In triangle $A B C$, angle $C$ is twice angle $A$ and $b=2 a$. Find the angles of this triangle.

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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 $60^{\circ}$.