Let ABC be an equilateral triangle and `AX, BY, CZ `be the altitudes. Then the right statement out of the four given responses is

Options

A. `AX = BY = CZ `
B. `AX != BY = CZ `
C. `AX = BY != CZ`
D. `AX != BY != CZ`

Answer

To solve the problem regarding the altitudes of an equilateral triangle ABC, we will follow these steps:

### Step 1: Understand the properties of an equilateral triangle
An equilateral triangle has all three sides equal and all three angles equal to 60 degrees. Let the length of each side be denoted as \( A \).

### Step 2: Define the altitudes
The altitudes \( AX \), \( BY \), and \( CZ \) are the perpendicular segments drawn from each vertex to the opposite side. In an equilateral triangle, these altitudes also bisect the opposite sides.

### Step 3: Calculate the length of the altitude
Using the Pythagorean theorem, we can calculate the length of the altitude from vertex \( A \) to side \( BC \):
- In triangle \( ABX \):
- Hypotenuse \( AB = A \)
- Base \( BX = \frac{A}{2} \) (since \( AX \) bisects \( BC \))

Using the Pythagorean theorem:
\[ AB^2 = AX^2 + BX^2 \]
\[ A^2 = AX^2 + \left(\frac{A}{2}\right)^2 \]
\[ A^2 = AX^2 + \frac{A^2}{4} \]
Subtracting \( \frac{A^2}{4} \) from both sides:
\[ A^2 - \frac{A^2}{4} = AX^2 \]
\[ \frac{3A^2}{4} = AX^2 \]
Taking the square root:
\[ AX = \frac{\sqrt{3}}{2} A \]

### Step 4: Repeat for the other altitudes
By symmetry, the calculations for altitudes \( BY \) and \( CZ \) will yield the same result:
- For altitude \( BY \):
\[ BY = \frac{\sqrt{3}}{2} A \]
- For altitude \( CZ \):
\[ CZ = \frac{\sqrt{3}}{2} A \]

### Step 5: Conclusion
Since all three altitudes are equal:
\[ AX = BY = CZ = \frac{\sqrt{3}}{2} A \]

### Final Answer
The correct statement is that all altitudes \( AX \), \( BY \), and \( CZ \) are equal.

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