Almost the Intercept Theorem Consider the following figure, with $|\text{AD}|=|\text{CE}|$.

If $|\text{AB}|=|\text{CB}|$, then $\text{AC}$ is parallell to $\text{DE}$ and $|\text{DE}|=\frac{|\text{AB}|}{|\text{AD}|}|\text{AC}|\le |AC|$, by the Intercept Theorem.
If we don't suppose anything, it seems true that we still always have $|\text{DE}|\le |\text{AC}|$. Does this result have a name ? How might it be proved ?
 A: (This feels like cracking walnuts with a sledgehammer...) From the Law of Cosines applied about $\angle B$ to find the lengths $DE$ and $AC$ in terms of $AB$, $CB$, and $d=AD=CE$ (using Mathematica), $DE\le AC\Leftrightarrow$
$$(AB+CB-d)d(\cos B-1)\le 0.$$
Since $d>0$, $d<AB$, and $d<CB$, the first two terms in the product are positive.  $\cos B\le 1$, so $\cos B-1\le 0$, so the inequality is true.
A: Here's an elementary geometry proof.  The left picture is the quadrilateral of relevance in your picture.  You want to show that given that $AD\cong CE$ and $\angle DAC+\angle ECA<180^\circ$ (since they are two of the three angles of $\triangle ABC$), we have $DE<AC$.

Since $\angle DAC+\angle ECA<180^\circ$, we can construct exterior to the quadrilateral $ACDE$ a segment $AF$ (in red) such that $AF$ is congruent and parallel to $CE$.  Now ACEF is a parallelogram since it has one pair of congruent and parallel opposite sides, so $AC\cong FE$, and it suffices to show that $DE<FE$.  Consider the triangle $\triangle DEF$ (in blue).  Since the angle opposite the greater side is greater, it suffices to show that $\angle FDE>\angle DFE$.  But this is easy since
$$
{\small \angle FDE=\angle FDA+\angle ADE=\angle AFD+\angle ADE=(\angle AFE+\angle DFE)+\angle ADE,}
$$
which is clearly greater than $\angle DFE$, as desired.  (Note that in the second step, we used that $\angle FDA=\angle AFD$ since $\triangle ADF$ is isosceles.)
