, 2021

Common remarks. We introduce some objects and establish some preliminary facts common for all solutions below.
Let $\Omega$ denote the circle $(ABCD)$, and let $\gamma_{B}$ and $\gamma_{D}$ denote the two circles from the problem statement (their centres are $O_{B}$ and $O_{D}$, respectively). Clearly, all three centres $O$, $O_{B}$, and $O_{D}$ are distinct.
Assume, without loss of generality, that $AB > BC$. Suppose that $AD > DC$, and let $H = AC \cap BD$. Then the rays $BB_{1}$ and $DD_{1}$ lie on one side of $BD$, as they contain the midpoints of the $\operatorname{arcs} ADC$ and $ABC$, respectively. However, if $BD_{1} \parallel DB_{1}$, then $B_{1}$ and $D_{1}$ should be separated by $H$. This contradiction shows that $AD < CD$.
Let $\gamma_{B}$ and $\gamma_{D}$ meet $\Omega$ again at $T_{B}$ and $T_{D}$, respectively. The common chord $BT_{B}$ of $\Omega$ and $\gamma_{B}$ is perpendicular to their line of

centres $O_{B}O$; likewise, $DT_{D} \perp O_{D}O$. Therefore, $O \in O_{B}O_{D} \Longleftrightarrow O_{B}O \parallel O_{D}O \Longleftrightarrow BT_{B} \parallel DT_{D}$, and the problem reduces to showing that
$$\begin{equation*} BT_{B} \parallel DT_{D}. \tag{1} \end{equation*}$$

Solution 1. Let the diagonals $AC$ and $BD$ cross at $H$. Consider the homothety $h$ centred at $H$ and mapping $B$ to $D$. Since $BD_{1} \parallel DB_{1}$, we have $h(D_{1}) = B_{1}$.
Let the tangents to $\Omega$ at $B$ and $D$ meet $AC$ at $L_{B}$ and $L_{D}$, respectively. We have
$$\angle L_{B}BB_{1} = \angle L_{B}BC + \angle CBB_{1} = \angle BAL_{B} + \angle B_{1}BA = \angle BB_{1}L_{B},$$
which means that the triangle $L_{B}BB_{1}$ is isosceles, $L_{B}B = L_{B}B_{1}$. The powers of $L_{B}$ with respect to $\Omega$ and $\gamma_{D}$ are $L_{B}B^{2}$ and $L_{B}B_{1}^{2}$, respectively; so they are equal, whence $L_{B}$ lies on the radical axis $T_{D}D$ of those two circles. Similarly, $L_{D}$ lies on the radical axis $T_{B}B$ of $\Omega$ and $\gamma_{B}$.
By the sine rule in the triangle $BHL_{B}$, we obtain
$$\begin{equation*} \frac{HL_{B}}{\sin \angle HBL_{B}} = \frac{BL_{B}}{\sin \angle BHL_{B}} = \frac{B_{1}L_{B}}{\sin \angle BHL_{B}}; \tag{2} \end{equation*}$$
similarly,
$$\begin{equation*} \frac{HL_{D}}{\sin \angle HDL_{D}} = \frac{DL_{D}}{\sin \angle DHL_{D}} = \frac{D_{1}L_{D}}{\sin \angle DHL_{D}}. \tag{3} \end{equation*}$$
Clearly, $\angle BHL_{B} = \angle DHL_{D}$. In the circle $\Omega$, tangent lines $BL_{B}$ and $DL_{D}$ form equal angles with the chord $BD$, so $\sin \angle HBL_{B} = \sin \angle HDL_{D}$ (this equality does not depend on the picture). Thus, dividing (2) by (3) we get
$$\frac{HL_{B}}{HL_{D}} = \frac{B_{1}L_{B}}{D_{1}L_{D}}, \quad \text{and hence} \quad \frac{HL_{B}}{HL_{D}} = \frac{HL_{B} - B_{1}L_{B}}{HL_{D} - D_{1}L_{D}} = \frac{HB_{1}}{HD_{1}}.$$
Since $h(D_{1}) = B_{1}$, the obtained relation yields $h(L_{D}) = L_{B}$, so $h$ maps the line $L_{D}B$ to $L_{B}D$, and these lines are parallel, as desired.

Solution 2. Let $BD_{1}$ and $T_{B}D_{1}$ meet $\Omega$ again at $X_{B}$ and $Y_{B}$, respectively. Then
$$\angle BD_{1}C = \angle BT_{B}D_{1} = \angle BT_{B}Y_{B} = \angle BX_{B}Y_{B},$$
which shows that $X_{B}Y_{B} \parallel AC$. Similarly, let $DB_{1}$ and $T_{D}B_{1}$ meet $\Omega$ again at $X_{D}$ and $Y_{D}$, respectively; then $X_{D}Y_{D} \parallel AC$.
Let $M_{D}$ and $M_{B}$ be the midpoints of the $\operatorname{arcs} ABC$ and $ADC$, respectively; then the points $D_{1}$ and $B_{1}$ lie on $DM_{D}$ and $BM_{B}$, respectively. Let $K$ be the midpoint of $AC$ (which lies on $M_{B}M_{D}$). Applying Pascal's theorem to $M_{D}DX_{D}X_{B}BM_{B}$, we obtain that the points $D_{1} = M_{D}D \cap X_{B}B$, $B_{1} = DX_{D} \cap BM_{B}$, and $X_{D}X_{B} \cap M_{B}M_{D}$ are collinear, which means that $X_{B}X_{D}$ passes through $K$. Due to symmetry, the diagonals of an isosceles trapezoid $X_{B}Y_{B}X_{D}Y_{D}$ cross at $K$.

Let $b$ and $d$ denote the distances from the lines $X_{B}Y_{B}$ and $X_{D}Y_{D}$, respectively, to $AC$. Then we get
$$\frac{X_{B}Y_{B}}{X_{D}Y_{D}} = \frac{b}{d} = \frac{D_{1}X_{B}}{B_{1}X_{D}},$$
where the second equation holds in view of $D_{1}X_{B} \parallel B_{1}X_{D}$. Therefore, the triangles $D_{1}X_{B}Y_{B}$ and $B_{1}X_{D}Y_{D}$ are similar. The triangles $D_{1}T_{B}B$ and $B_{1}T_{D}D$ are similar to them and hence to each other. Since $BD_{1} \parallel DB_{1}$, these triangles are also homothetical. This yields $BT_{B} \parallel DT_{D}$, as desired.