Is it possible ti find $∠X$ in below figure?

Question

I tried like below, but I get stuck on this problem. Is it possible to help me to find a clue?

$$∠A=20^\circ\\AN=MC,NB=AM \Longrightarrow AB=AC$$So$$∠B=∠C=\frac{180^\circ-20^\circ}{2}=80^\circ$$ Then i tried to draw a line between $M,B$ and it make $\frac{180^\circ-20^\circ}{2}$ by an isosceles triangle ... but I can't go furthere !

Answer 1

Let $K\in BN$ such that $KC=BC$.

Thus, $\measuredangle KCB=20^{\circ}$ and from here $$\measuredangle KCM=60^{\circ},$$ $$KM=MC=KC=BC$$ and $$\measuredangle AKM=40^{\circ}.$$ Now, let $N_1\in KA$ such that $KM=MN_1$.

Thus, $\measuredangle MN_1K=40^{\circ},$ which says $$AN_1=N_1M=KM=BC,$$ which gives $$N_1\equiv N$$ and $$\measuredangle NMC=180^{\circ}-20^{\circ}=160^{\circ}.$$