A Geometric Construction problem that's got me stumped since 1975 
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
 A: 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.
A: Let us consider the intersection of lines $O:=(e_1) \cap (e_2)$ as the origin, with line $(e_1)$ taken as $x$ axis, and line $(e_2)$ oriented by a unit vector $\vec{u_2}=(\cos\theta ,\sin \theta)$.
The pencil of lines issued from $A=(u,v)$ defines a natural projectivity: $$M \in Ox=(e_1) \ \to \ M' \in (e_2)$$
If $x$ is the abscissa of $M$, if $s$ is the abscissa of $M'$ on $(s_2)$ (meaning that $\vec{OM}=s \vec{u_2}$) a basic property of projectivities is the homographic correspondence:
$$s=h(x)=\frac{px+q}{x+r}\tag{1}$$
Particular correspondences $x=0 \leftrightarrow s=0,  \ \ \ \ \ x=1 \leftrightarrow s=\dfrac{1}{\cos \theta}, \ \ \ \ \ x=u \leftrightarrow s=\dfrac{u}{\cos \theta}$
show that relationship (1) can be written:
$$s=\dfrac{vx}{x \sin \theta +k}\tag{2}$$
where $k:=v \cos \theta - u \sin \theta.$
Remark: $k$ can be interpreted as the (signed) distance from $A$ to line $(e_2)$.
Now, let us take into account the existence of two pairs $M=B,M'=D$ and $M=C,M'=E$ with distances
$$\begin{cases}x_2−x_1&=&a&\ (i),\\ s_2−s_1&=&b& \ (ii) \end{cases}.\tag{3}$$
Taking into account (2) in (3 ii) gives:
$$
\dfrac{(x_2-x_1)kv}{(x_2 \sin \theta+k)(x_1 \sin \theta+k)}=b$$
Otherwise said, using (3 i):
$$
\dfrac{akv}{(x_2 \sin \theta+k)((x_2-a) \sin \theta+k)}=b$$
which is equivalent to :
$$(x_2 \sigma+k)((x_2-a) \sigma+k)=p \ \ \text{where} \ \ \begin{cases}\sigma&:=&\sin \theta \\ p&:=&kv\frac{a}{b}\end{cases}\tag{4}$$
(4) is a quadratic equation in variable $x_2$ with two solutions:
$$x_2,x'_2=\dfrac{1}{2 \sigma}\left(a\sigma-2k \pm \sqrt{\Delta}\right) \ \text{with} \ \Delta:=a^2\sigma^2+4p\tag{5}$$
Having two possible abscissas $x_2,x'_2$ we get, using (3 i), two possible values for $x_1, x'_1$ ; then two solutions for $s_2,s'_2$ using relationship (2) and finally values $s_1, s'_1$.
But, as one can see on the figure below, the second solution (corresponding to negative sign in (5)) is not exactly what the asker wants...
Of course, the existence of real solution(s) rests on the positivity of $\Delta$, which is always true in the case of a figure like the one given by the author of the question (but some more analysis is necessary).

Fig. 1: Case $a=2, b=1$ with $A(u,v)=(2 ; 2.5)$ and $\theta=\pi/5$.
It remains to provide an edge and compass construction of points $B,C,D,E$ based on formula (5).
Remark: some cases of relative positions of $M$ with respect to $(e_1)$ and $(e_2)$ and values of $a$ and $b$ give the impossibility of solutions. This is the case when $A$ is situated on the angle bissectors of $(e_1)$ and $(e_2)$.
A: I made a diagram showing my idea for perpendicular $x,y$  axes. I'm not yet sure what is a convenient length to mark the axes.  Given the fixed (positive) lengths $\alpha, \beta$  we can demand $ \alpha = 1$   or maybe $\frac{\alpha + \beta}{2}  = 1 $  I did indicate the points $x_1,x_2,x_3,x_4$  on my $x$ axis. The length "along"  one of the lines $e_1, e_2$  will now be expressed as $\sqrt{1+m^2}$  times the length along the $x$ axis.

I guess I'll put calculations below the diagram.. The point $A(\gamma, \delta)$  has a line to $B(x_1, mx_1).$  This must have the same slope as $A$ to $D(x_3, -mx_3).$  I get
$$  \frac{mx_1 - \delta}{x_1 - \gamma} =  \frac{mx_3 + \delta}{-x_3 + \gamma}  $$
Next, for the line $ACE$  I get
$$  \frac{mx_2 - \delta}{x_2 - \gamma} =  \frac{mx_4 + \delta}{-x_4 + \gamma}  $$
These are all expressed in Mobius transformations, which can readily be inverted
I did not draw in lengths $\alpha, \beta$ yet. From length $BC = \alpha$  I get
$$  |x_2 - x_1|  = \frac{\alpha}{\sqrt{1+m^2}}  $$
From length $DE = \beta$  I get
$$  |x_3 - x_4|  = \frac{\beta}{\sqrt{1+m^2}}  $$
Sunday 12:00 Pacific time. By inverting some Mobius transformations,
$$  x_1 = \frac{(\delta + m \gamma) x_3}{2mx_3 + (\delta - m \gamma)} $$
$$  x_2 = \frac{(\delta + m \gamma) x_4}{2mx_4 + (\delta - m \gamma)} $$
I put in extra parentheses to emphasize some hypotheses, namely $m \neq 0, \; \; $ $ \delta - m \gamma \neq 0, \; \; $ $ \delta + m \gamma \neq 0, \; \; $
I reached this:
$$  |2mx_3 + \delta - m \gamma| |2mx_4 + \delta - m \gamma| = \left( \frac{\beta}{\alpha}  \right) \left(  \delta^2 - m^2 \gamma^2  \right). $$
Together with
$  |x_3 - x_4|  = \frac{\beta}{\sqrt{1+m^2}}  $  this leads us to a finite number of $(x_3, x_4)$  pairs; well, we hope so. Note that, in my diagram, I depict $\left(  \delta^2 - m^2 \gamma^2  \right)  > 0 . \; \;  $  If we imagine a new $x_3x_4$  plane, without the absolute value signs we have  a line and a hyperbola. We expect that, if some  more inequalities are met, there is an intersection.
