SAUDI ARABIAN IMO Booklet 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 $O{x}^{\prime },O{y}^{\prime }$ are isogonal in angle $xOy$ then every pair of isogonal lines $Oa,Ob$ in angle $xOy$ are also isogonal in angle $x^{\prime }O{y}^{\prime }$. 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 $\angle DTE$. $(\ast )$

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 $AS\leftrightarrow A{x}^{\prime }$. Then $$(A{x}^{\prime },AK,AB,AC)=-1,$$ so $Ax\equiv A{x}^{\prime }$ 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 $(\ast )$ is true. $\square$