Point lies inside a triangle ABC with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ$ A triangle $ABC$ is given with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ.$ Point $M$ lies inside the triangle and $\measuredangle MAB=\measuredangle MBA=15^\circ.$ Find $\measuredangle BMC$.

I am not sure how to approach the problem. We can find some angles:$$\measuredangle ACB=180^\circ-45^\circ-30^\circ=105^\circ\\ \measuredangle CAM=45^\circ-15^\circ=30^\circ\\ \measuredangle MBC=30^\circ-15^\circ=15^\circ\\ \measuredangle AMB=180^\circ-2\cdot15^\circ=150^\circ.$$
I will be very grateful if we can see a solution without using trigonometry.

Please apply Trigonometric form of Ceva's theorem.
If $\angle BCM = x$,
$\sin \angle BAM \ \sin \angle ACM \sin CBM = \sin \angle MAC \ \sin \angle MCB \ \sin \angle MBA$
$\sin 15^0 \sin (105^0-x) \sin 15^0 = \sin 30^0 \sin x \sin 15^0$
$\cos (15^0-x) \sin 15^0 = 2 \sin 15^0 \cos 15^0 \sin x$
$\sin 30^0 \cos (15^0-x)  = \cos 15^0 \sin x$
$\sin (45^0-x) + \sin (15^0+x)  = \sin (15^0+x) + \sin(x-15^0)$
$45^0-x = x-15^0 \implies x = 30^0$
So $\angle BMC = 180^0 - 30^0 - 15^0 = 135^0$.
EDIT: For geometric solution, see the below construction that you can use. $AN$ is angle bisector of $\angle CAM$. You can prove $\angle ANM = \angle ANC = 60^0$ and $CN = NM$. So, $\angle MCB = 30^0$. $\therefore \angle BMC = 135^0$.

