Just use harmonic ez pz

It's well known that $DT\cap \odot{ABC}$ is the $Q$queue point and is also $\odot{ABC}\cap \odot{AFE}$.Its easy to see $-1=(EF, HQ)$ so $EFHQ$ is harmonic. Now projecting through $H$ onto $\odot{ABC}$ we get $-1=(QD, BC)$ so $QDBC$ is harmonic hence $Q, D, T$ collinear. $\blacksquare$

Projective Cookie. Let $AP \cap \omega=I$ and $DT \cap \omega =I'$ . Now let $AD \cap BC=J$ Now we have $(P,J;B,C)=-1$ . $(I,D;B,C) \overset{A}{=} (P,J;B,C)=-1$ . Now again we have $(I',D;B,C)=-1 \implies I\equiv I'$ and we are done $\blacksquare$. Edit - 600th post

Let $\odot AEF$ meet $\odot ABC$ at $Q$. $M$ - midpoint of $BC$. $Q \in AP$ ($\because P$ is radical center of $\odot ABC. \odot AEF, \odot BCEF$). $M \in QH$ and $\angle BQD = \angle BAD = \angle MQC$ (well-known). $\implies QD$ is symmedian in $\triangle BQC$, so $T \in QD$. $\blacksquare$

Somehow I missed the projective solution, but here goes complex number solution. Let $\odot(ABC)$ and $\odot(AFE)$ meet again at point $Q$. Now by radical axis on $\odot(BFEC)$, $\odot(AFE)$ and $\odot(ABC)$ we get that lines $AQ, FE, BC$ are concurrent at $P$, thus we are left to show that points $Q, D, T$ are collinear. We use $\odot(ABC)$ as a unit circle. Note that: $$ t = \frac{2bc}{b+c} \quad \text{and} \quad d=-\frac{bc}{a} $$To compute $Q$ we use the following well -known fact that points $Q,H,A'$, where $A'$ is antipode of $A$ with respect to $\odot(ABC)$, are collinear. This gives us that: $$ q = \frac{(2a+b+c)bc}{2bc+ab +ac} $$Now to finish the problem we need to check that: $$ \frac{q -d}{\overline{q}-\overline{d}} =\frac{d-t}{\overline{d}-\overline{t}} $$This is equivalent to: \begin{align*} -qd = \frac{\frac{bc}{a} + \frac{2bc}{b+c}}{\frac{a}{bc}+\frac{2}{b+c}} \\ \frac{(2a+b+c)(bc)^2}{(2bc+ca+ab)a} = \frac{\frac{bc(2a +b+c)}{a(b+c)}}{\frac{2bc+ab+ac}{bc(b+c)}} \\ \frac{(2a+b+c)(bc)^2}{(2bc+ca+ab)a}=\frac{(2a+b+c)(bc)^2}{(2bc+ca+ab)a} \end{align*}Thus we are done.

WhY tHiS g7 iS sO eAsY? It is known that radical axeses of $(AFE),(ABC),(EFBC)$ concurr at $P$.$AP \cap (ABC)$ at Queue point(call it $Q_a$). We have to prove $Q_a-D-T$.Call $AD \cap BC$ at $R$. We know that $(P,R;B,C)=-1$ from $A-pencil$.The $(ABC)$ intersects $AP,AB,AR,AC$ at $Q_a,B,D,C$ respectively. Thus,$Q_aBCD$ is harmonic quadrilateral.By other side $TD \cap (ABC) $ at $Q_a'$ and we know $Q_a'BCD$ is harmonic from tangency.From harmonic condition $Q_a'=Q_a$ PS.One can do it without diagram. .I don't know why i didn't solve it in 5 minutes (i tried lemmas like 45 minutes or something like that).

Just well-known facts... Proof outline: Since $P$ is the $A$-Ex point, it suffices to show that $DT$ passes through $Q$, the $A$-Queue point of $\Delta ABC$. But it is well-known that $(B,C;Q,D)=-1$, which finishes.

The problem is equivalent to proving that $QBDC$ is a harmonic quadrilateral where $Q=AP \cap (ABC) \cap (AFE)$. Let $K$ be foot of $A$ onto $BC$ and project from $K$ onto $(ABC)$. Then $(Q,D;BC)=(LA;BC)$ where $L$ is $QK \cap (ABC)$ but by APMO 2012 P4, we have that $(AL;BC)=-1$ so we are done.