Let $M,N$ be the midpoints of $AB,CD$ respectively, and let $P,Q$ be the feet of the perpendiculars from $M,N$ to $CD,AB$. We know that $MP=MA$, and $AB$ is tangent to the circle with diameter $CD$ iff $NQ=NC$, which is equivalent to the triangles $AMN,MNC$ having equal areas $(*)$. Since $AM=MB,\ NAM,NMB$ always have equal areas, so $(*)$ is, in turn, equivalent to $NMC,NMB$ having equal areas, i.e. $MN\|BC$ (because $B,C$ lie on the same side of the line $MN$), and this is just another way of saying that $BC\|AD$.

grobber wrote: We know that $MP=MA$, and $AB$ is tangent to the circle with diameter $CD$ iff $NQ=NC$, which is equivalent to the triangles $AMN,MNC$ having equal areas $(*)$. Sorry for reviving such an old thread, but I don't understand why that is equivalent to the triangles $AMN,MNC$ having equal areas. Thank you.

Assume that line $AB$ is tangent to the circle on diameter $CD$ and let the intersection point be $E$; similarly, define the intersection of line $CD$ and the circle on diameter $AB$ to the $F$. Also, let $O_1$ be the midpoint of $AB$ and $O_2$ be the midpoint of $CD$. By the definition of tangent lines, we have $\angle CFO_1=90^{\circ}$ and $\angle BEO_2=90^{\circ}.$ Claim 1: $EFO_2O_1$ is cyclic. Proof: Note that $\angle O_2FO_1=90^{\circ}=\angle O_2EO_1$. $\Box$ By Claim 1, we have $\angle FO_2E=\angle FO_1E$, which implies that the supplement angles are equal or $\angle CO_2E=\angle BO_1F. $ Claim 2: $ADFE$ is cyclic. Proof: Because a central angle equals twice an inscribed angle, we have $\angle CO_2E=2\angle CDE$ and $\angle BO_1F=2\angle BAF.$ But, $\angle CO_2E=\angle BO_1F $ so $$\angle FDE=\angle CDE=\angle BAF=\angle EAF,$$proving the desired. $\Box$ Claim 3: $BCFE$ is cyclic. Proof: This is proven similarly like Claim 2.$\Box$ By Claim 2 and 3, we have $$\angle DAB=\angle DAE=180^{\circ}-\angle DFE=\angle CFE=180^{\circ}-\angle CBE=180^{\circ}-\angle CBA,$$proving $AD\parallel BC$. $\blacksquare$ Now, assume that $AD\parallel BC$. Let $O_1, O_2$ be the midpoint of $AB, CD$ respectively and let $E$ be the perpendicular from $O_2$ to line $AB$ and $F$ be the point of tangency of the circle with diameter $AB$ and line $CD$. Claim 4: $EFO_2O_1$ is cyclic. Proof: By construction, $\angle O_2FO_1=90^{\circ}=\angle O_2EP_1$. $\Box$ Claim 5: $ADFE$ is cyclic. Proof: Since $AD\parallel BC$, $ABCD$ is a trapezoid with midline $O_1O_2$. Thus, $\angle O_2O_1B=\angle DAE.$ By Claim 4, $\angle EFO_2=\angle O_2O_1B.$ Hence, $$\angle DAE=\angle O_2O_1B=\angle O_2FE,$$proving the desired. $\Box$ By Claim 5, $\angle FAO_1=\angle O_2DE.$ By Claim 4, $FO_1A=\angle DO_2E$. By AA similarity, $\triangle O_2DE\sim \triangle O_1AF$. As $\triangle O_1AF$ is an isosceles triangle with base $AF$, $\triangle O_2DE$ must be an isosceles triangle with base $DE$. Hence, $DO_2=EO_2$ so $E$ lies on the circle with diameter $CD$. Thus, the line $AB$ is tangent to the circle with diameter $CD$ as $O_2E\perp AB$. $\blacksquare$

