We are given AB, obtuse ∠BAC (but we do not have point C.  I placed
C? at an arbitrary place along that side of the given
angle, and we are given the length of BC-AC, the green line segment below.
We want to draw ߡABC.



We extend CA so that AD = BC-AC 

 

We draw DB

   

We want to find the point on CD such that BD is the base
of an isosceles triangle.  We do this by constructing the
perpendicular bisector of BD, which will be the base of the
isosceles triangle.  We know that the perpendicular bisector of
the base of an isosceles triangle passes through its vertex.
That perpendicular bisector is the blue line:



The point where the blue line intersects CD is the desired
point C.  We draw the line from that point to B:



ߡCDB is isosceles, so BC = DC, CD - (BC-AC) = BC - (BC-AC) = AC.

So we have drawn ߡABC.

Edwin