Here's How to Crack It Let the tangents to $(ABCD)$ at $A$ and $C$ intersect at $K$ and let $H$ be the orthocenter of $\triangle{ABC}$, let $KD$ intersect $(ABCD)$ at $D$ and $V$, let $M$ be the midpoint of $AC$ and let $KA$ intersect $(ABCD)$ at $A$ and $R$. Let $T$ be the intersection of $(AOC)$ and $(AOD)$, for some well-know reasons $\angle{AVH}=90$ and $AV$, $BC$ and $KT$ concur at a point $W$ and $W, D$ and $R$ are colinear, now let $AD$ intersect $AK$ and $CK$ at $X_1$ and $X_2$, for some obvious reasons if $Y_1$ and $Y_2$ are the inverses of $X_1$ and $X_2$ wrt $(K,KA)$, then $Y_1Y_2$ passes through $W$, but clearly $WRMV$ is cyclic with diameter $WM$ and $(VMR)$ is the inverse of $(AOD)$ wrt $(K,KA)$ and hence the conclusion.

Another solution here: let the tangents at $a$ and $C$ to $\odot{ABC}$ meet at $T$, $M\in{BD\cap{AT}}$ and $N\in{BD\cap{CT}}$. $X\in{\odot{MNT}\cap{\odot{AOC}}}$ $L\in{XT\cap{BD}}$ $K\in{XO\cap{BD}}$ and finally $P\in{XT\cap{AC}}$. (i know it hurts seeing all of this, but i thought i'd just write them all from the begining). First notice that $MNT$ is isosceles $XO$ and $XT$ are bisectors of $\angle{CXT}$ and $\angle{NXM}$, also $m(\angle{TXO})=90$ since $TO$ is diameter in $\odot{AOC}$. Then we have $(N,M;L,K)=-1$ and also $(P,AC\cap{XO};A,C)=-1$ which gives $(TP,T(AC\cap{XO});TA,TC)=-1$ but intersecting this with $BD$ gives $(L,T(AC\cap{XO})\cap{BD};M,N)=-1$ hence $T(AC\cap{XO})\cap{BD}=K$, which is only possible if $OK$ passes through $AC\cap{BD}$, which is our main claim, proved. Now, using the power of point $K$ we get $XBOD$ cyclic, and we are left to prove that $\odot{MNT}$ and $\odot{BOD}$ are tangent at $X$, which is equivalent to $m(\angle{XNT})+m(\angle{DXO})=90$ but$\angle{XNT}\equiv{\angle{XCO}}$ and since $\angle{TCO}=90$ we are left to prove that $\angle{XCT}\equiv{\angle{XDO}}\iff{\angle{XOT}\equiv{\angle{XDO}}}\iff{\angle{TOB}\equiv{\angle{BDO}}\equiv{\angle{DBO}}}$ which is true since $OT\parallel{BD}$

ACGNmath wrote: Quadrilateral $ABCD$ with perpendicular diagonals is inscribed in a circle with centre $O$. The tangents to this circle at $A$ and $C$ together with line $BD$ form the triangle $\Delta$. Prove that the circumcircles of $BOD$ and $\Delta$ are tangent. Proposed by A. Kuznetsov Nice problem! Suppose tangents to $\odot(DAB)$ at $D$ and $B$ meet at point $X$. Let $\overline{XA}$ and $\overline{XC}$ meet $\odot(DOB)$ again at points $Y$ and $Z$ respectively. Let $S=\overline{AC} \cap \overline{BD}$. Let $W$ be the midpoint of $\overline{AC}$. Claim. $\odot(WSY)$ is tangent to line $\overline{BD}$. (Proof) Note that $\angle OWA=\angle OYA=90^{\circ}$ so $AOYW$ is cyclic. Thus $\angle AYW=\angle AOW$. Now we require $\angle SYX=\angle OAC$. Let $A'$ be the point symmetric to $A$ in line $\overline{BD}$. Let $R=\overline{XA} \cap \odot(ABCD)$ with $R \ne A$; note that $R$ lies on line $\overline{A'M}$. Observe that $\overline{SY} \parallel \overline{A'M}$ as it is the midline in $\triangle AMA'$. Thus $\angle SYX=\angle ARM$. Now $\overline{RM} \cap \odot(ABCD) \overset{\text{def}}{:=} L$ then $\overline{AL} \parallel \overline{BD}$ so $\angle ARM=\angle ACL=\angle OAC$ and we're done. $\blacksquare$ Finally, observe that $\odot(WSZ), \odot(WSY)$ are both tangent to line $\overline{BD}$. Hence $\odot(WYZ)$ touches line $\overline{BD}$ at point $S$. Now we invert at $O$; clearly $\triangle WYZ$ maps to the triangle formed by $\overline{BD}, \overline{AA}, \overline{CC}$; hence the inverse point $T$ of $S$ in $\odot(ABCD)$ is the tangency point of $\Delta$ and $\odot(BOD)$. $\blacksquare$ Remark. As always, "guessing" the tangency point is the hardest part of the proof. The intuition behind our candidate is as follows. We know $T \in \odot(BOD)$. Swapping $A$ and $C$ preserves $\Delta$ so whatever way we use to define $T$ given $\triangle DAB$ must also work if we used $\triangle CBD$ instead. Another idea is that we invert at $\odot(O)$ because (a) images of vertices of $\Delta$ are especially nice; (b) because "line-circle" tangency is convenient to prove as opposed to "circle-circle". The inverse of $T$ must lie on $\overline{BD}$. So we seek a point on $\overline{BD}$ symmetric with respect to $A$ and $C$; but depending on them. The most natural choice would then be $\overline{AC} \cap \overline{BD}$; which happens to work!

