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, we will use the Apollonius theorem, which relates the sides of a triangle to its medians.

### Step-by-Step Solution:

1. **Understanding the Medians**:
Let \( AD, BE, CF \) be the medians of triangle \( ABC \). The lengths of the sides opposite to vertices \( A, B, C \) are denoted as \( a = BC, b = CA, c = AB \).

2. **Applying Apollonius's Theorem**:
According to Apollonius's theorem, for any triangle, the square of the length of a median can be expressed in terms of the lengths of the sides of the triangle. The theorem states:
\[ m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4} \]
where \( m_a \) is the median from vertex \( A \) to side \( BC \) (which is \( AD \)), and similarly for the other medians.

3. **Calculating Each Median**:
- For median \( AD \):
\[ AD^2 = \frac{2b^2 + 2c^2 - a^2}{4} \]
- For median \( BE \):
\[ BE^2 = \frac{2c^2 + 2a^2 - b^2}{4} \]
- For median \( CF \):
\[ CF^2 = \frac{2a^2 + 2b^2 - c^2}{4} \]

4. **Summing the Squares of the Medians**:
Now, we sum the squares of the medians:
\[ AD^2 + BE^2 + CF^2 = \frac{(2b^2 + 2c^2 - a^2) + (2c^2 + 2a^2 - b^2) + (2a^2 + 2b^2 - c^2)}{4} \]
Simplifying this:
\[ = \frac{(2b^2 + 2c^2 - a^2 + 2c^2 + 2a^2 - b^2 + 2a^2 + 2b^2 - c^2)}{4} \]
\[ = \frac{(4a^2 + 4b^2 + 4c^2 - (a^2 + b^2 + c^2))}{4} \]
\[ = \frac{(3a^2 + 3b^2 + 3c^2)}{4} \]
\[ = \frac{3}{4}(a^2 + b^2 + c^2) \]

5. **Finding the Ratio**:
We need to evaluate the ratio:
\[ \frac{AD^2 + BE^2 + CF^2}{BC^2 + CA^2 + AB^2} \]
Since \( BC^2 + CA^2 + AB^2 = a^2 + b^2 + c^2 \), we have:
\[ \frac{AD^2 + BE^2 + CF^2}{BC^2 + CA^2 + AB^2} = \frac{\frac{3}{4}(a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} \]
\[ = \frac{3}{4} \]

### Final Answer:
Thus, the value of \( \frac{AD^2 + BE^2 + CF^2}{BC^2 + CA^2 + AB^2} \) is \( \frac{3}{4} \).

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