In the Following Statement, There Is an Issue of the Definition of a Triangle

A Sample Proof for Geometry 3050.

Walter Whiteley, November 2006.

In the following statement, there is an issue of the ‘definition’ of a triangle. We will return to that after the proof. Watch for the gap.

Theorem: The three angle bisectors of a triangle in the plane, or a small triangle in the sphere, meet in a point, the incenter of the triangle.

Proof: The angle bisector of angle BAC is the mirror along which one can fold the plane so that side BA lies on side AC. For any point D, a perpendicular DE to edge AB will fold onto (reflect onto) a perpendicular DF to side AC (see below).

Now, consider the special point P, at which the two angle bisectors to < BAC and <ACB intersect.

Drop a perpendicular from P to AB, called PE. This will fold onto an equal length perpendicular PF to AC, using the bisector of BAC. This, in turn, will fold onto a perpendicular PG to BC. This gives the equality of the two perpendiculars PE and PG.

Now focus on the substructure PEBG. We will add an ‘auxiliary line’ - the angle

bisector of EPG.

We can now apply ASA to the two triangles with the congruent sides PE and PG, and the corresponding angles congruent. The proof of ASA and the congruence generated, guarantees that the angle bisector of P goes through the original vertex B, and that this line also bisects the angle EBG (also called <ABC).
We conclude that the three angle bisectors meet in a common point P, as required.

Moreover, P has three equal perpendiculars to the three sides. As a result, if we draw a circle with center P and radius |PE|=|PF|=|PG|. A circle with this radius, will have the edges (perpendicular to the radii) as tangent to the circle. This circle is called the incircle of the triangle, and the point P is called the incenter. QED.

Alternate completion after the construction of P, and PE and PG.

Draw the edge EG. PEG is an isosceles triangle. By ITT and its corollary, the angle bisector of EPG is also the right bisector of EG.

Now continue to right bisector of EG on towards B. By ITT on PEG, we have equality of the angles PEG and PGE. This leaves the equal angles at BEG and BGE. This now gives the set up for the converse of ITT, which guarantees (as its corollary) that the right bisector of EG is also an angle bisector of angle EBG. We conclude that PB is the third angle bisector, as required.

Everything done so far, works whenever ITT, the Converse of ITT, or ASA apply. I suspect that your proofs of this on the sphere, require a small triangle – so this is the context I have used in the proof.

There is one special case that does not quite fit three proof: the first two angle bisectors might not meet in a point – in the plane! (On the sphere these lines must meet!)

However, in the plane, to be parallel, the angle bisectors would have to be bisector 180 degree angles – and no triangle has two interior angles of 180 degrees. Even a collinear triangle will have three angle bisectors meeting at the third interior vertex (and the incircle has radius 0)!

When we go on to generalize this discussion to the exterior angles of the triangle (the excenter), essentially the same proof works – with the very small gap for the case of collinear plane triangles. This is the case where there can be parallel exterior angle bisectors! Whether we continue to ‘see’ the three angle bisectors meeting in a point depends on whether you ‘see’ three parallel lines meeting at infinity!