Point lies inside a triangle ABC with $\measuredangle BAC=45^\circ$ and $\measuredangle ABC=30^\circ$

Question

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.

Answer 1

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, $\angle MCB = 30^0$. $\therefore \angle BMC = 135^0$.

Answer 2

I found a solution without trigonometry. The key is consider the midpoint of $BC$, which I have named $N$ in the picture below. Using the dashed lines and the observations that $|AD| = |CD| = |CN| = |DN| = |BN|$, it is easy to show that $\angle BAN = 15^{\circ}$. This means that $A$, $M$ and $N$ are colinear, and in addition that $\triangle NMB \sim \triangle NBA$. This gives the ratio $|NM| / |NB| = |NB| / |NA|$, and since $N$ was the midpoint of $BC$, we conclude that also $|NM| / |NC| = |NC| / |NA|$. This means that $\triangle NMC \sim \triangle NCA$, from which it follows that $\angle BMC = \angle BMN + \angle NMC = 30^{\circ} + 105^{\circ} = 135^{\circ}$, as desired.

A final remark: it is actually true that $|AM| = |AC|$, as follows easily from the above proof. However, I could not immediately find a solution that somehow exploits this fact, but it seems to me that it should be possible. Anyone up for it?