20.79 A quadrilateral is circumscribed around a circle, and the circle is divided into four arcs by the vertices of the quadrilateral. The sum of the angles of the corresponding segments with the sides of the quadrilateral as bases is 
(A) $180^{\circ}$.
(B) $540^{\circ}$.
(C) $360^{\circ}$.
(D) $450^{\circ}$.
(E) $1080^{\circ}$.
(The segments refer to those whose sides do not intersect the sides of the quadrilateral again)
(5th American High School Mathematics Examination, 1954)
[Solution] As shown in the figure, let the degree measures of the four arcs divided by the quadrilateral in the circle be $a, b, c, d$, and let the degree measures of the four segments be $x, y, z, t$, with the sum of the segment angles being $s$.
By $\angle x \stackrel{\mathrm{m}}{=} \frac{1}{2}(b+c+d)$, similarly, we can have the sum of the four angles:
$$
\begin{aligned}
s \stackrel{\mathrm{m}}{=} & \frac{1}{2}[(a+b+c)+(b+c+d)+(c+d+a) \\
& +(d+a+b)] \\
= & \frac{3}{2}(a+b+c+d),
\end{aligned}
$$

Thus, $s=360^{\circ} \cdot \frac{3}{2}=540^{\circ}$.
Therefore, the answer is $(B)$.