From a point O in the interior of a `DeltaABC` perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which one of the following is true ?

Options

A. `AF^(2)+BD^(2)+CE^(2)=AE^(2)+CD^(2)+BF^(2)`
B. `AB^(2)+BC^(2)=AC^(2)`
C. `AF^(2)+BD^(2)+CE^(2)=OA^(2)+OB^(2)+OC^(2)`
D. `AF^(2)+BD^(2)+CE^(2)=OD^(2)+OE^(2)+OF^(2)`

Answer

To solve the problem, we need to analyze the relationships between the segments created by drawing perpendiculars from point O to the sides of triangle ABC. We will derive some expressions and ultimately show that one of the options is true.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a triangle ABC.
- A point O is located inside the triangle.
- Perpendiculars OD, OE, and OF are drawn from point O to the sides BC, CA, and AB respectively.

2. **Using the Pythagorean Theorem**:
- Consider triangle OCE:
\[ OC^2 = OE^2 + EC^2 \]
- Consider triangle OBD:
\[ OB^2 = OD^2 + BD^2 \]
- Consider triangle OAF:
\[ OA^2 = OF^2 + AF^2 \]

3. **Adding the Equations**:
- Now, we add the three equations:
\[ OA^2 + OB^2 + OC^2 = (OF^2 + AF^2) + (OD^2 + BD^2) + (OE^2 + EC^2) \]
- This simplifies to:
\[ OA^2 + OB^2 + OC^2 = OF^2 + OD^2 + OE^2 + AF^2 + BD^2 + EC^2 \]

4. **Rearranging the Equation**:
- Rearranging gives us:
\[ AF^2 + BD^2 + EC^2 = OA^2 + OB^2 + OC^2 - (OF^2 + OD^2 + OE^2) \]
- We can label this as Equation (1).

5. **Considering Other Triangles**:
- Now, consider triangle OCD:
\[ OC^2 = OD^2 + CD^2 \]
- Consider triangle DBF:
\[ OB^2 = OF^2 + BF^2 \]
- Consider triangle OAE:
\[ OA^2 = OE^2 + AE^2 \]

6. **Adding These New Equations**:
- Adding these gives:
\[ OA^2 + OB^2 + OC^2 = (OE^2 + AE^2) + (OF^2 + BF^2) + (OD^2 + CD^2) \]
- This simplifies to:
\[ OA^2 + OB^2 + OC^2 = OE^2 + OF^2 + OD^2 + AE^2 + BF^2 + CD^2 \]

7. **Rearranging Again**:
- Rearranging gives:
\[ AE^2 + BF^2 + CD^2 = OA^2 + OB^2 + OC^2 - (OE^2 + OF^2 + OD^2) \]
- We can label this as Equation (2).

8. **Comparing Equations**:
- From Equation (1) and Equation (2), we can conclude that:
\[ AF^2 + BD^2 + EC^2 = AE^2 + BF^2 + CD^2 \]

### Conclusion:
Thus, the correct statement is that the sum of the squares of the segments created by the perpendiculars from point O to the sides of triangle ABC is equal to the sum of the squares of the segments created by the sides of the triangle.