Prove Diagonals of Parallelograms Bisect w/ Vector Method

[kanki]

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 \overrightarrow{AO}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}

 

[VietDao29]

Say you have a parallelogram ABCD, AB // CD, and AD // BC. O is the intersection of AC and BD.
You have:
\vec{OA} = \vec{OB} + \vec{BA} = \vec{OD} + \vec{DA}
\vec{OC} = \vec{OB} + \vec{BC} = \vec{OD} + \vec{DC}
\vec{OA} + \vec{OC} = 2\vec{OB} + \vec{BA} + \vec{BC}
= 2\vec{OB} + \vec{BD}
\vec{OA} + \vec{OC} will give you a vector which is on AC.
2\vec{OB} + \vec{BD} will give you a vector which is on BD.
The only vector that's both on AC and BD is \vec{0}.
So \vec{OA} + \vec{OC} = \vec{0} \Leftrightarrow 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: \vec{AO} = \frac{1}{2}(\vec{AB} + \vec{BC}), just state that O is the midpoint of AC, and \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})
Viet Dao,

 

[Yegor]

Probably, a little shorter than VietDao has, but similar.
\vec{AO}+\vec{OB}=\vec{AB}
\vec{AO}+\vec{OD}=\vec{AD}
2\vec{AO}+\vec{OB}+\vec{OD}=\vec{AC}
\vec{OB}+\vec{OD}=\vec{AC}-2\vec{AO}
Thus \vec{OB}+\vec{OD}=\vec{0} (for the reason, which explaned VietDao)
For me it also doesn't looks as simpliest way.

 

[kanki]

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?

 

[Yegor]

OB and OD are on the same line (they are parallel) (i believe it's given in definition of diagonal).

 

[OlderDan]

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

\vec{AC} = \vec{AB} + \vec{BC} = \vec{AO} + \vec{OC} = (1 + \alpha)\vec{AO}

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

Add the above

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}

\vec{AD} = \alpha\vec{AO} + \beta\vec{OD} = \alpha(\vec{AD} - \vec{OD}) + \beta\vec{OD} = \alpha\vec{AD} + (\beta - \alpha)\vec{OD}

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

\vec{OC} = \alpha\vec{AO} = \vec{AO}

\vec{BO} = \beta\vec{OD} = \vec{OD}