The Incenter of a Triangle
Consider any triangle ABC. Can you fit a circle inside it that
is tangent to all three sides? How can you construct such a circle?
Analysis
Let's analyze the situation to find more about such a circle, should it
exist. We'll call it the incircle of the triangle since it just fits in the triangle. Suppose its center is the point I. Draw lines
IA, IB, and IC to the vertices of the triangle.
Also draw perpendicular lines from I to the sides of the triangle; call
the feet of these perpendicular lines D, E, and F. What
can you say about this figure?
Look at the little triangles. Are any of them congruent? Well, yes. They're congruent in pairs, one pair for each vertex.
Triangles AEI and AFI are both right triangles with a common
hypotenuse AI and equal sides EI and FI, so they're congruent. That implies that their angles at A are equal, hence,
the line AI bisects the angle at A. Thus, the center of the
incircle lies on all three angle bisectors of the vertex angles. We can
use this information to construct the incenter and the incircle.
Synthesis
Now to construct the incenter and the incircle of a given triangle ABC
and to prove that the construction is correct. Since we don't yet know that
the three angle bisectors actually meet at a point, we can't start there. (We
only know that once we succeed in constructing the incenter, then the angle
bisectors will meet there.) Instead, we just take two angle bisectors. So,
take angle bisectors of angles A and B. They will meet at some point, call it I. (It's
easy to show they're not parallel.) Drop perpendiculars to each side to
get ID, IE, and IF. Then triangles AEI and AFI
are right triangles with a common hypotenuse AI and equal angles
EIA and FIA, and, hence, are congruent. (Note that the reason
they are congruent is not the same that we gave in the analysis.) Consequently,
line IE equals line IF. Similarly, using the fact that line
IB bisects the angle at B, we may conclude that triangles
BDI and BFI are congruent. Therefore,
line ID equals line IF. Thus, all three points D, E, and
F are equidistant from the point I. Therefore, the circle
with center I and any of those three lines as radius passes through
all three points D, E, and F. Furthermore, since the radii
ID, IE, and IF are perpendicular to the sides of the triangles,
those sides are tangent to the circle. Thus, the point I is the center
of the incircle DEF of the triangle.
David E. Joyce
Department of Mathematics and Computer Science
Clark University
Worcester, MA 01610
Email: djoyce@clarku.edu
My nonJava Homepage
and my
Java homepage