Using Geometer's Sketchpad, we can look at an example of the conjecture:

*The three angle bisectors
of the internal angles of a triangle are concurrent.*

Example:

The angle bisectors are: AD,
BE and FG. They are concurrent because the point *c* is on
all of the angle bisectors. Each angle bisector divides the opposite
side into two segments.

Proof:

For triangle ABC, first we construct
the angle bisectors of <ABC and <ACB. They intersect at
the point *l*.

Now construct perpendicular
lines from *l *to AC, from *l* to AB, and from *l*
to BC.

Triangle G*l*C is congruent
to triangle F*l*C by ASA. This gives us that *l*G =
*l*F.

Triangle E*l*B is congruent
to triangle F*l*B by ASA. This gives us that *l*E =
*l*F.

So we have, *l*G = *l*E
= *l*F. Construct a line from A to *l.*

Triangle AG*l *and triangle
AE*l* are both right triangles. The two triangles are congruent
because the hypoteneuse and the adjacent leg are equal. Therefore
<GA*l = < *EA*l*, such that A*l* is the angle
bisector. It is concurrent with the other two at the point *l.*