Given : A quad. ABCD in which, ∠A + ∠D = 90°
To Prove : AC2 + BD2 = AD2 + BC2.
Construction : Produce AB and DC intersect each other at E.
Proof : ∠A + ∠D = 90°
In triangle ADE
∠A + ∠D + ∠E = 180°
⇒ 90° + ∠E = 180°
⇒ ∠E = 180° – 90° = 90°
In right triangle BEC
BC2 = BE2 + CE2
(By Pythagoras Theorem)
AD2 = AF2 + DE2 …..(ii)
Adding equation (i) and (ii)
BC2 + AD2 = BE2 + AE2 + CE2 + DE2 …..(iii)
In right ∆AEC,
AC2 = AE2 + CE2 …(iv)
In right angled triangle BED
BD2 = BE2 + DE2 ……(v)
From equation (iv) and (v)
AC2 + BD2 = BE2 + AE2 + CE2 + DE2 …..(vi)
From equations (iii) and (iv)
BC2 + AD2 = AC2 + BD2
⇒ AC2 + BD2 = AD2 + BC2.