Let $M$ be the Miquel point of the three sides of $\Delta$ and $AC$, and let $X=AC\cap BD$. Since $MX$ bisects $\angle AMC$, hence $O,M,X$ are collinear. We get $OX\cdot OM=OB^2=OC^2$, thus $M$ lies on $(OBC)$. But $\Delta$ and $AOC$ are spirally similar about $M$ with a $90$ degree rotation, so $(\Delta)$ is orthogonal to $(AOC)$ which is in turn orthogonal to $(BOD)$.

DVDthe1st wrote: Let $M$ be the Miquel point of the three sides of $\Delta$ and $AC$, and let $X=AC\cap BD$. Since $MX$ bisects $\angle AMC$, hence $O,M,X$ are collinear. We get $OX\cdot OM=OB^2=OC^2$, thus $M$ lies on $(OBC)$. But $\Delta$ and $AOC$ are spirally similar about $M$ with a $90$ degree rotation, so $(\Delta)$ is orthogonal to $(AOC)$ which is in turn orthogonal to $(BOD)$. Why $MX$ bisect $\angle AMC$ ?

[asy][asy] size(10cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); pair X = extension(A, A+dir(90)*A, C, C+dir(90)*C); pair Y = extension(A, X, B, D); pair Z = extension (C, X, B, D); pair E = extension(A, C, B, D); draw(Y--Z--B); draw(A--X--Y); draw(C--Z--X); draw(circumcircle(B,O,D),red); draw(circumcircle(A,O,C),heavycyan); draw(circumcircle(X,Y,Z),red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); [/asy][/asy] $$\text{Figure 1: The Big Picture}$$ Above is the big picture. $E = AC \cap BD, X = AA \cap CC, Y = AX \cap BD, Z = CX \cap BD.$ Now we consider a smaller diagram: [asy][asy] size(10cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); pair E = extension(A, C, B, D); draw(circumcircle(B,O,D),red); path w = circumcircle(B,O,D); path ao = (10*A-9*O)--(10*O-9*A); pair G = OP(ao,w); path co = (10*C-9*O)--(10*O-9*C); pair H = IP(co, w); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); draw(A--O--G); draw(H--C--O); draw(H--G); [/asy][/asy] $$\text{Figure 2: A Small Configuration}$$ Denote $(BOD)=\omega$. Next, let $G=AO\cap\omega, H=CO\cap\omega$. Now we note that $\angle BAC=90^{\circ}-\angle DBA = 90^{\circ}-\frac{1}{2}\angle DOA = \angle OAD$. Hence $\angle BOC=\angle GOD$. Which means that $HG$ is parallel to $BD$. Next we consider a larger diagram: [asy][asy] size(10cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); pair E = extension(A, C, B, D); draw(circumcircle(B,O,D),red); path w = circumcircle(B,O,D); path ao = (10*A-9*O)--(10*O-9*A); pair G = OP(ao,w); path co = (10*C-9*O)--(10*O-9*C); pair H = IP(co, w); draw(A--O--G); draw(H--C--O); draw(H--G); path eg = (10*E-9*G)--(10*G-9*E); path eh = (10*E-9*H)--(10*H-9*E); pair P = IP(eg,w); pair Q = IP(eh, w); draw(G--P); draw(H--Q); pair foot = foot(O, B, D); path ofoot = (10*O-9*foot)--(10*foot-9*O); pair M = OP(ofoot,w); draw(A--Q--M, orange+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$M$", M, dir(M)); [/asy][/asy] $$\text{Figure 3: Extending the Configuration}$$ Let $P = GE \cap \omega, Q = HE \cap \omega$. Now note that $AE\cdot EC = BE\cdot ED = HE\cdot EQ$, hence $AQCH$ cyclic. Also note that $\angle AQH + \angle HQM = \angle ACH + \angle HOM = 180^{\circ}-\angle ACO + \angle HOM = 180^{\circ}$ Hence $A, Q, M$ collinear. $\angle BEP = \angle HGP = \angle HQP = \angle EQP$. Hence $(EQP)$ is tangent to $BD$ at $E$. [asy][asy] size(10cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); pair E = extension(A, C, B, D); draw(circumcircle(B,O,D),red); path w = circumcircle(B,O,D); path ao = (10*A-9*O)--(10*O-9*A); pair G = OP(ao,w); path co = (10*C-9*O)--(10*O-9*C); pair H = IP(co, w); draw(A--O--G); draw(H--C--O); draw(H--G); path eg = (10*E-9*G)--(10*G-9*E); path eh = (10*E-9*H)--(10*H-9*E); pair P = IP(eg,w); pair Q = IP(eh, w); draw(G--P); draw(H--Q); pair foot = foot(O, B, D); path ofoot = (10*O-9*foot)--(10*foot-9*O); pair M = OP(ofoot,w); draw(A--Q--M, orange+dashed); pair I = foot(O, A, C); draw(circumcircle(E, Q, P), green); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(240)); dot("$M$", M, dir(M)); dot("$I$", I, dir(I)); [/asy][/asy] $$\text{Figure 4: The Final Configuration}$$ Now let $I$ be the midpoint of $AC$. We remark that $\angle AIO = 90^{\circ}=\angle AQO$. Note that $\angle EQI = 180^{\circ}-\angle AQI-\angle EQM = 180^{\circ}-\angle AOI - \angle HOM = 90^{\circ}$. Similarly, we can show that $\angle EPI = 90^{\circ}$. Hence $E, Q, P, I$ are concyclic, with circumcircle tangent to $BD$ at $E$. Now invert the diagram with respect to $(O)$. This results in the first diagram, and we are done.

