. Let $ABC$ be a right triangle at $C$. We denote by $M$ the midpoint of $[AB]$ and by $H$ the foot of the altitude from $C$. Given that $CH=1$ and $CM=2$, determine the angle $\widehat{CAB}$.
. Let $\alpha=\widehat{B M C}$. We have $\sin \alpha=\frac{H C}{M C}=\frac{1}{2}$, so $\alpha=30^{\circ}$ or $\alpha=150^{\circ}$. Triangle $M B C$ is isosceles at $M$, so $\widehat{M B C}=\frac{180^{\circ}-\alpha}{2}$, and $\widehat{B A C}=90^{\circ}-\widehat{M B C}=\frac{\alpha}{2}$. Therefore, we have two possibilities: either $\widehat{M B C}=15^{\circ}$, or $\widehat{M B C}=75^{\circ}$. (Note that we switch from one possibility to the other by swapping $A$ and $B$).

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