## Simsons and 9-Point Circles in Cyclic Quadrilateral

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A Mathematical Droodle

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(When the simsons are shown, you may want to verify their definition. Click on them in turn, or click away from any to remove additional information.)

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Copyright © 1996-2018 Alexander Bogomolny

### Explanation

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Any quadrilateral ABCD defines four triangles: BCD, CDA, DAB, ABC. In each of the triangles one may consider associate remarkable points and lines. The above applet is specifically concerned with their 9-point circles and also the Simson lines of each of the points A, B, C, D with respect to the triangle formed by the other three. The applet purports to illustrate the fact that, provided the quadrilateral ABCD is cyclic, the eight aforementioned elements all concur in a point.

First let's prove the concurrence of the four 9-point circles. As we know, the quadrilateral N_{A}N_{B}N_{C}N_{D} formed by the 9-point centers of the four triangles BCD, CDA, DAB, ABC is similar to ABCD with the factor of 1/2. So that the distance from the circumcenter of N_{A}N_{B}N_{C}N_{D} to its vertices is R/2, where R is the circumradius of ABCD. We also know that in any triangle the radius of its 9-point circle equals half that of its circumradius. It therefore follows that the circles of radii R/2 centered at the points N_{A}, N_{B}, N_{C}, and N_{D} meet at the circumcenter of the quadrilateral N_{A}N_{B}N_{C}N_{D}. In other words, the 9-point circles of the triangles BCD, CDA, DAB, ABC meet at the circumcenter of N_{A}N_{B}N_{C}N_{D}.

Now let's tackle the simsons. As we showed elsewhere, the circumcenter of N_{A}N_{B}N_{C}N_{D} coincides with H, the center of homothety that maps ABCD onto H_{A}H_{B}H_{C}H_{D}, the quadrilateral formed by the orthocenters of the triangles BCD, CDA, DAB, ABC. The two quadrilaterals ABCD and H_{A}H_{B}H_{C}H_{D} are equal so that H is the midpoint of each of the segments AH_{A}, BH_{B}, CH_{C}, and DH_{D}. By a well known property of triangles [Honsberger, p. 43], the simson of A with respect to ΔBCD passes through the midpoint of AH_{A}, i.e., H. This of course applies to the other three simsons.

Note, that the result admits a natural generalization: if the quadrilateral is not cyclic, the simsons convert into circles, and the eight circles involved are still concurrent in a point.

### References

- R. Honsberger,
*Episodes in Nineteenth and Twentieth Century Euclidean Geometry*, MAA, 1995.

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Copyright © 1996-2018 Alexander Bogomolny

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