Consider the notations in the attached figure. $F$ is defined as the second intersection of circles $\odot(AOCP)$ and $\odot(PXY)$. Notice that $F$ is the center of spiral similarity sending segment $XY$ to segment $CA$, and moreover as triangles $PXY$ and $OAC$ are isosceles and similar, it follows that this spiral similarity centered at $F$ also sends $P$ to $O$. Also, $\angle PFO = 90^{\circ}$. $OP \perp CA$, hence $OP \parallel BD$. Thus $OP$ is a common tangent of circles $\odot(PXY)$ and $\odot(OBD)$. Then the midpoint of this segment lies on the radical axis of the aforementioned circles. Let $L$ be the midpoint of $JE$. We want to prove that $L$ lies on the radical axis of circles $\odot(PXY)$ and $\odot(OBD)$. Now, Desargues' Involution Theorem yields the existence of some involution on line $BD$, with pairs of points $(B,D)$ , $(X,Y)$ and $(E,E)$. However, by looking at our initial spiral similarity we get that : $\frac{YJ}{JX} = \frac{AE}{EC} = \frac{DE}{EB}$. This means that in the involution on line $BD$ we got earlier, $(J,J)$ is also a pair. As every involution is also an inversion, there is an inversion centered at $L$, with radius $LE^2 = LJ^2$, swapping $B$ with $D$ and $X$ with $Y$. Thus : $LB \cdot LD = LX \cdot LY$, so $L$ also lies on the radical axis of circles $\odot(PXY)$ and $\odot(OBD)$. So, $LK$ is the radical axis of $\odot(PXY)$ and $\odot(OBD)$, $K$ being the midpoint of $PO$. Now, $LK$ is also a median in triangle $FPO$, so it contains $F$. Because $KP = KF$, $KF$ is a tangent to $\odot(PXY)$ and since $KF$ is the radical axis of $\odot(PXY)$ and $\odot(OBD)$, we get that these two circles are tangent at point $F$ and the conclusion follows.

A different solution: Let $BD \cap AA = X, BD \cap CC = Y, AA \cap CC = Z, AO \cap \odot (ABCD) = A'$. Note that $OZ \perp AC \Rightarrow OZ \parallel BD \Rightarrow OZ$ is tangent to $\odot (BOD)$. Also, $\angle ZYX = \angle YZO = \angle AZO = \angle ZXY \Rightarrow ZX = ZY$. Now, $\triangle AA'C \sim \triangle ZOA \Rightarrow \frac{AA'}{AC} = \frac{OZ}{ZA} \Rightarrow \frac{2AO}{AC} = \frac{OZ}{ZA}$ And, $\triangle OAC \sim \triangle ZXY \Rightarrow \frac{ZX}{XY} = \frac{AO}{AC} = \frac{OZ}{2ZA} \Rightarrow 2ZA \cdot ZX = XY \cdot OZ$ Also, $XA+CY = (ZX+ZA)+(ZC-ZY) = 2ZA \Rightarrow (XA+YC) \cdot ZX = XY \cdot OZ \Rightarrow XA \cdot ZY+YC \cdot ZX = XY \cdot OZ \Rightarrow ZY \cdot \sqrt{XB \cdot XD} + ZX \cdot \sqrt{YB \cdot YD} - ZO \cdot XY = 0$ Thus, Using the aforementioned equality, and by the converse of Casey's Theorem on point circles $\odot (X), \odot (Y), \odot (Z)$ and circle $\odot (BOD)$, we get that $\odot (XYZ)$ is tangent to $\odot (BOD)$.

[asy][asy] size(20cm); pair O = origin; pair A = dir(120); pair B = dir(200); pair C = dir(240); pair D = dir(340); draw(unitcircle, cyan); draw(A--B--C--D--cycle,magenta); draw(A--C); draw(B--D); pair E = extension(A, C, B, D); draw(circumcircle(B,O,D), heavygreen); path w = circumcircle(B,O,D); path ao = (10*A-9*O)--(10*O-9*A); pair G = OP(ao,w); path co = (10*C-9*O)--(10*O-9*C); pair H = IP(co, w); draw(A--O--G); draw(H--C--O); draw(H--G); path eg = (10*E-9*G)--(10*G-9*E); path eh = (10*E-9*H)--(10*H-9*E); pair P = IP(eg,w); pair Q = IP(eh, w); draw(G--P); draw(H--Q); pair foot = foot(O, B, D); path ofoot = (10*O-9*foot)--(10*foot-9*O); pair M = OP(ofoot,w); pair I = foot(O, A, C); draw(circumcircle(A,C,H), red); draw(circumcircle(A, I, Q), orange); draw(circumcircle(E, Q, P), purple); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O$", O, dir(50)); dot("$E$", E, dir(E)); dot("$G$", G, dir(G)); dot("$H$", H, dir(H)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(240)); dot("$M$", M, dir(M)); dot("$I$", I, dir(I)); [/asy][/asy] INVERTED FIGURE (after inversion at O with circle (ABCD)) Here’s a sketch. Angle chasing gives $GH // AB$ Also by PoP, $AQCH$ is cyclic. Angle chasing gives $A,Q,M$ to be collinear. Also $A,I,Q,O$ is cyclic. Angle chasing gives $(EQP)$ to be tangential to BD. Also $E,Q,P,I$ are concyclic. In the original figure, after inversion, $ X \Rightarrow I$ $ Y \Rightarrow P$ $ (BOD) \Rightarrow BD$ $ Z \Rightarrow Q$. So, we are done.

