Prove that there cannot exist a triangle in which the trisectors of an angle also trisect the opposite side by using proportions.
Attempt I have started the proof by way of contradiction.
Suppose we have a triangle $\Delta BAC$ with trisectors $AD$ and $A$E so that $D$ and $E$ are points lying in side $BC$. I also supposed that $D$ and $E$ trisect $BC$ such that $BD=DE=EC$. I then looked at triangle $BAE$ and observed that $AD$ bisects $\angle BAE$ and that point $D$ is the midpoint of $BE$ therefore $BD$ is both a median and an angle bisector.
From this I get stuck and do not know how to apply proportions. I know we proceed similarly when looking at $\Delta DAC$.