Proving using vector

  • Thread starter kanki
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  • #1
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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]
 

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  • #2
VietDao29
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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,
 
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  • #3
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Probably, a little shorter than VietDao has, but similar.
[tex]\vec{AO}+\vec{OB}=\vec{AB}[/tex]
[tex]\vec{AO}+\vec{OD}=\vec{AD}[/tex]
[tex]2\vec{AO}+\vec{OB}+\vec{OD}=\vec{AC}[/tex]
[tex]\vec{OB}+\vec{OD}=\vec{AC}-2\vec{AO}[/tex]
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.
 
  • #4
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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?
 
  • #5
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OB and OD are on the same line (they are parallel) (i believe it's given in definition of diagonal).
 
  • #6
OlderDan
Science Advisor
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kanki said:
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?
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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