Prove that the area of a rhombus is half the product of its diagonals.

(N/A) Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ intersecting at point $O$.
By the properties of a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Thus,$AC \perp BD$ and $AO = OC = \frac{1}{2} AC$,$BO = OD = \frac{1}{2} BD$.
The area of the rhombus $ABCD$ is the sum of the areas of $\triangle ABC$ and $\triangle ADC$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BO$.
Area of $\triangle ADC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times OD$.
Total Area $= \frac{1}{2} \times AC \times BO + \frac{1}{2} \times AC \times OD = \frac{1}{2} \times AC \times (BO + OD)$.
Since $BO + OD = BD$,the Area $= \frac{1}{2} \times AC \times BD$.
Therefore,the area of a rhombus is half the product of its diagonals.