Consider a triangle, which we label ∆*ABC*, with circumcenter *O* and circumradius *R*=*AO*=*BO*=*CO*. Let us label the midpoints of the sides as *M*_{A}, *M*_{B} and *M*_{C}, so that *M*_{A} is the midpoint of *BC* (the side opposite *A*), and similarly, so that , and are the medians. Then , < and are segments of the perpendicular bisectors of the sides. Let us also label the feet of the altitudes from *A*, *B* and *C* as *H*_{A}, *H*_{B} and *H*_{C}, respectively, and let *H* be the orthocenter (the intersection of the altitudes , and ).

Let us construct the point *D* on the circumcircle diametrically opposed to *A*; that is to say, the point *D* such that *AD* is a diameter of the circumcenter. Then *AD*=2*R*, and *O* is the midpoint of *AD*.

Now, by Thales’ theorem, ∠*ABD* and ∠*ACD* are both right angles. Now, since *CD* and the altitude are both perpendicular to *AC*, they are parallel to each other. Similarly the altitude and segment *BD* are parallel, both being perpendicular to *AB*. Thus, the quadrilateral *BDCH* is a parallelogram.

Since the diagonals of a parallelogram bisect each other, we see that the midpoint *M*_{A} of *BC* is also the midpoint of *HD*.

Now, let *P*_{A} be the midpoint of the segment *AH*. Then is a midline of the triangle ∆*AHD*, and by the triangle midline theorem, and *P*_{A}M_{A}=½*AD*=*R*.

Now, let *N* be the intersection of and *HO*. By the midline-median bisection theorem proven in this post, we see that, as *HO* is the median of ∆*AHD* that crosses midline , *N* is the midpoint of both and *HO*. Thus, *NM*_{A}=*NP*_{A}=½*P*_{A}M_{A}=½*R*.

Now, consider the quadrilateral *HOM*_{A}H_{A}. Since and are both perpendicular to *BC*, *HOM*_{A}H_{A} is a right trapezoid. Letting *Q*_{A} be the midpoint of , we see then that *NQ* is the median (or midline) of trapezoid *HOM*_{A}H_{A}.

By the first of the three items proven here, we see that , and so . Thus, we see that is the perpendicular bisector of , and so, by the perpendicular bisector theorem, *NH*_{A}=*NM*_{A}, and so

*NH*_{A}=*NM*_{A}=*NP*_{A}=½*R*.

Constructing diameter *BE* of the circumcircle gives us parallelogram *CEAH*, by a similar argument as above. Letting *P*_{B} be the midpoint of the segment *BH*, analogous reasoning to the above shows that *NH*_{B}=*NM*_{B}=*NP*_{B}=½*R* as well. Lastly, diameter *CF*, and midpoint *P*_{C} of *CH* gives, by similar proof, that *NH*_{C}=*NM*_{C}=*NP*_{C}=½*R*. Thus, the nine points *M*_{A}, *M*_{B}, *M*_{C}, *H*_{A}, *H*_{B}, *H*_{C}, *P*_{A}, *P*_{B} and *M*_{C} are all equidistant from *N*.

Therefore, we see that for any triangle, the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments connecting the vertices to the orthocenter all lie on a single circle, which, for this reason, is usually known as the nine-point circle, and its center *N* the nine-point center. We also see that the radius of the the nine-point circle is half the radius of the circumcircle, and the nine-point center is the midpoint of the segment connecting the circumcenter and orthocenter. This latter tells us that for any non-equilateral triangle, the nine-point center lies on the Euler line (for an equilateral triangle, it coincides with the circumcenter, orthocenter, and centroid).