Line from Bisector of Side of Parallelogram to Vertex Trisects Diagonal

Theorem
Let $ABCD$ be a parallelogram.

Let $E$ be the midpoint of $AD$.

Then the point at which the line $BE$ meets $AC$ trisects $AC$.

Proof

 * TrisectorOfDiagonalOfParallelogram.png

Let the given intersection be at $F$.

We have that $E$ is the midpoint of $AD$.

Thus:

Since $\vec {AB} + \vec {BC}$ we have $\vec {AF} = n \paren {\vec {AB} + \vec {BC} }$ where $0 \le n \le 1$.

But:
 * $\vec {AB} + \vec {BF} = \vec {AF}$

That is:
 * $\vec {AB} + m \paren {\dfrac {\vec {BC} } 2 - \vec {AB} } = n \paren {\vec {AB} + \vec {BC} }$

That is:
 * $\paren {1 - m - n} \vec {AB} + \paren {\dfrac m 2 - n} \vec {AB} = 0$

These have a simultaneous solution because $\vec {AB}$ and $\vec {AD}$ are neither coincident nor parallel.

So:
 * $1 - m - n = 0, \dfrac m 2 - n = 0 \implies m = 2 n$

Hence $3n = 1$ and so:
 * $n = \dfrac 1 3$