17.56 Given that the three sides of triangle $P$ are $a, \sqrt{3} a, 2 a$, and the three sides of triangle $Q$ are $3 a, 2 \sqrt{3} a, \sqrt{3} a$, and triangle $R$ is a right triangle with one acute angle equal to $30^{\circ}$, let “c” denote similarity, and “ゆ” denote non-similarity. Then
(A) $P \sim Q$, but $Q \infty R, R \infty P$.
(B) $P \infty Q, Q \infty R$, but $R \backsim P$.
(C) $P \propto Q, Q \infty R, R \infty P$.
(D) $P \backsim Q \backsim R$.
(China Guangzhou, Wuhan, Fuzhou, and other five cities junior high school mathematics league, 1988)
[Solution] $\because \frac{a}{\sqrt{3} a}=\frac{\sqrt{3} a}{3 a}=\frac{2 a}{2 \sqrt{3} a}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} . \quad \therefore \quad P \cos Q$.
Also, $\because \sqrt{a^{2}+(\sqrt{3} a)^{2}}=2 a$, but $\frac{a}{2 a}=\frac{1}{2}$,
$\therefore$ Triangle $P$ is a right triangle with an internal angle of $30^{\circ}$, so $P \backsim R$. Thus, $\quad P \backsim Q \backsim R$.
Therefore, the answer is $(D)$.