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If O is the circumcentre and P the ortho...

If O is the circumcentre and P the orthocentre of `Delta ABC`, prove that `vec(OA)+ vec(OB) + vec(OC) =vec(OP)`.

Answer

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If O( vec0 ) is the circumcentre and O' the orthocentre of a triangle ABC, then prove that i. vec(OA)+vec(OB)+vec(OC)=vec(OO') ii. vec(O'A)+vec(O'B)+vec(O'C)=2vec(O'O) iii. vec(AO')+vec(O'B)+vec(O'C)=2 vec(AO)=vec(AP) where AP is the diameter through A of the circumcircle.

Knowledge Check

  • If S is the cirucmcentre, G the centroid, O the orthocentre of a triangle ABC, then vec(SA) + vec(SB) + vec(SC) is:

    A
    `3vec(SG)`
    B
    `vec(OS)`
    C
    `vec(SO)`
    D
    `vec(OG)`

  • If S is circumcentre, O is orthocentre of DeltaABC , then vec(SA)+vec(SB)+vec(SC) =

    A
    `vec(SO)`
    B
    `2vec(SO)`
    C
    `vec(OS)`
    D
    `2vec(OS)`

  • Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are vec(a) , vec(b) , vec(c ) and (vec(a) + vec(b) + vec(c ))/(4) respectively, then the position vector of the orthocentre of this triangle is

    A
    `vec(0)`
    B
    `-(vec(a)+vec(b)+vec(c))/(2)`
    C
    `vec(a) + vec(b)+vec( c) `
    D
    `(vec(a)+vec(b)+vec(c))/(2)`

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    If E is the intersection point of diagonals of parallelogram ABCD and vec( OA) + vec (OB) + vec (OC) + vec (OD) = xvec (OE) where O is origin then x=

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