Triangles PBQ and RBQ are congruent by the Side-Angle-Side Congruence Theorem since ÐPQB=ÐRQB (right angles), PQ=RQ and QB=QB.

Now for trisection, we must show that triangles RBQ and RBX are congruent. ÐRXB=ÐRQB (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 ÐPBQ, ÐQBR and ÐRBC are equal, and we have trisected the angle ÐABC.