Given isosceles triangle ABC (AB=BC) and points D, E (AD=CE), how do I prove that
$BD+BE > AB+BC$?
This is a task for middle school, so cosine theorem cannot be used (and I am not sure that it could help anyway).
I guess (but I can be wrong) that the triangle inequality must be applied somehow to solve it, but I do not see how.
On
This may not be what you're looking for because I had to invoke calculus. But the basic idea is elementary enough. What you need to prove is that for a triangle with a fixed height and fixed base (where you're allowed to vary the position of the vertex/apex only), an isosceles triangle has the minimum possible perimeter. (Can you see why this problem is equivalent to yours?)
If you let the triangle have a height $h$ and a base $b$, then you can draw a right triangle with those sides perpendicular to each other. Let this be a "reference" triangle. Now shift the apex so that the height now intersects the base at a point $\lambda b$ from the right angle (and the remaining length of the base is $(1-\lambda b)$. Note that any such triangle formed in this fashion by translating the apex has an invariant area ($=\frac 12 bh$).
The perimeter of the triangle can be computed using Pythagoras theorem to be $P = \sqrt{h^2 + \lambda^2b^2} + \sqrt{h^2 + (1-\lambda)^2b^2} + b$. Only the initial two terms are variable (and they are dependent on $\lambda$). By setting the derivative $\frac{dP}{d\lambda}$ to zero, you can show (after a bit of algebra) that $\lambda = \frac 12$ minimises the perimeter - and this occurs when the triangle is isosceles.
There might be a non-calculus way to do it, but I can't think of one at the moment. I'll think on this further (it's after midnight where I am, so it'll have to wait for the morrow).
On
The proof is a bit simpler if we consider, instead of $BD$, its symmetric $BF$ with respect to the altitude of $ABC$. We then need to prove that $BF+BE>2BC$, or $BE-BC>BC-BF$.
Let's construct then points $E'$ and $F'$ on ray $BC$, such that $BE'=BE$ and $BF'=BF$. We have then $BE-BC=BE'-BC=E'C$ and $BC-BF=BC-BF'=F'C$, so that the inequality to be proved can be rewritten as $E'C>F'C$.
This inequality can be proved by considering triangles $FCF'$ and $ECE'$. They have $FC=EC$ and $\angle FCF'=\angle ECE'$, but on the other hand $\angle FF'C>\angle EE'C$, because $\angle FF'C>90°$ while $\angle EE'C<90°$. That entails $\angle F'FC<\angle E'EC$
It follows that the parallel $EG$ to $FF'$ is inside $\angle E'EC$ and meets $CE'$ at $G$. Hence $E'C>GC=F'C$, as it was to be proved.
On
This might be suitable for middle school. It uses very little geometry even. Draw circle with center $B$ and radius $BD$, intersecting $AB$, $CB$, in $F$, $G$, and draw circle with radius $BA$, passing through $C$ and intersecting $EB$ at $H$. Draw a line through $F$, $G$ parallel to $AE$, and join $GH$.
$H$ lies above line $FG$. For considering $D$ and $C$ as pendulums suspended from $B$, then if $D$ in rising to $F$ moves through projected horizontal distance $DA$, then since $C$ is rising more steeply, and hence is less horizontal in its movement, it must rise higher than $F$ to move through a projected horizontal distance $CE=DA$.
Hence $$EH>CG=AF$$(Even the portion of $EH$ between the parallels is greater than $CG$.)
But $$EH=EB-CB$$ and $$AF=AB-DB$$Therefore$$EB-CB>AB-DB$$making$$EB+DB>AB+CB$$
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