## Gergonne and Soddy Lines Are Perpendicular

The applet below illustrates the fact that the Gergonne and Soddy lines in a scalene triangle are perpendicular.

In ΔABC, Gergonne line is the polar of the Gergonne point which is the intersection of the cevians joining the points of tangency of the incircle with the side lines and the opposite vertices of the triangle.

The Soddy line is the line through the Gergonne point Ge and the incenter I of ΔABC. In a scalene triangle I and Ge never coincide so that the Soddy line is uniquely defined.

### 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.

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Discussion

### Theorem

Wherever the Soddy line is uniquely defined (in particular, in a scalene triangle), it is perpendicular to Gergonne Line.

### 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 the proof we shall use the following notations:

• A, B, C - vertices of the given triangle,

• A1, B1, C1 - the points of tangency of the incircle with the side lines BC, AC, and AB, respectively,

• A2, B2, C2 - the points of intersection of the pairs of lines BC and B1C1, AC and A1C1, and AB and A1B1, that define the Gergonne line (these are known as Nobbs' points),

• Ge, I - Gergonne point and the incenter.

Our main tool in the proof will be the inversion in the incircle.

Under this inversion,

• the sides AB, BC, AC invert into three equal circles (Oc), (Oa), (Ob) whose diameters join I to the points C1, A1, B1, respectively,

• the lines A1B1, B1C1, A1C1 invert into circles (Oc), (Oa), (Ob) whose diameters join the vertices C, A, B to I. For example, the inversive image of A1B1 is a circle through I, A1, B1 and since IA1⊥BC and IB1⊥AC, the quadrilateral A1IB1C is cyclic and, moreover, CI is its diameter, again because the indicated segments form right (inscribed) angles,

• The Soddy line IGe inverts into itself,

• The Gergonne line inverts into circle (G) through I, A3, B3, C3, where, say, A3 is the second (besides I) intersection of (Oa) and (Oa); and, since (Oa) is the image of BC while (Oa) is the image of B1C1, A2 is the image of A3, etc.

The angles IA3A1 and IA3A are both right (since IA1 is a diameter of (Oa) and IA is a diameter of (Oa)) implying that AA3A1 is a straight line so that IA3⊥AA1. But then also IA3⊥AGe, i.e., ∠IA3Ge is right. Similarly, ∠IB3Ge = ∠IC3Ge = 90° which says that Ge lies on (G) and IGe serves as its diameter.

• IGe is perpendicular to (G) and, by the angle preservation property of inversion, their images - IGe and the Gergonne line - are also perpendicular.

### References

1. Zuming Feng, Why Are the Gergonne and Soddy Lines Perpendicular? A Synthetic Approach, Mathematics Magazine, v. 81, No. 3, June 2008, pp.211-214