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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.

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(1) and (2) can be solved

for $\mu$ and $\lambda$

by setting

$\lambda \vec{b}

=\vec{a}+\mu(\vec{c}-\vec{a})

$

and looking at the

two coordinates.

This gives two equations

in two unknowns,

and so can be solved.

Putting in the values for

$\lambda \vec{b}

$

gives the coordinates of the point

of intersection.

Similarly,

equation (3) gives

parameterizations for

the coordinates

of the line.

Looking at one of these,

say $x$,

you can solves for

the value of $t$

that gives that $x$ value.

That same value of $t$

should give the

$y$ coordinate of the

intersection.

I feel that there should be

a more purely vector approach,

but one does not occur to me

right now.

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\,.

$$

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