I am looking for an elegant solution of the following problem:
Construct $ABC$ with straightedge and compass, given $\widehat{A},r,b-c$.
By taking the lines $AB,AC$ as a skew reference system, the problem is clearly equivalent to finding a tangent to an ellipse that is parallel to a given line. As an alternative, by setting $s_a = p-a$ and so on, we get to know $r,s_a,s_b-s_c$, and since: $$ s_a\cdot s_b\cdot s_c = r^2\cdot (s_a+s_b+s_c) $$ $s_b$ and $s_c$ can be found by solving a second-degree equation.
Anyway, I am not really satisfied by these approaches. I am looking for something more in the spirit of these famous construction problems by John Horton Conway.
Ok, I found it. We may start with a circle $\Gamma$ having radius $r$ and centre $I$, then take a point $I_A$ on such circle and two points $B',C'$ on opposite sides of the tangent through $I_A$, such that $I_A B'-I_A C'=b-c$.
The tangent to $\Gamma$ through $B'$ and the tangent to $\Gamma$ through $C'$ meet at a candidate point $A'$ that lies on a line (the orange line in the picture above). On the other hand, $A$ must see $\Gamma$ under a known angle, hence the problem can be solved by simply intersecting a line and a circle.