Let $ABC $ a triangle with $ A_1$ the middle of the edge $[BC] $ and $BAA_1=30°$. $ Let D\in [AB] $ s.t. $CD=AB $.
I have to show that $AB=AA_1$.
This conclusion seems to be wrong because I can draw a triangle with $AB>AA_1$ but I am not sure if the circle with center $C $ and radius $ AB $ intersects the edge $[AB] $.