The diagonals of a rhombus $ABCD$ are $\vec{AC}$ and $\vec{BD}$, and they intersect at point $O$. Express the vectors that coincide with the sides of the rhombus in terms of the vectors $\vec{AC}$ and $\vec{BD}$.

1. Problem Description

The diagonals of a rhombus $ABCD$ are $\vec{AC}$ and $\vec{BD}$, and they intersect at point $O$. Express the vectors that coincide with the sides of the rhombus in terms of the vectors $\vec{AC}$ and $\vec{BD}$.

2. Solution Steps

Let $ABCD$ be the rhombus. The diagonals are $\vec{AC}$ and $\vec{BD}$, and they intersect at point $O$. We want to express the vectors $\vec{AB}$, $\vec{BC}$, $\vec{CD}$, and $\vec{DA}$ in terms of $\vec{AC}$ and $\vec{BD}$.

Since the diagonals of a rhombus bisect each other at right angles, we have:

$\vec{AO} = \frac{1}{2}\vec{AC}$

$\vec{OC} = -\vec{AO} = \frac{1}{2}\vec{AC}$

$\vec{BO} = \frac{1}{2}\vec{BD}$

$\vec{OD} = -\vec{BO} = \frac{1}{2}\vec{BD}$

Now, we can express the side vectors as follows:

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

$\vec{BC} = \vec{BO} + \vec{OC} = -\frac{1}{2}\vec{BD} + \frac{1}{2}\vec{AC} = \frac{1}{2}(\vec{AC} - \vec{BD})$

$\vec{CD} = \vec{CO} + \vec{OD} = -\frac{1}{2}\vec{AC} - \frac{1}{2}\vec{BD} = -\frac{1}{2}(\vec{AC} + \vec{BD})$

$\vec{DA} = \vec{DO} + \vec{OA} = \frac{1}{2}\vec{BD} + \frac{1}{2}\vec{AC} = \frac{1}{2}(\vec{AC} + \vec{BD})$

3. Final Answer

$\vec{AB} = \frac{1}{2}(\vec{BD} - \vec{AC})$

$\vec{BC} = \frac{1}{2}(\vec{AC} - \vec{BD})$

$\vec{CD} = -\frac{1}{2}(\vec{AC} + \vec{BD})$

$\vec{DA} = \frac{1}{2}(\vec{AC} + \vec{BD})$