# Find the angle $MBA$.

Triangle ABC has $\angle CAB=30 ^\circ$ and $\angle CBA=70^\circ$. Point $M$ lies inside triangle $ABC$ so that $\angle BAM=20^\circ= \angle ACM$. Find $\angle MBA$.

I've already drawn the diagram, but I can't get the angle. Also, I would greatly appreciate if someone could help me with the latex. Thanks!

• You can try Ceva's theorem. en.wikipedia.org/wiki/Ceva%27s_theorem For example in the picture given in the wikipedia page, notice $\dfrac{AF}{FB}=\dfrac{\sin \angle ACO}{\sin \angle BCO}$ – Fan May 14 '17 at 21:45
• Is $M$ a center of a circle? – JJR May 14 '17 at 22:03

Observe that $\angle \, BCM = 60^{\circ}$. On line $CM$ take point $O$ such that $BC = CO$ so that $M$ is between $C$ and $O$. Then triangle $BCO$ is isosceles with $\angle \, BCO = 60^{\circ}$. Therefore $BCO$ is in fact equilateral with $$BO = CO = BC \,\, \text{ and } \,\, \angle \, BOC = \angle \, BCO = \angle \, CBO = 60^{\circ}$$ Since $$\angle \, BAC = 30^{\circ} = \frac{1}{2} \, \angle \, BOC$$ point $O$ is in fact the center of the circle superscribed around the triangle $ABC$ so $$AO = BO = CO = BC$$

Let $D$ be the point with properties $\angle \, DCM = 20^{\circ} = \angle \, ACM$ and $AC = DC$. The triangle $ACD$ is isosceles with $CM$ its angle bisector. Therefore, $CM$ is the orthogonal bisector of segment $AD$. Since $O$ is on $CM$, triangles $ACO$ and $DCO$ are congruent so $$\angle \, CAO = \angle \, CDO = 20^{\circ}$$ At the same time, triangles $ACM$ and $DCM$ are also congruent, so $$\angle \, CDM = \angle \, CAM = 10^{\circ}$$ which means that $DM$ is the angle bisector of $\angle \, CDO$.

If you calculate the angles (which is straight-forward) you find that $$\angle \, BNC = 80^{\circ} = \angle \, DNO$$ Furthermore, again a simple angle chase yields $\angle \, DON = 80^{\circ}$. Thus $$\angle \, DON = 80^{\circ} = \angle \, DNO$$ so triangle $DNO$ is isosceles with $DN = DO$. however, $DM$ is the angle bisector of $angle \, NDO = \angle \, CDO$ so $CM$ is the orthogonal bisector of segment $NO$. Thus, $MN = MO$. However, $\angle \, NOM = \angle\, BOC = 60^{\circ}$ so triangle $MNO$ is equilateral. Hence $$MN = MO = NO$$

Finally, we can conclude that triangles $MBO$ and $NCO$ are congruent because $MO = NO, \,\, BO = CO$ and $\angle \, BOM = \angle \, CON = 60^{\circ}$. Consequently, $$\angle \, MBO = \angle \, NCO = 20^{\circ}$$ Therefore $$\angle \, MBA = \angle \, MBO + \angle \, OBA = 20^{\circ} + 10^{\circ} = 30^{\circ}$$

• Thanks for the non trig solution! This is the one I was looking for! :D – user406996 May 19 '17 at 2:20

Let MBA=x. Since AM, CM and BM are collinear, we use the sine form of Ceva's Theorem. $$sinACMsinBAMsinCBM=sinCAMsinABMsinBCM$$ $$sin20\sin20\sin(70-x)=sin10\ sinx\sin60$$ You can obtain all these angles by drawing the triangle and then angle chasing. Notice that x must be a multiple of 10 since problem makers don't want to be assholes and force you to do tedious calculations. A rough drawing gives that x must be 20 or 30, and x=30 solves the above equation found by using Ceva's Theorem. Thus MBA is 30.

To make it rigorous, notice \begin{align} \sin 10^{\circ}\sin 60^{\circ}&=(\cos 50^{\circ}-\cos 70^{\circ})/2\\ &=(\sin40^{\circ}-\sin 20^{\circ})/2\\ &=\frac{\sin 20^{\circ}}{2}(2\cos 20^{\circ}-1)\\ &=\sin 20^{\circ}(\cos 20^{\circ}-\sin 30^{\circ})\\ &=\sin 20^{\circ}(\sin 70^{\circ}-\sin 30^{\circ})\\ &=2\sin 20^{\circ}\sin 20^{\circ}\cos 50^{\circ}. \end{align}

We can see $x=30^{\circ}$.

• Good job, but how do you know where to go at each step? You could've taken a wrong turn in the middle steps and failed to arrive at the final result. – Display name May 14 '17 at 23:10
• @Displayname It is true. Luckily there are only two main steps here, and your goal is to create two $20^\circ$ from $10^\circ$ and $60^\circ$. So actually options are limited. – Fan May 14 '17 at 23:14