# Gergonne and Medial Triangles Are Orthologic

What is this about?

A Mathematical Droodle

What if applet does not run? |

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

Copyright © 1996-2017 Alexander BogomolnyThe applet suggests the following statement and one of its proofs:

In ΔABC, let M_{a}, M_{b}, M_{c} be the midpoints of sides BC, AC, and AB. Let A', B', C' be the points where the incircle touches those sides. (ΔA'B'C' is known as Gergonne or contact triangle of ΔABC.) Then the perpendiculars from the midpoints M_{a}, M_{b}, M_{c} onto the opposite sides of ΔA'B'C' are concurrent.

What if applet does not run? |

According to R. Honsberger, the problem has been submitted by Bulgaria to an IMO, but was left unused. It was discussed and solved by J. T. Groenman, The Netherlands, in *Crux Mathematicorum*, 1987, 75. Here's Groenman's solution:

The center of the incircle of a triangle lies at the intersection of its angle bisectors. The incenter is thus naturally equidistant from the sides of the triangle. It follows that the tangents AB', AC' from vertex A of ΔABC to its incircle are equal and ΔAB'C' is isosceles. The bisector of angle A is perpendicular to B'C', hence parallel to the perpendicular from M_{a}. The latter, being parallel to the angle bisector in ΔABC, serves as the angle bisector of angle M_{b}M_{a}M_{c} in the medial triangle ΔM_{a}M_{b}M_{c}, since the two triangles are homothetic. The same holds for the other two perpendiculars. The three intersect as being angle bisectors of ΔM_{a}M_{b}M_{c}.

The fact just proven means that the Gergonne and medial triangles are orthologic, which, as we know from Maxwell's theorem, is a symmetric relationship. We conclude that the perpendiculars from the vertices of Gergonne triangle A'B'C' onto the opposite sides of ΔM_{a}M_{b}M_{c} also concur. But this fact is obvious. Indeed the sides of the medial triangle are parallel to those of ΔABC, so that the perpendiculars at the vertices of ΔA'B'C' where the incircle of ΔABC touches its sides meet at the incenter of ΔABC, naturally. Going backwards, we get another proof of the Bulgarian problem.

### References

- R. Honsberger,
*From Erdös To Kiev*, MAA, 1996, pp. 99-102

### Maxwell Theorem and orthologic triangles

- Maxwell's Theorem
- Gergonne and Medial Triangles Are Orthologic
- Orthologic Triangles in a Quadrilateral
- Pedal Triangle and Isogonal Conjugacy
- Projective Proof of Maxwell's Theorem
- Maxwell's Theorem by Way of Trigonometric Ceva

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

Copyright © 1996-2017 Alexander Bogomolny62058961 |