66th Belarusian Mathematical Olympiad

By condition, the perimeter $P(A{B}_2{C}_1)$ of the triangle $A{B}_2{C}_1$ is equal to the perimeter $P(BC{B}_2{C}_1)$ of the quadrilateral $BC{B}_2{C}_1$, i.e., (see the Fig.)



$$P(A{B}_2{C}_1)=A{B}_2+B_2{C}_1+C_{1}A=BC+C{B}_2+B_2{C}_1+C_{1}B=P(BC{B}_2{C}_1),$$ so, taking into account $C_{1}A=C_{1}B$, we have $$A{B}_2=BC+C{B}_2=BC+(AC-A{B}_2).$$ Therefore, $A{B}_2=(AC+BC)/2$. Since $B_1{C}_1$ is the midline of the triangle $ABC$, we have $B_1{C}_1=BC/2$ whence $$B_1{B}_2=A{B}_2-A{B}_1=(AC+BC)/2-AC/2=BC/2=B_1{C}_1.$$ Therefore, the triangle $C_1{B}_1{B}_2$ is isosceles and $\angle B_1{B}_2{C}_1=\angle B_1{C}_1{B}_2$. Further, $A_1{C}_1$ is the midline of the triangle $ABC$, then $A_1{C}_1\parallel AC$ and $\angle B_1{B}_2{C}_1=\angle B_2{C}_1{A}_1$. Thus $\angle B_1{C}_1{B}_2=\angle B_2{C}_1{A}_1$, and so $C_1{B}_2$ is the bisector of the angle $C_1$ of the triangle $A_1{B}_1{C}_1$. In the same way, we obtain that $B_1{C}_3$ is the bisector of the angle $B_1$ of the triangle $A_1{B}_1{C}_1$, and $A_1{C}_2$ is the bisector of the angle $A_1$ of the triangle $A_1{B}_1{C}_1$. Therefore, the segments $A_1{C}_2$, $B_1{C}_3$, $C_1{B}_2$ are the bisectors of the angles of the triangle $A_1{B}_1{C}_1$ and, hence, they are concurrent.