Loci of Equi-angular Points

Given a line segment AB, the loci of points P such that the angle 
APB has a constant value are circular arcs that pass through the 
points A and B.  As a special case of this we have the well-known
fact that if APB is a right angle, then the locus of points is the 
circle with the diameter AB.  In general, if AB is any chord of a 
circle, the angle APB for any point P on the circle is constant
except at the singular points A and B themselves.  (Note that if the
chord is not a diameter of the circle, the constant angle will be
different on either side of the chord.)  

This is equivalent to the fact that for any three points A, B, and 
C on a circle, the angle ABC depends only on the distance from A to 
C. Of course, if O is the center of the circle, it follows that the 
angle AOC depends only on the angle ABC (and vice versa).  Another 
interesting equivalent statement is:  If a particle travels in a 
circular path at constant speed, then its angular velocity about 
any fixed point on the circle is constant.

For a simple proof of the equi-angular property of circles and chords,
consider the figure below.

            

We've labeled the angles taking advantage of the fact that the triangles
with one vertex at the central point O and the other two vertices on
the circle are isosceles.  Since the sum of the interior angles of a
triangle is pi, we see immediately from the triangle ABP that

                2a + 2b + 2c  =  pi

and therefore
                                       pi
         angle APB   =    b + c   =   ---  -  a
                                       2

which is constant for every point P.  Strictly speaking we've only
shonw this when ABP contains the point O, but it's not hard to show
that it also applies even if O is outside ABP, as long as P is
on the larger of the two arcs of the circle cut by the segment AB.

On the other hand, for point such as Q, on the the smaller of the
two arcs cut by AB, we can examine the triangles OAQ and OBQ to
give the relations

                    2a + 2d + f  =  pi
 
                    2a + 2e + g  =  pi

Adding these gives

               2(2a+d+e) + (f+g)  =  2pi

Also, from the triangle AOB we have 2a + (f+g) = pi, so we can
substitute for (f+g) into the preceding equation to give

                                        pi
         angle AQB   =    2a+d+e   =   ---  +  a
                                        2

which again is constant for all points Q on the smaller arc.  Notice
that the line tangent to the circle at A makes an angle of pi/2 with
the line OA, and so it makes the angles  pi/2 + a  and  pi/2 - a 
with the line AB.  This stands to reason, because the lines AP and
AQ approach this tangent line (from opposite sides) as P and Q 
approach A.

A trigonometric proof of this equi-angular property in the xy plane 
goes like this:  Let T denote the point (0,1) at the top of a unit 
circle, and let P denote another point on the circle with coordinates 
(x,y).  If 'a' is the clockwise angle between the horizontal and the 
line TP, then the equation of the line is simply 

                       y = 1 - x tan(a)

Inserting this into the circle's equation x^2 + y^2 = 1 and solving for
x gives 
                 x  =  2tan(a)/(1+tan^2(a))  =  sin(2a)

and so y = cos(2a).  This shows that the angular velocity about the 
point T is always exactly 1/2 the angular velocity about the center of 
the circle.  Thus, when P sweeps out a given angle about the point T, 
it sweeps out exactly twice that angle about the center, regardless of 
where on the circle point P happens to be relative to T.  This proves
that the length of the chord is fixed for a given swept angle about T.

One immediate conseqence of this is the fact that if a quadrilateral is
inscribed in a circle, and we construct the bisectors of the angles at
the vertices, the intersections of these bisector lines with the circle
form a rectangle.  This is illustrated for the quadrilateral shown in
red inscribed in the circle below.

                

The dotted red lines bisect the interior angles of the quadrilateral, 
and the points where these lines meet the circle comprise the vertices 
of the green rectangle.  To prove that it is, in fact, a rectangle,
consider the figure below.

                

The blue lines are radii from the center O of the circle to the 
vertices of the original quadrilateral.  Now, as the red dotted line
through vertex D rotates at constant angular speed from the line DA 
to the line DC, the green line OP from the center of the circle to the 
point of intersection of the red dotted line with the circle rotates
at twice the angular speed.  Hence the line DP bisects the angle ADC
precisely when the radii OP bisects the angle AOC.  Likewise the 
bisector of the interior angle ABC strikes the circle at the same 
points as does a radii from the center bisecting the angles AOC on
the in the other direction.  Similarly the other two points of interest
fall along the radii from the center that bisect the angle BOD in
both directions.  Hence each diagonal of the resulting figure is a
diameter of the circle, so the figure is a rectangle.