## Experimentation with Dynamic Geometry Software: An Example

Computers open new horizons for experimentation. A dynamic sketch of a triangle and its three meadians shows convincingly that the medians concur in a point. It is often tempting to assert that fact without proof. Building on visual acceptance is easy, more attractive and probably suits better the temperament of a great number of students. I think that for many students such an approach might even be a better pedagogy than accepting proven facts only. I am certain that any generalization in this repsect may cause more harm than good, but in any event, students must be aware that reliance on visual perception alone may be misleading.

Here's an example I borrowed from Paul Yiu's lecture at the 7^{th} Statewide Conference organized by Florida Higher Education Consortium, November 12, 1998.

Let I be the incenter of ΔABC. Consider three triangles AIB, BIC, and CIA. In each of the three triangles pass a straight line through its incenter and circumcenter. (In the applet, I_{C} and O_{C} stand for the incenter and circumcenter of ΔAIB, and similarly for the other 4 points.)

Doodling with the applet one may form (at least) two conjectures. (A working geometer would probably dismiss one of them as a known result. I still carry on.)

Circumcenters of triangles AIB, BIC, and CIA all lie on the circumcircle of triangle ABC.

The three lines through the incircle and the circumcircle of triangles AIB, BIC, and CIA meet at a point.

In the applet, by checking the *Show calculations* you may obtain the (approximate) sum of the **Dist**ances of the circumcenters O_{A}, O_{B}, O_{C} to the circumcircle of ΔABC and the (approximate) **Area** of the triangle formed by the above three lines.

What if applet does not run? |

As you move the vertices of the triangle, you may surmise that one conjecture is probably right whereas the other conjecture is probably wrong. You may want to formally prove the former.

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

Copyright © 1996-2018 Alexander Bogomolny

66201887