Question

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

AB² + CD² = AD² + BC²

Answer 1

In right-angled triangle ΔAOB, applying Pythagoras’ theorem gives:

AB² = OA² + OB²

In right-angled triangle ΔCOD, applying Pythagoras’ theorem gives:

CD² = OC² + OD²

Adding the two equations together:

AB² + CD² = (OA² + OB²) + (OC² + OD²)

AB² + CD² = OA² + OD² + OB² + OC²

Since OA² + OD² = AD² (from right-angled ΔAOD) and OB² + OC² = BC² (from right-angled ΔBOC), substituting these yields:

AB² + CD² = AD² + BC²