Triangles $\mathrm{PBQ}$ and $\mathrm{RBQ}$ are congruent by the Side-Angle-Side Congruence Theorem since $\angle \mathrm{PQB}=\angle \mathrm{RQB}$ (right angles), $\mathrm{PQ}=\mathrm{RQ}$ and $\mathrm{QB}=\mathrm{QB}$.

Now for trisection, we must show that triangles $\mathrm{RBQ}$ and $\mathrm{RBX}$ are congruent. $\angle \mathrm{RXB}=\angle \mathrm{RQB}$ (right angles again), $\mathrm{RQ}=\mathrm{RX}$ (since it's the width of the horizontal leg of the carpenter's square), and $\mathrm{RB}=\mathrm{RB}$, so these two triangles are congruent, and so all three triangles are congruent. So the three angles $\angle \mathrm{PBQ}$, $\angle \mathrm{QBR}$ and $\angle \mathrm{RBC}$ are equal, and we have trisected the angle $\angle \mathrm{ABC}$.