hint

lemma : let $MNPQ$ a trapezoid s.t $MN\parallel PQ$ and $MP\cap NQ=R$ .Let $K$ point of the circle $(NPR)$ then the circle $(MKN)$ is tangent to $MP$ at $M$ iff the line $NK$ bisects $MQ$. proof : let $L$ the point of intersection of $NK$ with $PQ$; $PM$ tangent to $(MNK) $ $\iff \angle KMP=\angle MNK = \angle NLP=\angle KLP\iff LKMP$ is cyclic $\iff \angle MLK=\angle MPK\iff \angle MLN=\angle KPR =\angle KNR=\angle QNL$ since $MN\parallel PQ $ it's equivalent to $MNQL$ is parallelogram which leads to the result . back to the problem : applying the lemma we deduce that the chords cut the side $AD$ at its midpoints RH HAS

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only lemma proving TST

Linearity of a power of a point for the win!! haha Define $P(X,\omega_1,\omega_2)=P(X,\omega_1)-P(X,\omega_2)$. Thus, $P(X,\omega_1,\omega_2)$ is linear. $$PC\cdot P(A,\omega_1,\omega_2)+AP\cdot P(C,\omega_1,\omega_2)=AC\cdot P(P,\omega_1,\omega_2)$$$$PD\cdot P(B,\omega_1,\omega_2)+PB\cdot P(D,\omega_1,\omega_2)=BD\cdot P(P,\omega_1,\omega_2)$$Notice that $P(B,\omega_1,\omega_2)=-BD^2$ and $P(C,\omega_1,\omega_2)=AC^2$ and $P(P,\omega_1,\omega_2)=PA^2-PD^2$, therefore, $$PC\cdot P(A,\omega_1,\omega_2)=AC\cdot (PA^2-PD^2)-AP\cdot AC^2$$$$PB\cdot P(D,\omega_1,\omega_2)=BD\cdot (PA^2-PD^2)+PD\cdot BD^2$$Let $M$ be the midpoint of $AD$. We have $$P(M,\omega_1,\omega_2)=\frac{1}{2} (P(A,\omega_1,\omega_2)+P(D,\omega_1,\omega_2))$$I claim this equality: $$\frac{AC\cdot (PA^2-PD^2)-AP\cdot AC^2}{PC}=-\frac{BD\cdot (PA^2-PD^2)+PD\cdot BD^2}{PB}$$$$\frac{AC\cdot PD^2+AP\cdot AC\cdot PC}{PC}=\frac{AC\cdot PD^2+AP\cdot AC^2-AC\cdot PA^2}{PC}=\frac{-AC\cdot (PA^2-PD^2)+AP\cdot AC^2}{PC}=\frac{BD\cdot (PA^2-PD^2)+PD\cdot BD^2}{PB}=\frac{BD\cdot PA^2+PD\cdot BD\cdot PB}{PB}$$$$PB\cdot (AC\cdot PD^2+AP\cdot AC\cdot PC)=PC\cdot (BD\cdot PA^2+PD\cdot BD\cdot PB)$$$$PB\cdot AC\cdot PD^2+PB\cdot AP\cdot AC\cdot PC=PC\cdot BD\cdot PA^2+PC\cdot PD\cdot BD\cdot PB$$$$PB\cdot AC\cdot PD^2-PC\cdot PD\cdot BD\cdot PB+PB\cdot AP\cdot AC\cdot PC-PC\cdot BD\cdot PA^2=0$$Since $\frac{PA}{PC}=\frac{PB}{PD}$, we have that $$PB\cdot PD\cdot (AC\cdot PD-BD\cdot PC)+AP\cdot PC(AC\cdot PB-PA\cdot BD)=0$$$$PB\cdot PD\cdot (AC\cdot PD-(PB+PD)\cdot PC)+AP\cdot PC(AC\cdot PB-PA\cdot (PB+PD))=0$$$$PB\cdot PD\cdot (AC\cdot PD-PB\cdot PC-PD\cdot PC)+AP\cdot PC(AC\cdot PB-PA\cdot PB-PA\cdot PD)=0$$$$PB\cdot PD\cdot (AP\cdot PD-PB\cdot PC)+AP\cdot PC(CP\cdot PB-PA\cdot PD)=0$$Hence, we have showed that $M$ lies on the radical axis of $\omega_1$ and $\omega_2$. Now, redefine $P(X,\omega_1,\omega_3)=(X,\omega_1)-P(X,\omega_3)$. $P(X,\omega_1,\omega_3)$ is linear. $$PC\cdot P(A,\omega_1,\omega_3)+AP\cdot P(C,\omega_1,\omega_3)=AC\cdot P(P,\omega_1,\omega_3)$$$$PD\cdot P(B,\omega_1,\omega_3)+PB\cdot P(D,\omega_1,\omega_3)=BD\cdot P(P,\omega_1,\omega_3)$$Notice that $P(B,\omega_1,\omega_3)=0$ and $P(C,\omega_1,\omega_3)=AC^2$ and $P(P,\omega_1,\omega_3)=PA^2$, therefore, $$PC\cdot P(A,\omega_1,\omega_3)=AC\cdot PA^2-AP\cdot AC^2=AC\cdot AP\cdot (-PC)\implies P(A,\omega_1,\omega_3)=-AC\cdot AP$$$$PB\cdot P(D,\omega_1,\omega_3)=BD\cdot PA^2\implies P(D,\omega_1,\omega_3)=\frac{BD\cdot PA^2}{PB}$$We have $$P(M,\omega_1,\omega_3)=\frac{1}{2} (P(A,\omega_1,\omega_3)+P(D,\omega_1,\omega_3))$$$$BD\cdot PA=PB\cdot AC\Longleftrightarrow (PD+PB)\cdot PA=PB\cdot (PA+PC)\Longleftrightarrow PA\cdot PD=PB\cdot PC$$Hence, $P(M,\omega_1,\omega_3)=0$, i.e $M$ lies on the radical axis of $\omega_1,\omega_3$. By radical axis theorem, $M$ also lies on the radical axis of $\omega_2,\omega_3$. We are done.

