In the triangle $ABC$, $BM$ is altitude and $E$ is in that segment such that $CE$ is angle bisector. Also, the angle $EAM = 30º$ and the angle $MCB = 20º$. Find the value of $ABM$.
My problem with this exercise is that it has to be done without trigonometry. I did it with trigonometry and I got that the answer is $40º$ (if i am not mistaken). Any hints are appreciated.
On
An alternative ending to another answer, borrowing the figure from that answer and starting at the point where it has been proved that $\triangle BOE'$ is equilateral:
Since $\triangle AEE'$ is equilateral, $E'A \cong E'E$. Also, $E'B \cong E'O$ and $\angle AE'B \cong \angle EE'O$, so by side-angle-side, $\triangle AE'B \cong \triangle EE'O$ and therefore $\angle ABM \cong \angle ABE' \cong \angle EOE'$. But $$ \angle EOE' = 180^\circ - 80^\circ - 60^\circ = 40^\circ. $$
Therefore $\angle ABM = 40^\circ.$
In my experience, when angles are nice multiples of $10^\circ$ or $\,20^\circ$, looking for a suitable equilateral $\triangle$ will usually lead you to a solution.
Solution : Take $E'$ symmetric to $E$ wrt to $BC$. So $\angle E'AM=\angle EAM=30^\circ$, $\angle E'CM=\angle ECM=10^\circ$. We see $\triangle EAE'$ is equilateral and $\angle BCE' = 30^\circ$.
One great thing about an angle of a $\triangle$ being equal to $30^\circ$ is that its opposite side subtends $60^\circ$ at the circumcenter, forming an equilateral $\triangle$. So let $O$ be the circumcenter of $\triangle BCE'$. In figure, $\triangle BOE'$ has $\angle BOE'=2\angle BCE'=60^\circ$ and $OB,OE'$ are equal circumradii $ \Rightarrow \triangle BOE'$ is equilateral.
Now a property about altitude and circumradius from same vertex is, they make equal angles with adjacent sides as can be seen in the following diagram.
Thus $CO$ makes with side $CB$ same angle altitude $CM$ makes with side $CE'$ (in first diagram). $\angle OCB = \angle MCE' = 10^\circ = \angle BCE$. This means $O$ lies on $EC\,!$
Now we focus on the colored $\triangle OEE'$ and $\triangle BAE'$. With $OE'=BE'$, $\angle OE'E = 60^\circ = \angle BE'A$, $EE'=AE'$, the two triangles are congruent and $$\angle ABM = \angle ABE' = \angle EOE' = 180^\circ - 80^\circ - 60^\circ = 40^\circ \quad \square$$