Let $X = AC \cap BD$, $B_2 = (O_B) \cap (ABCD)$ and $D_2 = (O_D) \cap (ABCD)$. Let the tangent to $(ABCD)$ at $B$ and $D$ intersect $AC$ at $P$ and $Q$ respectively. Claim 1: $D, D_2, P$ colinear and $PB = PB_1$

Similarly, $B, B_2, Q$ colinear and $QD = QD_1$. Let $M$ be the midpoint of $\overarc{ABC}$ and $N$ its antipode. And let $P_1 = MB \cap AC, \: Q_1 = ND \cap AC$. Claim 2: $BP_1DD_1$ and $DQ_1BB_1$ are cyclic

Since $X \in BD$, which is the radical axis of $(BP_1DD_1)$ and $(DQ_1BB_1)$, \[XD_1 \cdot XP_1 = XB_1 \cdot XQ_1 \iff \frac{XP_1}{XB_1} = \frac{XQ_1}{XD_1}\] Notice $P, Q$ are midpoints of $B_1P_1, D_1Q_1$. So, \[ \frac{XP_1}{XB_1} = \frac{XQ_1}{XD_1} \rightarrow \frac{B_1P_1}{XB_1} = \frac{D_1Q_1}{XD_1} \rightarrow \frac{B_1P}{XB_1} = \frac{D_1Q}{XD_1} \rightarrow \frac{XP}{XB_1} = \frac{XQ}{XD_1}\] Finally, \[ \frac{XP}{XQ} = \frac{XB_1}{XD_1} = \frac{XD}{XB} \]Hence, $BQ || DP$ which immediately gives $O, O_B, O_D$ colinear.

Let the circle with center $O_B$ which passes through $B$ and is tangent to $AC$ at $D_1$ be $\gamma_B$, the circle with center $O_D$ which passes through $D$ and is tangent to $AC$ at $B_1$ be $\gamma_D$, and the circumcircle of $ABCD$ be $\gamma$. Moreover, let $P \neq B$ be the other intersection point of $\gamma_B$ and $\gamma$ and $Q \neq D$ be the other interserction point of $\gamma_D$ and $\gamma$. Thus, we just neeed to prove that $PB \parallel DQ$. Let the tangents to $\gamma$ at $B$ and $D$ meet $AC$ at $X$ and $Y$, respectively. Moreover let $AC$ and $BD$ meet at $T$. We proceed with some Claims: Claim 1: $X \in DQ$ and $Y \in PB$. Proof: Note that $\angle YDD_1=\angle YDA+\angle ADD_1=\angle ACD+\angle D_1DC=\angle YD_1D,$ and so $YD=YD_1$, hence $Y$ belongs to the radical axis of $\gamma$ and $\gamma_B$, which is $PB$. Similarly $X \in DQ$ $\blacksquare$ Claim 2: $\angle YDT+\angle TBX=180^\circ$. Proof: Note that $\angle YDT+\angle TBX=\angle YDA+\angle ADT+\angle TBC+\angle CBX=$ $\angle ACD+\angle ADB+\angle DBC+\angle BDC=(\angle ACD+\angle DAC)+(\angle ADB+\angle BDC)=$ $\angle ACD+\angle DAC+\angle ADC=180^\circ,$ as desired $\blacksquare$ To the problem, apply the Law of Sines in triangles $YDT$ and $BTX$ to obtain $\frac{YT}{\sin \angle YDT}=\frac{YD}{\sin \angle YTD}$ and $\frac{XT}{\sin \angle TBX}=\frac{BX}{\sin \angle BTX}$ Dividing these two and using Claim 2 we have that $\frac{YT}{XT}=\frac{YD}{BX}=\frac{YD_1}{XB_1}$ Thus, $\frac{YT}{YD_1}=\frac{XT}{XB_1},$ implying that $\frac{YT}{TD_1}=\frac{XT}{TB_1}$, and so $\frac{YT}{XT}=\frac{TD_1}{TB_1}=\frac{TB}{TD},$ where the last equality follows since $BD_1 \parallel DB_1$. Hence by the converse of Thales theorem, $YB \parallel DX$, i.e. $PB \parallel DQ$, which is what we wanted to prove.

