Is there a point $H$ such that $\frac{AH \cdot DM}{HD^2} = \frac{BH \cdot EN}{HE^2} = \frac{CH \cdot FP}{HF^2}$?

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$ 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 correct? $ $ \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.