$20.76 \triangle ABC$ is inscribed in a semicircle with radius $r$, with base $AB$ on the diameter $AB$, and point $C$ not coinciding with $A, B$. Let $s=AC+BC$, for all possible positions of $C$, we have
(A) $s^{2} \leqslant 8 r^{2}$.
(B) $s^{2}=8 r^{2}$.
(C) $s^{2} \geqslant 8 r^{2}$.
(D) $s^{2} \leqslant 4 r^{2}$.
(E) $s^{2}=4 r^{2}$.
(12th American High School Mathematics Examination, 1961)
[Solution] Let $B C=a, A C=b$. Since $A B$ is the diameter of $\odot O$, then $\angle A C B=90^{\circ}$. By the Pythagorean theorem,
$$
a^{2}+b^{2}=4 r^{2} \text {. }
$$

Also, $s=a+b$, then
$$
s^{2}=a^{2}+b^{2}+2 a b=4 r^{2}+2 a b \text {, }
$$

and $2 a b \leqslant a^{2}+b^{2}=4 r^{2}$, with equality holding if and only if $a=b=\sqrt{2} r$.
Therefore,
$$
s^{2} \leqslant 8 r^{2}
$$

Hence, the answer is $(A)$.