$7 \cdot 49$ A circle with radius $r$ has chords $AB, CD$ of lengths 10 and 7, respectively. The extensions of $AB$ and $CD$ meet at a point $P$ outside the circle. If $\angle APD=60^{\circ}, BP=8$, then $r^{2}$ is
(A) 70.
(B) 71.
(C) 72.
(D) 73.
(E) 74.
(43rd American High School Mathematics Examination, 1992)
[Solution] Let $PD = x$. By the secant theorem, we have
$$
8 \cdot 18 = x \cdot (x + 7),
$$

which gives $x = 9$ (the negative value has been discarded).
Connect $BC$. In $\triangle PBC$, by the cosine rule, we have
$$
\begin{aligned}
BC^2 & = PC^2 + PB^2 - 2 \cdot PC \cdot PB \cdot \cos 60^\circ \\
& = 16^2 + 8^2 - 16 \cdot 8 = 192.
\end{aligned}
$$

Connect $AC$. In $\triangle APC$, by the cosine rule again, we have
$$
\begin{aligned}
AC^2 & = PA^2 + PC^2 - 2 \cdot PA \cdot PC \cdot \cos 60^\circ \\
& = 18^2 + 16^2 - 16 \cdot 18 = 292.
\end{aligned}
$$

Thus, $AC^2 = BC^2 + AB^2$, so
$$
\angle ABC = 90^\circ.
$$

Therefore, $AC$ is the diameter of the circle, and thus $r^2 = \frac{AC^2}{4} = \frac{292}{4} = 73$. Hence, the answer is $(D)$.