Conference dypss1::brain_bogglers

Why isn't this solved yet?   ;-)

 Problem: Given a triangle with two equal angle bisectors then prove
 the angles are equal.

The Diagram:

      |                 \       
      |                  \'
      |                  '\A   
      |                 '  \           <EBC = x
      |                '    \          <ABC = 2x
      |               '      \           BE = 1
      |              '        \          DC = 1
      |             '          \
      |            '            \
      |           '              \
      |        K '                \
     H+---------+-----------------.\E
      |        '               .    \
      |       '             .        \
      |    D +           .            \
      |     '|   +    .                \
      |    ' |     .  +                 \
      |   '  |  .          +             \
      |  '  .|                    +       \
      | '.   |                          +  \
    --+------+------------------------------+--
     B|      J                              C
      |<-d-->|

Construction:

 1. Label the origin (point 0,0) B.  
 2. From B draw line BE one unit long at angle x to point E.
 3. From B draw line AB at angle 2x. Point A is on this line somewhere
    (but point A is indefinite).
 4. From point E draw a line parallel to the X axis back to the Y axis
    at point H. Line HE intersects line AB at point K.
 5. Pick a point D somewhere on line segment BK.
 6. From point D draw a line one unit long to intersect the X axis at
    point C (to the right of E).
 7. From point D draw a line perpendicular to the X axis at point J.
    Call the length of line segment BJ "d".
    
Observations:

 1. Line BE has a slope of tan(x).
 2. Line AB has a slope of tan(2x).
 3. Point E is at coordinate (cos(x), sin(x)).
 4. Point D is at coordinate (d, d tan(2x)).
 5. The length of segment JC = sqrt(1 - (d tan(2x))^2).
 
                                       - sin(x)
 6. The slope of line CE (and AC) = ---------------
                                    d + JC - cos(x)

Conclusions:

 1. The slope of line CE may be derived from angle "x" and distance "d".

 2. <DCB ("y") bisects <ACB ("2y") when d equals a value such that
 
    tan(y)  = (d tan(2x)) / sqrt(1 - (d tan(2x))^2)
     AND
    tan(2y) = sin(x) / (d + JC - cos(x))
    
    The unique(1) solution for these equations is
    
        d = sin(x)/tan(2x)   or (d tan(2x)) = sin(x)
        
    tan(y) = sin(x) / sqrt(1 - sin(x)^2) = sin(x)/cos(x) = tan(x)
        
    tan(2y) = sin(x) / ( sin(x)/tan(2x) + cos(x) - cos(x) )
            = sin(x) / ( sin(x)/tan(2x) )
            = tan(2x)

 3. When point D is at point K then CD and BE are equal angle bisectors.
    Further, angle x = angle y and triangle ABC is isosceles.

 4. When point D is not at point K then line CD does not bisect <ACB
    and therefore it is not an angle bisector.

So, does this prove it or am I restating the obvious?
    
Chuck
(who presses the lever a lot just to get pellets)

1. Where point D lies along line segment AB is constrained by two 
   factors:

   a. <DCB must be less than 45 degrees.  Otherwise 2*<DCB is more
      than 90 degrees and <DCB as an angle bisector does not make
      sense.  This limits distance (d tan(2x)) < sin(45).

   b. BC must be greater than cos(x). Otherwise the slope of AC
      goes positive and the triangle ABC loses its form. This is
      only a problem when x > 30 degrees or so.