In an isosceles `DeltaABC` right angled at B, then:

Options

A. `AB^2=2AC^2`
B. `AB^2=AC^2`
C. `AB^2=(AC^2)/4`
D. `AC^2=2AB^2`

Answer

To solve the problem regarding the isosceles triangle \( \Delta ABC \) which is right-angled at \( B \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Triangle**:
We have an isosceles triangle \( \Delta ABC \) with a right angle at \( B \). This means that \( AB = BC \) (the two sides opposite the equal angles).

2. **Labeling the Sides**:
Let \( AB = BC = x \). Since \( \Delta ABC \) is right-angled at \( B \), the hypotenuse \( AC \) will be opposite the right angle.

3. **Applying the Pythagorean Theorem**:
According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse \( AC \) is equal to the sum of the squares of the other two sides:
\[ AC^2 = AB^2 + BC^2 \]

4. **Substituting the Values**:
Since \( AB = x \) and \( BC = x \), we can substitute these values into the equation:
\[ AC^2 = x^2 + x^2 \]

5. **Simplifying the Equation**:
This simplifies to:
\[ AC^2 = 2x^2 \]

6. **Finding the Relationship**:
We can express \( AC \) in terms of \( x \):
\[ AC = \sqrt{2x^2} = x\sqrt{2} \]

### Final Answer:
Thus, the relationship between the sides of the triangle is:
\[ AC^2 = 2 \cdot AB^2 \]