Let the tangents to $(ABCD)$ at $B$ and $D$ intersect $\overline{AC}$ at $E$ and $F$ respectively. Let $T = \overline{AC} \cap \overline{BD}$. Then we have $EB = EB_1$ and $FC = FC_1$. Therefore, \[ \frac{TE}{EB_1} = \frac{TE}{EB} = \frac{\sin \angle TBE}{\sin \angle T} = \frac{\sin \angle C}{\sin \angle T} = \frac{\sin \angle A}{\sin \angle T} = \frac{\sin \angle TDF}{\sin \angle T} = \frac{TF}{FD} = \frac{TF}{FD_1}. \]Thus, \[\frac{TB_1}{TE} = \frac{TD_1}{TF} \Longrightarrow \frac{TB}{TD} = \frac{TD_1}{TB_1} = \frac{TF}{TE}. \]and hence $\overline{BF} \parallel \overline{DE}$. Now let $\overline{BF}$ and $\overline{DE}$ intersect $(ABCD)$ at $B'$ and $D'$ respectively. Then $ED \cdot ED' = EB^2 = EB_1^2$, so $(DB_1D')$ is tangent to $\overline{AC}$ at $B_1$. Similarly, $(BD_1B')$ is also tangent to $\overline{AC}$ at $D_1$. Since $\overline{BF} \parallel \overline{DE}$, we see that segment $BB'$ and $DD'$ share the same perpendicular bisector, say $\ell$. Then as $O$, $O_B$ and $O_D$ all lie on $\ell$, we are done.

Let $\omega_B$ and $\omega_D$ be the circles in the problem with centers $O_B$ and $O_D$ respectively. I initially thought that $\overline{BD_1}||\overline{DB_1}$ meant that I should look at the insimillicenter of $\omega_B$ and $\omega_D$, and went down a 1.5hr rabbit hole with Monge. Now that I have finished Eromanga Sensei, I have seen the light, and this took only an extra 20-30 minutes. Let $T_B$ be on $\overline{AC}$ such that $\overline{BT_B}$ is tangent to $(ABCD)$. Let $M_B=\overline{BB_1}\cap (ABCD)$. Let $B'$ be on $(ABCD)$ such that $\overline{BB'}||\overline{AC}$. Define $T_D,M_D,D'$ similarly. Let $X=\overline{BD}\cap\overline{AC}$. Note that by symmetry, $\overline{BD},\overline{B'D'},\overline{M_BM_D}$ concur. $$B(X,P;T_D,D_1)=(X,P;T_D,D_1)\overset{D}{=}(B,D';D,M_D)=(D,B';B,M_B)\overset{B}{=}(X,P;T_B,B_1)=D(X,P;T_B,B_1).$$$\overline{BX}||\overline{DX},\overline{BP}||\overline{DP},$ and $\overline{BD_1}||\overline{DB_1}$, so $\overline{BT_D}||\overline{DT_B}$. $T_DD=T_DD_1$, so the radical axis of $\omega_B,(ABCD)$ is $\overline{BT_D}$. Similarly, the radical axis of $\omega_D,(ABCD)$ is $\overline{DT_B}$, so $O_B-O-O_D$ as desired.

Let $\omega, \omega_B,\omega_C$ denote $\odot (ABCD)$ and circles centered at $O_B,O_D$ respectively. Let $AC$ meet tangent to $\omega$ at $B,D$ and $BD$ at $X,Y,Z$, and $\omega$ meet $\omega_B ,\omega_C$ again at $B_2,D_2.$ Note that there exist a homothety $\varphi :BO_BD_1\stackrel{Z}{\mapsto} DO_DB_1.$ Inversion wrt $X$ which fixes $B,B_1$ also fixes $\omega ,\omega_D$ and so swaps $D,D_2,$ in particular $X\in DD_2$ and analogously $Y\in BB_2.$ Thus $$(ACD_2B)_{\omega}\stackrel{X}{=}(CADB)_{\omega}=(ACBD)_{\omega}\stackrel{Y}{=}(CAB_2D)_{\omega}\implies Z\in B_2D_2\implies \varphi (B_2)=D_2,$$which yields $BB_2\parallel DD_2.$ Therefore $O,O_B,O_D$ lie on common perpendicular bisector of segments $BB_2,DD_2.$

