12. $A B C$ is a triangle with $A C=B C$ and $A \widehat{C} B<60^{\circ}$. Let $A^{\prime}$ and $B^{\prime}$ be points on sides $B C$ and $A C$ respectively such that $A A^{\prime}=B B^{\prime}=A B$. Let $C^{\prime}$ be the intersection of $A A^{\prime}$ and $B B^{\prime}$. Given that $A C^{\prime}=A B^{\prime}$ and $B C^{\prime}=B A^{\prime}$, what is the measure in degrees of the angle $A \widehat{C} B$?
12. The answer is 36. Indeed, triangles $C A B$ and $B A B^{\prime}$ are isosceles with a common angle adjacent to the base, so they are similar. Similarly, $B A B^{\prime}$ and $A B^{\prime} C^{\prime}$ are similar. Reapplying the same reasoning to triangles $A B A^{\prime}$ and $B A^{\prime} C^{\prime}$, it is found that all five such triangles are similar. Therefore, $A \widehat{C} B = B^{\prime} \widehat{A} C^{\prime} = A^{\prime} \widehat{A} B = A \widehat{B} B^{\prime} = C^{\prime} \widehat{B} A^{\prime}$; moreover, the sum of these five angles is equivalent to the sum of the interior angles of $A B C$, and thus $=180^{\circ}$. Therefore, $A \widehat{C} B = \frac{180^{\circ}}{5} = 36^{\circ}$.

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