If `AD, BE and CF` are the medians of a `DeltaA B C ,t h e n(A D^2+B E^2+C F^2):(B C^2+C A^2+A B^2)` is equal to

Options

A. `4:3`
B. `3:2`
C. 3:4
D. 2:3

Answer

To solve the problem, we need to find the ratio of the sum of the squares of the medians of triangle \( ABC \) to the sum of the squares of its sides. Let's denote the sides of the triangle as follows:

- \( a = BC \)
- \( b = CA \)
- \( c = AB \)

The medians from the vertices \( A, B, \) and \( C \) to the opposite sides \( BC, CA, \) and \( AB \) are denoted as \( AD, BE, \) and \( CF \) respectively.

### Step 1: Write the formulas for the lengths of the medians

The lengths of the medians can be calculated using the following formulas:

- \( AD = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2} \)
- \( BE = \frac{1}{2} \sqrt{2c^2 + 2a^2 - b^2} \)
- \( CF = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2} \)

### Step 2: Calculate the squares of the medians

Now we will square the lengths of the medians:

- \( AD^2 = \left(\frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}\right)^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2) \)
- \( BE^2 = \left(\frac{1}{2} \sqrt{2c^2 + 2a^2 - b^2}\right)^2 = \frac{1}{4}(2c^2 + 2a^2 - b^2) \)
- \( CF^2 = \left(\frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}\right)^2 = \frac{1}{4}(2a^2 + 2b^2 - c^2) \)

### Step 3: Sum the squares of the medians

Now, we will sum the squares of the medians:

\[ AD^2 + BE^2 + CF^2 = \frac{1}{4} \left( (2b^2 + 2c^2 - a^2) + (2c^2 + 2a^2 - b^2) + (2a^2 + 2b^2 - c^2) \right) \]

Combining the terms:

\[ = \frac{1}{4} \left( 4a^2 + 4b^2 + 4c^2 - (a^2 + b^2 + c^2) \right) = \frac{1}{4} \left( 3a^2 + 3b^2 + 3c^2 \right) = \frac{3}{4} (a^2 + b^2 + c^2) \]

### Step 4: Calculate the sum of the squares of the sides

The sum of the squares of the sides is:

\[ BC^2 + CA^2 + AB^2 = a^2 + b^2 + c^2 \]

### Step 5: Find the ratio

Now we can find the ratio of the sum of the squares of the medians to the sum of the squares of the sides:

\[ \frac{AD^2 + BE^2 + CF^2}{BC^2 + CA^2 + AB^2} = \frac{\frac{3}{4}(a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} \]

This simplifies to:

\[ \frac{3}{4} \]

### Final Answer

Thus, the ratio \( (AD^2 + BE^2 + CF^2) : (BC^2 + CA^2 + AB^2) \) is equal to \( \frac{3}{4} \).

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