Let ABC be an equilateral triangle. Let `BE bot CA` meeting CA at E, then `(AB^2+BC^2+CA^2)` is equal to :

Options

A. `2BE^2`
B. `3BE^2`
C. `4BE^2`
D. `6BE^2`

Answer

To solve the problem, we need to find the value of \( AB^2 + BC^2 + CA^2 \) for an equilateral triangle \( ABC \) where \( BE \) is perpendicular to \( CA \) and meets \( CA \) at point \( E \).

### Step-by-step Solution:

1. **Identify the sides of the equilateral triangle**:
Let the length of each side of the equilateral triangle \( ABC \) be \( a \). Therefore, we have:
\[ AB = a, \quad BC = a, \quad CA = a \]

2. **Calculate \( AB^2 + BC^2 + CA^2 \)**:
Since all sides are equal, we can write:
\[ AB^2 + BC^2 + CA^2 = a^2 + a^2 + a^2 = 3a^2 \]

3. **Find the length of \( BE \)**:
In an equilateral triangle, the height \( h \) can be calculated using the formula:
\[ h = \frac{\sqrt{3}}{2} a \]
Since \( BE \) is the height from point \( B \) to line \( CA \), we have:
\[ BE = h = \frac{\sqrt{3}}{2} a \]

4. **Calculate \( BE^2 \)**:
Now, we can find \( BE^2 \):
\[ BE^2 = \left(\frac{\sqrt{3}}{2} a\right)^2 = \frac{3}{4} a^2 \]

5. **Relate \( AB^2 + BC^2 + CA^2 \) to \( BE^2 \)**:
We know that:
\[ AB^2 + BC^2 + CA^2 = 3a^2 \]
We can express this in terms of \( BE^2 \):
\[ 4BE^2 = 4 \left(\frac{3}{4} a^2\right) = 3a^2 \]

6. **Final Result**:
Therefore, we conclude that:
\[ AB^2 + BC^2 + CA^2 = 4BE^2 \]

### Final Answer:
\[ AB^2 + BC^2 + CA^2 = 4BE^2 \]