|  | The problem is nearly equivalent to this (point c has been moved up |de|):
      a+
       |\
       | \
       |  \
      b+   \
       | \  \
       |   \ \
       |     \\
      c+-------+e
            
It's more useful to label the edges than the vertices, so let
	A = |ab|
	B = |bc|
	X = |ce|
	G = angle aeb
We are given A, B, and G, and want to determine X.
By the law of cosines, we have
	A� = |be|� + |ae|� - 2 |be| |ae| cos(G)
	A� = (B�+X�) + ((A+B)�+X�) - 2 cos(G) sqrt((B�+X�)((A+B)�+X�))
	(B�+X�) + ((A+B)�+X�) - A� = 2 cos(G) sqrt((B�+X�)((A+B)�+X�))
Squaring both sides, and further manipulations yields...
	   4      2
	T X  + T X  + T  = 0
	 4      2      0
	T  = 4 sin�(G)
	 4
	T  = 8 B (A+B) sin�(G) - 4 A� cos�(G)
	 2
	T  = 4 B� (A+B)� sin�(G)
	 0
Use the quadratic formula, take a sqquare root, and you have:
            A�cos�G - 2B(A+B)sin�G � AcosG sqrt(A�cos�G - 4B(A+B)sin�G) 
X = sqrt( --------------------------------------------------------------- )
                                    2sin�G
Nasty, but not hideous.
Note that we've introduced some extra roots.  In the final sqrt, a negative
is easily understood as the `mirror' point on ce to the left of abc.  But
what of the � from the quadratic formula?
 | 
|  | I drew a figure much like that in .1, and then applied the rule for tangent of 
sum of two angles to the two angles with vertex at e.
I got the quadratic equation
	x^2 * Tan(G)/(B-A) + X * (1-B/(B-A)) + B * Tan(G) = 0
I did not bother to solve it, since I had gone far enough to see that the result
would indeed be nasty.  Anyway. my algebra usually has errors in it.
>Note that we've introduced some extra roots.  In the final sqrt, a negative
>is easily understood as the `mirror' point on ce to the left of abc.  But
>what of the � from the quadratic formula?
My opinion is that the problem is in general undetermined.  Visualize the 
triangle as you let x slide from 0 to infinity, keeping A and B constant.  The
angle G starts at 0, increases to some maximum, and then decreases to zero.  
For a given angle G, there will be two solutions provided the given G is below
the maximum.  At the maximum, the discriminant is zero, so the two answers are
the same.
So you could ask another question: what is the value of the angle G which 
allows you to compute a unique value for X?
 |