Any ideas? The official solution?

Please,how to define ,exactly,the points A1and B1?

Anyone can explain how we define A1 and B1? Draw a figure?

Weird problem. We do not need $AH_1,BH_2$ to be altitudes, let alone $\angle MCB=\angle NCA=30^{\circ}.$ Let $\triangle {ABC}$ and $\triangle DEF$ be two any perspective triangles and define $X \equiv BF \cap CE,$ $Y \equiv CD \cap AF,$ $Z\equiv BD \cap AE.$ Then $\triangle {ABC}$ and $\triangle XYZ$ are perspective. Consider an homology sending the perspectrix of $\triangle {ABC},\triangle DEF$ to the line at infinity $\Longrightarrow$ $\triangle {ABC}$ and $\triangle DEF$ become homothetic $\Longrightarrow$ $\frac{ZD}{ZB}=\frac{DE}{AB}=\frac{DF}{AC}=\frac{YD}{YC}$ $\Longrightarrow$ $BC \parallel YZ$ and similarly $CA \parallel ZX,$ $AB \parallel XY$ $\Longrightarrow$ $\triangle {ABC}$ and $\triangle XYZ$ are homothetic. Thus back in the original figure, $\triangle {ABC},$ $\triangle XYZ$ and $\triangle DEF$ are perspectrix with common perspectrix even.