# Four Concurrent Lines in a Cyclic Quadrilateral

What are they?

A Mathematical Droodle

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Copyright © 1996-2017 Alexander Bogomolny

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Let ABCD be a cyclic quadrilateral. Let O be the center of its circumscribed circle, K its center of gravity, i.e. the point of intersection of two lines joining the midpoints of the opposite sides. Let T be symmetric to O in K. We are going to show that the four lines, each drawn through the midpoint of one of the sides perpendicular to the opposite side, all pass through T. The line in a quadrilateral drawn from a midpoint of a side perpendicular to the opposite side is called a *maltitude*. In a cyclic quadrilateral maltidues are concurrent. Point T of concurrency is known as the *anticenter* of the quadrilateral ABCD. (This looks very much as a generalization of Brahmagupta's theorem. It also admits the following 3D analogue [Cofman, Problem 90]: the six planes (the lines as well) through the midpoints of the edges of a tetrahedron perpendicular to the opposite edges are concurrent. The proof below works with virtually no modification also in this case. Point T in that case is known as the *Monge point* of the tetrahedron.)

We need only consider one of these lines. So, let P be the midpoint of AB, R the midpoint of CD.

Here's a couple of related problems:

### Problem 1

Let the extensions of the opposite sides AD and BC of the cyclic quadrilateral meet at a point U. Prove that UT, where T is the anticenter of ABCD, is perpendicular to SQ.

### Problem 2

Let O_{AB} and O_{CD} be reflections of the circumcenter of the cyclic quadrilateral in AB and CD, respectively. Prove that the anticenter T of ABCD lies on O_{AB}O_{CD}.

### References

- R. Honsberger,
*Episodes in Nineteenth and Twentieth Century Euclidean Geometry*, MAA, 1995. - J. Cofman,
*Numbers and Shapes Revisited*, Clarendon Press, 1995 - D. Wells,
*Curious and Interesting Geometry*, Penguin Books, 1991

### Solution to Problem 1

In ΔUSQ the two altitudes SH and QF intersect in T, the anticenter of ABCD. T is therefore the orthocenter of ΔUSQ, such that UT must be the third altitude.

### Solution to Problem 2

P, the midpoint of AB is, by construction, the midpoint of OO_{AB}, whereas R, the midpoint of CD, is, by construction, the midpoint of OO_{CD}. In ΔOO_{AB}O_{CD}, the midline PR is parallel to the base O_{AB}O_{CD}.

Now, on one hand, the center of gravity K lies on PR by Varignon's theorem. As we saw, K bisects OT. Therefore, T lies on O_{AB}O_{CD} (Make a drawing.)

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