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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.

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

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  • $\begingroup$ 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 $X$. $\endgroup$
    – conditionalMethod
    Oct 25, 2019 at 16:00
  • $\begingroup$ @conditionalMethod The nine point circle will also pass through M, D, X. Therefore the two circles are actually identical. $\endgroup$
    – Mick
    Oct 25, 2019 at 16:03
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    $\begingroup$ 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? $\endgroup$
    – conditionalMethod
    Oct 25, 2019 at 16:06
  • $\begingroup$ @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. $\endgroup$
    – Mick
    Oct 25, 2019 at 17:15
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    $\begingroup$ 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? $\endgroup$
    – conditionalMethod
    Oct 25, 2019 at 17:17

1 Answer 1

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Besides M and D, need one more point (out of the remaining 7 points of the nine-point circle) to fix the circle. If the circle drawn cuts AD at X, then H, the ortho-center should be another point on and between AD such that AX = XH.

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