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Proving using vector

  1. Jul 26, 2005 #1
    I'm stucked here...
    How to prove that the diagonals of parallelogram bisect each other using vector method?
    Let's say for ABCD which is parallelogram, AB//DC, AD//BC, O is the point of intersection of the diagonals.
    Can i straight away apply that [itex]\overrightarrow{AO}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}[/itex]
  2. jcsd
  3. Jul 26, 2005 #2


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    Say you have a parallelogram ABCD, AB // CD, and AD // BC. O is the intersection of AC and BD.
    You have:
    [itex]\vec{OA} = \vec{OB} + \vec{BA} = \vec{OD} + \vec{DA}[/itex]
    [itex]\vec{OC} = \vec{OB} + \vec{BC} = \vec{OD} + \vec{DC}[/itex]
    [itex]\vec{OA} + \vec{OC} = 2\vec{OB} + \vec{BA} + \vec{BC}[/itex]
    [itex] = 2\vec{OB} + \vec{BD}[/itex]
    [itex]\vec{OA} + \vec{OC}[/itex] will give you a vector which is on AC.
    [itex]2\vec{OB} + \vec{BD}[/itex] will give you a vector which is on BD.
    The only vector that's both on AC and BD is [itex]\vec{0}[/itex].
    So [itex]\vec{OA} + \vec{OC} = \vec{0} \Leftrightarrow[/itex] O is the midpoint of AC.
    From there, you can easily prove O is the midpoint of DB.
    Hope someone will come up with another shorter proof.
    To use: [itex]\vec{AO} = \frac{1}{2}(\vec{AB} + \vec{BC})[/itex], just state that O is the midpoint of AC, and [itex]\vec{AC} = \vec{AB} + \vec{BC} \Leftrightarrow \frac{1}{2} \vec{AC} = \frac{1}{2} (\vec{AB} + \vec{BC}) \Leftrightarrow \vec{AO} = \frac{1}{2}(\vec{AB} + \vec{BC})[/itex]
    Viet Dao,
    Last edited: Jul 26, 2005
  4. Jul 26, 2005 #3
    Probably, a little shorter than VietDao has, but similar.
    Thus [tex]\vec{OB}+\vec{OD}=\vec{0}[/tex] (for the reason, which explaned VietDao)
    For me it also doesn't looks as simpliest way.
  5. Jul 26, 2005 #4
    But, aren't we have to assume that magnitude of OB and OD is not the same? And so AO is not half of AC.
    So how to prove that OB + OD = AC - 2AO = 0?
  6. Jul 26, 2005 #5
    OB and OD are on the same line (they are parallel) (i believe it's given in definition of diagonal).
  7. Jul 26, 2005 #6


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    Assume only that the two segments of each diagonal are parallel, not equal

    [tex] \vec{AC} = \vec{AB} + \vec{BC} = \vec{AO} + \vec{OC} = (1 + \alpha)\vec{AO} [/tex]

    [tex] \vec{BD} = \vec{BC} + \vec{CD} = \vec{BC} - \vec{AB} = \vec{BO} + \vec{OD} = (1 + \beta)\vec{OD} [/tex]

    Add the above

    [tex] 2\vec{BC} = 2\vec{AD} = (1 + \alpha)\vec{AO} + (1 + \beta)\vec{OD} = \vec{AO} + \vec{OD} + \alpha\vec{AO} + \beta\vec{OD} = \vec{AD} + \alpha\vec{AO} + \beta\vec{OD} [/tex]

    [tex] \vec{AD} = \alpha\vec{AO} + \beta\vec{OD} = \alpha(\vec{AD} - \vec{OD}) + \beta\vec{OD} = \alpha\vec{AD} + (\beta - \alpha)\vec{OD} [/tex]

    Since [tex] \vec{AD} [/tex] and [tex] \vec{OD} [/tex] are not parallel, the only possible combination of coefficients on the right hand side is [tex] \alpha = \beta = 1 [/tex]

    [tex] \vec{OC} = \alpha\vec{AO} = \vec{AO} [/tex]

    [tex] \vec{BO} = \beta\vec{OD} = \vec{OD} [/tex]
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