, 2019

Denote $\phi = \widehat{XAB} = \widehat{YAC}$, $\alpha = \widehat{CAX} = \widehat{BAY}$. Then, because the quadrilaterals $ABSK$ and $ACTL$ are cyclic, we have
$$\widehat{BSK} + \widehat{BAK} = 180^\circ = \widehat{BSK} + \phi = \widehat{LAC} + \widehat{LTC} = \widehat{LTC} + \phi,$$
so, due to the 90-degree angles formed, we have $\widehat{KSL} = \widehat{KTL}$. Thus, $KLST$ is cyclic.

Figure 6: G6
Consider $M$ to be the midpoint of $BC$ and $K'$ to be the symmetric point of $K$ with respect to $M$. Then, $BKCK'$ is a parallelogram, and so $BK \parallel CK'$. But $BK \parallel CT$, because they are both perpendicular to $AX$. So, $K'$ lies on $CT$ and, as $\widehat{KTK'} = 90^\circ$ and $M$ is the midpoint of $KK'$, $MK = MT$. In a similar way, we have that $MS = ML$. Thus, the center of $(KLST)$ is $M$.
Consider $D$ to be the foot of altitude from $A$ to $BC$. Then, $D$ belongs in both $(ABKS)$ and $(ACLT)$. So,
$$\widehat{ADT} + \widehat{ACT} = 180^\circ = \widehat{ABS} + \widehat{ADS} = \widehat{ADT} + 90^\circ - \alpha = \widehat{ADS} + 90^\circ - \alpha,$$

and $\overline{AD}$ is the bisector of $\widehat{SDT}$.
Because $DM$ is perpendicular to $AD$, $DM$ is the external bisector of this angle, and, as $MS = MT$, it follows that $DMST$ is cyclic. In a similar way, we have that $DMLK$ is also cyclic.
So, we have that $ST$, $KL$ and $DM$ are the radical axes of these three circles, $(KLST)$, $(DMST)$, $(DMKL)$. These lines are, therefore, concurrent, and we have proved the desired result. □

We continue after proving that $M$ is the center of $(KLST)$. If $D$ is the foot of perpendicular from $A$ to $BC$, then $ASDKB$ is cyclic, as well as $ATDLC$. The radical axes of those two circles and $(KLST)$ are concurrent, thus $KS$ and $LT$ intersect on point $Q \in AD$. So, if $P$ is the intersection point of $KL$ and $TS$, due to Brokard's theorem, $AQ$ is perpendicular to $MP$. This is, of course, equivalent to proving that $P$ belongs on $BC$. □