I guess you meant $\omega_2$ has diameter $CD$. Let $T=EF\cap BC$ and $Y=AD\cap EF$. Since $AD$ is an angle bisector and $E,F$ are the projections of $D$ on $AB,AC$, it follows that $AED$ and $AEF$ are congruent, and thus $DY\perp EF$. Furthermore, let $Y'$ be the intersection of the common internal tangent to $\omega_1$ and $\omega_2$ with $EF$, and $Z$ the point at infinity of $\ell$. Now consider the inversion $\Phi$ with center $Y'$ and radius squared $Y'D^2=Y'Q\cdot Y'E=Y'F\cdot Y'Q$ (which leaves fixed $\omega_1$ and $\omega_2$), which preserves cross ratios. Therefore, applying $\Phi$, we get $(Q,P;Y,T)=(F,E;T,Y)$.By projecting through $A$ this is equal to $(C,B;T,D)=(B,C;D,T)$. Therefore, there must exist a point to which to project trough in order to obtain $(B,C;D,T)=(Q,P;Y,T)$. In other words, $BQ,CP,YD$ concur, which of course is equivalent to $BQ,CP$ and $AD$ concurring, as wanted.

Pretty funny Let $X_{\infty}$ be the point at infinity along $\overrightarrow{DA}$. Check that $\angle X_{\infty}PQ = \angle BPD = 90^{\circ}$, $\angle X_{\infty}QP = \angle CQD = 90^{\circ}$, and\[\angle BDP = \angle AEP = \angle AFQ = \angle CDQ\]so Jacobi finishes.

Does there is any other sullotion using angel chase? I have prove that if $(ABD)$ intersects ω_2 at $T$ then $B,T,Q$ are collinear. Now if we prove that $(A,D/,B,T)=-1$ we are done. Edit:Two below if we prove this then $G,M$are isogonal to the triangle $ABC$ where $M$ midpoint to $AD$

P2nisic wrote: Does there is any other sullotion using angel chase? I have prove that if $(ABD)$ intersects ω_2 at $T$ then $B,T,Q$ are collinear. Now if we prove that $(A,D,B,T)=-1$ we are done. hello, excuse me explain that why prove that $(A,D,B,T)=-1$ we are done.

Another solution I guess... WLOG let $AB\leq AC$. $\angle DEA=\angle DFA=90^{\circ}$ thus $AEDF$ is cyclic. Let $B'$ and $C'$ be $BP\cap AC$ and $CQ\cap AB$, respectively. $\angle DBB'=\angle DEF=\angle DAC=\angle DAB'$ implying that $DBAB'$ is cyclic. Furthermore, since $AD$ bisects angle $\angle BAB'$ and $DP$ is perpendicular to $BB'$, we have that $P$ is the midpoint of $BB'$. Similarly we deduce that $AC'DC$ is cyclic and that $Q$ is the midpoint of $CC'$. Now, let $G=AC\cap DP$, $I=AB\cap DQ$, and $H=BC\cap GI$. Note that $\angle GDI=180^{\circ}-\angle PDQ=180^{\circ}-(180^{\circ}-\angle DPQ -\angle DQP)=\angle ABC +\angle ACB = 180^{\circ} -\angle BAC =\angle GAI$, and this implies that $GADI$ is cyclic. Since $Q$ is the midpoint of $CC'$ and $DQ\perp CC'$ we have that $ID$ is the angle bisector of $\angle AIC$, implying that $D$ is the incenter of $\bigtriangleup AIC$. Also note that $\angle CGD=\angle AGD=\angle AID=\angle CID$ and also $\angle GCD=\angle ICD$ implies that $CG=CI$ by congruence. This implies that $H$ is the midpoint of $GI$ and also that $CH\perp GI$. Since $\angle DEI=90^{\circ}=\angle DHI$, we have that $DEHI$ is cyclic. Furthermore, $\angle DEH +\angle DEF=180^{\circ}-\angle DIH +\angle DAF=\angle DAG +\angle DAF=180^{\circ}$. Hence, $H$, $E$, and $F$ are collinear. Now we are ready to finish the problem. Observe that $\bigtriangleup ABC$ and $\bigtriangleup DQP$ are perspective axially ($AB\cap DQ=I$, $AC\cap DP=G$, and $BC\cap QP=H$ are collinear). So, by Desargue's Theorem, they are perspective centrally $\Rightarrow AD$, $BQ$, and $CP$ are concurrent. $QED$

$\text{solved by using trigonometric ratio}$