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.

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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.

Proof

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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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 (O_c), (O_a), (O_b) whose diameters join I to the points C₁, A₁, B₁, respectively,
the lines A₁B₁, B₁C₁, A₁C₁ invert into circles (O^c), (O^a), (O^b) 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 (O_a) and (O^a); and, since (O_a) is the image of BC while (O^a) 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 (O_a) and IA is a diameter of (O^a)) 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

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

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