In the figure,if $AD \perp BC$,prove that $AB^2 + CD^2 = BD^2 + AC^2$.

(N/A) Given: In $\Delta ABC$,$AD \perp BC$.
To prove: $AB^2 + CD^2 = BD^2 + AC^2$.
Proof:
In right-angled $\Delta ADB$,by Pythagoras theorem:
$AB^2 = AD^2 + BD^2$ --- $(1)$
In right-angled $\Delta ADC$,by Pythagoras theorem:
$AC^2 = AD^2 + CD^2$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$AB^2 - AC^2 = (AD^2 + BD^2) - (AD^2 + CD^2)$
$AB^2 - AC^2 = AD^2 + BD^2 - AD^2 - CD^2$
$AB^2 - AC^2 = BD^2 - CD^2$
Rearranging the terms,we get:
$AB^2 + CD^2 = BD^2 + AC^2$
Hence proved.