The purpose of this investigation is to explore pedal triangles. Particularly, what is the significance if the pedal point is the same as the incenter of a given triangle?
I claim that points of tangency between the incircle and original triangle are the vertices of the pedal triangle when their centers are the same point.
To begin with let's discuss incircles, shall we?
Given ĆABC, the incenter is found at the intersection of the angle bisectors.
Using 'I' as the center of a circle, the shortest distance from 'I' to each of the edges of ĆABC will be a perpendicular line through 'I' and each edge. These three points where the perpendicular lines meet will are all equidistant from 'I' and hence will all be radii of the incircle.
What does this have to do with pedal triangles however?
To create a pedal triangle, choose an arbitrary point and draw in the lines perpendicular to each edge (or the line which the edge is on). Where these perpendicular lines intersect you will find the vertices of the pedal triangle. But wait, the moment of triumph has almost passed us by. Didn't we get the three points where the incircle will meet the edges of ĆABC in the exact same manner, by drawing perpendiculars from 'I' to the edges?
In fact, this is a special case of a pedal triangle, sometimes called the contact triangle.
If you would like to experiment with pedal triangles, press HERE.