In a quadrilateral ABCD, with unequal sides if the diagonals AC and BD intersect at right angles, then

Options

A. `AB^(2)+BC^(2)=CD^(2)+DA^(2)`
B. `AB^(2)+CD^(2)=BC^(2)+DA^(2)`
C. `AB^(2)+AD^(2)=BC^(2)+CD^(2)`
D. `AB^(2)+BC^(2)=2(CD^(2)+DA^(2))`

Answer

To solve the problem step by step, we will use the properties of triangles and the Pythagorean theorem.

### Step-by-Step Solution:

1. **Identify the Quadrilateral and Diagonals**:
We have a quadrilateral ABCD with diagonals AC and BD intersecting at point M. We know that these diagonals intersect at right angles.

2. **Apply Pythagorean Theorem in Triangle CDM**:
In triangle CDM, we can apply the Pythagorean theorem:
\[ CD^2 = DM^2 + CM^2 \quad \text{(Equation 1)} \]

3. **Apply Pythagorean Theorem in Triangle ADM**:
In triangle ADM, we also apply the Pythagorean theorem:
\[ AD^2 = AM^2 + DM^2 \quad \text{(Equation 2)} \]

4. **Apply Pythagorean Theorem in Triangle ABM**:
In triangle ABM, we use the Pythagorean theorem:
\[ AB^2 = AM^2 + BM^2 \quad \text{(Equation 3)} \]

5. **Apply Pythagorean Theorem in Triangle BCM**:
In triangle BCM, we apply the Pythagorean theorem:
\[ BC^2 = BM^2 + CM^2 \quad \text{(Equation 4)} \]

6. **Add All Four Equations**:
Now, we add all four equations together:
\[ (DM^2 + CM^2) + (AM^2 + DM^2) + (AM^2 + BM^2) + (BM^2 + CM^2) = CD^2 + AD^2 + AB^2 + BC^2 \]

This simplifies to:
\[ 2DM^2 + 2CM^2 + 2AM^2 + 2BM^2 = CD^2 + AD^2 + AB^2 + BC^2 \]

7. **Factor Out the Common Terms**:
We can factor out the 2 from the left side:
\[ 2(DM^2 + CM^2 + AM^2 + BM^2) = CD^2 + AD^2 + AB^2 + BC^2 \]

8. **Divide Both Sides by 2**:
Dividing both sides by 2 gives:
\[ DM^2 + CM^2 + AM^2 + BM^2 = \frac{1}{2}(CD^2 + AD^2 + AB^2 + BC^2) \]

9. **Rearranging the Equation**:
Rearranging the terms leads us to:
\[ AB^2 + CD^2 = AD^2 + BC^2 \]

### Conclusion:
Thus, we conclude that in quadrilateral ABCD, if the diagonals AC and BD intersect at right angles, then:
\[ AB^2 + CD^2 = AD^2 + BC^2 \]