I'm no good in geometry so I will just convert it into an Algebraic problem by defining a coordinate system.
Without loss of generality we can assume that the length of AB is 2.
Place the coordinate system such that
A = (-1 , 0)
B = (1 , 0)
The length of AC is just 2 sin(beta)/sin(gamma) by the law of sines and likewise the length of BC is 2 sin(alpha)/sin(gamma).
The equation of the line AC is
(x,y) = (-1 , 0) + t (cos(alpha), sin(alpha))
By knowing the length of AC we get that
C = (-1 + 2 cos(alpha) sin(beta) /sin(gamma) , 2 sin(alpha) sin(beta) /sin(gamma))
By Thales's theorem B1 lies on the circle
x^2 + y^2 = 1
and it lies on the line AC so pluging the line in this circle equation solving the equation for t gives
t = 2 cos(alpha)
which means
B1 = (-1 + 2 cos^2(alpha) , 2 sin(alpha) cos(alpha))
The equation for the line BC is
(x,y) = (1 , 0) + t (-cos(beta), sin(beta))
again the point A1 will lie both on this line and the circle
x^2 + y^2 = 1
so this time
t = 2 cos(beta)
and
A1 = (1 - 2 cos^2(beta), 2 sin(beta) cos(beta))
So all the relevant points are known.
A = (-1 , 0)
B = (1 , 0)
C = (-1 + 2 cos(alpha) sin(beta) /sin(gamma) , 2 sin(alpha) sin(beta) /sin(gamma))
A1 = (1 - 2 cos^2(beta), 2 sin(beta) cos(beta))
B1 = (-1 + 2 cos^2(alpha) , 2 sin(alpha) cos(alpha))
The lengths between any of those points can now be calculated but they almost all have long and ugly expressions. I can't simplify that shit but if you put random angles alpha and beta and then set gamma = 180°- alpha - beta then it works out numerically.
