3. On a circle with center $M$ lie points $A, B$, and $C$. The reflection of $M$ in the line $A B$ lies inside triangle $A B C$ and is the intersection of the angle bisectors of angle $A$ and angle $B$. (The angle bisector of an angle is the line that divides the angle into two equal angles.)

Determine the size of angle $C$.
## 3. Version for Class 4 and Lower

The reflection of point $M$ in the line $A B$ is called $I$. We define $\alpha=\angle C A I$ and $\beta=\angle C B I$. Since $A I$ is the bisector of $\angle C A B$, we see that $\angle I A B=\alpha$. Because $I$ is the reflection of $M$ in the line $A B$, we see that $\angle B A M=\alpha$. Triangle $A M C$ is isosceles with vertex angle $M$, because $|A M|=|C M|$. Therefore, we see that $\angle M C A=\angle C A M=3 \alpha$. Similarly, we see that $\angle I B A=\angle A B M=\beta$ and $\angle M C B=3 \beta$. The sum of the angles in triangle $A B C$ is now $2 \alpha+(3 \alpha+3 \beta)+2 \beta=180^{\circ}$. From this, we deduce that $\alpha+\beta=\frac{180^{\circ}}{5}=36^{\circ}$ and thus that $\angle A C B=3 \alpha+3 \beta=3 \cdot 36^{\circ}=108^{\circ}$.

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