Let ABC be an equilateral triangel. Let `BE_|_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 \( CA \) 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:
\[ AB = BC = CA = a \]

2. **Calculate \( AB^2 + BC^2 + CA^2 \)**:
Since all sides are equal, we can express the sum of squares as:
\[ AB^2 + BC^2 + CA^2 = a^2 + a^2 + a^2 = 3a^2 \]

3. **Determine the length of \( BE \)**:
In triangle \( BEC \), where \( BE \) is perpendicular to \( CA \), we can use the properties of a 30-60-90 triangle. The angle \( \angle ABC \) is \( 60^\circ \). Thus, we can use the sine function:
\[ \sin(60^\circ) = \frac{BE}{BC} \]
Substituting the known values:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad BC = a \]
Therefore:
\[ \frac{\sqrt{3}}{2} = \frac{BE}{a} \implies BE = a \cdot \frac{\sqrt{3}}{2} \]

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

5. **Relate \( AB^2 + BC^2 + CA^2 \) to \( BE^2 \)**:
We have already calculated \( AB^2 + BC^2 + CA^2 = 3a^2 \) and \( BE^2 = \frac{3a^2}{4} \). Now, we can express \( 3a^2 \) in terms of \( BE^2 \):
\[ 3a^2 = 4 \cdot BE^2 \]
Thus, we conclude:
\[ AB^2 + BC^2 + CA^2 = 4 \cdot BE^2 \]

### Final Result:
The value of \( AB^2 + BC^2 + CA^2 \) is equal to \( 4 \cdot BE^2 \).

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