Let $P$ be the intersection of $AB$ and $CD$, $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively, and $E$ and $F$ be the projections of $M$ and $N$ into $CD$ and $AB$, respectively. First, assume that $BC\parallel AD$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -6.2845153948328, xmax = 4.129678684630141, ymin = -4.282213752758109, ymax = 4.287504536064257; /* image dimensions */ /* draw figures *//* special point *//* special point *//* special point */ draw((-5.94,-3.32)--(-2.84,-3.3), linewidth(1.6)); /* special point */ draw((-4.58,0.76)--(-3.3095855735832895,0.7681962221059143), linewidth(1.6)); /* special point *//* special point *//* special point *//* special point */ draw((-5.26,-1.28)--(-3.101919142755515,-1.030896077312692), linewidth(1.6)); draw((-3.0747927867916447,-1.2659018889470428)--(-5.037249845363277,-0.611749536089832), linewidth(1.6)); /* special point */ draw((-5.94,-3.32)--(-3.6356530151564,3.5930409545307973), linewidth(1.2)); draw(shift((-5.26,-1.28))*xscale(2.150348808914498)*yscale(2.150348808914498)*arc((0,0),1,-108.43494882292201,71.56505117707798), linewidth(1.6)); draw(shift((-3.0747927867916447,-1.2659018889470428))*xscale(2.047604155621539)*yscale(2.047604155621539)*arc((0,0),1,96.58442303520167,276.58442303520167), linewidth(1.6)); draw((-3.6356530151564,3.5930409545307973)--(-2.84,-3.3), linewidth(1.6)); /* dots and labels */ label("$A$", (-4.525169256067944,0.839753635097643), NE * labelscalefactor); label("$B$", (-5.8872436860794455,-3.2322813796242427), NE * labelscalefactor); label("$C$", (-2.7800113926157066,-3.218093104311623), NE * labelscalefactor); label("$D$", (-3.2482244779321605,0.8539419104102628), NE * labelscalefactor); label("$M$", (-5.206206471073695,-1.1891697346069898), NE * labelscalefactor); label("$N$", (-3.0212120729302434,-1.1749814592943701), NE * labelscalefactor); label("$E$", (-3.049588623555483,-0.9479690542924532), NE * labelscalefactor); label("$F$", (-4.979194066071777,-0.5223207949138589), NE * labelscalefactor); label("$P$", (-3.574554810122416,3.677408697621605), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By power of a point, we have that $PE^2=PA\cdot PB$. However, note that $$\dfrac{PE}{PF}=\dfrac{PM}{PN}=\dfrac{PA}{PC}=\dfrac{PB}{PD}$$Then,$$\dfrac{PF^2}{PC\cdot PD}=\dfrac{PF^2}{PC\cdot PD}\cdot\dfrac{PE^2}{PF^2}\cdot\dfrac{PC}{PA}\cdot\dfrac{PD}{PB}=\dfrac{PE^2}{PA\cdot PB}=1$$Thus, $ PF^2=PC\cdot PD$, so the circle with diameter $CD$ is tangent to $AB$. Now, assume that the circle with diameter $CD$ is tangent to $AB$. Let $D’$ be the point in $CD$ such that $AD’\parallel BC$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -6.2845153948328, xmax = 4.129678684630141, ymin = -4.282213752758109, ymax = 4.287504536064257; /* image dimensions */ /* draw figures *//* special point *//* special point *//* special point */ draw((-4.58,0.76)--(-5.94,-3.32), linewidth(1.6)); draw((-5.94,-3.32)--(-2.84,-3.3), linewidth(1.6)); /* special point */ draw((-4.58,0.76)--(-3.3095855735832895,0.7681962221059143), linewidth(1.6) + linetype("2 2")); draw((-3.3095855735832895,0.7681962221059143)--(-2.84,-3.3), linewidth(1.6)); draw(shift((-5.26,-1.28))*xscale(2.150348808914498)*yscale(2.150348808914498)*arc((0,0),1,-108.43494882292201,71.56505117707798), linewidth(1.6)); draw(shift((-3.0747927867916447,-1.2659018889470428))*xscale(2.047604155621539)*yscale(2.047604155621539)*arc((0,0),1,96.58442303520167,276.58442303520167), linewidth(1.6) + linetype("2 2")); /* special point */ draw((-4.58,0.76)--(-3.3949432789346017,1.507682087883365), linewidth(1.6)); draw((-3.3949432789346017,1.507682087883365)--(-3.3095855735832895,0.7681962221059143), linewidth(1.6)); draw(shift((-3.1174716394673005,-0.8961589560583174))*xscale(2.4198021148943787)*yscale(2.4198021148943787)*arc((0,0),1,96.58442303520167,276.58442303520167), linewidth(1.6)); /* dots and labels */ label("$A$", (-4.525169256067944,0.839753635097643), NE * labelscalefactor); label("$B$", (-5.8872436860794455,-3.2322813796242427), NE * labelscalefactor); label("$C$", (-2.7800113926157066,-3.218093104311623), NE * labelscalefactor); label("$D'$", (-3.2482244779321605,0.8539419104102628), NE * labelscalefactor); label("$D$", (-3.333354129807879,1.591732226666493), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] As we already proved, the circle with diameter $CD’$ must be tangent to $AB$. However, if $D’\ne D$, this circle would contain or be contained by the circle with diameter $CD$, and, hence, it would be impossible for both circles to be tangent to $AB$. Thus, $D=D’$, so $BC\parallel AD$. We are done. $\square$