$$P$$ is a point in the plane of parallel lines $$l$$, $$m$$. Explain how you locate a point $$Q$$ on $$l$$ and a point $$R$$ on $$m$$ such that $$\overline{PQ}$$ and $$\overline{PR}$$ are equal in length and perpendicular to each other. Discuss only the case when $$P$$ lies between $$l$$ and $$m$$.

Draw the common perpendicular through $$P$$ meeting $$l$$ at $$L$$ and $$m$$ at $$M$$, and let $$PL = a$$, $$PM = b$$. On the same side of line $$LM$$, mark $$Q$$ on $$l$$ with $$LQ = b$$ and $$R$$ on $$m$$ with $$MR = a$$. The right triangles $$PLQ$$ and $$RMP$$ have legs $$a$$ and $$b$$, so they are congruent, giving $$PQ = PR$$ and $$\angle LPQ = \angle MRP$$. Since $$\angle MRP + \angle MPR = 90^\circ$$ and $$L, P, M$$ are collinear, $$\angle QPR = 180^\circ - (\angle LPQ + \angle MPR) = 90^\circ$$. Equivalently, $$R$$ is where the image of $$l$$ under a $$90^\circ$$ rotation about $$P$$ meets $$m$$; doing the construction on the other side of $$LM$$ gives the second (mirror-image) solution.