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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 ?

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Here is a quick intuitive argument that $|DE|<|AC|$. This certainly holds at least sometimes! If we can get to $|DE|> |AC|$, then on the way we can get to $|DE|=|AC|$. But then $ACDE$ is a parallelogram, so the angles at $A$ and $C$ add up to $180^\circ$, ridiculous. – André Nicolas Jan 30 '12 at 22:56
up vote 2 down vote accepted

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 $DE<AC$.

enter image description here

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.)

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(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.

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Yeah, I hoped to find an elementary proof before getting to this extent ;) I didn't know you could use Mathematica to do that kind of formal manipulation, it must be very helpful and time-saving! – Klaus Jan 30 '12 at 23:17
@Klaus: I'm still thinking about it and hoping for a more geometric proof... this didn't feel as bad as cracking walnuts with a steamroller (that would probably have been a proof with coordinates), but it didn't feel elegant. Using Mathematica meant that I could type in the raw formulas from the Law of Cosines with the appropriate inequality symbol between them, tell it to simplify, and get the expression I put in my answer as the result, rather than having to do the algebraic work myself. – Isaac Jan 30 '12 at 23:19
A dream comes true, I'll definitely have to learn that! Thanks for the idea. – Klaus Jan 30 '12 at 23:21

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