Yay! My first problem on the shortlist This problem is a proposal by Jafet Baca and me. The problem statement, which originally asked to prove $BD_1\parallel DB_1$ if and only if $O$ lied on $O_BO_D$, was mine. Jafet solved this G5 and I later found a solution for the converse using moving points.

Let $\odot(O_B)$ and $\odot(O_D)$ meet $\odot(ABCD)$ again at $X$ and $Y$. The key claim is that $BX\parallel DY$, which obviously implies the problem. To prove the claim, let the external bisector meet $AC$ at $B_2$ and $D_2$. Let $M_B$ and $M_D$ be the midpoints of $B_1B_2$ and $D_1D_2$, recalling the well-known fact that $BM_B$ and $DM_D$ are tangent to $\odot(ABCD)$. Therefore, by radical center theorem, $B,X,M_D$ and $D,Y,M_B$ are colinear. Now, let $T=AC\cap BD$. Perform an inversion at $T$ that fixes $\odot(ABCD)$. Notice that $\odot(DD_1D_2)$ is orthogonal to $\odot(ABCD)$, so it must map to a circle through $B$ orthogonal to $\odot(ABCD)$, which is $\odot(BB_1B_2)$. Therefore, the inversion swaps $\{B_1,D_2\}$ and $\{B_2,D_1\}$. Since $BD_1\parallel DB_1$, it follows that $\odot(TDB_2)$ and $\odot(TBD_2)$ are tangent, implying that $DB_2\parallel BD_2$. Thus, $DM_B\parallel BM_D$, done.

Let $\omega$ be the circumcircle of $ABCD$, $G \neq B = BD_1 \cap \omega$, $H \neq D = DB_1 \cap \omega$. Along the $DB_1$ 'axis', the distance between $O_D$ and $O$ is given by $B_1H/2$, as $O$ must lie on the perpendicular bisector of $DH$ and $O_D$ must lie on the perpendicular bisector of $DB_1$. We can write $d(O_D, O) = B_1H/2$(note the ordering of points on the left-hand-side). Similarly, $d(O, O_B) = GD_1/2$. Notice that $O$ is between $O_B$ and $O_D$. It suffices to find another 'axis' where the distance satisfy $d_1(O_D, O): d_1(O, O_B) = HB_1: D_1G$. Consider the 'axis' $AC$ and let $M$ be the midpoint of the segment. We have $OM // O_DB_1 // O_BD_1$. It suffices to show $B_1M: MD_1 = HB_1: D_1G$ and that $M$ is between $B_1$ and $D_1$, which by intercept theorem, means it suffices to show $G, H, M$ are collinear. Let $M_1 \neq B = BB_1 \cap \omega$ and $M_2 \neq D = DD_1 \cap \omega$. By Pascal's theorem, $B_1 = HD \cap BM_1$, $D_1 = GB \cap DM_2$, and $X = M_1M_2 \cap GH$ are collinear. $X$ must be on $AC$, so $X = M_1M_2 \cap AC = M$ and conclusion follows.

Trivially $\triangle BO_BD_1\sim \triangle DO_DB_1$. Let $M,N$ be midpoints of arcs $ADC, ABC$ respectively, and draw line $XY$ through $O$ parallel to both $BO_B, DO_D$. Since $MN \parallel O_BD_1 \parallel O_DB_1$ we get $\triangle BO_BD_1, \triangle DO_DB_1, \triangle XON, \triangle YOM$ are all homothetic. Let $BX \cap ND = E, CY \cap MB = F$. By homothety, $O_B, E, O$ collinear, $O_D, F, O$ collinear, and by Pascal's on $YDNMBX$ we get $F,O,E$ collinear, thus $O_D, O, O_B$ collinear as desired.

