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

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 $$\Delta DAC$$.

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: 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.