Euler's Line

Kate Berryman

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Take any triangle ABC.
Construct the circumcenter, centroid, and orthocenter.
These three points lie on the Euler line.
Proof: We know that PY is perpendicular to AC, since it is the perpendicular bisector. We know that BO is perpendicular to AC because it is the altitude of ABC.
Construct a line segment LM.
Construct a line segment MN. No we know that triangles MLY and BMN are similar triangles. So we have that 2MY = MB. We also know that MN/LM =BN/ LY = 2/1.

Now, we construct triangles LMZ and CMN. So we have 2CM = ZM. Furthermore, we have that CM = 2GM. We already know that NM = 2LM. Therefore, LMZ and CMN are similar triangles.

So L, M, and N are colinear.

Center of the Nine Point Circle and the Exeter Point

Also on the Euler line lie the center of the 9 point circle and the Exeter line. We construct the 9 point circle by using the points R, P, X, Q, Y, and Z. The other three points are the midpoints of the line segments from the orthocenter to the vertices. These points are labeled L, M, and N.
A circle passes through all 9 of these points. The center of the 9 point circle lies on Euler's line.
The Exeter point is found by constructing a triangle with points X, Y, and Z.
Where this triangle intersects the medians construct a point on all three sides.
Then construct a perpendicular line through each of these points through to the opposite side. Construct a point where these lines are concurrent.
Construct a circumcircle and tangents at each vertex and construct points at the intersection of the tangent lines.
Now, construct segments from each vertex to through the midpoint on the opposite side. Construct a point where this line intersects the circumcircle. In the picture to the right these points are L, M, and N. Connect these points to construct a triangle.
Construct lines from the vertices of the triangle made by the tangent lines through points L, M, and N.
Where these points concur is the Exeter Point.

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