12.56*. In triangle $ABC$, angle $C$ is twice as large as angle $A$ and $b=2a$. Find the angles of this triangle.
12.56. Let $CD$ be the bisector. Then $BD = ac/(a+b)$. On the other hand, $\triangle BDC \sim \triangle BCA$, so $BD:BC = BC:BA$, i.e., $BD = a^2/c$. Therefore, $c^2 = a(a+b) = 3a^2$. The sides of triangle $ABC$ are $a, 2a$, and $\sqrt{3}a$, so its angles are 30, 90, and $60^{\circ}$.