I am looking for a way to prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. However, I would like to know whether it is possible to prove without vectors, Pythagoras theorem, cosine theorem, and trigonometric functions. Thus, I'd like to know the more elementary proof of this property.
P.S.: I know the easy way to prove this via the cosine theorem:
Consider parallelogram $ABCD$ with $BC\parallel AD$, $AB\parallel CD$.
In triangle $ABD$, by cosine theorem:
$BD^2 = AB^2+AD^2-2\cdot AB\cdot AD\cdot\cos\angle BAD$.
In triangle $ADC$, by cosine theorem:
$AC^2=CD^2+AD^2-2\cdot CD\cdot AD\cdot\cos\angle ADC$
Summing up these two equalities we get:
$AC^2+BD^2=CD^2+AD^2+AB^2+AD^2-2\cdot CD\cdot AD\cdot\cos\angle ADC-2\cdot AB\cdot AD\cdot\cos\angle BAD$
By properties of parallelogram, $AB = CD$, and $\cos\angle BAD=\cos(180^\circ - \angle ADC) = -\cos\angle ADC$
Then we have:
$AC^2+BD^2=2AB^2+2AD^2+2\cdot AB\cdot AD\cdot\cos\angle ADC-2\cdot AB\cdot AD\cdot\cos\angle ADC$
It follows that:
$AC^2+BD^2=2(AB^2+AD^2)$
QED.