Can be trivialized by complex numbers. $k = b + c - p, l = c + a - p, m = a + b - p$ The midpoint of $AK$ is $\frac{a + b + c - p}{2}$. Notice this is also the midpoint of $BL$ and $CM$, then we are done.

By simple angle chasing, you can find that $ALKB$ is a parallelogram, so if $AK\cap BL={X}$, then $X$ is the midpoint of the segments $AK$ and $BL$, and same goes for the quadrangle $AMKC$. It is a parallelogram too, so $CM$ passes through $X$, the midpoint of $AK$.

This point is the center of homotety that takes the two congruent triangles $KLM$ and $ABC$.