Question

Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (you may assume that the trapezium is not a parallelogram).

Answer 1

Let $O$ be the origin of reference. Let the position vectors of $A$ and $B$ be $\vec{a}$ and $\vec{b}$ respectively.

Since, $BC\Vert OA,\overrightarrow{BC}=\alpha \overrightarrow{OA}=\alpha \overrightarrow{a}$ for some constant $\alpha$,
Equation of $OC$ is $\overrightarrow{r}=t(\overrightarrow{b}+\alpha \overrightarrow{a})$ and
equation of $AB$ is $\overrightarrow{r}=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a})$
Let $P$ be the point of intersection of $OC$ and $AB$. Then, at point $P,t(\overrightarrow{b}+\alpha \overrightarrow{a})=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a})$ for some values of $t$ and $\lambda$.
$\Rightarrow (t\alpha -1+\lambda )\overrightarrow{a}=(\lambda -t)\overrightarrow{b}<br/>$
Since, $\overrightarrow{a}$ and $\overrightarrow{b}$ are non-parallel vectors, we must have
$t\alpha -1+\lambda =0$ and $\lambda =t$
$\Rightarrow t=1/(\alpha +1)$
Thus, position vector of $P$ is
$\overrightarrow{r_1}=\frac{1}{\alpha +1}(\vec{b}+\alpha \overrightarrow{a})$
Equation of $MN$ is
$\overrightarrow{r}=\frac{1}{2}\overrightarrow{a}+k\left[\overrightarrow{b}+\frac{1}{2}(\alpha -1)\overrightarrow{a}\right].....$(i)
For $k=1/(\alpha +1)\{$ which is the coefficient of $\overrightarrow{b}$ in $\overrightarrow{r_1}\}$, we get
$\overrightarrow{r}=\frac{1}{2}\overrightarrow{a}+\frac{1}{\alpha +1}\left[\overrightarrow{b}+\frac{1}{2}(\alpha -1)\overrightarrow{a}\right]$
$=\frac{1}{(\alpha +1)}\overrightarrow{b}+\frac{1}{2}(\alpha -1)\cdot \frac{1}{\alpha +1}\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}$
$=\frac{1}{(\alpha +1)}\overrightarrow{b}+\frac{1}{2(\alpha +1)}(\alpha -1+\alpha +1)\overrightarrow{a}$
$=\frac{1}{(\alpha +1)}(\overrightarrow{b}+\alpha \overrightarrow{a})={\overrightarrow{r}}_1$
$\Rightarrow P$ lies on $MN$.