Proof of the Nine-Point Circle

The Theorum

Given triangle ABC, Let D be the midpoint of side AB. Let E be the midpoint of side BC, and let F be the midpoint of side CA. (After connecting the midpoints D,E, & F, the medial triangle DEF is formed.)

Let point J be the foot of the altitude from vertex C to side AB. Let point K be the foot of the altitude from vertex A to side BC. Let point L be the foot of the altitude from vertex B to side CA. (After the feet of the altitudes are connected, the orthic triangle JKL is formed.)

If H is the orthocenter, let point R be the midpoint of the segment AH. Let S be the midpoint of the segment BH, and let T be the midpoint of the segment CH. (After connecting points R,S, & T, the orthocenter mid-segment triangle is formed.)

We must prove that points D,E,F,J,K,L,R,S, & T all lie on the same circle, ehich forms the nine point circle of this triangle.

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The Proof

We have:

DE parallel to RT

How?? DE is parallel to AC, because DE joins the midpoints of the two sides AB & BC of triangle ABC.

RT is parallel to AC, because RT joins the midpoints of the two sides AH & CH of triangle AHC.

Therefore DE is parallel to RT because they are both parallel to AC.

We have:

DR parallel to ET

How? DR is parallel to BH, because DR joins the midpoints of sides AB & AH of triangle ABH.

ET is parallel to BH, because ET joins the midpoints of sides CH & CB of triangle CHB.

Therefore DR is parallel to ET.

We have:

DR is perpendicular to AC

How? From the proof of DR is parallel to BH, we have that DR is parallel to BL, since BL is a segment that passes through H.

Our theorum states that BL is the altitude of the vertex B and the side AC, so from the definition of altitude, we conclude that AL is perpendicular to AC.

From the proof of DE parallel to RT, we have that DE is parallel to AC.

Therefore we have DR is perpendicular to AC.

Now, we have that, when the points are connected, DERT forms a rectangle. Similarly, DFTS forms a rectangle, as well.

We can see that the segment DT lies on the diagonal of both of the rectangles, passes through the center of the circle and connects points on the circle. Therefore DT is the diameter of the circle.

We now have that D,E,F,R,S, & T are all points on the circle with the diameter DT. Sincle DT is the diagonal of both rectangles, the six points at the vertices of the rectangles are equidistance from hte midpoint of DT.

Since segment CJ is an altitude of the triangle ABC, J is on the circle with diameter DT.

The same for points L and K. They are both the feet of the altitudes of different sides of triangle ABC.

So, L and K will lie on the circle, as well.

Angle TJD is a right angle. Since the diameter DT subtends the right angle at J, J must be on the circle.

This argument can be similarly made for points K and L.

Finally we have the proof of the nine point circle.

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