# Droz-Farny Line Theorem

### 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.

 What if applet does not run?

In 1899, Arnold Droz-Farny (1865-1912), a Swiss science and mathematics teacher, published without proof the following theorem:

If two perpendicular straight lines are drawn through the orthocenter of a triangle, they intercept a segment on each of the three sidelines. The midpoints of the three segments are collinear.

A synthetic proof of the theorem along with a bibliography and a short bibliographical note on Droz-Farny has recently appeared in the Forum Geometricorum. The theorem is curious, but the proof is absolutely remarkable in its simple elegance.

### Proof (J.-L. Ayme, 2004)

The statement of the theorem is quite obvious in two cases: when the given triangle is right and, for an arbitrary triangle, when one of the given lines coincides with one of the altitudes of the triangle. In the following we shall ignore the two possibilities.

Consider ΔABC. Assume the given lines meet the side AB of the triangle in points Z and Z', the side BC in points X, X', and the side AC in points Y and Y'. Let Ma, Mb, Mc denote the midpoints of segments XX', YY', and ZZ', respectively. The Droz-Farny theorem states that the points Ma, Mb, Mc are collinear.

Let CABC denote the circumcircle of ΔABC. A similar notation will be used for other triangles, e.g. CXX'H will denote the circumcircle of ΔXX'H, etc.

The reflection Ka of the orthocenter H in BC lies on CABC. Since the two given lines are perpendicular, the circle CXX'H has XX' as its diameter (and Ma as its center.) The circle is symmetric in XX', hence in BC. So that the reflection Ka of the orthocenter H in BC lies on CXX'H and serves as a common point of the latter with the circumcircle CABC. Similar considerations apply to points Kb and Kc and circles CYY'H and CZZ'H.

The reflections of line XYZ in the sides of ΔABC meet on CABC in, say, point Q. The crucial step in the proof to observe we now can apply Miquel's Pivot theorem to each of the three triangles: XYQ, YZQ, and XZQ.

Indeed, we may think of points Ka, Kb, and H as lying on the sidelines XQ, YQ, and, respectively, XY of ΔXYQ. Circle CXX'H passes through X, Ka, and H. Circle CYY'H passes through Y, Kb, and H. And, finally, circle CABC passes through Ka, Ka, and Q. According to the Pivot theorem the three circles have a common point, say, M. Thus, besides sharing point Ka, circles CXX'H and CABC meet at M, while circles CYY'H and CABC meet in M and Kb.

If we started with, say, ΔYZQ, the Pivot theorem would lead to a point N common to circles CYY'H, CZZ'H, and CABC. But circles meet only in two points of which one is Kb. We are bound to conclude that N = M. Therefore, all three circles CXX'H,CYY'H, CZZ'H meet in M and H, which means they are coaxal. In particular, their centers Ma, Mb, and Mb are collinear.

Note: The collinearity holds when the three points Ma, Mb, and Mb are defined by the same linear combination of the points of intersection: Ma = tX + (1-t)X', Mb = tY + (1-t)Y', Mc = tZ + (1-t)Z', with the same t. The analytic proof for the general case can be found on a seperate page.

### References

1. J.-L. Ayme, A Purely Synthetic Proof of the Droz-Farny Line Theorem, Forum Geometricorum, Volume 4 (2004) 219-224 