TITLE: A Geometric Construction problem that's got me stumped since 1975
QUESTION [9 upvotes]: Hello, I'm a retired civil engineer from Greece. Ever since I was a student I have really liked maths. I had to solve this problem during my first year at NTUA, back in 1975! I couldn't solve it then and my question remains: what is the solution? The problem is as follows:
Let there be two lines (e1) and (e2) and a point A beyond them. Both lines and the point A are on the same plane. How can we construct a line (x1) starting from A which intersects the line (e1) on point B and the line (e2) on point D and similarly a line (x2) starting from A which intersects the line (e1) on point C and the line (e2) on point E so that BC = a cm and DE = b cm, where the lengths a,b are given.
TIA

REPLY [7 votes]: Define $p:=a/2$ and $q:=b/2$ for some slightly-cleaner arithmetic.
Let the lines meet at $O$, and complete parallelogram $APOQ$, with $P$ on the $p$-line and $Q$ on the $q$-line.

Take $\lambda$ and $\mu$ such that $|AQ|=\lambda p$ and $|AP|= \mu q$.
Traveling from $P$ in the direction of $O$ by distance $p\sqrt{1+\lambda\mu}$ gives $P'$, the midpoint of the target segment on the $p$-line; likewise, traveling from $Q$ in the direction of $O$ by distance $q\sqrt{1+\lambda\mu}$ gives midpoint $Q'$ on the $q$-line. $\square$

To see that the desired property holds, it helps to consider $p$ and $q$ to be vectors (as shown in the figure), and to take $A$ as the origin. Then one endpoint of the $p$-segment and the corresponding endpoint of the $q$ segment have position vectors
$$
P_\star := p \left(\sqrt{1+\lambda\mu} + 1\right) + \mu q \qquad\qquad Q_\star := \lambda p + q \left(\sqrt{1+\lambda\mu} - 1\right)
$$
Now, the endpoints are collinear with $A$ (our origin) if these are scalar multiples, so we need only observe
$$
\left(\sqrt{1+\lambda\mu}-1\right) P_\star = \mu Q_\star
$$
Likewise, the other pair of endpoints are collinear, as well.
But, as @JeanMarie's answer indicates, there's more to say: A second solution exists, and it arises here by reversing the directions of vectors $p$ and $q$, so that the "traveling" done is away from $O$:


Converting to a fully straightedge-and-compass construction is technically pretty straightforward; the hard part is constructing the scale factor $\sqrt{1+\lambda\mu}$, but this isn't actually "hard" (given a unit segment). I suspect there's a construction that naturally leverages the geometry of the figure without requiring too many "side computations", but I haven't given it much thought.