In this solution we will convert this to an incircle problem. Let the tangents to $(O)$ at $A$ and $C$ meet at $Z$ Let the tangents to $(O)$ at $D$ and $C$ meet at $X$ Let the tangents to $(O)$ at $A$ and $D$ meet at $Y$ Let $BD$ meet $ZC$ and $ZA$ at $P$ and $Q$ Let $BD$ intersect $AC$ at $M$ Let $(AZCO)$ intersect $(XYZ)$ at $N$ We will prove that $N$ is the desired point of tangency By the sharky-devil theorem $O, M, N$ collinear and $DN$ is angle bisector of $\angle XNY$ $\angle MNC=\angle ONC=\angle OZC=\angle MPC$ $\implies MPNC$ cyclic $\angle ZNP=\angle ZNC-\angle PNC=180-\angle ZAC-\angle PMC=90-\angle ZAC=\angle AZO=\angle ZQP$ $\implies ZPNQ$ cyclic $MO\cdot MN=MA\cdot MC=MB\cdot MD$ $\implies DOBN$ cyclic $\angle OCM=\angle CNM$ $\implies OM\cdot ON=OC^2=OD^2$ $\implies \Delta ODM \sim \Delta ODN$ $\implies \angle OMD=\angle ODN$ $\angle ZQN+\angle ODN=\angle NPC+ \angle OMD=\angle NMC+\angle PMN=90=\angle ZNO$ $\implies (ZPNQ)$ and $(DOBN)$ are tangent as desired

Solved with Aayuu Let the $A,C$ tangents meet $BD$ at $Y,Z$ and intersect each other at $X$. Let $AC \cap BD = P$ and finally, let $M$ be the miquel point of quadrilateral $ABCD$ which is well known to lie on $(BOD)$ and $(AOC)$ and on line $OP$. See that $\angle AMP = \angle AMO = \angle ACO = 90 - \angle ADC = 90 - \angle PAY = \angle AYP$ so $M \in (APY)$ and since it lies on $(AXC)$ too, it must be the miquel point of $PYXC$ and so lies on $(XYZ)$. To show tangency, it suffices to show $90 = \angle XMO = \angle XZM + \angle MDO$. But $\angle XZM + \angle MDO = \angle MYA + \angle MDO = 90 - \angle OPD + \angle MDO = 90$ since $OP.OM = OD^2$, so we are done. $\blacksquare$

We perform a normal inversion with respect to $(ABCD)$ with center $O$. Let $E=(AOC)\cap (BOD)$ and $P^*$ is the midpoint of $AC$. That inversion swaps $\{AC, (AOC)\}, \{PC, (P^*OC)\}$. The points $Q, E, C, Q^*, E^*$ are concyclic by the obvious reason. Notice $(P^*OC)$ is tangent to $PC$. Now \begin{align*} \angle P^*Q^*O+\angle OQ^*E^*&=\angle P^*CO+\angle QCE^*=\angle PCO=90^\circ \end{align*}Means $\angle P^*Q^*E^*$ is right angle and $P^*E^*$ is the diameter of $\triangle P^*Q^*E^*$. Similarly $\angle P^*R^*E^*=90^\circ$. So $P^*Q^*E^*R^*$ is cyclic. The line $BD$ is tangent to the cyclic quadrilateral $P^*Q^*E^*R^*$ at $E^*$. So the cyclic quadrilateral $PQER$ is tangent to $(BOD)$ at $E$.

Let $AA\cap CC=P, AA\cap BD=Q, CC\cap BD=R$ and $p,q,r$ be the power of $P,Q,R$ to $(BOC)$ respectively . Now $BD$ is radical axis of $(ABCD),(BOC)$ so $q=QA,r=RC$. Also, $OP$ is tangent to $(BOC)$ so $p=OP$. Notice that $PQ/QR=OP/(2PA)$ and $PQ=PR,$ $$PR\cdot q+PQ\cdot r=PR\cdot QA+PQ\cdot RC=PQ\cdot (QA+RC)=2PQ\cdot PA=QR\cdot OP=QR\cdot p.$$Thus from the converse of Casey's theorem, $(PQR)$ and $(BOC)$ are tangent to each other.

