EDIT: There was originally a typo in this question. Now that it has been edited, I can go ahead and write a correct proof. First, note that $\angle AEC=\angle ADC=\angle YTD$ by a combination of cyclic quads and parallel lines, so quadrilateral $AEYT$ must be cyclic. Therefore $\angle AYX=\angle AEB=\angle APB$ and $BP\parallel XY$. This means that triangles $\triangle AXY$ and $\triangle ABP$ are homothetic with center of homothety $A$, implying that their circumcircles are homothetic. Therefore, since the circumcircle of $\triangle ABP$ is simply $(O)$, circles $(O)$ and $\omega$ are internally tangent, as desired. $\blacksquare$ EDIT2: Oops, I forgot the case where $X$ and $Y$ are outside $(O)$. I'm sure the solution is very similar, although several changes need to be made as a result of configuration issues.

Thanks , typo edited.

As above, $A,T,E,Y$ are concyclic, hence $\angle AYT=\angle AET$, let $\alpha$ being the common angle value. Seeing that $\angle XAO=\angle BAO=90^\circ-\alpha$ we infer that the circumcenter of $\Delta AXY$ belongs to $(AO$, so being done. Best regards, sunken rock

@sunken : rock very very beautiful!!!!!! can any body solve with menelaus i tried but i didn't success

Let $AY$ meet the circumcircle of $ABCDE$ at $F$.clearly $ATYE$ is cyclic but $\angle AEB=\angle AFB$, hence we get $\angle AYX=\angle AFB$,now if we draw a tangent $MA$ to $\omega $ at $A$ then we get $\angle MAX=\angle AYX=\angle AFB$,so $MA$ is also tangent to the circumcircle of $ABCDE$,hence the result.

djmathman wrote: EDIT: There was originally a typo in this question. Now that it has been edited, I can go ahead and write a correct proof. First, note that $\angle AEC=\angle ADC=\angle YTD$ by a combination of cyclic quads and parallel lines, so quadrilateral $AEYT$ must be cyclic. Therefore $\angle AYX=\angle AEB=\angle APB$ and $BP\parallel XY$. This means that triangles $\triangle AXY$ and $\triangle ABP$ are homothetic with center of homothety $A$, implying that their circumcircles are homothetic. Therefore, since the circumcircle of $\triangle ABP$ is simply $(O)$, circles $(O)$ and $\omega$ are internally tangent, as desired. $\blacksquare$ EDIT2: Oops, I forgot the case where $X$ and $Y$ are outside $(O)$. I'm sure the solution is very similar, although several changes need to be made as a result of configuration issues. What's P?

Oops. I forgot to mention that $P$ is the second intersection point of $\omega$ and $AY$.

From $A$,draw a tangent $AL$ to circle $(O)$. $\angle{TAE}=\angle{DAE}=\angle{DCE}=\angle{CYX}=\angle{CYT} \Rightarrow T,A,Y,E$ concyclic.So $\angle{LAB}=\angle{AEB}=\angle{AET}=\angle{AYT}=\angle{AYX}$ or $AL$ is also tangent to $\omega$.The result follows..