prove that $MN \parallel BC$ in an equilateral triangle

Question

$\Delta ABC$ is equilateral with $M$ and $N$ being interior points. if

$\angle MAB=\angle MBA=40^{\circ}$

$\angle NAB=20^{\circ}$ and $\angle NBA=30^{\circ}$. Prove that $MN \parallel BC$

from the given angles we have

$\angle MBN=30$, $\angle MAN=20$

$\angle ANB=130$, $reflex(\angle ANB)=230$

Hence $\angle AMB=100$

after this can i have any HINT:

Answer 1

First $\angle MBN = 40-30=10°$ and $\angle AMB = 80°$ because $\angle MAB = \angle MBA = 40°$

HINT : Trace $MC$ and $NC$ and the key word is symmetry