$H$ is a point in non-isoceles triangle $\triangle ABC$. The intersections of $AH$ and $BC$, $BH$ and $CA$, $CH$ and $AB$ are respectively $D$, $E$, $F$. $AD$, $BE$ and $CF$ cuts $(A, B, C)$ respectively at $M$, $N$ and $P$. Is there a point $H$ such that the following equality is satisfied? $$\large \frac{AH \cdot DM}{HD^2} = \frac{BH \cdot EN}{HE^2} = \frac{CH \cdot FP}{HF^2}$$
If there is not, prove why.
If there is, illustrate how to put down point $H$.
Of course, point $H$ should be one of the triangle centres identified in the Encyclopedia of Triangle Centers. But I don't which one it is.
Unfortunately, this point is probably not in ETC. If you consider the triangle with $B=(0,0)$, $C=(6,0)$, $(BC,CA,AC)=(6,9,13)$ and $A$ above the $x$ axis you will find that the point you are looking for has coordinates equal to (approximately) $(5.13198,1.12946)$ and $6$, $9$, $13$ search doesn't find the value $1.1294$.(https://faculty.evansville.edu/ck6/encyclopedia/Search_6_9_13.html)