Let $X,Y$ be the midpoint of $AB, CD$ respectively and segments $XU, YV$ the radii of circles with diameters $AB,CD$ respectively. Then $X,Y,V,U$ are concyclic. From here by angle chasing we get $A,V,U,D$ are concyclic and $B,V,U,C$ are concyclic. Radical axis theorem for circles $AVUD, AUB, CDV$ gives concurrency of $AU, DV$ and $k$, where $k$ is the radical axis of circles with diameters $AB,CD$. Name that point $M$. Radical axis theorem for circles $BVUC, AUB, CDV$ gives concurrency of $BU, CV$ and $k$. Name that point $N$. Both $M$ and $N$ lie on line $k$, so $MN$ is the same line as $k$ (they are different points). Pappus theorem for line $CD$ with point $U$ and line $AB$ with point $V$ gives collinearity of $M,N$ and $AC \cap BD.\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -6.033901452545057, xmax = 6.771620019775647, ymin = -1.3198145537394828, ymax = 8.561845591696793; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen fuqqzz = rgb(0.9568627450980393,0,0.6); /* draw figures */ draw(shift((-0.39204752638678153,-0.3683718726733888))*xscale(4.633058672856573)*yscale(4.633058672856573)*arc((0,0),1,-0.07064827432254935,179.92935172567746), linewidth(0.7) + zzttff); draw((-5.025102677192676,-0.36265910430001536)--(4.241007624419113,-0.3740846410467622), linewidth(0.7) + zzttff); draw((-0.39204752638678153,-0.3683718726733888)--(-3.1496184397628713,3.354669266480686), linewidth(0.7) + zzttff); draw((xmin, 0.1030908588047777*xmin + 1.7048540384711446)--(xmax, 0.1030908588047777*xmax + 1.7048540384711446), linewidth(0.7) + red); /* line */ draw(circle((-0.6681508130778658,2.309880010104485), 2.6924461317777277), linewidth(0.7) + green); draw((-5.025102677192676,-0.36265910430001536)--(-3.1496184397628713,3.354669266480686), linewidth(0.7)); draw((-5.248092045148656,1.8003781478995706)--(-0.9508580636735964,-0.3676828337864753), linewidth(0.7)); draw((-3.1496184397628713,3.354669266480686)--(4.241007624419113,-0.3740846410467622), linewidth(0.7)); draw((-0.9508580636735964,-0.3676828337864753)--(3.359583845610758,8.17588563786515), linewidth(0.7)); draw(circle((-2.9863703837049282,0.9405282675319206), 2.4196542502208946), linewidth(0.7) + linetype("2 2") + blue); draw(shift((-0.944254099768949,4.98813189288236))*xscale(5.355818798161678)*yscale(5.355818798161678)*arc((0,0),1,-143.47350314793462,36.526496852065385), linewidth(0.7) + fuqqzz); draw((-5.248092045148656,1.8003781478995706)--(3.359583845610758,8.17588563786515), linewidth(0.7) + fuqqzz); draw(circle((1.6500687575491988,3.6792317526770493), 4.810648396992589), linewidth(0.7) + linetype("2 2") + blue); draw((-0.9442540997689493,4.98813189288236)--(-0.9508580636735964,-0.3676828337864753), linewidth(0.7) + fuqqzz); draw((-5.248092045148656,1.8003781478995706)--(4.241007624419113,-0.3740846410467622), linewidth(0.7) + linetype("4 4") + red); draw((-5.025102677192676,-0.36265910430001536)--(3.359583845610758,8.17588563786515), linewidth(0.7) + linetype("4 4") + red); /* dots and labels */ dot((-5.025102677192676,-0.36265910430001536),linewidth(3pt) + dotstyle); label("$A$", (-5.3,-00.9), NE * labelscalefactor); dot((4.241007624419113,-0.3740846410467622),linewidth(3pt) + dotstyle); label("$B$", (4.293771437670248,-0.8), NE * labelscalefactor); dot((-0.39204752638678153,-0.3683718726733888),linewidth(3pt) + dotstyle); label("$X$", (-0.3844066853447462,-0.8), NE * labelscalefactor); dot((-3.1496184397628713,3.354669266480686),linewidth(3pt) + dotstyle); label("$U$", (-3.6,3.497123089873386), NE * labelscalefactor); dot((-0.9508580636735964,-0.3676828337864753),linewidth(3pt) + dotstyle); label("$V$", (-1.2,-0.9), NE * labelscalefactor); dot((-0.9442540997689493,4.98813189288236),linewidth(3pt) + dotstyle); label("$Y$", (-1.0583814996774148,5.073034788092411), NE * labelscalefactor); dot((3.359583845610758,8.17588563786515),linewidth(3pt) + dotstyle); label("$C$", (3.4017459481123042,8.1), NE * labelscalefactor); dot((-5.248092045148656,1.8003781478995706),linewidth(3pt) + dotstyle); label("$D$", (-5.7,1.8518316313554102), NE * labelscalefactor); dot((-4.200465229183971,1.2718244706149617),linewidth(3pt) + dotstyle); label("$M$", (-4.4,0.7), NE * labelscalefactor); dot((0.09998872574750177,1.71516196207925),linewidth(3pt) + dotstyle); label("$N$", (0.05,1.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Nice solution, but can you give a full proof of how $A, V, U, D$ and $B, V, U, C$ are concyclic?

$\angle YVX=90^\circ=\angle YUX\implies \ X,Y,V,U\text{ are concyclic}\implies \angle UBV=\frac12\angle UXV=\frac12\angle UYV=\angle UCV\implies$ $B,C,V,U\text{ are concyclic}$ $X,Y,V,U\text{ are concyclic}\implies \angle UAV=90^\circ-\frac12\angle VXU=90^\circ-\frac12\angle VYU=\angle UDV\implies\ A,D,V,U\text{ are concyclic}$

Nice. Thanks.

Let circles tangency points to $AB,CD$ be $Q$ and $P$ and let $M,N$ be midpoints of $AB,CD$. $X=AP\cap DQ, Y=CQ\cap BP$. $\angle MPN=\angle NQM=90 \implies PNQM$ is cyclic $\implies \angle PCQ=\frac{\angle PNQ}{2}=\frac{\angle PMA}{2}=\angle PBQ \implies PQBC$ is cyclic. So $Y$ (Similarly $X$) lies on radical axis of circles which are given in statemet. Applying Pappus to $D-P-C$ and $A-Q-B$ gives that $AC\cap BD \in XY$, which gives desired result.