Given, quadrilateral ABCD, in which
∠A+∠D=90∘ To prove :
AC2+BD2=AD2+BC2 Construction : Produce AB and CD to meet at E.
Also, join AC and BD.
Proof :
In
ΔAED,
∠A+∠D=90∘ [given] ∴∠E=180∘−(∠A+∠D)=90∘ [because sum of all angles in a triangle=180∘] Then, by using Pythagoras theorem,
AD2=AE2+DE2 ..(i) In ΔBEC,
by using Pythagoras theorem,
BC2=BE2+EC2...(ii) On adding equations (i) and (ii), we get,
AD2+BC2=AE2+DE2+BE2+CE2 ...(iii) In ΔAEC,
by using Pythagoras theorem,
AC2=AE2+CE2 ...(iv) & in DeltaBED,
by using Pythagoras theorem,
BD2=BE2+DE2 ...(v) On adding both equations, we get,
AC2+BD2=AE2+CE2+BE2+DE2 ...(vi) From Eq. (iii) and (vi), we get,
AC2+BD2=AD2+BC2