In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC² + BD² = AD² + BC².

Given,

Quadrilateral ABCD,

In which ∠A + ∠D = 90°

Construct produce AB and CD to meet at E.

Also, join AC and BD.

in ∆AED,

∠A + ∠D = 90° [given]

∴∠E = 180°- (∠A + ∠D) = 90° [∵ sum of angles of a triangle = 180°]

Then,

By Pythagoras theorem,

AD² = AE² + DE²

In ∆BEC,

by Pythagoras theorem,

BC² = BE² + EF²

On adding both equations, we get,

AD² + BC² = AE² + DE² + BE² + CE²

In ∆AEC,

by Pythagoras theorem,

AC² = AE² + CE²

And in ∆BED,

by Pythagoras theorem,

BD² = BE² + DE²

On adding both equations, we get,

AC² + BD² = AE² + CE² + BE² + DE² ….(i)

From Equation (i) and (ii),

AC² + BD² = AD² + BC²

Hence proved.