In the given figure, diagonals AC and BD intersect at right angle. Show that :

AB² + CD² = AD² + BC²

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle AOB,

⇒ AB² = OB² + OA² ……….(1)

In right angle triangle COD,

⇒ CD² = OD² + OC² ……….(2)

In right angle triangle AOD,

⇒ AD² = OD² + OA² ……….(3)

In right angle triangle BOC,

⇒ BC² = OB² + OC² ……….(4)

Adding equations (1) and (2), we get :

⇒ AB² + CD² = OB² + OA² + OD² + OC²

⇒ AB² + CD² = (OA² + OD²) + (OC² + OB²)

Substituting value from (3) and (4) in above equation, we get :

⇒ AB² + CD² = AD² + BC².

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