Constructing a cyclic quadrilateral of given sides. Suppose we are given sides $a,b,c,d$. We need to construct a cyclic quadrilateral with the given sides. How can we do that?
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 A: Here's a consolidated construction of the angle $\alpha$ between edges of length $a$ and $d$. As noted in Wikipedia's "Cyclic Quadrilateral" article, the angle satisfies
$$\tan\frac{\alpha}{2} = \sqrt{\frac{s-a}{s-b}\frac{s-d}{s-c}} \tag{$\star$}$$
where $s := \frac{1}{2}(a+b+c+d)$ is the semi-perimeter.

In the diagram, the circle about $O$ through $U$ and $V$ has unit radius, and $S$ is easily constructed such that $|OS| = s$. Circles of radius $a$, $b$, $c$, $d$ about $S$ give points $S_a$, $S_b$, $S_c$, $S_d$ on $OS$ such that $S_x = s-x$.
We connect $S_b$ and $S_c$ (corresponding to the denominators of the fractions in $(\star)$) to $U$ and $V$, respectively. Parallels to the connecting lines, passing through $S_a$ and $S_d$ (corresponding to the numerators), determine points $P$ and $Q$ such that
$$|OP| = p := \frac{s-a}{s-b} \qquad |OQ| = q := \frac{s-d}{s-c}$$
The semicircle with diameter $PQ$ meets $\overleftrightarrow{OS}$ at $T$ such that
$$|OT| = \sqrt{pq} = \tan\frac{\alpha}{2} = \tan \angle OUT = \tan \angle OVT$$ 
Therefore, 
$\angle UTV = 180^\circ - \alpha$. From there, finding $\alpha$ itself is trivial. (Of course, $180^\circ - \alpha$ itself is the angle between edges of length $b$ and $c$ in the quadrilateral.) This is enough information to complete the construction of the entire quadrilateral.
