The theorem we are going to prove is the existence of the nine point circle, which is a circle created using nine important points of a triangle. Those nine points are the midpoint of each side, the feet of each altitude, and the midpoints of the segments connecting the orthocenter with each vertex.
Proof 1:
This shows the triangle with points ABC as the vertices, R, Q, and P as the feet of the altitudes BA, AC, and CA respectively and the orthocenter, H. We will also add midpoints to the legs of the triangle.
For segments BC, AC and AB, the midpoints are D, E and F respectively. We will also create midpoints of the segments connecting the orthocenter and the vertices. For the segments HA, HB and HC, the midpoints are L, M and N respectively.
Next we will draw the segments for FENM: FE, EN, NM and NF and the segments for FLND: FL, LN, ND and DF.
Due to the theorem stating that the line joining two midpoints of a side of triangle is parallel to the third side (See Mathematical Background page), using triangle ABC, FE is parallel to BC. Using triangle BHC, MN parallel BC and by the transitivity of parallel lines, FE is parallel to MN.
By triangle CHA, EN is parallel to AH and by triangle BHA, FM is parallel to AH. By transitivity, FM is parallel to EN.
FEMN is a rectangle. To get this fact we use the perpindicular bisectors. AP is perpendicular to BC and since BC is parallel to FE, AP is perpendicular to FE. By the same logic FE is perpendicular MN. Since FM is parallel to EN and is parallel to AH, FM is perpendicular to BC. By the definition of a rectangle, FEMN is one.
FLND is a rectangle using the same logic as above.
DL, EM, and FN are diameters of the same circle. This proof is on the Mathematical Background Page.
DP is perpendicular to LP because AH is perpendicular BC. So P is on the circle (See Mathematical Background: Perpendicular on the circle). BQ is perpendicular to EQ and CR is perpendicular FR. So points R, Q, and P are on the circle.
Proof 2:
Construct triangle ABC; the orthocenter, point H; the circumcenter, point O; and the altitude from vertex A. Construct the circle in which ABC is inscribed in. We can do that because we know three non colinear points will create a circle (See Mathematical Background page),. Next we construct point K on the circle, colinear to AP. OK is the circumradius.
The midpoint of BC is A'. The midpoint of AH is L'. By the midpoint theorem on the mathematical background page, then L(AH)=2*L(OA'). Since L' bisects AH so L(L'H)=L(OA'). Therefore OH and L'A' bisect eachother at point N.
The nine points of the circle are A', B', C', P, Q, R, L', M', and N'. We can see from this (See Mathematical Background page: Perpendicular on the Circle)) proof that A', P, and L' are on the circle. We can use the same logic and the altitudes from the other vertices of the triangle to see that the other six points also lie on the same circle. (Pedoe)
Proof 3:
First we constructed triangle ABC and the midpoints of sides AB, BC, and CA to be J, D and F respectively. H is the base point of the altitude AH. Connect points J and F and by the midpoint triangle parallel theorem, JF is parallel to BC.
The HDFJ is a trapezoid because it is a quadrilateral with one pair of parallel sides, JF and HD. We can say it is an isosceles trapezoid because L(DF)=1/2L(AB) because it is the midline of triangle ABC(See Mathematical Background page), and so triangle ABC and triangle FDC are similar triangles with a scaling factor of 1/2 and we know that because HJ is the radius of the Thales Circle having the diameter AB, L(HJ)=1/2L(AB).
The trapezoid is therefore a quadrilateral in a circle H is on the circle circumscribing the triangle DFJ. We repeat this same method twice using the other vertices as starting points. Therefore all the midpoints of the sides of the triangle and the feet of the altitudes are on the circle.
Now using triangle ABC we will look at the altitudes and at triangle BOC where O is the orthocenter. H is the base of the altitude from A. Triangle OBC has the same altitude bases as triangle ABC. So by part one of this proof, the circumscribed circle passing through the altitude base triangle of triangle ABC passes through the the midpoints of OB and OC which in triangle ABC are the midpoints of the segments connecting the orthocenter to the vertices. We can repeat this process with triangle ALC or triangle ABK to find that the midpoint of OA is also on the circle. Now we have that all of the midpoints of the segments connecting each vertex to the orthocenter are on the circle, and the proof is complete. (Dorrie)