Try it

We have triangle $ABC$ with largest side $AB$. We thus also know that $\angle ACB$ is the largest angle of triangle $ABC$.

Draw a linesegment in the triangle and call the endpoints $X$ and $Y$. If $X$ and/or $Y$ are not on the perimeter of triangle $ABC$ we can make a longer linesegment by extending $XY$ until both points are.

If $XY$ is parallel to $AB$ then we have that $ABC$ and $XYC$ are like triangles. Since $|XC| \leq |AC|$, we then also have $|XY| \leq |AB|$.

If $XY$ is not parallel to $AB$, suppose that $X$ is the point closest to $AB$. Otherwise, mirror your image.

Draw a line parallel to $AB$ through $X$, intersecting triangle $ABC$ in a second point $Z$.

We have that $\angle YXZ \leq \angle CAB$.

By construction, $\angle YZX = \angle CBA$.

This gives the following:

$\angle XYZ = 180 - \angle YXZ - \angle YZX \geq 180 - \angle CAB - \angle CBA = \angle ACB$

So it follows that

$\angle XYZ \geq \angle ACB \geq \angle CAB \geq \angle YXZ$ and

$\angle XYZ \geq \angle ACB \geq \angle CBA = \angle YZX$.

So in triangle $XYZ$ side $XZ$ is the longest.

Thus we have $|XY| \leq |XZ| \leq |AB|$ (the last inequality follows from the third paragraph).

I'm not entirely sure but here is my guess:

Given any triangle with one distinct greatest side length $A$ and two shorter side lengths $B$ and $C$. Draw a line $D$ from a corner that intersects side length $B$ or $C$, say that we intersect length $C$, the length is split into two lengths and call the length that is furthest from the greatest length $E$. Now given a constant angle opposite from the greatest length $A$. We will prove that length $D$ cannot under any circumstance be greater than length $A$ as we will assume that $E > C$. In order to prove that $D < A$, we will assume that $D > A$ in which we will contradict ourselves to prove that this is not possible.

Firstly using the cosine rule we can come up with:

$A=\sqrt{B^{2}+C^{2}-2BCcos(\theta)}$ and $D=\sqrt{B^{2}+E^{2}-2BEcos(\theta)}$

As we are assuming $A < D$

$\sqrt{B^{2}+C^{2}-2BCcos(\theta)} < \sqrt{B^{2}+E^{2}-2BEcos(\theta)}$

Let $2Bcos(\theta)$ be a constant $k$ and rearrange the equation:

$B^{2}+C^{2}-Ck < B^{2}+E^{2}-Ek$

$C(C-k) < E(E-k)$ where it is obvious that if we know $C > E$, hence $C(C-k) \neq < E(E-k)$ therefore we contradict ourselves and hence $A > D$. (proved)

I hope this helped.

The simple observation that for drawing the line inside a triangle which could be greater than the greatest side must be parallel to the greatest side and hence triangle being a closed figure the required line will not be greater than the largest side. a simple but effective one .. !! :)