$$P$$ is a point not lying on a line $$l$$. Explain how you construct an equilateral triangle $$PQR$$ such that $$\overline{QR} \subset l$$.

Drop the perpendicular $$PM$$ from $$P$$ to $$l$$. At $$P$$ construct an angle of $$60^\circ$$ (using an equilateral triangle on $$PM$$) and bisect it, so as to draw rays making $$30^\circ$$ with $$PM$$ on either side of it, and let these rays meet $$l$$ at $$Q$$ and $$R$$. The right triangles $$PMQ$$ and $$PMR$$ are congruent (common side $$PM$$, right angles at $$M$$, equal angles of $$30^\circ$$ at $$P$$), so $$PQ = PR$$. Also $$\angle QPR = 30^\circ + 30^\circ = 60^\circ$$, so the isosceles triangle $$PQR$$ is equilateral, and its side $$\overline{QR}$$ lies on $$l$$.