Triangle $ABC$ lies in a regular decagon as shown in the figure.
What is the ratio of the area of the triangle to the area of the entire decagon? 
Write the answer as a fraction of integers.
[img]https://1.bp.blogspot.com/-Ld_-4u-VQ5o/Xzb-KxPX0wI/AAAAAAAAMWg/-qPtaI_04CQ3vvVc1wDTj3SoonocpAzBQCLcBGAsYHQ/s0/2007%2BMohr%2Bp1.png[/img]
1. **Identify the center and symmetry:**
   Let \( O \) be the center of the regular decagon. Since \( O \) is the center, it is equidistant from all vertices of the decagon. Therefore, \( OA = OB = OC \).

2. **Divide the decagon into triangles:**
   The regular decagon can be divided into 10 congruent isosceles triangles, each with a central angle of \( \frac{360^\circ}{10} = 36^\circ \).

3. **Area of one triangle:**
   Each of these triangles has an area that is \( \frac{1}{10} \) of the area of the decagon. Let the area of the entire decagon be \( A \). Then the area of one of these triangles is \( \frac{A}{10} \).

4. **Area of triangle \( AOC \):**
   Since \( \triangle AOC \) is one of these 10 triangles, its area is \( \frac{A}{10} \).

5. **Area of triangle \( ABC \):**
   Notice that \( \triangle ABC \) consists of two of these isosceles triangles, \( \triangle AOC \) and \( \triangle BOC \). Therefore, the area of \( \triangle ABC \) is:
   \[
   [ABC] = [AOC] + [BOC] = 2 \times \frac{A}{10} = \frac{2A}{10} = \frac{A}{5}
   \]

6. **Ratio of areas:**
   The ratio of the area of \( \triangle ABC \) to the area of the entire decagon is:
   \[
   \frac{[ABC]}{A} = \frac{\frac{A}{5}}{A} = \frac{1}{5}
   \]

The final answer is \(\boxed{\frac{1}{5}}\).