Oo very cute but hard problem.. Thanks to L567( for spoiling the tangency point). [asy][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12.3cm); 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 = -4.8, xmax = 17.5, ymin = -11.9, ymax = 3.2; /* image dimensions */ pen wrwrwr = rgb(0.4,0.4,0.4); draw((2.1,-2.8)--(3.6,1.3)--(10.2,-2.9)--cycle, linewidth(0.4) + wrwrwr); draw((2.1,-2.8)--(3.5,-5.2)--(10.2,-2.9)--cycle, linewidth(0.4) + wrwrwr); /* draw figures */ draw((2.1,-2.8)--(3.6,1.3), linewidth(0.4) + wrwrwr); draw((3.6,1.3)--(10.2,-2.9), linewidth(0.4) + wrwrwr); draw((10.2,-2.9)--(2.1,-2.8), linewidth(0.4) + wrwrwr); draw(circle((6.2,-2), 4.1), linewidth(0.4) + wrwrwr); draw((2.1,-2.8)--(3.5,-5.2), linewidth(0.4) + wrwrwr); draw((3.5,-5.2)--(10.2,-2.9), linewidth(0.4) + wrwrwr); draw((10.2,-2.9)--(2.1,-2.8), linewidth(0.4) + wrwrwr); draw(circle((-0.4,-3), 1.1), linewidth(0.4) + wrwrwr); draw(circle((6.1,-11.7), 9.7), linewidth(0.4) + wrwrwr); draw((0.2,-3.9)--(6.2,-2), linewidth(0.4) + wrwrwr); draw((0.2,-3.9)--(3.5,-5.2), linewidth(0.4) + wrwrwr); draw((3.6,1.3)--(3.5,-5.2), linewidth(0.4) + wrwrwr); draw(circle((2.9,-2), 3.3), linewidth(0.4) + wrwrwr); draw(circle((4.5,-6.8), 5.1), linewidth(0.4) + wrwrwr); draw((0.2,-3.9)--(2.9,-2), linewidth(0.4) + wrwrwr); draw((2.9,-2)--(6.2,-2), linewidth(0.4) + wrwrwr); draw((6.1,-11.7)--(6.2,-2), linewidth(0.4) + wrwrwr); draw((-1.5,-2.8)--(2.1,-2.8), linewidth(0.4) + wrwrwr); draw((-1.5,-2.8)--(3.6,1.3), linewidth(0.4) + wrwrwr); draw((-0.4,-1.9)--(3.5,-5.2), linewidth(0.4) + wrwrwr); /* dots and labels */ dot((2.1,-2.8),linewidth(5pt) + dotstyle); label("$A$", (2.2,-2.7), NE * labelscalefactor); dot((3.6,1.3),linewidth(5pt) + dotstyle); label("$B$", (3.6,1.4), NE * labelscalefactor); dot((10.2,-2.9),linewidth(5pt) + dotstyle); label("$C$", (10.3,-2.7), NE * labelscalefactor); dot((3.5,-5.2),linewidth(4pt) + dotstyle); label("$D$", (3.6,-5.1), NE * labelscalefactor); dot((6.2,-2),linewidth(4pt) + dotstyle); label("$O$", (6.2,-1.8), NE * labelscalefactor); dot((-0.4,-1.9),linewidth(4pt) + dotstyle); label("$E$", (-0.4,-1.8), NE * labelscalefactor); dot((-1.5,-2.8),linewidth(4pt) + dotstyle); label("$F$", (-1.5,-2.7), NE * labelscalefactor); dot((0.7,-2.8),linewidth(4pt) + dotstyle); label("$G$", (0.7,-2.7), NE * labelscalefactor); dot((0.2,-3.9),linewidth(4pt) + dotstyle); label("$M$", (0.3,-3.8), NE * labelscalefactor); dot((3.5,-2.8),linewidth(4pt) + dotstyle); label("$P$", (3.6,-2.7), NE * labelscalefactor); dot((-0.4,-3),linewidth(4pt) + dotstyle); label("$O_1$", (-0.4,-2.9), NE * labelscalefactor); dot((6.1,-11.7),linewidth(4pt) + dotstyle); label("$O_3$", (6.1,-11.6), NE * labelscalefactor); dot((2.9,-2),linewidth(4pt) + dotstyle); label("$O_2$", (2.9,-1.8), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] We introduce the miquel point, $M$ for quadrilateral $ABCD$. Define $O_2$ as the center of $BOD$ and $O_3$ as the center of $COA$. Claim: $O_2MO_3O$ cyclic quadrilateral Proof: Note that $O_2O_3$ is the perpendicular bisector of $OM$. Note that $$OO_2||AC,O_3O||BD, \text{ as } BD\perp AC\implies O_2O\perp OO_3\implies O_2MO_3O\text{ cyclic} $$ Now, we define $M'=(EFG)\cap (BODE)$. Claim: $M'=M$ i.e $O-P-M',P=BD\cap AC.$ Proof: Note that $M'$ is the spiral center taking $E\rightarrow B, F\rightarrow B, G\rightarrow D.$ But note that $$\angle EM'O=90\implies \angle FM'B=90=\angle GMD$$$$\implies GM'DP, FM'PB\text{ cyclic}$$$$\implies \angle PM'D=\angle PGD=\angle EFG=\angle EM'G=\angle OM'D\implies O-P-M'$$$$\implies M'=M.$$ Define $O_1$ as the center of $(EFG).$ Then note that $$\angle O_1MO_2=90.$$But $$ \angle O_2MO_3=90\implies O_1-M-O_3.$$So $M$ is the tangency point.

