Construct parallel through a triangle satisfying a sum condition I would like to draw, using the classical compass and rule methods, some points $D$ and $E$ given a triangle $\Delta ABC$ such that $BD + EC = DE$ and $DE$ is parallel to $BC$, as in the following picture:

It is clear that if $x$ is as in the picture, then, by Thales,
$$ \frac{x}{f(x)} = \frac{AC}{BC} \implies f(x) = \frac{BC}{AC} x,$$
and,
$$ \frac{x}{g(x)} = \frac{AC}{AB}\implies g(x) = \frac{AB}{AC}x, $$
so the condition $BD + EC = DE$ gives me
$$ x = \frac{AB(AB+AC)}{(AB+BC+CA)} $$
I know how to construct then the segment $x$ if I could use a segment whose length was $1$, but I cannot. I would be very pleased if you give me some hint or idea. Thanks in advance.
 A: Simply note that the line ($\overleftrightarrow{EF}$) must pass through the incenter ($I$) of the triangle.

A: This construction works by first assuming we have a solution segment $DE$ and then deducing that the intersection (which we call $X$ here) of the bisectors to triangle $ABC$ from vertices $A$ and $B$ lies on segment $DE,$ and from there the parallel may be constructed. The initial assumption of existence of a solution follows by continuity, on considering the various parallels to side $AB$ as they move up from just above side $AB$ to just below vertex $C.$
Construct a segment $DX$ on $DE$ having length $AD,$ and another segment $EY$ also on $DE$ having length $BE.$ Then since $AD+BE=DE$ also we have $DX+YE=DE.$ If here $X \neq Y$ there would either be an overlap or a gap on the left side of the last equation, so in fact $X=Y$ and we can call both of these $X.$
Then angle $DAX$ equals angle $DXA,$ and the latter is equal to angle $XAB$ since $DE$ is parallel to $AB.$ Similar reasoning on the other side gives the equal angles $EXB,\ XBE,\ XBA.$ We thus find that the segments $AX$ and $BX$ are angle bisectors of triangle $ABC$ at vertices $A,B,$ making $X$ constructible (it is the incenter, by the way). Then with $X$ constructed we can draw the parallel $DE$ to $AB$ passing through $X.$
Note we have shown that if the construction is possible, then the point $X$ above is the incenter. In one way it is clear by continuity there must be some such parallel, but if we simply define this parallel in terms of the incenter which we call $X$ here, we can check it determines the parallel $DE$ which has the required sum property.
A: Part I - Analysis:
Let $\displaystyle r=\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$. Observe that
\begin{align}
BD&=AB\cdot\left(1-\frac{AD}{AB}\right)=AB\cdot(1-r),\\
EC&=AC\cdot\left(1-\frac{AE}{AC}\right)=AC\cdot(1-r).
\end{align}
So if $BD+EC=DE$, then we have
\begin{align}
&&&(AB+AC)\cdot(1-r)=BD+EC=DE=r\cdot BC\\
&\Longleftrightarrow&& \frac{1-r}{r}=\frac{BC}{AB+AC}\\
&\Longleftrightarrow&& \frac{1}{r}
=1+\frac{BC}{AB+AC}
=\frac{AB+AC+BC}{AB+AC}\\
&\Longleftrightarrow&& r=\frac{AB+AC}{AB+AC+BC}.
\end{align}

Part II - Drawing:
Consider the following figure.

To find the point $E$, first draw the points $F$ and $G$ on the line $AB$ 
so that $BF=BC$ and $AG=AC$. Next, draw a parallel line to $FC$ passing through $B$, and intersect with $GC$ at $H$. Then it is clear that
$$\frac{GH}{GC}=\frac{GA+AB}{GA+AB+BF}=\frac{AB+AC}{AB+AC+BC}=r.$$
Finally, draw a parallel line to $AB$ passing through $H$, 
and intersect with $AC$ at $E$. Then such $E$ is the desired point on $AC$.
The way of finding the point $D$ on $AB$ is just draw a parallel
line to $BC$ passing through $E$, and intersect with $AB$ at $D$.
A: $$\begin{align}
BD+EC & = (BA+AC) - (DA+AE) \\
& = (BA+AC) - (BA+AC)x/AC \\
& = (BA+AC)(1 - x/AC) \\
& = (BA+AC)(AC - x)/AC
\end{align}$$
and
$$DE = BC\,x/AC$$
and you need to get
$$BD+EC = DE$$
This is equivalent to
$$(BA+AC)(AC - x) = BC \,x$$
which in turn is
$$(AC-x)/x = BC/(BA+AC)$$
$$EC/AE = BC/(BA+AC)$$
So, extend the $BC$ line past the $B$ end to some $Q$ point, so that $BQ = BA+AC$ length. Then you'll have
$$EC/AE = BC/BQ$$
Now it's enough to draw a line $m$ through $B$, parallel to $AQ$ – the intersection of $m$ and $AC$ is the point $E$ you seek.
