In a quadrilateral ABCD, it is given that `AB |\|CD` and the diagonals `AC and BD` are perpendicular to each other. Show that `AD.BC >= AB. CD`.

Explanation

Let AC and BD intersect at O (origin)
According to the question `veca.vecb =0`
Also AB||CD
`(muveca-lambdavecb)||(veca-vecb)`
`Rightarrow mu = lambda`
`Let AD. BC ge AB. CD`
`Rightarrow |lambdavecb-veca||muveca-vecb||ge|(vecb-veca)||(lambdavecb - muveca)|`
`or (lambda^(2)b^(2)+a^(2))(mu^(2)a^(2)+b^(2))ge(b^(2)+a^(2))(lambda^(2)b^(2)+mu^(2)a^(2))`
`or lamda^(2)mu^(2)a^(2)b^(2)+a^(2)b^(2)gelamda^(2)a^(2)b^(2)+mu^(2)b^(2)a^(2)`
`or (lamda^92)-1) (mu^(2)-1)ge0`
`or (lamda^(2)-1)^(2)ge0` which is true.
Hence, AD. BC `ge` AB. CD