From Jim Wilson’s
Assignments: “Of a triangle,
given two vertices A and B, and the angle at the third vertex C (the angle
opposite side AB). What is the locus of the point C? See Script # 18 in
Assignment 5.

(Note: Think of this as turning a fixed angle so that its sides
always rest on the two endpoints of the segment AB. Remember that the size of
the angle is fixed but the sides can move along A and B simultaneously.)”

By exploration (with or without technology), we may arrive at a
picture like the following:

Such an exploration leads us to believe that the locus of the
vertex point, C, will be the arc of a circle. Using GSP we can develop a more elegant and convincing
locus. If we start with a segment
AB and an angle at the third vertex C as the assignment suggests, we can create
an arbitrary point, Drag, on the circle whose center is A and radius is AB. This creates an angle .

Since the sum
of the interior angles of any triangle is 180^{o}, (or a straight
angle), we can extend the initial side of the given angle whose vertex is C to
a line. Now we can add angle and the angle
whose vertex is C to discover the third angle of the triangle, .

We can now construct angle , using vertex B and side AB.

The
intersection of rays emanating from A and B not containing the segment AB is
one possible position of our vertex point C. By using the locus capabilities of GSP, we can construct the
path of C as the point Drag is moved.

It
appears that our locus is an arc of a circle. If this is the case, we should be able to find the center of
the circle. Since a circle is the
set of all points equidistant from a fixed point, O, we must have that O lies
on the perpendicular bisector of AB.
(This perpendicular bisector is the set of all points equidistant from A
and B). For construction purposes,
we can build in a special way
so that it is 90^{o}, (any acute angle will work, but 90^{o} is
easiest to construct). The angle is now the
complement of the fixed angle whose vertex is C. The complement of the angle whose vertex is C can now be
constructed at point B using AB as the initial side. The center, O, is the intersection of the perpendicular bisector
of AB and the terminal side of angle . (Similar
triangles give that this intersection lies on the perpendicular bisector of segment
AC’)

By
exploration, we discover that the locus contains the minor arc of the circle
when the angle whose vertex is C is obtuse. A semicircle results when C is a right angle.

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