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

$$H$$ 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.