It's easy to prove the following 2 claims Claim 1 If 2 diagonals intersect at a vertex of the pentagon they can't be both bisectors Claim 2 If there are at least 3 inspectors, there are 2 diagonal inspectors that intersect in a vertex of the Pentagon Using the claims there are at most 2 bisectors.Example is easy to make

i think its quite trivial to prove that no. of bisector diagnols $\leq 2$ by forming equations with perimeter. Then just show a case that 2 works and we are done.

Here is an example: $$A\left(\frac{4}{3\sqrt5},\frac23\right),\,B\left(-\frac{2}{\sqrt5},1\right),\,C\left(-\frac{2}{\sqrt5},-1\right),\,D\left(\frac{4}{3\sqrt5},-\frac23\right),\,E\left(\frac{3}{\sqrt5},0\right)$$Then $AC$ and $BD$ are bisectors.