angle bisector and relationships between lengths

Question

in triangle $ABC$, bisector of ∠A and segment $BC$ meets in $D$. $AD+AC=BC$, and $AD+AB=CD$. if we let ∠C=x°, how much is x?

approach i tried to use $AB:AC=BD:DC$, but it was hard to find the solution.

Answer 1

I invite you to fill in the missing details of this path.

First produce $AC$ to $E$ so that $AE \cong AD$ (HP implies then that $\triangle EBC$ is isosceles). Similarly produce $AD$ to $F$ so that $AF \cong AB$. Let finally $G \in AC$, so that $AG \cong AF \cong AB$.

1. Show that the isosceles triangles $\triangle AFB$ and $\triangle AFG$ are congruent (SAS criterion).
2. Prove that $\triangle AFE \cong \triangle ADB$ (again SAS).
3. Conclude that $\triangle FEB$ and $\triangle GDB$ are isoceles and congruent. As a result $\triangle FGB$ is equilateral.
4. Using the fact that $\triangle EBC$ is isoceles we can write therefore $$2(2\measuredangle GBC + 60^\circ) + x = 180^\circ.$$
5. Exterior Angle Theorem on $\triangle GBC$ yields $$\measuredangle GBC + x = 30^\circ.$$
6. Deduce, from 4. and 5., $$\boxed{x = 20^\circ}.$$