Let tangent to $ABCD$ at $BD$ meet $AC$ at $Y,X$. Claim $: BY = B_1Y$ and $DX = D_1X$. Proof $:$ Note that $\angle BB_1Y = \angle BCA + \angle CBB_1 = \angle BCA + \angle ABB_1 = \angle YBB_1$. with same approach $DX = D_1X$. Claim $: BX$ and $DY$ are Radical Axis of $ABCD$ with $\omega_B$ and $\omega_D$. Proof $:$ Note that $XD^2 = XD_1^2$ so $X$ has same power wrt $ABCD$ and $\omega_B$ so $BX$ is Radical Axis. we have same approach for $DY$. Note that $OO_B \perp BX$ and $OO_D \perp DY$ so we need to prove $DY || BX$. Let $BD$ meet $AC$ at $K$. Note that $BD_1 || DB_1 \implies \frac{TB_1}{TD_1} = \frac{TD}{TB}$ so we need to prove $\frac{TB_1}{TD_1} = \frac{TY}{TX}$ or $\frac{TY}{B_1Y} = \frac{TX}{D_1X}$. $\frac{TY}{B_1Y} = \frac{TY}{BY} = \frac{\sin{YBT}}{\sin{BTY}} = \frac{\sin{BCD}}{\sin{BTY}} = \frac{\sin{BAD}}{\sin{BTY}} = \frac{\sin{TDX}}{\sin{DTX}} = \frac{TX}{DX} = \frac{TX}{D_1X}$.

My favorite problem of Shortlist! Let us denote some points: $AC\cap BD = P$, $BB\cap AC = B'$, $DD\cap AC = D'$ and $B'D\cap \omega = D_2$, $BD'\cap \omega = B_2$. $\angle{D'DD_1}=\angle{D'DC}+\angle{D_1DC} = \angle{CAD}+\angle{D_1DA} = \angle{CD_1D}$ so $D'D_1 = D'D$, similarly $B'B = B'B_1$ Now since $B'B_1^{2} = B'B^{2} = B'D_2 * B'D$, $\omega_{D_2B_1D}$ and $\omega_{BD_1B_2}$ is tangent to $AC$. If we write sinus: $\frac{BB'}{B'P} = \frac{\sin BPB'}{\sin B'BP}$ and $\frac{DD'}{D'P} = \frac{\sin DPD'}{\sin D'DP}$ so $\frac{BB'}{B'P} = \frac{DD'}{D'P}$ Now if we use statement given to us: $\frac{PB_1}{PD_1} = \frac{PD}{PB} \implies \frac{PB_1 - B'B_1}{PD' - D'D_1} = \frac{PB_1 - B'B}{PD' - D'D} = \frac{PB'}{PD'}$ so $BB_2||DD_2$. We can see that 2 radical axis of $\omega$,$\omega_{D_2B_1D}$ and $\omega_{BD_1B_2}$ are parallel hence the last one also must be parallel which implies that all of its center lies on one line.

