In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB² + CD² = AD² + BC².

By pythagoras theorem,

In right angle triangle AOB,

AB² = OB² + OA² …….(i)

In right angle triangle COD,

CD² = OC² + OD² …….(ii)

In right angle triangle AOD,

AD² = AO² + OD² …….(iii)

In right angle triangle BOC,

BC² = OB² + OC² …….(iv)

Adding (i), (ii) we get,

AB² + CD² = OB² + OA² + OC² + OD²

AB² + CD² = (OA² + OD²) + (OC² + OB²)

Substituting value from (iii) and (iv) in above equation we get,

AB² + CD² = AD² + BC².

Hence, proved that AB² + CD² = AD² + BC².