45th Mongolian Mathematical Olympiad

Denote $BC\cap A{A}_1=A_2$, $AC\cap B{B}_1=B_2$, $AB\cap C{C}_1=C_2$. From well known property we have: $\frac{A{B}_2}{B_{2}C}=\frac{S_{AB{B}_1}}{S_{BC{B}_1}}$. In another way: $$S_{AB{B}_1}=\frac{AB\cdot A{B}_1\cdot \frac{1}{2}\sin (\angle A+\angle AMC-90^{\circ })}{BC\cdot B_{1}C\cdot \frac{1}{2}\sin (\angle C+\angle AMC-90^{\circ })},$$ here $$\angle AC{B}_1=\angle CA{B}_1=\frac{180^{\circ }-\angle A{B}_{1}C}{2}=\frac{180^{\circ }-(360^{\circ }-2\angle AMC)}{2}=$$ $$=\angle AMC-90^{\circ }.$$ If we observe that $A{B}_1=B_{1}C$, so by easy calculation we get following: $$\begin{array}{rl}\frac{A{B}_2}{B_{2}C}& =\frac{AB(\sin (\angle A+\angle AMC)\cos 90^{\circ }+\sin 90^{\circ }\cos (\angle A+\angle AMC))}{BC(\sin (\angle C+\angle AMC)\cos 90^{\circ }+\sin 90^{\circ }\cos (\angle C+\angle AMC))}=\\ & =\frac{AB\cos (\angle A+\angle AMC)}{BC\cos (\angle C+\angle AMC)}=\frac{AB\cos (\angle A+2\angle B)}{BC\cos (\angle C+2\angle B)}\end{array}(1)$$

(here because of $M$ is circumcenter of $ABC$) By similar way, we can get $$\frac{C{A}_2}{A_{2}B}=\frac{CA\cdot \cos (\angle C+2\angle A)}{AB\cdot \cos (\angle B+2\angle A)},(2)$$ and $$\frac{B{C}_2}{C_{2}A}=\frac{BC\cdot \cos (\angle B+2\angle C)}{CA\cdot \cos (\angle A+2\angle C)}.(3)$$

From (1), (2) and (3) we getting that $$\begin{array}{rl}\frac{A{B}_2}{B_{2}C}\cdot \frac{C{A}_2}{A_{2}B}\cdot \frac{B{C}_2}{C_{2}A}& =\frac{AB\cos (\angle A+2\angle B)}{BC\cos (\angle C+2\angle B)}\cdot \\ & \frac{CA\cos (\angle C+2\angle A)}{AB\cos (\angle B+2\angle A)}\cdot \frac{BC\cos (\angle B+2\angle C)}{CA\cos (\angle A+2\angle C)}=\end{array}$$ $$=\frac{\cos (\angle A+2\angle C)}{\cos (\angle C+2\angle B)}\cdot \frac{\cos (\angle C+2\angle A)}{\cos (\angle B+2\angle A)}\cdot \frac{\cos (\angle B+2\angle C)}{\cos (\angle A+2\angle C)}=1.$$

Then by Menelaus' theorem the lines $A{A}_1,B{B}_1,C{C}_1$ passing same point.