It's nothing but a limiting case of the following configuration: $\triangle ABC$ is scalene with incirce $(I,r)$ and B-excircle $(I_b)$ touching its sidelines $BC,CA$ at $M,N.$ Then $MN$ cuts the A-altitude at a point $X,$ such that $AX=r.$ This is an old problem posted many times before, e.g Inradius and altitude and elsewhere.

Can somebody post a solution using complex numbers just for learning purposes?

Let $O$ be the midpoint of $KM$. Let the perpendicular bisector of $KM$ intersect $KL$ at $N$. Let $CN$ intersect $AB$ at $Q$. Note that $ON\parallel MC\parallel KQ$. Thus, $\angle CMN=\angle MNO=\angle KNO=\angle QKN=\angle BKL=\angle BLK$ $=\angle CLN$ i.e. $CNLM$ is a cyclic quad. But $\angle MCN=180^{\circ}-\angle MLN=90^{\circ}$. Thus $OQ\perp AB$. But since $MKQC$ is a trapezoid, and $ON$ is the perpendicular bisector of $KM$, and $ON\parallel KQ$, $N$ must be the midpoint of $CQ$.