Let $H_AH_C\cap AC=X$ and $BX\cap (ABC)=\{B, D\}$. By simple angle-chasing, one can show that $\angle KXL=\angle BAC-\angle BCA$ and $\angle KLX=\angle BAC+\dfrac{\angle ABC}2$. Hence, $|XK|=|XL|$..........(1) We have $|XD|\cdot|XB|=|XA|\cdot |XC|=|XH_A|\cdot |XH_C|$, hence $D\in (BH_CHH_A)$ and $\angle ADH=90^\circ$. Then, $M\in DH$ and $HM\perp BX$. Thus, $LT\perp BX$. Also, $BT\perp XL$. So, $T$ is the orthocenter of $\triangle BXL$ and $XT\perp BL$. Using this and (1), one can get the desired result.