Let ABC be an arbitrary triangle, and P be any point inside. Let `d_1, d_2` and `d_3` denote the perpendicular distancea from P to the sides BC, CA and AB respectively. Let `h_1, h_2` and `h_3` denote, reapectively, the length of the altitudse from A, B and C to the opposite sides of the triangle, then `d_1/h_1 + d_2/h_2 + d_3/h_3` is

Options

A. A)`sqrt 3`
B. B)3
C. C)1
D. D)`pi`

Answer

To solve the problem, we need to analyze the relationship between the perpendicular distances from point P inside triangle ABC to the sides of the triangle and the altitudes from the vertices to the opposite sides.

### Step-by-Step Solution:

1. **Understanding the Triangle and Points**:
- Let triangle ABC be any arbitrary triangle.
- Let P be a point inside triangle ABC.
- Let \( d_1, d_2, d_3 \) be the perpendicular distances from point P to the sides BC, CA, and AB respectively.
- Let \( h_1, h_2, h_3 \) be the lengths of the altitudes from vertices A, B, and C to the opposite sides.

2. **Using the Area of the Triangle**:
- The area of triangle ABC can be expressed in two ways:
- Using the altitude from A:
\[ \text{Area} = \frac{1}{2} \times BC \times h_1 \]
- Using the perpendicular distance from point P to side BC:
\[ \text{Area} = \frac{1}{2} \times BC \times d_1 \]

3. **Setting Up the Ratios**:
- From the area expressions, we can equate them:
\[ \frac{1}{2} \times BC \times h_1 = \frac{1}{2} \times BC \times d_1 \]
- This simplifies to:
\[ d_1 = \frac{h_1}{k_1} \]
- Where \( k_1 \) is a constant that represents the ratio of the distances.

4. **Repeating for Other Sides**:
- Similarly, we can set up the equations for \( d_2 \) and \( d_3 \):
\[ d_2 = \frac{h_2}{k_2} \]
\[ d_3 = \frac{h_3}{k_3} \]

5. **Summing the Ratios**:
- Now we can express the sum:
\[ \frac{d_1}{h_1} + \frac{d_2}{h_2} + \frac{d_3}{h_3} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} \]

6. **Using the Property of the Triangle**:
- For any point P inside triangle ABC, the sum of the ratios \( \frac{d_1}{h_1} + \frac{d_2}{h_2} + \frac{d_3}{h_3} \) will always equal 1.
- This is a known property of triangles and can be proven using the concept of areas or by considering the centroid of the triangle.

7. **Conclusion**:
- Therefore, we conclude that:
\[ \frac{d_1}{h_1} + \frac{d_2}{h_2} + \frac{d_3}{h_3} = 1 \]

### Final Answer:
The value of \( \frac{d_1}{h_1} + \frac{d_2}{h_2} + \frac{d_3}{h_3} \) is **1**.