Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

(A) Let $ABCD$ be a parallelogram.
Let us draw perpendicular $DE$ on the extended side $AB$,and $AF$ on side $DC$.
Applying Pythagoras theorem in $\triangle DEA$,we obtain:
$DE^{2} + EA^{2} = DA^{2} \dots (i)$
Applying Pythagoras theorem in $\triangle DEB$,we obtain:
$DE^{2} + EB^{2} = DB^{2}$
$DE^{2} + (EA + AB)^{2} = DB^{2}$
$(DE^{2} + EA^{2}) + AB^{2} + 2EA \times AB = DB^{2}$
$DA^{2} + AB^{2} + 2EA \times AB = DB^{2} \dots (ii)$
Applying Pythagoras theorem in $\triangle AFC$,we obtain:
$AC^{2} = AF^{2} + FC^{2}$
$AC^{2} = AF^{2} + (DC - FD)^{2}$
$AC^{2} = AF^{2} + DC^{2} + FD^{2} - 2DC \times FD$
$AC^{2} = (AF^{2} + FD^{2}) + DC^{2} - 2DC \times FD$
$AC^{2} = AD^{2} + DC^{2} - 2DC \times FD \dots (iii)$
Since $ABCD$ is a parallelogram,$AB = CD \dots (iv)$ and $BC = AD \dots (v)$.
In $\triangle DEA$ and $\triangle AFD$:
$\angle DEA = \angle AFD = 90^{\circ}$
$\angle EAD = \angle ADF$ (since $EA \parallel DF$)
$AD = AD$ (Common side)
Therefore,$\triangle EAD \cong \triangle FDA$ ($AAS$ congruence criterion).
This implies $EA = DF \dots (vi)$.
Adding equations $(ii)$ and $(iii)$:
$DB^{2} + AC^{2} = DA^{2} + AB^{2} + 2EA \times AB + AD^{2} + DC^{2} - 2DC \times FD$
Using $AB = DC$ and $EA = DF$:
$DB^{2} + AC^{2} = AD^{2} + AB^{2} + 2EA \times AB + AD^{2} + AB^{2} - 2AB \times EA$
$DB^{2} + AC^{2} = 2AD^{2} + 2AB^{2}$
Since $AD = BC$ and $AB = CD$:
$AC^{2} + BD^{2} = AB^{2} + BC^{2} + CD^{2} + DA^{2}$.