1.

If I is the centre of a circle inscribed in a DeltaABC, then abs(bar(BC))bar(IA)+abs(bar(CA))bar(IB)+abs(bar(AB))bar(IC)=

Answer»

`bar(0)`
`bar(IA)+bar(IB)+bar(IC)`
`(bar(IA)+bar(IB)+bar(IC))/3`
`2(bar(IA)+bar(IB)+bar(IC))`

ANSWER :A


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