $A B E$ is an equilateral triangle. Hence, $\angle A B C=120^{\circ}$. Using cosine rule, $|A C|=\sqrt{20-16 \cos 120^{\circ}}=2 \sqrt{7}$. Therefore, by the extended law of sines, the circumdiameter $=\frac{|A C|}{\sin 120^{\circ}}=\frac{2 \sqrt{7}}{\sqrt{3} / 2}$. This implies that the circumradius equals $2 \sqrt{7 / 3}$ or $2 \sqrt{21} / 3$.