Solution: Let $\omega$ be the circumcircle of $ABCD$ and let ${\omega}_B$ and ${\omega}_D$ be the circles with center $O_B$ passing through $B$ and with center $O_D$ passing through $D,$ respectively. Further let $R$ and $S$ be the second intersections of ${\omega}_B$ and ${\omega}_C$ with $\omega,$ respectively. Now consider the negative homothety $\Phi$ that sends ${\omega}_B$ to ${\omega}_D.$ Hence, from definition we have that $\Phi$ sends $D_1$ to $B_1.$ Hence $\Phi$ must send $B$ to $D$ since we are assuming $\overline{BD_1} \parallel \overline{DB_1}.$ So if we denote by $Q:= \overline{BD}\cap \overline{AC} $ we have that $Q$ is the center of $\Phi.$ Now we prove the following claim wich solves the problem. Claim: $\Phi$ sends $R$ to $S.$ First we will show that the Claim solves the problem. Indeed, if $\Phi$ sends $R$ to $S$ then we must have that $\overline{BR}\parallel \overline{CS}$ which implies that $BRDS$ is an isosceles trapezoid and hence all points $O,O_b$ and $O_d$ must lie on the common perpendicular bisector of $BR$ and $CS.$ Now it remains to prove the Claim. Proof: It is sufficient to show that $R$ and $S$ concurr at $Q,$ i.e., that $\overline{RS},\overline{BD}$ and $\overline{AC}$ concurr. We then define the point $X:=\overline{BR} \cap \overline{AC}.$ We claim that $\overline{XD}$ is tangent to $\omega;$ indeed consider the circle ${\omega '}_B$ tangent to $\omega$ at $D$ and to $\overline{AC}$ at $D_1$ which exists since $DD_1$ passes through the midpoint of one of the arcs $AC$ of $\omega.$ So we have that $X$ is the radical center of the circles ${\omega}_B,{\omega '}_B$ and $\omega.$ Similarly, define $Y:=\overline{AC} \cap \overline{BS}$ and analogoulsy we show that $\overline{YB}$ is tangent to $\omega.$ Now to establish the claim, apply Pascal's Theorem to the degenerate hexagon $BBRSDD$ to conclude that $\overline{RS} \cap \overline{BD}$ lies on $\overline{AC}.\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17.712256548248305cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.004925918694425, xmax = 8.707330629553882, ymin = 1.2667477155254272, ymax = 13.323357989704114; /* image dimensions */ pen wqwqwq = rgb(0.3764705882352941,0.3764705882352941,0.3764705882352941); pen qqffff = rgb(0.,1.,1.); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen zzttqq = rgb(0.6,0.2,0.); pen dcrutc = rgb(0.8627450980392157,0.0784313725490196,0.23529411764705882); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen xfqqff = rgb(0.4980392156862745,0.,1.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffqqff = rgb(1.,0.,1.); /* draw figures */ draw(circle((-5.354835856181882,6.494664424417499), 2.327399836985479), linewidth(1.2) + wqwqwq); draw(circle((-7.382178480474679,6.209117717934945), 1.264844707878139), linewidth(1.2) + qqffff); draw(circle((-4.022918684103314,6.728907360916935), 1.9789904404767618), linewidth(1.2) + xdxdff); draw((xmin, 6.615384615384636*xmin + 46.58257467700062)--(xmax, 6.615384615384636*xmax + 46.58257467700062), linewidth(1.2) + zzttqq); /* line */ draw(circle((-6.7553786539055425,7.141953635421922), 0.7845118141729529), linewidth(1.2) + dcrutc); draw(circle((-4.7402070805265115,5.809749948175339), 1.4071409757786508), linewidth(1.2) + dcrutc); draw((-7.467512233804722,7.4710805880662114)--(-3.8003961516937834,4.7624672051866765), linewidth(1.2) + yqqqyq); draw((-7.467512233804722,7.4710805880662114)--(-3.0535795870448217,6.146800104664222), linewidth(1.2) + blue); draw((-3.8003961516937834,4.7624672051866765)--(-7.656092125318946,6.842528744170774), linewidth(1.2) + xfqqff); draw((-5.156256264524813,12.471956311682519)--(-7.467512233804722,7.4710805880662114), linewidth(1.2) + qqwuqq); draw((-6.717327373159708,2.144870516097806)--(-3.8003961516937834,4.7624672051866765), linewidth(1.2) + qqwuqq); draw((-7.619705208172238,5.958794272594481)--(-3.0535795870448217,6.146800104664222), linewidth(1.2) + xfqqff); draw((-6.717327373159708,2.144870516097806)--(-7.619705208172238,5.958794272594481), linewidth(1.2) + qqwuqq); draw((-7.656092125318946,6.842528744170774)--(-3.182011836969981,7.328706259540812), linewidth(1.2) + blue); draw((-5.156256264524813,12.471956311682519)--(-3.182011836969981,7.328706259540812), linewidth(1.2) + qqwuqq); draw((-7.11561808980492,4.972680217220363)--(-4.484668585942924,8.65327483924573), linewidth(1.2) + linetype("4 4") + ffqqff); /* dots and labels */ dot((-5.712155119565556,8.794471578336058),linewidth(3.pt) + dotstyle); label("$A$", (-6.089418343302069,8.961057181410371), NE * labelscalefactor); dot((-3.8003961516937834,4.7624672051866765),linewidth(3.pt) + dotstyle); label("$D$", (-3.6123329596980356,4.2480186196859755), NE * labelscalefactor); dot((-6.375948230621498,4.403224843658272),linewidth(3.pt) + dotstyle); label("$C$", (-6.220945000838566,4.4672297155801335), NE * labelscalefactor); dot((-7.467512233804722,7.4710805880662114),linewidth(3.pt) + dotstyle); label("$B$", (-7.777343781687118,7.689632825224255), NE * labelscalefactor); dot((-3.0535795870448217,6.146800104664222),linewidth(3.pt) + dotstyle); label("$M$", (-2.779330795300219,6.067470715607486), NE * labelscalefactor); dot((-7.656092125318946,6.842528744170774),linewidth(3.pt) + dotstyle); label("$N$", (-8.215765973475442,6.615498455342881), NE * labelscalefactor); dot((-6.131541710963054,6.020067973706444),linewidth(3.pt) + dotstyle); label("$D_1$", (-6.133260562480901,5.47560075669326), NE * labelscalefactor); dot((-5.979679222339047,7.024696744603731),linewidth(3.pt) + dotstyle); label("$B_1$", (-6.505919425500976,7.360816181383019), NE * labelscalefactor); dot((-6.072327091145308,6.411795458654618),linewidth(3.pt) + dotstyle); label("$Q$", (-5.84828613781849,6.3305240306804755), NE * labelscalefactor); dot((-7.11561808980492,4.972680217220363),linewidth(3.pt) + dotstyle); label("$R$", (-7.273158261130544,5.4536796471038445), NE * labelscalefactor); dot((-4.484668585942924,8.65327483924573),linewidth(3.pt) + dotstyle); label("$S$", (-4.489177343274684,8.873372743052709), NE * labelscalefactor); dot((-6.717327373159708,2.144870516097806),linewidth(3.pt) + dotstyle); label("$X$", (-7.229316041951712,2.1655132086914746), NE * labelscalefactor); dot((-5.156256264524813,12.471956311682519),linewidth(3.pt) + dotstyle); label("$Y$", (-4.927599535063009,12.380750277359235), NE * labelscalefactor); dot((-4.332833306710579,17.919215878761307),linewidth(3.pt) + dotstyle); label("$Y'$", (-9.004925918694425,13.323357989704114), NE * labelscalefactor); dot((-7.303113035356361,-1.7303269415108327),linewidth(3.pt) + dotstyle); label("$X'$", (-9.004925918694425,13.323357989704114), NE * labelscalefactor); dot((-7.619705208172238,5.958794272594481),linewidth(3.pt) + dotstyle); label("$D'$", (-8.193844863886026,5.62904852381917), NE * labelscalefactor); dot((-3.182011836969981,7.328706259540812),linewidth(3.pt) + dotstyle); label("$B'$", (-2.9327785624261327,7.40465840056185), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $N$ be the midpoint of arc $ABC$. Let $AC\cap BD=E$ and $NB\cap AC=F$. Define $l$ be the radical axis of $(BB_1F)$ and point $D$. By simple angle chasing, $BD_1DF$ is cyclic and segments $OB,O_DB_1$ are tangent to $(BB_1F)$. Now $OB=OD,O_DB_1=O_DD$ hence $O,O_D$ are lie on $l$. Also we get $\measuredangle EFD=\measuredangle D_1BD=\measuredangle B_1DE\Rightarrow\triangle EDB_1\sim\triangle EFD\Rightarrow ED^2=EB_1\cdot EF$ thus $E$ lies on $l$. From the above, $O,E,O_D$ are collinear. Similarly $O,E,O_B$ are collinear so we have the desired result.

