If AD, BE and CF are the medians of a `Delta ABC,` then evaluate `(AD^(2)+BE^(2)+CF^(2)): (BC^(2) +CA^(2) +AB^(2)).`

Answer

To solve the problem of evaluating the ratio \((AD^2 + BE^2 + CF^2) : (BC^2 + CA^2 + AB^2)\), where \(AD\), \(BE\), and \(CF\) are the medians of triangle \(ABC\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the sides of the triangle**:
Let \(a = BC\), \(b = CA\), and \(c = AB\).

2. **Use the formula for the length of the medians**:
The lengths of the medians can be expressed using the following formulas:
- \(AD = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}\)
- \(BE = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}\)
- \(CF = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}\)

3. **Square the lengths of the medians**:
- \(AD^2 = \left(\frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}\right)^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2)\)
- \(BE^2 = \left(\frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}\right)^2 = \frac{1}{4}(2a^2 + 2c^2 - b^2)\)
- \(CF^2 = \left(\frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}\right)^2 = \frac{1}{4}(2a^2 + 2b^2 - c^2)\)

4. **Combine the squared medians**:
\[ AD^2 + BE^2 + CF^2 = \frac{1}{4} \left( (2b^2 + 2c^2 - a^2) + (2a^2 + 2c^2 - b^2) + (2a^2 + 2b^2 - c^2) \right) \]
Simplifying this expression:
\[ = \frac{1}{4} \left( 4a^2 + 4b^2 + 4c^2 - (a^2 + b^2 + c^2) \right) \]
\[ = \frac{1}{4} \left( 3a^2 + 3b^2 + 3c^2 \right) = \frac{3}{4} (a^2 + b^2 + c^2) \]

5. **Calculate the sum of the squares of the sides**:
\[ BC^2 + CA^2 + AB^2 = a^2 + b^2 + c^2 \]

6. **Form the ratio**:
\[ \frac{AD^2 + BE^2 + CF^2}{BC^2 + CA^2 + AB^2} = \frac{\frac{3}{4}(a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} \]
Simplifying this gives:
\[ = \frac{3}{4} \]

### Final Result:
Thus, the required ratio is:
\[ (AD^2 + BE^2 + CF^2) : (BC^2 + CA^2 + AB^2) = \frac{3}{4} \]