## 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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Copyright © 1996-2018 Alexander Bogomolny

### Theorem

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

### Proof

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In the proof we shall use the following notations:

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

A

_{1}, B_{1}, C_{1}- the points of tangency of the incircle with the side lines BC, AC, and AB, respectively,A

_{2}, B_{2}, C_{2}- the points of intersection of the pairs of lines BC and B_{1}C_{1}, AC and A_{1}C_{1}, and AB and A_{1}B_{1}, 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 C_{1}, A_{1}, B_{1}, respectively,the lines A

_{1}B_{1}, B_{1}C_{1}, A_{1}C_{1}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 A_{1}B_{1}is a circle through I, A_{1}, B_{1}and since IA_{1}⊥BC and IB_{1}⊥AC, the quadrilateral A_{1}IB_{1}C 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, A

_{3}, B_{3}, C_{3}, where, say, A_{3}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 B_{1}C_{1}, A_{2}is the image of A_{3}, etc.The angles IA

_{3}A_{1}and IA_{3}A are both right (since IA_{1}is a diameter of (O_{a}) and IA is a diameter of (O^{a})) implying that AA_{3}A_{1}is a straight line so that IA_{3}⊥AA_{1}. But then also IA_{3}⊥AGe, i.e., ∠IA_{3}Ge is right. Similarly,∠IB which says that Ge lies on (G) and IGe serves as its diameter._{3}Ge = ∠IC_{3}Ge = 90°- 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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny