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If ` vec a , vec b , vec ca n d vec d` are the position vectors of the vertices of a cyclic quadrilateral `A B C D ,` prove that `(| vec axx vec b+ vec bxx vec d+ vec d xx vec a|)/((vec b- vec a).(vec d- vec a)) + (| vec bxx vec c+ vec cxx vec d+ vec d xx vec b|)/((vec b- vec c).( vec d- vec c))=0dot`

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`(|vecaxxvecb+vecb xxvecd+vecd xxveca|)/((vecb-veca).(vecd-veca))=(|(veca-vecd)xx(vecb-veca)|)/((vecb-veca).(vecd-veca))`
`= (|veca-vecd||vecb-veca|sinA)/(|vecb-veca||vecd-veca|cosA)`
`tanA`
`Also (|vecbxxvecc+veccxxvecd+vecd xxvecb|)/((vecb-vecc).(vecd-vecc))=(|(vecb-vecc)xx(vecc-vecd)|)/((vecb-veca).(vecd-vecc))`
`=(|vecb-vecc||vecc-vecd|sinC)/(|vecb-vecc|.|vecd-vecc|cosA)`
`tan C`
As it is a cycle quadrilateral, we have
`A = 180^(@)-C`
`tan A=tan(180^(@)-C)`
`tan A + tan C =0`
`(|vecaxx vecb+vecbxxvecd+vecd xxveca|)/((vecb-veca).(vecd-veca))+(|vecbxxvecc+veccxxvecd+vecd xxvecb|)/((vecb-vecc).(vecd-vecc))`
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