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If veca,vecb,vec c,vecd are the position...

If `veca,vecb,vec c,vecd` are the position vectors of the verticles of a cyclic quadrilateral ABCd prove that
`(|vecaxxvecb+vecbxxvecd+vecd xxveca|)/((vecb-veca).(vecd-veca))+(|vecbxxvec c+veccxxvecd+vecd xxvecb|)/((vecb-vecc).(vecd-vecc))=0`

Text Solution

Verified by Experts

Consider `(|vecaxxvecb+vecbxxvecd+vecd xxveca|)/((vecb-veca).(vecd-veca))`
We have,
`(veca-vecd)xx(vecb-veca)`
`=vecaxxvecb-vecaxxveca-vecd xxvecb+vecd xxveca`
`=vecaxxvecb+vecbxxvecd+vecd xxveca`
`:. (|vecaxxvecb+vecbxxvecd+vecd xxveca|)/((vecb-veca).(vecd-veca))=(|(veca-vecd)xx(vecb-veca)|)/((vecb-veca).(vecd-veca))`
`=(|veca-vecd||vecb-veca| sinA)/(|vecb-veca||vecd-veca|cos A)=tanA`
Again, `(|vecbxxvec c+vecaxxvecd+vecd xxvecb|)/((vecb-vecc).(vecd-vecc))=(|(vecb-vecc)xx(vecc-vecd)|)/((vecb-vecc):(vecd-vecc))`
`=(|vecb-vecc||vecc-vecdsinC)/(|vecb-vecc||vecd-vecc|cosC)=tanC`
in a cyclic quadrilateral `A+C=pi rArrA=pi-C`
`rArr tanA=tan(pi-C)`
`rArr tanA=-tanC`
`rArr tanA+tanC=0`
`rArr (|vecaxxvecb+vecbxxvecd+vecd xxveca|)/((vecb-veca).(vecd-veca))+(|vecbxxvec c+veccxxvecd+vecd xxvecd|)/((vecb-vecc).(vecd-vecc))=0`
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