Interactive · 1 question
Let's say we have a triangle ABC (and it's incenter), and we want to construct its incircle. How can we do this?
Right! Since the shortest distance between a point and a line is the perpendicular, we just need to construct the perpendicular to one of the sides! We construct the perpendicular (the circle centered around D doesn't necessarily have to pass through A):
Now we find the perpendicular bisector of AE, which passes through D.
Finally, we have the perpendicular (the intersections of the line through D and the segment AC).
Getting rid of all of the clutter...
Since we have the incenter and the inradius (DM), we can finally draw the incircle!