Dear Martin, I propose vectors, it is extremely easy.

vec(AB) = (x1-x0, y1-y0) , vec(BC) = (x2-x1, y2-y1) 

In the plane, the vectors orthogonal to vec(AB) are k(y1-y0, -(x1-x0)) , where k is a real.

With the condition AB=2BC we get k=2 or k=-2. Hence we have 2 convenient points C, C´. The coordinates of C are computed from the system of equations

•  x2-x1 = 1/2(y1-y0)
•  y2-y1 = -1/2(x1-x0)

for C´ the constants are in order -1/2 , 1/2.

Take a unit vector perpendicular to AB. This only needs the coordinates of A and B.  Multiply this unit vector by half the distance between A and B and add it to point B.

Number of isosceles triangles drawn on the longest side = infinitely large.

Let BC = 2a  be the longest side. Draw its perpendicular bisector (say LMN) through the mid point M of AB. Choose any point (say  A) on LMN and complete the isosceles triangle ABC. Since base BC of the triangle is given longest no side AB (=AC) should be larger than 2a.  

Thank you all! I appreciate your very helpful inputs! I completely forgot vectors being an option. I guess I want to look at high school maths again. :-)