In a rhombus $ABCD$,prove that $AC^{2} + BD^{2} = 4AB^{2}$.

(N/A) Let the diagonals $AC$ and $BD$ of the rhombus $ABCD$ intersect at point $O$.
Since the diagonals of a rhombus bisect each other at right angles $(90^{\circ})$,we have $AO = OC = \frac{AC}{2}$ and $BO = OD = \frac{BD}{2}$.
In the right-angled triangle $\triangle AOB$,by the Pythagoras theorem:
$AB^{2} = AO^{2} + BO^{2}$
Substitute the values of $AO$ and $BO$:
$AB^{2} = (\frac{AC}{2})^{2} + (\frac{BD}{2})^{2}$
$AB^{2} = \frac{AC^{2}}{4} + \frac{BD^{2}}{4}$
$AB^{2} = \frac{AC^{2} + BD^{2}}{4}$
Multiplying both sides by $4$,we get:
$4AB^{2} = AC^{2} + BD^{2}$
Hence,it is proved that $AC^{2} + BD^{2} = 4AB^{2}$.