Given a parallelogram constructed on vectors $\vec{AB}$ and $\vec{AD}$, express the vectors $\vec{OA}$, $\vec{OB}$, $\vec{OC}$, and $\vec{OD}$ in terms of $\vec{AB}$ and $\vec{AD}$, where $O$ is the intersection point of the diagonals $AC$ and $BD$.

1. Problem Description

Given a parallelogram constructed on vectors $\vec{AB}$ and $\vec{AD}$, express the vectors $\vec{OA}$, $\vec{OB}$, $\vec{OC}$, and $\vec{OD}$ in terms of $\vec{AB}$ and $\vec{AD}$, where $O$ is the intersection point of the diagonals $AC$ and $BD$.

2. Solution Steps

Let the parallelogram be $ABCD$.

Since $O$ is the intersection of diagonals of a parallelogram, it is the midpoint of both diagonals.

$\vec{OA} = \vec{OA}$ (Trivial)

$\vec{OB} = \vec{OA} + \vec{AB}$

$\vec{OC} = \vec{OA} + \vec{AC}$

$\vec{OD} = \vec{OA} + \vec{AD}$

Since $O$ is the midpoint of $AC$, we have $\vec{AO} = \frac{1}{2} \vec{AC}$, so $\vec{OA} = -\frac{1}{2} \vec{AC}$.

Also, $\vec{AC} = \vec{AB} + \vec{BC} = \vec{AB} + \vec{AD}$.

So, $\vec{OA} = -\frac{1}{2} (\vec{AB} + \vec{AD})$

Since $O$ is the midpoint of $BD$, we have $\vec{BO} = \frac{1}{2} \vec{BD}$, so $\vec{OB} = \vec{OD} + \vec{DB} = \vec{OD} - \vec{BD}$.

Also, $\vec{BD} = \vec{AD} - \vec{AB}$.

$\vec{OB} = \vec{OA} + \vec{AB} = -\frac{1}{2} (\vec{AB} + \vec{AD}) + \vec{AB} = \frac{1}{2} \vec{AB} - \frac{1}{2} \vec{AD}$

$\vec{OC} = \vec{OB} + \vec{BC} = \frac{1}{2} \vec{AB} - \frac{1}{2} \vec{AD} + \vec{AD} = \frac{1}{2} \vec{AB} + \frac{1}{2} \vec{AD}$

Or, $\vec{OC} = - \vec{OA} = \frac{1}{2} (\vec{AB} + \vec{AD})$.

$\vec{OD} = \vec{OA} + \vec{AD} = -\frac{1}{2} (\vec{AB} + \vec{AD}) + \vec{AD} = -\frac{1}{2} \vec{AB} + \frac{1}{2} \vec{AD}$

3. Final Answer

$\vec{OA} = -\frac{1}{2} \vec{AB} - \frac{1}{2} \vec{AD}$

$\vec{OB} = \frac{1}{2} \vec{AB} - \frac{1}{2} \vec{AD}$

$\vec{OC} = \frac{1}{2} \vec{AB} + \frac{1}{2} \vec{AD}$

$\vec{OD} = -\frac{1}{2} \vec{AB} + \frac{1}{2} \vec{AD}$