Prove,by vector method,that the point of intersection of the diagonals of the trapezium lies on the line passing through the mid-points of the parallel sides.
My Attempt:
Let the trapezium be $OABC$ and that the O is a origin and the position vectors of $A,B,C$ be $\vec{a},\vec{b},\vec{c}$.Then the equation of $OB$ diagonal is $\vec{r}=\vec{0}+\lambda \vec{b}…………….(1)$
And equation of $AC$ diagonal is $\vec{r}=\vec{a}+\mu(\vec{c}-\vec{a})…….(2)$
And the equation of the line joining the mid points of $OA$ i.e.$\frac{\vec{a}}{2}$ and $BC$ i.e. $\frac{\vec{b}+\vec{c}}{2}$ is $\vec{r}=\frac{\vec{a}}{2}+t(\frac{\vec{b}+\vec{c}}{2}-\frac{\vec{a}}{2})………(3)$.
Here $\lambda,\mu,t$ are scalars.
I do not know how to solve $(1)$ and $(2)$ and put into $(3)$ to prove the desired result.
Please help me.Thanks.
Best Answer
You're missing an additional piece of information: There are two parallel sides to the trapezium. Express this by assuming that $\vec b -\vec c = k\vec a$ for some $k$. You now can solve for $$\vec b = \vec c + k\vec a\,.\tag4$$ Put (4) into the equation (1) = (2) to find $$\mu = \lambda = \frac1{k+1}\,.$$ You then check that with this choice of $\lambda$ and $\mu$, the resulting $\vec r$ satisfies (3). Note that with (4), equation (3) becomes $$ \vec r = t\vec c + \left(\frac {t(k-1)+1}2\right)\vec a\,. $$