# Proving using vector

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

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VietDao29
Homework Helper
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,

Last edited:
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.

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?

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

OlderDan
Homework Helper
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

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

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

$$2\vec{BC} = 2\vec{AD} = (1 + \alpha)\vec{AO} + (1 + \beta)\vec{OD} = \vec{AO} + \vec{OD} + \alpha\vec{AO} + \beta\vec{OD} = \vec{AD} + \alpha\vec{AO} + \beta\vec{OD}$$

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

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

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

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