Let $AT\cap DC=P$ and $BT\cap DC=Q$. Let $X,Y,Z,W$ be the feet of the altitudes from $A,B,K,L$ to $CD$. We need to prove that $|KZ|=|LW|$ and we have $|AX|=|BY|$. Then, it suffices to prove that $\frac{|AX|}{|KZ|}\overset{?}{=}\frac{|BY|}{|LW|}\Leftrightarrow \frac{|AP|}{|PK|}\overset{?}{=}\frac{|BQ|}{|QL|}\Leftrightarrow \frac{|AP|^2}{|DP||PC|}\overset{?}{=}\frac{|BQ|^2}{|DQ||QC|}.$ By the law of Sines, we get $\frac{|AP|^2}{|DP||PC|}=\frac{\sin(ADX)\sin(ACX)}{\sin(DAP)\sin(CAP)}.$ By comparing the law of Sines in the triangles $ADT$ and $ACT$, we get $\frac{1}{\sin(DAP)\sin(CAP)}=\frac{|AD||AC|}{\sin(ATD)\sin(ATC)|DT||TC|}.$ As $|AD||AC|=\frac{|AX|^2}{\sin(ADX)\sin(ACX)}$, we get $\frac{|AP|^2}{|DP||PC|}=\frac{|AX|^2}{\sin(ATD)\sin(ATC)|DT||TC|}$.....(1) Similarly, one can find that $\frac{|BQ|^2}{|DQ||QC|}=\frac{|BY|^2}{\sin(BTC)\sin(BTD)|DT||TC|}$.....(2) Since $|AX|=|BY|$ and $\angle ATD=\angle BTC$, the RHS of (1) and (2) are equal and we get the desired result. @below my bad, thanks for the heads-up

There were two T in the solution,can you fix it pls

Let $(ADT) \cap (BCT) = T'$ and $DL \cap CK = X$. We know that $\angle DAT = \angle DT'T, \angle CBT = \angle CT'T$. Then $\angle DT'C = \alpha + \beta$ where $\angle DAT = \alpha$ and $\angle CBT = \beta$. Because points $A,D,C,K$ and points $B,C,D,L $ lie on a circle, $\angle DCX = \angle DAT = \alpha$ and $\angle CDX = \angle CBT = \beta$, which means that the points $X,D,C,T'$ are cyclic. From these two observations, we get $\angle DT'X = \alpha = \angle DAT$, so $X,T',T$ are collinear. Then we proceed with some calculations, $$\frac{XC}{XD} = \frac{\sin \beta}{\sin \alpha} \; \; \;\;\; \frac{CK}{CT} = \frac{\sin CTK}{\sin ADC} \; \;\;\; \;\frac{DL}{DT} = \frac{\sin LTD}{\sin BCD}$$From the latter two equations, we get $$\frac{CK}{DL} = \frac{CT}{DT}\cdot \frac{\sin BCD}{\sin ADC} = \frac{\sin \beta}{\sin \alpha}\cdot \frac{BC}{AD} \cdot \frac{\sin BCD}{\sin ADC}$$$\frac{BC}{AD} = \frac{\sin DAB}{\sin CBA} = \frac{\sin ACD}{\sin BCD}$ which gives us that $$\frac{XC}{XD} = \frac{\sin \beta}{\sin \alpha} = \frac{CK}{DL}$$which is sufficient to show the lines $CD$ and $KL$ are parallel.

I spent so much time looking for a synthetic solution only to find a thread filled with trig Let $(TBD)$ intersect $\overline{BC}$ at $P$ and $(TCA)$ intersect $\overline{AD}$ at $Q$. Then $\triangle DLT \sim \triangle DCP$, so by spiral similarity, $\triangle DLC \sim \triangle DTP$. Similarly, $\triangle CKD \sim \triangle CTQ$. Therefore, it suffices to show that \[ \frac{d(T, \overline{PD})}{PD} = \frac{d(T, \overline{CQ})}{CQ}, \]as this would imply that $L$ and $K$ have the same distance from line $\overline{CD}$. Let $E = \overline{PC} \cap \overline{QD}$, and $F = \overline{PD} \cap \overline{CQ}$. Note that $\angle DPC = 180^\circ - \angle DTB = 180^\circ - \angle ATC = \angle DQC$, so $PQCD$ is cyclic and by Reim's theorem, $PQAB$ is also cyclic. Therefore, the radical axis of $(PDT)$ and $(QCT)$ passes through $E$ and $F$, so $E$, $F$, $T$ are collinear. Finally, observe that $\triangle EPD \sim \triangle EQC$, so \[ \frac{d(T, \overline{PD})}{d(T, \overline{CQ})} = \frac{d(E, \overline{PD})}{d(E, \overline{CQ})} = \frac{PD}{CQ}, \]as desired.