Let $A B C D$ be a trapezoid such that $(A B)$ is parallel to $(C D), A B=3 C D=3 D A$ and $\widehat{A D C}=120^{\circ}$. Determine the angle $\widehat{C B A}$ in degrees.
Answer: 30. We can assume that $CD=1$ and $AB=3$. Let $M$ and $N$ be the points on $[AB]$ such that $AM=MN=NB=1$. Then $AMCD$ is a parallelogram because $AM=DC$ and $(AM) \|(DC)$. Furthermore, $DA=DC$ so it is a rhombus, and thus $(DM)$ is the bisector of $\widehat{ADC}$. We deduce that $ADM$ is equilateral, so $\widehat{MAD}=60^{\circ}$. Since $MNCD$ is a parallelogram, we also have $\widehat{NMC}=60^{\circ}$. Moreover, $NM=NB=1$ and $NC=MD=1$, so $MBC$ is a right triangle at $C$ (and $N$ is the center of its circumcircle), therefore $\widehat{CBA}=180^{\circ}-\widehat{BMC}-\widehat{MCB}=30^{\circ}$.