## Five points on a circle, centroids and perpendiculars

Suppose you have different points on a circle. For each triangle formed by three of these points, consider the line passing through its centroid, perpendicular to the line determined by the other two.

*Proof:* Pick one point among the five points on the circle, and consider the homothety of center and ratio . This homothety maps the centroid of a triangle on the midpoint of . Thus, if are the remaining vertices, the line passing through the centroid of , perpendicular on is mapped to a line passing through the midpoint of , perpendicular on .

Now, we have the following result: Suppose you have points on a circle. Draw the lines which pass through the midpoint of a segment determined by these, perpendicular to the line determined by the other two. Then these lines are concurrent. Furthermore, if is the point of concurrency, the center of the circle, and the centroid of , then is the midpoint of .

To prove this secondary result, note that if we take the midpoints of and , for example, and consider the intersection point of the perpendicular from to and the perpendicular from to , then is a parallelogram. Thus, the midpoint of is the same as the midpoint of , which is the centroid of .

Thus, among the lines, there are families of six lines which are concurrent. Each such two families have at least two common lines, which means the intersection point is unique.

Take a look at this MathOverflow discussion, where you can find more informations and generalizations. Here you can find a Geogebra interactive picture.