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.


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