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 $AC$.

Let the given intersection be at $F$.

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

Thus:

$\ds \vec {AB} + \vec {BE}$

$=$

$\ds \frac {\vec {AD} } 2$

$\ds \leadsto \ \ $

$\ds \vec {BE}$

$=$

$\ds \frac {\vec {AD} } 2 - \vec {AB}$

$\ds \leadsto \ \ $

$\ds \vec {AF}$

$=$

$\ds m \paren {\frac {\vec {AD} } 2 - \vec {BE} }$

for some $m$ such that $0 \le m \le 1$

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:

$n = \dfrac 1 3$

$\blacksquare$

1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $69$