1. **State the problem:** We have two chords AC and BD intersecting at point O inside a circle. We need to find the ratio of the areas of triangles AOB and COD. 2. **Recall properties:** When two chords intersect inside a circle, the products of the segments are equal: $$AO \times OC = BO \times OD$$. 3. **Express areas:** The area of a triangle is $$\frac{1}{2} \times \text{base} \times \text{height}$$. Here, triangles AOB and COD share the point O and lie inside the circle. 4. **Use the fact that triangles AOB and COD share the same angle at O:** The angles \(\angle AOB\) and \(\angle COD\) are vertical angles and thus equal. 5. **Area ratio using sides and sine of included angle:** $$\frac{\text{area of } \triangle AOB}{\text{area of } \triangle COD} = \frac{\frac{1}{2} AO \times BO \times \sin \theta}{\frac{1}{2} CO \times DO \times \sin \theta} = \frac{AO \times BO}{CO \times DO}$$ 6. **Use chord segment product equality:** From step 2, $$AO \times OC = BO \times OD$$, rearranged as $$\frac{AO \times BO}{CO \times DO} = 1$$. 7. **Conclusion:** The ratio of the areas is $$1:1$$. **Final answer:** $$\text{area of } \triangle AOB : \text{area of } \triangle COD = 1 : 1$$