Show that $m(\angle ABM)=30^{\circ}$.

Question

Let $ABC$ a triangle and $M$ a point in the interior of the triangle such that $$m(\angle BAM)=20^{\circ}, m(\angle MAC)=60^{\circ}, m(\angle ACM)=20^{\circ}, m(\angle MCB)=10^{\circ}.$$ I have to show that $m(\angle ABM)=30^{\circ}$.

I constructed a point $P$ on $CM$ s.t. $m(\angle AMC)=20^{\circ}$. Then $\triangle APC$ is isosceles. Also $AP=PM$. I denote $AM\cap BC=\{N\}$. Now I am stuck.

I need a synthetic proof.

Answer 1

Consider point $X$ on $MC$ such that $\angle XAC=20^\circ$. Easy angle chasing reveals that triangles $AXC$ and $AMX$ are isosceles with $AX=XC$ and $AM=MX$. Also observe that $\angle CXA =2\angle CBA$ which along with $AX=XC$ implies that $X$ is circumcenter of $ABC$. Hence $AXB$ is isosceles and $\angle AXB=2\angle ACB=60^\circ$. This shows that $AXB$ is equilateral. Using $AM=MX$ we see that $MB$ bisects angle $XBA$ which means that $\angle MBA=30^\circ$.

Answer 2

Let $\measuredangle ABM=x$.

Thus, by Cheva $$\sin(70^{\circ}-x)\sin20^{\circ}\sin20^{\circ}=\sin{x}\sin60^{\circ}\sin10^{\circ}.$$ But $\sin(70^{\circ}-x)\sin20^{\circ}\sin20^{\circ}$ decreases and $\sin{x}\sin60^{\circ}\sin10^{\circ}$ increases,

which says that our equation has one root maximum.

Id est, it's enough to prove that: $$\sin40^{\circ}\sin^220^{\circ}=\sin30^{\circ}\sin60^{\circ}\sin10^{\circ}$$ or $$\sin40^{\circ}(1-\cos40^{\circ})=\sin60^{\circ}\sin10^{\circ}$$ or $$\sin40^{\circ}=\frac{1}{2}\sin80^{\circ}+\sin60^{\circ}\sin10^{\circ}$$ or $$\cos50^{\circ}=\cos60^{\circ}\cos10^{\circ}+ \sin60^{\circ}\sin10^{\circ},$$ which is obvious.