[Math] Prove the Median of a Trapezoid Bisects Both Diagonals

Question

I am trying to prove that the median of a trapezoid bisects both of the trapezoid's diagonals using basic vector operations (addition, subtraction, etc.). I have tried to do this by labeling the two parallel sides of the trapezoid as $\mathbf{A}$ and $k\mathbf{A}$, where $k$ is some constant. I then labeled one of the non-parallel sides as $\mathbf{B}$ and used vector addition to obtain a formula for the fourth side. Does anyone have a solution using this method? Thanks!

Answer 1

Using your set-up, let the parallel edges be $OC=k\underline a$ and $AB=\underline a$, and let $OS=\underline b$.

Let the median intersect the diagonal $OB$ at $D$. The idea is to write the vector $OD$ in two different ways and make them equal. We have $$OD=\lambda(\underline a+\underline b)=\tfrac {1}{2}\underline b+\mu\underline a$$

Now since $\underline a$ and $\underline b$ are not parallel, we have $$\lambda=\mu=\tfrac {1}{2}$$

Hence diagonal $OB$ is bisected, and similarly with the other diagonal