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In a quadrillateral ABCD, it is given th...

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

Text Solution

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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
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