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Diagonals AC and BD of square ABCD inter...

Diagonals AC and BD of `square ABCD` intersect each other at point P. Show that `ar(triangle APB)xxar(triangle CPD)=ar(triangle APD)xxar(triangle BPC)`

Text Solution

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`AM _|_ BD & CN _|_ BD `
`L.H.S.= area(/_APB)xxarea(/_CPD)`
`=1/2xx(BPxxAM)xx1/2(PDxxCN)`
`=1/2(BPxxCN)xx1/2(PDxxAM)`
`=area(/_BPC)xxarea(/_APD)`
`L.H.S=R.H.S`
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