Let the diagonals of the kite be $d_1$ and $d_2$. The diagonals of a kite intersect at right angles, and one diagonal ($d_2$) is bisected. Let the segments of $d_1$ be $x$ and $y$, so $d_1 = x + y$.

The kite is divided into two triangles with the same base $d_2$ and heights $x$ and $y$. The area of the kite is the sum of the areas of these two triangles:

$$\frac{1}{2} \times d_2 \times x + \frac{1}{2} \times d_2 \times y = \frac{1}{2} \times d_2 \times (x + y) = \frac{1}{2} \times d_1 \times d_2$$

∴ The area of the kite is $\frac{1}{2} \times d_1 \times d_2$.

The diagonals divide the kite into four right-angled triangles. The two triangles on either side of segment $y$ are congruent, and the two on either side of $x$ are congruent.

The area of the two triangles with height $x$ is:

$$2 \times \left[\frac{1}{2} \times \frac{d_2}{2} \times x\right] = \frac{1}{2} \times d_2 \times x$$

The area of the two triangles with height $y$ is:

$$2 \times \left[\frac{1}{2} \times \frac{d_2}{2} \times y\right] = \frac{1}{2} \times d_2 \times y$$

Total area of the kite:

$$\frac{1}{2} \times d_2 \times x + \frac{1}{2} \times d_2 \times y = \frac{1}{2} \times d_2 \times (x + y) = \frac{1}{2} \times d_1 \times d_2$$

∴ The area of the kite is $\frac{1}{2} \times d_1 \times d_2$.