If ABCD is a rhombus, then :

Options

A. `AC^2+BD^2=4AB^2`
B. `AC^2+BD^2=AB^2`
C. `AC^2+BD^2=2AB^2`
D. `2(AC^2+BD^2)=3AB^2`

Answer

To solve the problem, we need to analyze the properties of a rhombus and its diagonals. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding the Rhombus**:
A rhombus is a type of quadrilateral where all four sides are of equal length. Let the length of each side be denoted as \( a \).

2. **Labeling the Rhombus**:
Let's label the vertices of the rhombus as \( A, B, C, \) and \( D \). The diagonals of the rhombus are \( AC \) and \( BD \).

3. **Properties of Diagonals**:
The diagonals of a rhombus bisect each other at right angles. Let the lengths of the diagonals be \( AC = d_1 \) and \( BD = d_2 \).

4. **Using the Pythagorean Theorem**:
Since the diagonals bisect each other, we can form right triangles. Each triangle formed by the diagonals will have legs of lengths \( \frac{d_1}{2} \) and \( \frac{d_2}{2} \), and the hypotenuse will be the side of the rhombus, which is \( a \).
Therefore, we can write:
\[ a^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 \]

5. **Expanding the Equation**:
Expanding the equation gives us:
\[ a^2 = \frac{d_1^2}{4} + \frac{d_2^2}{4} \]
Multiplying through by 4 to eliminate the fractions:
\[ 4a^2 = d_1^2 + d_2^2 \]

6. **Final Result**:
Rearranging the equation, we find:
\[ d_1^2 + d_2^2 = 4a^2 \]
This shows that the sum of the squares of the diagonals is equal to four times the square of the side length of the rhombus.

### Conclusion:
Thus, if \( ABCD \) is a rhombus, we have shown that:
\[ AC^2 + BD^2 = 4AB^2 \]