SAMC

Let $A{A}_1\cap C{C}_1=\{M\}$. We have $△AO{A}_1\equiv △CO{C}_1$ (S.A.S.), then $\widehat{A{A}_{1}O}\equiv \widehat{M{A}_{1}O}\equiv \widehat{C{C}_{1}O}$. It follows that $A_1{C}_{1}MO$ is cyclic, hence $\widehat{A_{1}MC}\equiv \widehat{A_{1}O{C}_1}=90^{\circ }$, that is $A_{1}M\perp C{C}_1$.

The quadrilateral $A_1{B}_1{C}_{1}M$ is cyclic, then $\widehat{A_{1}M{B}_1}\equiv \widehat{A_1{C}_1{B}_1}=45^{\circ }$.

The quadrilateral $AOCM$ is cyclic, hence we have $\widehat{OMC}\equiv \widehat{OAC}=45^{\circ }$ and $\widehat{AMO}\equiv \widehat{ACO}=45^{\circ }$.

The quadrilateral $OBMC$ is cyclic, since $\widehat{OMC}\equiv \widehat{OBC}=45^{\circ }$. It follows $\widehat{OMB}=90^{\circ }$.



We have $$\widehat{A_{1}MB}=\widehat{OMB}-\widehat{AMO}=90^{\circ }-45^{\circ }=45^{\circ },$$

hence $\widehat{A_{1}M{B}_1}=\widehat{AMB}=45^{\circ }$, that is points $M,B,B_1$ are collinear. It follows that the lines $A{A}_1$, $B{B}_1$, $C{C}_1$ are concurrent.

Solution 2:

We use complex coordinates. Assume that the origin of the complex plane is at $O$ and we have $A(1)$, $B(1+i)$, $C(i)$. If $A_1(a_1)$, then $C_1(a_{1}i)$ and $B_1(a_1(1+i))$. As in the previous solution, consider $A{A}_1\cap C{C}_1=\{M\}$, and $M(m)$. The points $A_1,A,M$ are collinear if and only if $$\frac{m-a_1}{1-a_1}\in {\mathbb{R}}^{\ast }$$ That is equivalent to $$\begin{array}{c}\frac{m-a_1}{1-a_1}=\frac{\bar{m}-{\bar{a}}_1}{1-{\bar{a}}_1}.\end{array}\text{\hspace{1em}(1)}$$ The points $C_1,C,M$ are collinear if and only if $$\frac{m-a_{1}i}{i-a_{1}i}\in {\mathbb{R}}^{\ast }$$ hence $$\begin{array}{c}\frac{m-a_{1}i}{i-a_{1}i}=\frac{\bar{m}+{\bar{a}}_{1}i}{-i+{\bar{a}}_{1}i}\end{array}\text{\hspace{1em}(2)}$$ From (1) and (2) we get $$m=\frac{1}{2}\left[-\frac{(1+i){\bar{a}}_1(1-a_1)}{1-{\bar{a}}_1}+(1+i)a_1\right]=\frac{1+i}{2}\cdot \frac{a_1-{\bar{a}}_1}{1-{\bar{a}}_1}.$$ We have only to show that the points $B_1,B,M$ are collinear. That is $$\frac{m-a_1(1+i)}{1+i-a_1(1+i)}\in {\mathbb{R}}^{\ast }$$ Indeed, $$\begin{array}{c}\frac{m-a_1(1+i)}{1+i-a_1(1+i)}=\frac{\frac{1}{2}\cdot \frac{a_1-{\bar{a}}_1}{1-{\bar{a}}_1}-a_1}{1-a_1}\\ =-\frac{a_1+{\bar{a}}_1-a_1{\bar{a}}_1}{2(1-a_1)(1-{\bar{a}}_1)}\in {\mathbb{R}}^{\ast }\end{array}$$