Construction of a triangle with ruler and compass

Question

Triangle constructions with ruler and compass can be very difficult. I have the following problem. To construct a triangle given: the difference of side lengths b-c, angle A and angle B.

Answer 1

It is not that difficult to construct a triangle, given the angles at the vertices $A$ and $B$, and the measure of $b-c$. Without loss of generality, we assume that $b\gt c$. The case where $b = c\space$ is easy and, therefore, we do not discuss that in our answer. For brevity, we let the angles at the vertices $A$ and $B$ equal $\space\omega\space$ and $\space\beta\space$ respectively

We start the construction by drawing a line segment $PQ$, which is longer than $b-c$. Then, we mark points $B$ and $D$ so that $BD=b-c\space$ as shown in the diagram. Now, we need to construct the segment $BR$ so that $\measuredangle PBR=\beta$. After drawing the line $BS$ to make $\measuredangle SBD = \omega$, we construct its angle bisector $BT$. Next, the perpendicular from $D$ to $BT$ is dropped and extended to meet $BR$ at $C$. Drawing the line parallel to $BS$ through $C$ to cut $PQ$ at $A$ completes our construction.

Devising a construction is one thing and proving it is another. We would like to leave the job of compiling a sound proof of the above described construction to OP.

We add the following diagram at OP's request to chase angles. It clearly shows that $\triangle ADC$ is isosceles.