Let $W$ be the point of intersection of $(ABC)$ with the circle centered at $O_B$ and let $Y$ be the point of intersection of $(ABC)$ with the circle centered at $O_D$. Let the tangents at $B$ and $D$ meet $AC$ at $X$ and $Z$. Note that \[\angle CD_1D = \angle CAD+\angle ADD_1 = \angle CDZ + \angle CDD_1 = \angle DZD_1.\]Therefore, $ZB\cdot ZW = ZD^2=ZD_1^2$ which implies that $Z$ lies on $BW$. Similarly, $X$ lies on $DY$. Pascal's on $DDBBWY$ shows that $BD$ and $YW$ intersect on $AC$. Let $T$ be the point of intersection of $O_BO_D$ and $AC$. Note that \[\angle O_BD_1B=90^\circ - \angle BD_1B_1 = 90^\circ - \angle DB_1D_1 = \angle O_DB_1D\]which implies $\triangle O_BD_1B\sim \triangle O_DB_1D$. Thus, \[\frac{BD_1}{DB_1}=\frac{O_BD_1}{O_DB_1}=\frac{D_1T}{B_1T}\]which implies $\triangle BD_1T\sim \triangle DB_1T$, so $BD$ also intersects $AC$ at $T$. Since $BD$ and $YW$ intersect on $AC$, $YW$ also intersects $AC$ at $T$. Consider $\mathcal{H}$ the homothety centered at $T$ that sends $O_B$ to $O_D$. $\triangle B_1DY$ is taken to $\triangle D_1BW$, so $BW\parallel DY$. Clearly, $OO_B\perp BW$ and $OO_D\perp DY$ so $OO_B\parallel OO_D$ as desired.

