Question

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

Answer 1

Given: Quadrilateral ABCD, in which ∠A + ∠D = 90°

To prove: AC² + BD² = AD² + BC²

Construct: Produce AB and CD to meet at E

Also join AC and BD

Proof: 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 + EC²

On adding both equations, we get

AD² + BC² = AE² + DE² + BE² + CE²  ........(i)

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  ........(ii)

From Eqs. (i) and (ii)

AC² + BD² = AD² + BC²

Hence proved.