Go from A to D in three equal steps Given two parallel lines $r$ and $s$, line $p$, perpendicular to both, and points $A$ and $D$ on different sides of $p$ with respect to the parallel lines, how can I prove the existence of two points, $B$ and $C$, respectively on $r$ and $s$, such that segments $AB$, $BC$ and $CD$ are congruent?
I'd prefer a constructive proof, for it to be a more general solution useful to solve this question too. (also, geometric proofs are fun)
Here are two examples I constructed in reverse (I fitted the lines to the points and not viceversa):



Let $a,x,d,y$ the distances shown in the picture and $t=\overline{AB}=\overline{BC}=\overline{CD}$

Then we have
$$a^2+x^2=t^2\tag{1}$$
$$d^2+y^2=t^2\tag{2}$$
$$(x-y)^2+e^2=t^2\tag{3}$$
If we add $a^2d^2$ to  $(3)$ and reorder the terms we get
$$x^2+a^2-2xy+y^2+b^2+e^2=t^2+a^2+d^2$$
an further
$$2xy=t^2+e^2-a^2-d^2\tag{4}$$
if we substitute by $(1)$ and $(2)$.
Squaring $(5)$ and substituting again by $(1)$ and $(2)$ gives
$$4(t^2-a^2)(t^2-d^2)=(t^2+e^2-a^2-d^2)^2\tag{5}$$
$(5)$ is an equation of degree $4$ in $t$ but quadratic in $t^2$:
$$3t^4-2t^2(a^2+d^2+e^2)+(2a^2d^2+2a^2e^2+2d^2e^2)-(a^4+d^4+e^4)=0$$
 So it can be easily solved.
The discriminant $\Delta$  of the quadratic equation satisfies
$$\frac{1}{8}\Delta=(a^2-d^2)^2+(a^2-e^2)^2+(d^2-e^2)^2\ge0$$
For the second picture we could start with the equations 
$$a^2+x^2=t^2$$
$$d^2+y^2=t^2$$
$$(x+y)^2+e^2=t^2\tag{6}$$
but we would get the same solutions because the equations differ only in the sign of $y$ in $(6)$
