In triangle ABC, D and E are the midpoints of AB and AC respectively. Express AD and AE in terms of a and b. Find DE and show that DE = (1)/(2)BC.

Step 1: Express $\vec{AD}$ and $\vec{AE}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
Given that $D$ is the midpoint of $AB$, we have:

$$\vec{AD} = \frac{1}{2}\vec{AB} = \frac{1}{2}\mathbf{a}$$

Given that $E$ is the midpoint of $AC$, we have:

$$\vec{AE} = \frac{1}{2}\vec{AC} = \frac{1}{2}\mathbf{b}$$

Step 2: Find $\vec{DE}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ (part i).
To find $\vec{DE}$, we can use the vector addition rule:

$$\vec{DE} = \vec{DA} + \vec{AE}$$

Since $\vec{DA} = -\vec{AD}$, we substitute the expressions from Step 1:

$$\vec{DE} = -\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$$ $$\vec{DE} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$$