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

Explanation

Let the P.V.s of the points A, B, C and D be `vecO, B(vecb), D(vecd) and C(vecd + tvecb)`
For any point `vecr` on `vec(AC) and vec(BD), vecr = lamda(vecd + t vecb) and vecr = (1-mu) vecb + mu vecd`, respectively.
For the point of intersection, say T, compare the coefficients.
`lamda = mu, tlamda =1-mu=1-lamda or (t+1)lamda =1`
`therefore lamda = (1)/(t+1) = mu`

Therefore, `vecr` (position vector of T) = `(vecd + tvecb)/(t+1)" "`(i) ltbgt Let R and S be the midpoints of the parallel sides AB and DC, then R is `(b)/(2)` and S is `d+ t(b)/(2)`.
Let T divide SR in the ratio `m:1`.
Position vector of T is `(m(vecb)/(2)+ vecd+ t(vecb)/(2))/(m+1)` , which is equivalent to `(vecd+ tvecb)/(t+1)`.
Comparing coefficients of `vecb and vecd`.
`(1)/(m+1) = (1)/(t+1) and (m+t)/(2(m+1))= (t)/(t+1)`
From the first relation, `m= t` which satisfies the second relation. Hence proved.