|
|
|
|
|
|
|
|
Four Concurrent Lines in a Cyclic Quadrilateral: What are they?
|
| Buy this applet What if applet does not run? |
| Buy this applet What if applet does not run? |
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. 
OKR = TKP,
TPK = ORK.
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 OAB and OCD be reflections of the circumcenter of the cyclic quadrilateral in AB and CD, respectively. Prove that the anticenter T of ABCD lies on OABOCD. |
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 USQ the two altitudes SH and QF intersect in T, the anticenter of ABCD. T is therefore the orthocenter of USQ, such that UT must be the third altitude.
Solution to Problem 2
P, the midpoint of AB is, by construction, the midpoint of OOAB, whereas R, the midpoint of CD, is, by construction, the midpoint of OOCD. In OOABOCD, the midline PR is parallel to the base OABOCD.
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 OABOCD (Make a drawing.)
| 33058196 |  |
