Find the minimum length of the fence

Consider a triangle, and draw a fence.

Now, $AB = c, BC = a, CA = b$ , and also consider $DC = x$ and $EC = y$

As the fence divides the triangle into 2 equal halves, so area of $\Delta ABC$ is two times area $\Delta CDE$

So, $\frac{1}{2}$ $(a.b.sinC)$ = 2. $\frac{1}{2}$ $(x.y.sinC)$

$\Rightarrow xy = \frac{1}{2}ab$

Now, let the length of the fence is $DE =$ $l$

So, $l^2 = x^2 + y^2 - 2x.y.cosC$

$\Rightarrow l^2 = x^2 + \frac{a^2b^2}{4x^2} - ab.cosC$

$\Rightarrow l^2 = (x - \frac{ab}{2x})^2 + ab - abcosC$

$\Rightarrow l^2 = (x - \frac{ab}{2x})^2 + 2a.b.sin^2(\frac{C}{2}$ )

Also, we know, $\Delta = \frac{1}{2}a.b.sinC \Rightarrow \frac{2\Delta }{sinC} = ab$

So, the above equation becomes, $l^2 = (x - \frac{ab}{2x})^2 + 2\Delta tan\frac{C}{2}$

Taking, $C$ as the common point we get $l_{min} = \sqrt{2\Delta tan\frac{C}{2}}$

We also know $\angle A < \angle B < \angle C$ , which means $tan\frac{A}{2} < tan\frac{B}{2} < tan\frac{C}{2}$

Thus, the shortest possible fence for this triangle which will make the area of two parts equal, has the common point at A and the shortest possible length of the fence is $\sqrt{2\Delta tan\frac{A}{2}}$