Let $M$ be the midpoint of segment $AB$. Since $MX$ and $MY$ are midlines of $\triangle BAC$ and $\triangle BAD$ respectively $\Longrightarrow \angle CAD=\angle XMY$. Clearly $AB$ bisects $XY$ at $N$. From power of point and AM-GM inequality: $$NX^2=NB\times NA< \dfrac{(NA+NB)^2}{4}=NM^2\Longrightarrow NX<NM$$Hence $M$ is outside of circle with diameter $XY$. So $\angle XMY=\angle CAD<90^\circ$

Does this work? By radical axis $AB$ intersects $XY$ at $M$, the midpoint of the segment. A homothety at $B$ sends $X\to C$, $Y\to D$ and $M\to N$, so $N$ is the midpoint of $CD$. Hence $AB$ bisects $CD$ too. Now note that the angle inequality is equivalent to $NA>\frac{CD}{2}$ (consider the circle with diameter $CD$). Since $MB=MN$, $$ NA>\frac{CD}{2} \iff AM+MB > XY$$But by power of a point, $MX^2=MY^2=MB\cdot MA$ so $BM=\frac{MY^2}{MA}$. The inequality becomes \begin{align*}AM+\frac{MY^2}{MA} &> XY \\ \iff AM^2+MY^2&>2\cdot MY\cdot MA \\ \iff (AM-MY)^2&>0 \end{align*}If $AM=MY$, then $BM=\frac{MY^2}{MA}=AM$ so $B=A$, contradiction. Hence this final inequality is true, completing the proof.

Denote by $M,N$ and $O_1,O_2$ the midpoints of $AB$ and $XY$ and the centers of $w_1$ and $w_2$ respectively. As mentioned above, since $MX$ and $MY$ are midlines, we are required to prove $\angle{XMY}<90^{\circ}$. It is well known that the $AB$ intersects $XY$ at $N$. Now notice that $O_1O_2$ is the perpendicular bisector of $AB$ hence $O_1,M,O_2$ are collinear as well as having $\angle{NMO_1}=90^{\circ}$. Since $XY$ is the common tangent of $w_1$ and $w_2$, we notice that $\angle{O_1XN}=\angle{O_2YN}=90^{\circ}$, and hence we have $MO_1XN$ and $MO_2YN$ are $cyclic$. Now we find that $$180^{\circ}-\angle{O_1NO_2}=\angle{O_1NX}+\angle{O_2NY}=\angle{O_1MX}+\angle{O_2MY}=180^{\circ}-\angle{XMY}\implies\angle{XMY}=\angle{O_1NO_2}$$Also note that $MO_1<XO_1$ so we have $\angle{O_1XM}<\angle{O_1MX}$, using the similar result, adding them up, and noting the $cyclic$ quadrilaterals once again, we find that $$180^{\circ}-\angle{XMY}>\angle{O_1NO_2}\implies 180^{\circ}>2\angle{XMY}$$Which is exactly what we want! $$Q.E.D$$

let)$T, S : O_1O_2\cap w_1, w_2$ $\angle CAD=\angle XMY =180-\angle XMO_1-\angle YMO_2$ $\le 180-\angle XTO_1-\angle YSO_2$ $=180-(90-1/2\angle XO_1O_2)-(90-1/2\angle YO_2O_1) $ $=1/2(180)=90$ Equivalent condition is $S=M=T$, which means $A=B$, so contradiction. as a result, problem holds. easy problem! P.S sorry i used your picture, @cycle

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Let AB meet XY and CD at M,N. note that instead we can prove NA > NC. Let NC = x, NB = z and AB = y. we have MX^2 = MB.MA so x^2/4 = z^2/4 + zy/2 so x^2 = z^2 + 2zy. Assume NA ≤ NC so y + z ≤ x or y^2 + z^2 + 2zy ≤ x^2 which is true only when y = 0 and it's not true so NA > NC. we're Done.