Redacted

Let $M$ be the midpoint of $AD$, which we claim is the desired intersection point. Define the linear function $f(X)=\mathbb{P}(X,w_1)-\mathbb{P}(X,w_3)$. Then \begin{align*} f(A)&=AP\cdot AC\\ f(B)&=0\\ f(C)&=-AC^2 \end{align*}By the linearity, $$\frac{f(A)-f(B)}{AB}=\frac{f(D)-f(C)}{CD}$$Solving for $f(D)$ gives, $$f(D)=AC\cdot AP\left(\frac{CD}{AB}-\frac{AC}{AP}\right)=AC\cdot AP\left(\frac{CD}{AB}-\frac{CP}{AP}-1\right)=-AC\cdot AP$$Thus $f(M)=0$ so $M$ lies on the common chord of $w_1$ and $w_3$. By symmetry, we are finished.

Attachments:

totally same with above. Let $f(Q)=\mathbb{P}(Q,\omega_1)-\mathbb{P}(Q,\omega_3)$. Let $DB\cap \omega_1$ at $T$. $f(A)=-AP \cdot AC$,$f(B)=0$,$f(C)=AC^2$. We have to compute $f(D)$. From angle chasing $ATCD$ is cyclic. $f(D)=(DT*DB) - (DP * DB)$ $\implies $$f(D)=DB *TP=AP *AC$. Let radical axis of $\omega_1$ and $\omega_3$ intersects $AD$ at $W$. So, $f(A)\frac{WD}{AD}+f(D)\frac{AW}{AD}=f(W)=0$. Substuting gives us $WD=AW$ which implies desired point is the midpoint of $AD$. Doing same thing to $\omega_2$ and $\omega_3$ gives us desired result.