Thomas sent us a very clear explanation of his solution.

Construct a perpendicular from $R$ to a point $X$ on $BC$.

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

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