, 2023

Since $BCED$ is cyclic, we have two similar triangles $ABC$ and $AED$. On the other hand
$$\angle TBC = \angle TDE = \angle KED \text{ and } \angle TCB = \angle TED = \angle KDE.$$
Thus two points $T, K$ are corresponding in the above pair of similar triangles, it can be deduced that
$$\angle TAB = \angle KAE \text{ and } \angle TAC = \angle KAB$$
implies that $AT, AK$ are symmetric through $\ell$.

Lemma: if two rays $Ox', Oy'$ are isogonal in angle $xOy$ then every pair of isogonal lines $Oa, Ob$ in angle $xOy$ are also isogonal in angle $x'Oy'$. We will use this idea to solve the problem. Let $J$ be the center of $(TPQ)$ and $H$ be the projection of $T$ onto $PQ$. Then, according to the symmetry of the altitude and the line joining the center, we immediately have $TJ, TH$ isogonal in $\angle PTQ$. On the other hand,
$$\angle PTD = \angle ACT = \angle ABT = \angle QTE$$
so lines $TP, TQ$ are isogonal in angle $\angle DTE$ and then $TJ, TH$ are also isogonal in $\angle DTE$. To prove three points $J, T, O$ are collinear, we will show that lines $TO, TH$ are isogonal in $\angle DTE$. $(*)$

Denote $S$ as the intersection of $DE, BC$, then by applying Brocard's theorem to the completed quadrilateral $BCED, AS$ we have $TO \perp AS$. Draw $Ax \perp TH$ then $Ax \parallel XY$. Notice that $AXKY$ is a parallelogram, so $AK$ passes through the midpoint of $XY$, leads to
$$(Ax, AK, AB, AC) = -1.$$
Furthermore, $A(ST, BC) = -1$ so consider the symmetry over the line $\ell$, we have $AB \leftrightarrow AC$, $AT \leftrightarrow AK$ and take $AS \leftrightarrow Ax'$. Then
$$(Ax', AK, AB, AC) = -1,$$
so $Ax \equiv Ax'$ or $AS, Ax$ is symmetric about $\ell$. Through $T$, draw lines $Tb, Tc$ perpendicular to $AB, AC$, then two sets of four lines
$$(AS, Ax, AB, AC) \text{ and } (TO, TH, Tb, TC)$$
are perpendicular respectively. This implies that $TO, TH$ are isogonal in $\angle bTC$. Finally, since $\angle ADT = \angle AET$, we have $Tb, Tc$ are also isogonal in $\angle DTE$ and deduce to $(*)$ is true. $\Box$