In a quadrilateral $ABCD$,$\angle A + \angle D = 90^{\circ}$. Prove that $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $\angle A + \angle D = 90^{\circ}$.
To prove: $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Construction: Produce $AB$ and $DC$ to meet at point $E$.
Proof: In $\triangle AED$,$\angle A + \angle D = 90^{\circ}$ (given).
Since the sum of angles in a triangle is $180^{\circ}$,we have $\angle E = 180^{\circ} - (\angle A + \angle D) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
In right-angled $\triangle AED$,by Pythagoras theorem: $AD^{2} = AE^{2} + DE^{2}$ ... $(i)$
In right-angled $\triangle BEC$,by Pythagoras theorem: $BC^{2} = BE^{2} + CE^{2}$ ... $(ii)$
Adding $(i)$ and $(ii)$,we get: $AD^{2} + BC^{2} = AE^{2} + DE^{2} + BE^{2} + CE^{2}$ ... $(iii)$
In right-angled $\triangle AEC$,by Pythagoras theorem: $AC^{2} = AE^{2} + CE^{2}$ ... $(iv)$
In right-angled $\triangle BED$,by Pythagoras theorem: $BD^{2} = BE^{2} + DE^{2}$ ... $(v)$
Adding $(iv)$ and $(v)$,we get: $AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2}$ ... $(vi)$
Comparing $(iii)$ and $(vi)$,we get: $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Hence proved.