Here is a solution I think is different from the ones posted above: Let the circumcircle of $ABCD$ be $\omega$, let$\omega_B$ be the circle centered at $O_B$ passing through $D_1$ and define $\omega_D$ similarly. Let $\omega_B$ and $\omega_D$ intersect $\omega$ at $B_2$ and $D_2$ respectively. Let $DD_2$ and the tangent to $\omega$ at $B$ intersect at $D_3$. By angle-chase it follows $D_3B_1=D_3B$ and so by power of point $D_3$ with respect to $\omega_B$ and $\omega$ it follows that $D_3$ lies on $AC$. Similarly, let $B_3$ be the intersection of $BB_2$ and the tangent to $\omega$ at $D$. Then $B_3$ lies on $AC$. Now let $AC\cap BD=P$. By Pascal's on $BB_2DBDD_2$ it follows that $B_3, D_3$ and the intersection of $BD$ and $B_2D_2$ are collinear. However, $B_3D_3\equiv AC$, so $BD\cap B_2D_2=P$. Now we'll do the rest by angle-chase and applying Thales' a few times: We have \[\angle DD_2B_1 = \angle DB_1D_1=\angle BD_1B_1=\angle BB_2D_2\]Let the radius of $\omega_B$ be $r_B$ the radius of $\omega_D$ be $r_D$. Then by sine law and Thales' \[\frac{DP}{PB}=\frac{B_1P}{PD_1}=\frac{B_1D}{D_1B}=\frac{r_D}{r_B}=\frac{O_DB_1}{O_BD_1}\]Also $O_BD_1\perp AC$ and $O_DB_1\perp AC$, so $O_BD_1\parallel O_DB_1$ and by the above ratios $P$ lies on $O_BO_D$. Now $\angle O_BB_2P=\angle O_BPB_2=\angle O_DPD_2=\angle O_DD_2P$ and so $O_BB_2\parallel O_DD_2$ and by Thales' \[\frac{D_2P}{B_2P}=\frac{r_D}{r_B}=\frac{DP}{PB}\]and so $DD_2\parallel BB_2$. Now $OO_D$ is the perpendicular bisector of $DD_2$ which coincides with the perpendicular bisector of $BB_2$ which is $OO_B$. So $OO_B\equiv OO_D$. $\square$