Let $X = AA\cap BD$, $Y = CC\cap BD$, $Z = AA\cap CC$, $E = AC\cap BD$. We claim the tangency point is the miquel point $M$ of $ABCD$. First, we have $$\angle MAX = \angle MAB - \angle BAX = \angle MDC - \angle BDA = 180^\circ - \angle MCD - \angle CBD = \angle YCD - \angle MCD = \angle MCY$$ and $$\frac{AX}{CY} = \frac{AE}{CE} = \frac{BA}{BC}\cdot\frac{DA}{DC} = \frac{MA}{MC}$$ so $\triangle MAX\sim\triangle MCY$ and hence $M$ is the spiral center mapping $AX\to CY$. In particular, $M$ lies on $(XYZ)$. Now since $M$ lies on the line through $AB\cap CD$ and $AD\cap BC$ by pascal, we have $\angle OMZ = 90^\circ$. To prove the tangency, it suffices to show $$\angle MDO + \angle MYZ = \angle OMZ = 90^\circ.$$ We have $$\angle MYZ = 180^\circ - \angle MEC = \angle AEM$$ and if $W = BB\cap DD$, $$\angle MDO = \angle MWO = 90^\circ - \angle WOM = 90^\circ - \angle AEM$$ as desired.

Very clean and beautiful problem statement, a great hard 1/4 or standard 2/5. First, we look at images of points under inversion about $(ABCD)$. Let $X = \overline{AA} \cap \overline{BD}$, $Y = \overline{CC} \cap \overline{BD}$, and $P = \overline{AA} \cap \overline{CC}$. Additionally, denote by $E, F, G$ the images of $P, X, Y$ under inversion about $(ABCD)$. Furthermore, set $M = (AOC) \cap (BOD)$. Furthermore let $T = \overline{AC} \cap \overline{BD}$ be the image of $M$. Claim. $MATFX$ is cyclic (and so is its symmetric variant). Proof. Note that $(MAT)$ is orthogonal to $(ABCD)$, so it suffices to show $X$ lies on it. But $\angle ATX = 90^\circ$ and $\overline{OA}$ is tangent to $(MAT)$ by orthogonality, so $\overline{AX}$ is a diameter and the result follows. $\blacksquare$ Now, I claim that $(EFG)$ is tangent to $\overline{BD}$ at $T$, which will imply the original problem. It suffices to show $\angle TFE = \angle TGE = 90^\circ$. To do so, note $$\angle TFE = \angle TFO + \angle EFO = \angle CAP + \angle EAO = 90^\circ.$$The other angle follows similarly.

Invert first about circle $O,$ then about the circle with diameter $AC.$ Denote by $C',D',O'$ the images of $C,D,O,$ and $M$ the midpoint of $AC.$ Let $X$ be the center of circle $AB'CD',$ and let $E,F$ be the intersections of circle $B'D'O'$ with lines $AO'$ and $CO'.$ First, since $M$ is on the polar of $O'$ with respect to circle $X,$ we have that $M,O'$ are images under an inversion at circle $X,$ so circles $B'D'M$ and $B'D'O'$ are images of each other. Thus, since $XM$ is tangent to circle $B'D'M$ we have that $XO'$ is tangent to circle $B'D'O'.$ Since $XO'$ is an angle bisector of lines $AO'$ and $CO',$ we get $EO'=FO'.$ From here, we may let $P$ be the midpoint of $EF.$ A boring trig bash tells us that $PO'$ has the same length as the diameter of circle $B'D'M,$ so $EF$ is tangent to $B'D'M$ at the antipode of $M,$ and inverting back gives us our desired result.

Define the points $R = BD \cap AA$, $S = BD \cap CC$, and $T = AA \cap CC$. Inverting about circle $O$, we note $T$ maps to the midpoint of $AC$, which we denote as $M$. $R$ maps to $(AOM) \cap (BOD)$, which we denote as $P$. $S$ maps to $(COM) \cap (BOD)$, which we denote as $Q$. Let $T = AC \cap BD$. It is clear that $APXR$ and $CQXS$ are cyclic from the right angles. Thus \[\measuredangle MPX = \measuredangle MPR + \measuredangle RPX = \measuredangle MAO + \measuredangle RAX = \measuredangle RAO = 90,\] and analogously for $Q$, so $MPXQ$ is cyclic, with its circumcircle tangent to $BD$. Inverting back, we get the desired. $\blacksquare$

Solved with kingu. Really nice problem which simply dies to inversion. Let $G$ be the intersection of the tangents to $(ABCD)$ at $A$ and $C$. Let $F$ and $H$ be the intersections of $BD$ and the tangents to $(ABCD)$ at $A$ and $C$ respectively. We define $X= (AOC) \cap \overline{OE}$. Our goal is to show that this is the desired tangency point. Consider the inversion $\Omega$ centered at $O$ with radius $OA$. It is clear that $(ABCD)$ is fixed under this inversion. Further, $G$ maps to the midpoint of $AC$ - $M$. Now, we do the proof in three parts. Claim : $X$ lies on $(BOD)$. Proof : Note that since $AOCG$ is clearly cyclic, $X$ maps to $\overline{AC} \cap \overline{OE}$ which is simply $E$. But, since $E$ lies on $\overline{BD}$, it is clear that $X$ lies on $(BOD)$ as desired. Claim : $X$ lies on $(FGH)$. Proof : Let $H^*$ and $F^*$ be the inverses of $H$ and $F$ respectively under $\Omega$. It is clear that $H^* = (BOD) \cap (MOC)$ and $F^* = (BOD) \cap (MOA)$. Now, note that since $O-H-H^*$ under inversion, \[\measuredangle CH^* H = \measuredangle CH^* O = \measuredangle CMO = \measuredangle = 90^\circ = \measuredangle CEH \]Thus, $H^* HCE$ must be cyclic. Now, this allows us to see that, \[\measuredangle MH^* E = \measuredangle MH^* C + \measuredangle CH^* E = \measuredangle MOC = \measuredangle CHE = \measuredangle MOC + 90 + \measuredangle HCE = 90^\circ\]Similarly, we can show that $\measuredangle EF^* M = 90^\circ$. Thus, $MF^* E H^*$ is cyclic, which implies the claim. Now, note that circle $(MF^* E H^*)$ is tangent to $\overline{BD}$ at $E$ since $ME \perp BD$ and the center of $(MF^* E H^*)$ lies on $ME$ since we showed $\measuredangle EF^* M = 90^\circ$. Thus, we must have that the circumcircles of $BOD$ and $\Delta$ are tangent as desired.

