Let `O A C B` be a parallelogram with `O` at the origin and`O C` a diagonal. Let `D` be the midpoint of `O Adot` using vector methods prove that `B Da n dC O` intersect in the same ratio. Determine this ratio.

Answer

`2:1`

Explanation

OACB is a parallelogram with O as origin. Let with respect to O, position vectors of A and B be `veca and vecb`, respectively. Then P.V. of C is `veca + vecb`.
Also D is the midpoint of OA, therefore, the positions vector of D is `veca//2`

CO and BD intersect each other at P.
Let P divide CO in the ratio `lamda : 1` and BD in the ratio `mu: 1`. Then by section theorem, positioin vector of point P dividing CO in ratio `lamda : 1` is
`" "(lamda xx 0 +1 xx (veca + vecb))/(lamda +1) = (veca + vecb)/(lamda +1)" "`(i)
and position vector of point P dividing BD in the ratio `mu :1` is
`(mu((veca)/(2))+1 (vecb))/(mu+1) = (muveca + 2vecb)/(2(mu+1))" "` (ii)
As (i) and (ii) represent the position vector of the same point, hence,
`" "(veca + vecb)/(lamda +1)= (muveca + 2 vecb)/( 2(mu+1))`
Equating the coefficients of `veca and vecb`, we get
`" "(1)/(lamda +1) = (mu)/(2(mu+1))" "` (iii)
`" " = (1)/(mu+1)" "` (iv)
From (iv) we get `lamda = mu`, i.e., P divides CO and BD in the same ratio.
Putting `lamda =mu ` in Eq. (iii), we get `mu =2`
Thus, the required ratio is `2:1`.