Do Equal Bisectors Imply Isosceles?

It seems intuitively obvious that if the diagonals to the opposite 
sides of a triangle are equal in length, then the triangle is 
isosceles.  However, this proposition is not totally trivial to
prove, and in fact it's arguably false without some additional 
restrictions.  For example, consider a triangle ABC in a cartesian 
coordinate system, taking the unit segment from (0,0) to (1,0) on 
the x axis as the base of the triangle, we have


                 y|      C
                  |      /\
                  |     /   \
                  |    /      \ D
                  |  E/         \
                  |  /            \
                  | /  2a       2b  \
              ____|/__________________\B_____
                 A|0                  1      x

Let 2a and 2b denote the angles at verticies A and B respectively.
The bisector of angle A strikes the line BC at point D, and the
bisector of angle B strikes the line AC at point E.

Notice that if we set a = 60 degrees  and  b = 165.867 degrees the
two bisectors are both of length 0.9003 units, and yet the triangle
is NOT isosceles.  Of course, in this case we have turned the
triangle "inside out", so to speak, and the clockwise ordering of
the points around the triangle is ACB instead of ABC.  However,
nothing in the statement of the problem (or in Euclid's Elements)
excludes this range of possibilities, so the proposition is 
strictly false.

In general, the angle bisectors strike the opposite sides at the 
points [Lcos(a),Lsin(a)] and [1-Lcos(b),Lsin(b)] where L is the 
length of the bisectors.  Thus we have

 tan(2a) = Lsin(b)/(1-Lcos(b))     tan(2b) = Lsin(a)/(1-Lcos(a))

Equating the values of 1/L given by these two equations gives

     cos(a) + sin(a)/tan(2b)  =  cos(b) + sin(b)/tan(2a)

which, in terms of just the parameters A=cos(a) and B=cos(b) can
be written as
                                          ______________
(A-B)[1 - A^2 - B^2 + AB + 2(AB)^2 - 2AB /(1-A^2)(1-B^2) ] = 0

Obviously A=B is a solution, corresponding to the isosceles cases, 
but this still leaves the other possible roots of the right hand
factor, i.e., values of A,B such that
                                      ______________
 1 - A^2 - B^2 + AB + 2(AB)^2 =  2AB /(1-A^2)(1-B^2)         (1)

Since A and B are cosines they must both be in the range -1
to +1, but this leaves infinitely many non-isosceles solutions,
such as the one noted above.  However, it can be shown that all
the real non-trivial roots of this equation correspond to cases 
where either angle a or b exceeds 90 degrees, so they represent 
"inside-out" solutions.  If we restrict ourselves to just cases 
where point "C" is above the x axis, then the only solutions are 
with A=B, which represent isosceles triangles.

To formally prove that none of the other solutions of (1) is 
"proper", notice that we can always orient the triangle so that 
point C is above the x axis, which implies that the half-angles 
"a" and "b" are both in the range 0 to pi/2.  Since the cosines 
of angles in this range are strictly positive, we know that 
A,B > 0.  Now observe that the above equation can be written 
in the form

     /  ______________________     \2
    (  /1 - A^2 - B^2 + (AB)^2 - AB )  +  AB  =  0         (2)
     \                             /

which is clearly impossible if A and B are both positive.

Incidentally, it's certainly true that the non-isosceles 
"solutions" of this problem don't satisfy the _intended_ 
conditions.  The formal ambiguity (if there is any) hinges 
on the question of whether the term "bisector of the angle 
ABC" is defined well enough to distinguish between the two 
"principal bisectors" of the lines AB and BC.  We know the 
"bisector" must pass through point B, and the angles between 
the bisector and the lines AB and BC must be equal, but these 
conditions are satisfied by both of the principal bisectors.  

When we refer to _the_ bisector of angle ABC it's clearly 
our intention to specify a line that passes "between" the 
points A and C.  Unfortunately many historical treatments 
of plane geometry (including Euclid's) neglected to define 
the notion of "betweeness", let alone impose this condition 
on the definition of "bisector".  As a result, on a purely 
formal basis (as with crunching out answers via trig formulas) 
it's possible for some unintended "solutions" to appear.

By the way, the quantity under the square root in (2) is the 
product of 1-A^2 and 1-B^2, and since A and B are just the 
cosines of a and b, these terms are just the squares of the 
sines.  Thus the above equation can be written in the nice 
form
     /                            \2
    ( cos(a)cos(b) +- sin(a)sin(b) )  +  cos(a)cos(b)  =  0
     \                            /

which, using the addition rule for cosines, reduces to the 
even nicer form

             cos^2(a+-b) + cos(a)cos(b) = 0

Together with the "isosceles factor", cos(a)-cos(b), this 
completely describes all sets of three co-planar lines L1,L2,L3 
such that the distance from the intersection of L1,L2 along an 
angle-bisector of those lines to the line L3 equals the distance 
from the intersection of L2,L3 along an angle-bisector of those 
lines to the line L1.