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*and

_{B}*M*, so that

_{C}*M*is the midpoint of

_{A}*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*and

_{B}*H*, respectively, and let

_{C}*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*=½

_{A}M_{A}*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*=½

_{A}*P*=½

_{A}M_{A}*R*.

Now, consider the quadrilateral *HOM _{A}H_{A}*. Since and are both perpendicular to

*BC*,

*HOM*is a right trapezoid. Letting

_{A}H_{A}*Q*be the midpoint of , we see then that

_{A}*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*, and so

_{A}*NH*=

_{A}*NM*=

_{A}*NP*=½

_{A}*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*=½

_{B}*R*as well. Lastly, diameter

*CF*, and midpoint

*P*of

_{C}*CH*gives, by similar proof, that

*NH*=

_{C}*NM*=

_{C}*NP*=½

_{C}*R*. Thus, the nine points

*M*,

_{A}*M*,

_{B}*M*,

_{C}*H*,

_{A}*H*,

_{B}*H*,

_{C}*P*,

_{A}*P*and

_{B}*M*are all equidistant from

_{C}*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).