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

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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.

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