Let $\omega_B,\omega_D$ be the circles from problem statement and let $\Omega=(ABCD)$ Also let $Y=\omega_B \cap \Omega, Z=\omega_D \cap \Omega, X=BD\cap AC$ Since $BB_1$ is angle bisector of $\angle ABC$ it is well known that there exists a circle $\gamma$ which is tangent to $AC$ at $B_1$ and to $\Omega$ at $B$ Using radical axis theorem on $\Omega, \omega_D, \gamma$ we get that $DZ, AC$ and tangent at $B$ to $\Omega$ are concurrent, at $Q$ Similarly $BY, AC$ and tangent at $D$ to $\Omega$ are concurrent at $P$ Now $(A,C;B,Y)\stackrel{B}{=}(A,C;Q,P)\stackrel{D}{=}(A,C;Z,D)\stackrel{X}{=}(C,A;ZX\cap\Omega,B)=(A,C;B,ZX\cap\Omega)$ Hence $Y=ZX\cap\Omega$ which means that $Z,X,Y$ are collinear Because $BD_1\parallel DB_1$ we have $\frac{XD_1}{XB_1}=\frac{XB}{XD}$ Using this and the fact that $B_1D_1$ is internal common tangent of $\omega_B$ and $\omega_D$ we get that $X$ is their insimilicenter Then a homothety at $X$ which sends $\omega_B$ to $\omega_D$ sends $B$ to $D$; $D_1$ to $B_1$; $Y$ to $Z$ (since $Z,X,Y$ collinear) Hence triangles $\Delta D_1BY$ and $\Delta B_1DZ$ are homothetic $\implies BY\parallel DZ$ We have $OO_B\perp BY, OO_D\perp DZ$ so this means $O_D, O, O_B$ collinear

How do you construct this: Let $ABCD$ be a cyclic quadrilateral with circumcircle $k$ and $E$ is a point on $BC$. Construct $F$ on $BC$ such that if $G\ne A$ is the intersection of $k$ and $AF$, then $FD \parallel EG$. Use only straightedge (ruler without markings) and compass. If there's a nice construction it can lead to a potential solution to the problem.

Here’s a sketch of this: Let $DD\cap AC=T$. There exists a circle tangent to $(O)$ at $D$ and $AC$ at $D_1$ since angle bisector. Radical axis on $(BD_1E), \omega, (ABCD)$ gives $DE$ passes through $T$. Repeating this on $B$ and pascal on $DDBBEF$ gives $AC, BD, EF$ concur at what must be the insimilicenter of the two triangles by the parallel condition. But this implies that $BE \parallel DF$ so the three circumcenters lie on the common perpendicular bisector of $BE, DF$.

Let $(O_B)\cap (ABC)=X,(O_C)\cap (ABC)=Y,XB_1\cap (ABC)=P,YD_1\cap (ABC)=Q$ and $DB_1\cap (ABC)=U, BD_1\cap (ABC)=V,NB_1\cap (ABC)=S,SU\cap AC=K$. We have $OO_B\perp DX$ and $OO_D\perp BY$ so it's enough to prove $XD\parallel YB$. Since $NK.NX=NA^2=NB_1.NS$ we get that $K,X,B_1,S$ are concyclic. \[\measuredangle NXP=\measuredangle KXB_1=\measuredangle KSB_1=\measuredangle USN\]Thus, $NU=NP$. Similarily $NV=NQ$ hence $UP\parallel AC\parallel VQ$. \[\measuredangle PXD=\measuredangle B_1XD=\measuredangle CB_1D=\measuredangle B_1D_1B=\measuredangle D_1YB=\measuredangle QYB\]Hence $BP\parallel DQ$. $AC,MN,UV,PQ$ pass through the midpoint of $AC$. Let $W$ be the midpoint of $AC$. \[\frac{WD_1}{WB_1}=\frac{WV}{WU}=\frac{WQ}{WP}\]So we conclude that $D_1Q\parallel PB_1$ which is equavilent to $XP\parallel YQ$. Pascal at $DXPBYQ$ yields $DX\cap BY,XP_{\infty},PB_{\infty}$ are collinear thus, $DX\parallel BY$ as desired.$\blacksquare$