Prove that in a parallelogram,the sum of squares of all the sides is equal to the sum of squares of the diagonals.

(N/A) Let $ABCD$ be a parallelogram with diagonals $AC$ and $BD$ intersecting at $O$.
By the Parallelogram Law,we know that the sum of the squares of the four sides is equal to the sum of the squares of the two diagonals.
Let the sides be $AB$,$BC$,$CD$,and $DA$. Since $ABCD$ is a parallelogram,$AB = CD$ and $BC = DA$.
According to Apollonius Theorem in $\triangle ABC$ and $\triangle ADC$ with median $BO$ (where $O$ is the midpoint of $AC$):
$AB^2 + BC^2 = 2(BO^2 + AO^2)$
$AD^2 + CD^2 = 2(DO^2 + AO^2)$
Since $O$ is the midpoint of $BD$,$BO = DO$.
Adding the two equations:
$AB^2 + BC^2 + CD^2 + DA^2 = 2(BO^2 + AO^2 + DO^2 + AO^2) = 2(2BO^2 + 2AO^2) = 4BO^2 + 4AO^2$.
Since $BD = 2BO$ and $AC = 2AO$,then $BD^2 = 4BO^2$ and $AC^2 = 4AO^2$.
Therefore,$AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$.