Let $AA_{1} \cap CC_{1} = \{M\}$. We have $\triangle AOA_{1} \equiv \triangle COC_{1}$ (S.A.S.), then $\widehat{AA_{1}O} \equiv \widehat{MA_{1}O} \equiv \widehat{CC_{1}O}$. It follows that $A_{1}C_{1}MO$ is cyclic, hence $\widehat{A_{1}MC} \equiv \widehat{A_{1}OC_{1}} = 90^{\circ}$, that is $A_{1}M \perp CC_{1}$.

The quadrilateral $A_{1}B_{1}C_{1}M$ is cyclic, then $\widehat{A_{1}MB_{1}} \equiv \widehat{A_{1}C_{1}B_{1}} = 45^{\circ}$.

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 $\widehat{A_{1}MB_{1}} = \widehat{AMB} = 45^{\circ}$, that is points $M, B, B_{1}$ are collinear. It follows that the lines $AA_{1}$, $BB_{1}$, $CC_{1}$ 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 $AA_{1} \cap CC_{1} = \{M\}$, and $M(m)$. The points $A_{1}, A, M$ are collinear if and only if
$$\frac{m-a_{1}}{1-a_{1}} \in \mathbb{R}^{*}$$
That is equivalent to
$$\begin{equation*} \frac{m-a_{1}}{1-a_{1}} = \frac{\bar{m}-\bar{a}_{1}}{1-\bar{a}_{1}} . \tag{1} \end{equation*}$$
The points $C_{1}, C, M$ are collinear if and only if
$$\frac{m-a_{1}i}{i-a_{1}i} \in \mathbb{R}^{*}$$
hence
$$\begin{equation*} \frac{m-a_{1}i}{i-a_{1}i} = \frac{\bar{m}+\bar{a}_{1}i}{-i+\bar{a}_{1}i} \tag{2} \end{equation*}$$
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
$$\frac{m-a_{1}(1+i)}{1+i-a_{1}(1+i)} \in \mathbb{R}^{*}$$
Indeed,
$$\begin{gathered} \frac{m-a_{1}(1+i)}{1+i-a_{1}(1+i)} = \frac{\frac{1}{2} \cdot \frac{a_{1}-\bar{a}_{1}}{1-\bar{a}_{1}} - a_{1}}{1-a_{1}} \\ = -\frac{a_{1}+\bar{a}_{1}-a_{1}\bar{a}_{1}}{2(1-a_{1})(1-\bar{a}_{1})} \in \mathbb{R}^{*} \end{gathered}$$