Let circles $\odot(BEC)$ and $\odot(ADC)$ intersect at $C,P.$ Then $P$ is the center of the rotation taking the oriented segments $\overline{BD}$ and $\overline{EA}$ into each other $\Longrightarrow$ $PB=PE$ $\Longrightarrow$ $CP$ bisects $\angle ECB$ internally, i.e. radical axis of $\odot(BEC)$ and $\odot(ADC)$ is the internal bisector of $\angle ACB$ $\Longrightarrow$ $CP$ is the perpendicular bisector of $\overline {KL}$ and the conclusion follows.

we consider it in $xOy$ system,with x-axis the bisector of $\angle ACB$. let $D(d,kd),E(e,-ke),A(e+c,-k(e+c)),B(d+c),k(d+c))$ it suffices to prove that $x_S=x_T$ and it's easy to prove: $x_S=x_T=\frac{(k^2+1)(d+e+c)}{4}$.

My contribution: The circumcenters of triangles $\triangle ABC, \triangle ACD, \triangle BCE, \triangle CDE$ are the vertices of a rhombus! Best regards, sunken rock