Find angle without trigonometry

Question

I solved the following problem using the sine law. Desired value is $\angle MAC=10°$. Can you find a geometric solution?

Answer 1

Maybe you can use the following. Consider a regular 9-gon $I_1I_2\ldots I_9$ with center $O$, and find your picture by setting $A=O$, $B=I_3$ and $C=I_5$. Then note that $M$ is the intersection of diagonals $I_1I_5$ and $I_3I_6$.

Answer 2

Let's go another way around.

Let $M'$ be such that $\angle M'CA = \angle M'AC = 10^{\circ}$ and let $D$ be a reflection of $C$ across line $AM'$. Then $AM' = CM' = DM'$ and a triangle $ABD$ is equlateral (since $AD = AC = AB$ and $\angle BAD = 60^{\circ}$). So $ABM'$ and $DBM'$ are congruent so $\angle DBM' = 30^{\circ}$

Now, since $A$ is a center of circle around through $B,D$ and $C$ we have also $$\angle DBC = {1\over 2}\angle DAC = 10^{\circ}$$ so $\angle CBM' = 20^{\circ}$ and thus $M'=M$.