# Concurrency in the Intouch Triangle

What is this about?

A Mathematical Droodle

What if applet does not run? |

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

Copyright © 1996-2017 Alexander Bogomolny

What if applet does not run? |

The applet may suggest the following statement [Savchev, pp. 141-141]:

In ΔABC, the points A', B', C' of tangency of the incircle with the sides ΔABC form the *intouch* ΔA'B'C'. From the midpoints A_{m}, B_{m}, C_{m} of the sides of the latter drop perpendiculars A_{m}A_{h}, B_{m}B_{h}, C_{m}C_{h} on the opposite sides of ΔABC. The three lines A_{m}A_{h}, B_{m}B_{h}, C_{m}C_{h} are concurrent.

The proof is remarkably short and reminds one of that employed to prove the concurrency of maltitudes in an inscribed quadrilateral.

Recollect that the medians in a triangle are divided in ratio 1:2 by the center of the triangle. We apply this fact to the intouch triangle A'B'C'. (I'll use G' to denote its center, although the point is not named in the applet.) This means that points A', B', C' can be obtained from points A_{m}, B_{m}, C_{m} by a homothety with coefficient -2 and center G'. A homothety maps straight lines onto parallel lines. Therefore, say, A'A_{m} is mapped onto a line through A' parallel to A'A_{m} and thus perpendicular to BC. Since BC is tangent to the incircle of ΔABC, that line passes through its incenter I. The same holds of the images of B'B_{m} and C'C_{m}. The three images therefore concur (in the incenter I of ΔABC.) Since homothety is a reversible transformation, the lines themselves are also concurrent. Furthermore, the point of concurrency S and the incenter I are collinear with G' and

### References

- S. Savchev, T. Andreescu,
*Mathematical Miniatures*, MAA, 2003

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

Copyright © 1996-2017 Alexander Bogomolny61254426 |