let $AQ$ intersect $(ABC)$ again at $F$ claim 1: $\measuredangle AOD= \measuredangle DOS$. Proof: It suffices to show $AOD\cong SOD$. Note that $\measuredangle ADO=\measuredangle ESO=\measuredangle SEO=\measuredangle SDO$ and $\measuredangle DAO=\measuredangle OED=\measuredangle OSD$. This implies $AOD\cong SOD$ claim 2: $SF\parallel BC$ proof: by claim 1, $\measuredangle AQD=\measuredangle AOD= \measuredangle DOS=\measuredangle DES=\measuredangle AFS$. to finish, consider the homothety mapping $PQ$ to $SF$, This maps $(APQ)$ to $(ABC)$, which implies that they are tangent at $A$

Let $F=OD \cap BS$. Firstly, since $OA=OS$, $\measuredangle OAD = \measuredangle AEO = \measuredangle DSO$, and $\measuredangle ADO = \measuredangle EDO = \measuredangle ESO = \measuredangle OES = \measuredangle ODS$, we have $\Delta ADO \stackrel{-}{\cong} \Delta SDO$. It follows that $\measuredangle AOD = \measuredangle DOS$, so $\measuredangle AOF = \measuredangle DOS = \measuredangle AES = \measuredangle ABF$, so we have that $A,B,O,F$ are concyclic. Next, note that $\measuredangle QAB+ \measuredangle BAO = \measuredangle QAO = \measuredangle BDO = \measuredangle BFD + \measuredangle DBF = \measuredangle BAO + \measuredangle CAP$, so $AB,AC$ are isogonal in $\angle PAQ$. Extend $AB,AC$ to meet $(APQ)$ at $B',C'$, then it follows that $BC//B'C'$. Then we may take a homothety centered at $A$ taking $(ABC)$ to $(AB'C')$, so the two circles are tangent.