Tuymaada 2018 [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.9064240992493cm); 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 = -247.90374708817487, xmax = 231.0026770110744, ymin = -60.907577424272496, ymax = 213.95772936508536; /* image dimensions */ pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); pen zzffff = rgb(0.6,1.,1.); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen ffttww = rgb(1.,0.2,0.4); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen wwzzff = rgb(0.4,0.6,1.); /* draw figures */ draw(circle((-2.520434060887578,2.5734323754399524), 53.57768597454855), linewidth(1.) + ffcqcb); draw((-52.20454218520992,22.624812235139977)--(42.450367652169426,31.69722724616801), linewidth(1.) + ffcqcb); draw(circle((51.53274545315879,124.50296489253384), 65.43989162003422), linewidth(1.) + zzffff); draw(circle((48.841827430532,7.496365089185839), 51.597646962017876), linewidth(1.) + zzffff); draw((-2.520434060887578,2.5734323754399524)--(50.028156533291515,59.080372254943896), linewidth(1.) + afeeee); draw((-4.877087266520244,27.161019740653998)--(-1.405537984204764,19.223233421990987), linewidth(1.) + ffttww); draw((22.812867954764343,29.8150264460206)--(-1.405537984204764,19.223233421990987), linewidth(1.) + ffttww); draw((-4.877087266520244,27.161019740653998)--(10.73469725235309,42.283773753821364), linewidth(1.) + ffttww); draw((10.73469725235309,42.283773753821364)--(22.812867954764343,29.8150264460206), linewidth(1.) + ffttww); draw(circle((-8.064523620234793,60.416387832445906), 58.108040967178), linewidth(1.) + dtsfsf); draw(circle((30.820139267697137,49.655762419115284), 21.39558827966574), linewidth(1.) + eqeqeq); draw((-52.20454218520992,22.624812235139977)--(8.972773200180354,174.21240239757438), linewidth(1.) + wwzzff); draw((8.972773200180354,174.21240239757438)--(42.450367652169426,31.69722724616801), linewidth(1.) + wwzzff); draw((8.972773200180354,174.21240239757438)--(29.53841590307158,-40.35439744829668), linewidth(1.) + ffcqcb); draw(circle((8.967890344122047,28.488023093337336), 13.908427047557144), linewidth(1.) + dotted + ffttww); draw((-13.608613179582,118.25934328945185)--(42.450367652169426,31.69722724616801), linewidth(1.) + wwzzff); draw((-2.520434060887578,2.5734323754399524)--(-13.608613179582,118.25934328945185), linewidth(1.) + eqeqeq); /* dots and labels */ dot((-2.520434060887578,2.5734323754399524),dotstyle); label("$O$", (-0.8622290228931107,6.9656041602161745), NE * labelscalefactor); dot((-52.20454218520992,22.624812235139977),dotstyle); label("$C$", (-50.607790681089774,26.779514312209763), NE * labelscalefactor); dot((42.450367652169426,31.69722724616801),dotstyle); label("$A$", (44.246034514624206,36.054110553568464), NE * labelscalefactor); dot((22.812867954764343,29.8150264460206),dotstyle); label("$E$", (24.432124362630617,33.946247771441485), NE * labelscalefactor); dot((20.80062641772244,50.8092756099098),linewidth(4.pt) + dotstyle); label("$D$", (22.32426158050364,54.18173047986047), NE * labelscalefactor); dot((29.53841590307158,-40.35439744829668),linewidth(4.pt) + dotstyle); label("$B$", (31.177285265436947,-36.877941708024956), NE * labelscalefactor); dot((-13.608613179582,118.25934328945185),linewidth(4.pt) + dotstyle); label("$H$", (-11.823115489953393,121.63333950792375), NE * labelscalefactor); dot((19.18991088322482,67.61429759206256),linewidth(4.pt) + dotstyle); label("$F$", (21.059543911227454,71.0446327368763), NE * labelscalefactor); dot((8.972773200180354,174.21240239757438),linewidth(4.pt) + dotstyle); label("$G$", (10.520230000592568,177.70248951250136), NE * labelscalefactor); dot((50.028156533291515,59.080372254943896),linewidth(4.pt) + dotstyle); label("$T$", (51.834340530281324,62.61318160836838), NE * labelscalefactor); dot((-4.877087266520244,27.161019740653998),linewidth(4.pt) + dotstyle); label("$K$", (-3.3916643614454833,30.57366732003832), NE * labelscalefactor); dot((10.73469725235309,42.283773753821364),linewidth(4.pt) + dotstyle); label("$L$", (12.628092782719545,45.75027935135256), NE * labelscalefactor); dot((-1.405537984204764,19.223233421990987),linewidth(4.pt) + dotstyle); label("$M$", (0.40248864638307585,22.563788747955808), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $H=AA \cap CC$, $G= CC \cap BD$ and $F= AA \cap BD$. Let $E$ be the intersection of diagonals of $ABCD$. We begin by inverting around $(ABCD)$, let $L$ and $M$ be the inverses of $F$ and $G$, we know that $L$ and $M $lie on $(BOD)$. Let $K$ be the inverse of $H$, then $K$ lies on $AC$, finally let $T$ be the inverse of $E$. We show that $T$ is the desired tangency point. We can easily get that $FLEAT$ is cyclic, hence by easy angle chase we show that $\angle KLE = KME=90$, hence $(KLME)$ is cyclic and tangent to $BD$, inverting back and we are done.

