$19 \cdot 12$ As shown in the figure, if $a, b$, and $d$ are the side length, the shortest diagonal, and the longest diagonal of a regular nonagon, respectively, then
(A) $d=a+b$.
(B) $d^{2}=a^{2}+b^{2}$.
(C) $d^{2}=a^{2}+a b+b^{2}$.
(D) $b=\frac{a+d}{2}$.
(E) $b^{2}=a d$.
(28th American High School Mathematics Examination, 1977)
[Solution] As shown in the figure, $P_{1} P_{2} \cdots P_{9}$ is a regular nonagon; and
$P_{1} P_{2}=a, \quad P_{2} P_{4}=b, \quad P_{1} P_{5}=d$.
Also, $Q$ and $R$ are on $P_{1} P_{5}$, and $P_{2} Q \perp P_{1} P_{5}$, $P_{4} R \perp P_{1} P_{5}$.
Since $P_{2} P_{3}=P_{3} P_{4}$ and each interior angle of the regular nonagon is
$$
\left[\frac{180(9-2)}{9}\right]^{\circ}=140^{\circ} \text {, }
$$

we can get $\angle P_{3} P_{2} P_{4}=20^{\circ}$.
Therefore, $\angle P_{1} P_{2} Q=30^{\circ}, \quad P_{1} Q=\frac{a}{2}$.
Similarly, $P_{5} R=\frac{a}{2}$.
Thus, $P_{1} P_{5}=P_{1} Q+P_{2} P_{4}+P_{5} R$, which means $d=a+b$. Hence, the answer is $(A)$.