Sufficient conditions to use a “nine-point circle” to determine the ortho-center.

In $$\triangle ABC$$, M is the midpoint of BC and A(X)(H)D is an altitude. Of course, a circle (actually is the “nine-point circle”) can be drawn through M, D, X.

If, in addition, AX = XH, can I conclude that H is the orthocenter? If that is not sufficient, what else do I need?

• Do you know for sure that the circle is the nine-point circle? If so, then its intersection with $AH$ is half way between $A$ and the orthocenter. So, yes. That would be $H$. Note, however, that there are many circles that pass through $D$ and $M$ and intersect $AH$ somewhere that can be called $X$ and then an $H$ drawn on $AH$ at double the distance from $A$ to $X$. Oct 25, 2019 at 16:00
• @conditionalMethod The nine point circle will also pass through M, D, X. Therefore the two circles are actually identical.
– Mick
Oct 25, 2019 at 16:03
• You haven't said who is $X$, until you have said who is the circle. So, what is the information that is actually known in the drawing of the figure? Oct 25, 2019 at 16:06
• @conditionalMethod The notation that I use in "A(X)(H)D is an altitude" means X, and also H, are points on the altitude AD. Sorry for the confusion.
– Mick
Oct 25, 2019 at 17:15
• I understood that. But still I am not sure if you have answered my question. Is it known/given that the circle is the nine-point circle? Oct 25, 2019 at 17:17