If P denotes the perimeter and S denotes the sum of the distances of a point within a triangle from its angular points, then

Options

A. `P gt S`
B. `P le S`
C. `S lt P`
D. `S le P`

Answer

To solve the problem, we need to establish a relationship between the perimeter \( P \) of a triangle and the sum \( S \) of the distances from a point inside the triangle to its vertices (angular points).

### Step-by-Step Solution

1. **Define the Triangle and Points**:
Let \( \triangle ABC \) be a triangle with vertices \( A \), \( B \), and \( C \). Let \( O \) be a point inside the triangle.

2. **Understand the Distances**:
The distances from point \( O \) to the vertices are denoted as:
- \( OA \) (distance from \( O \) to \( A \))
- \( OB \) (distance from \( O \) to \( B \))
- \( OC \) (distance from \( O \) to \( C \))

Therefore, the sum of the distances \( S \) can be expressed as:
\[ S = OA + OB + OC \]

3. **Use the Triangle Inequality**:
According to the triangle inequality, for any triangle, the sum of the lengths of any two sides must be greater than the length of the third side. We can apply this to the triangles formed by point \( O \) and the vertices of triangle \( ABC \).

- For triangle \( OAB \):
\[ OA + OB > AB \]
- For triangle \( OBC \):
\[ OB + OC > BC \]
- For triangle \( OCA \):
\[ OC + OA > CA \]

4. **Combine the Inequalities**:
Adding these inequalities together gives:
\[ (OA + OB) + (OB + OC) + (OC + OA) > AB + BC + CA \]
Simplifying this, we get:
\[ 2(OA + OB + OC) > AB + BC + CA \]
Which can be rewritten as:
\[ 2S > P \]
Where \( P = AB + BC + CA \) is the perimeter of triangle \( ABC \).

5. **Final Result**:
Dividing both sides by 2, we find:
\[ S > \frac{P}{2} \]

### Conclusion
From our derivation, we can conclude that the sum of the distances from a point inside a triangle to its vertices is always less than the perimeter of the triangle. Thus, we can state:
\[ P > S \]