$ $

Let $\overline{AA} \cap \overline{CC} = T, \overline{AA} \cap BD = X,$ and $\overline{CC} \cap BD = Y.$ Also let $AC$ and $BD$ intersect at $K.$ Now consider inverting with respect to $(ABCD)$. This sends $T$ to the midpoint $M$ of $AC.$ We will show that it sends the circumcircle of $\Delta$ to the circle with diameter $MK,$ which finishes the problem since $(BOD)$ inverts to $BD.$ It suffices to prove that $X$ is sent to a point on this circle, because similar logic will hold for $Y.$ The inverse of $\overline{AA}$ is the circle with diameter $OA.$ Additionally, if $V = \overline{BB} \cap \overline{DD},$ then $BD$ is sent to the circle with diameter $OY.$ Therefore, $X$ is sent to the intersection of these two circles, which is the foot of the altitude from $O$ to $AV.$ Taking a homothety of scale $2$ at $A,$ the problem reduces to showing that if the reflection of $A$ across $BC$ is $A'$ and $AV$ intersects $(ABCD)$ at point $W,$ then $\angle CWA' = 90^\circ.$ However, it is well-known that $W$ is the Humpty point of $\triangle BA'D,$ which finishes.

Let $AC \cap BD = E$, $AA \cap BD = P$, $CC \cap BD = Q$, $AA \cap CC = R$. Also let $X$ be the Miquel point of the lines $AA,CC,AC,BD$. We claim that $X$ is also the Miquel point of $ABCD$. Trivially $X$ lies on $(ACR)$, and since $\measuredangle OAR = 90^\circ = \measuredangle OCR$ we also have $O$ on $(ACR)$, so $(ACROX)$ are concyclic. Note that $RP=RQ$ since $RA=RC$ and $AC \perp BD$. Since $(AEPX)$ and $(CEQX)$ are concyclic, $\measuredangle AXE = \measuredangle APE = \measuredangle RPQ = \measuredangle PQR = \measuredangle EQC = \measuredangle EXC$, so $XE$ is an angle bisector of $\angle AXC$. $XE$ must pass through an arc midpoint of $AC$ on $(ACROX)$, which is either $R$ or $O$. But by definition, $X,E,R$ cannot be collinear, so $XE$ passes through $O$. The inverted image of $X$ wrt $(ABCD)$ must thus lie on lines $AC$ and $OE$, so it must be $E$. This implies that $X$ is the Miquel point of $ABCD$. Next we claim that $(BOD)$ and $(XPQR)$ are tangent at $X$. Firstly, since $E$ lies on $BD$, inversion about $(ABCD)$ shows that $X$ lies on $(BOD)$. It suffices to show that $\measuredangle OXR = \measuredangle OBX + \measuredangle XPR$. Clearly $\measuredangle OXR = 90^\circ$. Again by inversion about $(ABCD)$ we see that $\measuredangle OBX = - \measuredangle OEB$, then $\measuredangle OBX + \measuredangle XPR = \measuredangle BEO + \measuredangle XPA = \measuredangle PEX + \measuredangle XEA = \measuredangle PEA = 90^\circ$, so we are done. $\square$

We claim that the tangency point is the second intersection of $(AOC)$ and $(BOD)$ (incidentally the Miquel point, though this is not used). Let this point be $E$. Let $T$ be where the tangents of $(ABCD)$ at $A$ and $C$ meet; let $X$ be where the tangent to $(ABCD)$ at $A$ meets line $BD$; let $Y$ be where the tangent to $(ABCD)$ at $C$ meets line $BD$. Invert the problem at $O$ and refer to points' images by their original names (except for $O$). Note that point $X$ is now the $A$-dumpty point of $\triangle ABD$, and point $Y$ is now the $C$-dumpty point of $\triangle CDB$. To show that $(TXY)$ is tangent to $\overline{BD}$ at $E$, it suffices to show that $TE$ is a diameter of $(TXY)$. Claim: We have $\angle TYE = 90^{\circ}$. Proof: Let $E'$, $T'$ and $Y'$ be the images of their respective points under a homothety of factor $2$ centered at $C$. Then, $E'$ is the reflection of $C$ across $\overline{BD}$; $T'$ is the orthocenter of $\triangle C'BD$; $Y'$ is the harmonic conjugate of $C$ in $\triangle CDB$, so it is the $C'$-humpty point in $\triangle C'BD$. Then, we obviously have $\angle T'Y'E' = 90^{\circ}$ so we're done.