Proving that $\angle MAC = 30^{\circ}$

Question

In $\Delta ABC$, $M$ is an interior point such that,

$MB=MA$,

$MC=CB$,

$\angle CBA = 2 \angle BAC$.

Prove that $\angle MAC = 30^\circ$

I was able to solve this question using trigonometry here's my solution I was however unable to find any synthetic solution and require some aid for that.

Answer 1

First construct the $\triangle{AMD}\cong\triangle{BMC}$.

Let $\angle CBA = 2x, \angle BAC = x, \angle MAB=\angle MBA=a$

And let the intersection of $\overline{AC}$ and $\overline {BD}$ be $O$

By looking at the diagram below you can easily find that $\angle CAD=x$ and $\angle MAC=x-a$.

As $\triangle{ABD}\cong\triangle{ABC}$ you can find out that quadrilateral $ABCD$ is an isosceles trapezoid a.k.a trapezoid inscribed in a circle

Then you can easily find $\angle ACD = x$ and $AD=DC=MD=MC$.

Therefore $\triangle {MCD}$ is equilateral and $\angle CMD=60$.

As The sum of the angles around a single point is equal to 360 degrees you get $\angle CMD=180-4x+4a$

By both of the above equations you get,

$180-4x+4a=60$

By simplifying, $x-a=30=\angle MAC$

*Sorry for my bad handwriting if you need any further clarification please ask in the comments.

Thank you for sharing such an amazing problem.