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 $$\alpha = \beta = 1$$

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

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