Since $B'$ is on the bisector of angle $\angle B$, it is equidistant from lines $AB$ and $BC$. Construct circle $\omega_{B}$ with centre $B'$ which is tangent to $AC'$. The line $BC$ is tanget to it. Analogously line $BC$ is tangent to circle $\omega_{C}$ with centre at $C'$ which is tangent to $AC$. Then common tangent to circles $\omega_{B}$ and $\omega_{C}$ will be the line $BC$. We know that lines $AB'$, $AC'$ and $BC$, so we can find the vertices of the triangle