Geometry

To find the area of triangle BID, we can use the formula for the area of a triangle given the length of one side and the lengths of the two adjacent sides to an angle. The formula is:

$$\text{Area} = \frac{1}{2} \times \text{side} \times \text{adjacent side} \times \sin(\text{angle})$$

Given:
- $DI = 3$
- $BD = 4$
- $BI = 6$

We need to find the angle at vertex B.

We know that the angle bisectors of a triangle divide the opposite side into segments that are proportional to the adjacent sides. Therefore, from the given information, we can set up the following proportions:

$$\frac{DI}{BD} = \frac{DI}{DI + IB} = \frac{3}{4}$$

$$\frac{3}{4} = \frac{3}{3 + IB}$$

$$3 + IB = 4$$

$$IB = 1$$

Now, we can find $\sin(\angle B)$ using the Law of Sines in triangle BDI:

$$\sin(\angle B) = \frac{DI}{BI} = \frac{3}{6} = \frac{1}{2}$$

Now, we can use the formula for the area of triangle BID:

$$\text{Area} = \frac{1}{2} \times BD \times DI \times \sin(\angle B)$$

$$\text{Area} = \frac{1}{2} \times 4 \times 3 \times \frac{1}{2}$$

$$\text{Area} = 6 \, \text{square units}$$

So, the area of triangle BID is $6 \, \text{square units}$.