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, $B C \| O A, \overrightarrow{ B C }=\alpha \overrightarrow{ O A }=\alpha \overrightarrow{ a }$ for some constant $\alpha$,
Equation of $O C$ is $\overrightarrow{ r }=t(\overrightarrow{ b }+\alpha \overrightarrow{ a })$ and
equation of $A B$ is $\overrightarrow{ r }=\overrightarrow{ a }+\lambda(\overrightarrow{ b }-\overrightarrow{ a })$
Let $P$ be the point of intersection of $O C$ and $A B$. 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 }
$
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 $M N$ is
$\overrightarrow{ r }=\frac{1}{2} \overrightarrow{ a }+k\left[\overrightarrow{ b }+\frac{1}{2}(\alpha-1) \overrightarrow{ a }\right] .....$(i)
For $k=1 /(\alpha+1)\left\{\right.$ which is the coefficient of $\overrightarrow{ b }$ in $\left.\overrightarrow{ r _{1}}\right\}$, 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 $M N$.