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In a quadrilateral ABCD, with unequal sides if the diagonals AC and BD intersect at right angles, then
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- AB² + BC² = CD² + DA²
- AB² + CD² = BC² + DA²
- AB² + AD² = BC² + CD²
- AB² + BC² = 2(CD² + DA²)
- AB² + BC² = CD² + DA²
Correct Option: B
On the basis of question we draw a figure of quadrilateral ABCD
| OB² + OC² = BC² OC² + OD² = CD² OD² + OA² = AD² OA² + OB² = AB² |  | Pythagoras theorem |
∴ 2 (OB² + OA² + OD² + OC²) = AB² + BC² + CD² + DA²
⇒ 2(AB² + CD²) = AB² + BC² + CD² + DA²
⇒ AB² + CD² = BC² + DA²
