Impossible to have a triangle where trisectors of an angle trisect the opposite side

Question

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 $\Delta DAC$.

Answer 1

It is enough to show (through the sine theorem, for instance) that in the following configuration the red segments are longer than the blue ones:

Answer 2

You have a $\triangle{ABC}$. Let us say that side $BC$ is trisected by points $D, E$ such as $BD=DE=EC$. Note that $A_{\triangle{ACE}}=A_{\triangle{AED}}=A_{\triangle{ADB}}$ (equal bases, same height). Using area of a triangle formula this leads to $AC=AD$, $AE=AB$. But then $AE$ is a median and height of $\triangle{ACD}$ and $AD$ is a median and height of $\triangle{AED}$. We have two perpediculars from one point $A$ to side $BC$ that go to two distinct points. This is impossible.