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Explanation The applet suggests what is known as Cantor's theorem (M. B. Cantor, 1829-1920):

Perpendiculars from the midpoints of a triangle to the opposite sides of its tangential triangle are concurrent. Furthermore, they meet at the center of the nine-point circle.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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(Tangential triangle is formed by the tangents to the circumcircle at the vertices of the given triangle. The lines in the theorem are known as Cantor's lines and their common point is sometimes referred to as the Cantor's point.)

In complex variables (or affine geometry, where we can add points) the proof is truly simple. Let the origin is set at the circumcenter O of ΔABC. Let zA, zB and zC be the complex numbers corresponding to its vertices. The midpoints of the sides are given as (side centroids or barycenters) mA = (zB + zC)/2, etc. The center of the nine-point circle is located at N = (zA + zB + zC)/2. Therefore, the line mAN = N - mA = zA/2 is parallel to the circumradius zA drawn to the vertex A. It is thus perpendicular to the tangent to the circumcircle at A. We see that Cantor's perpendicular from mA passes through N, the nine-point center. The same holds for the other two lines.

### Remark

Cantor actually proved a more general statement. For any inscribed n-gon we may consider n lines, one per vertex. Pick a vertex p and the centroid g of the remaining vertices. From g draw the perpendicular to the tangent at p. Cantor's theorem asserts that all n lines always concur. The point of concurrence is known as Cantor's point of the polygon.

### References

1. R. Honsberger, Mathematical Chestnuts from Around the World, MAA, 2001, pp 241-244 