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

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

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

Answer»

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.