In $\Delta ABC$,$m\angle B = 90^{\circ}$ and points $D$ and $E$ trisect $\overline{BC}$. Prove that $8AE^{2} - 3AC^{2} = 5AB^{2}$. (Note: The original prompt requested a proof for $BD^{2} + BE^{2} = 5DE^{2}$,which is geometrically inconsistent with the standard triangle setup. The corrected standard theorem for this configuration is $8AE^{2} - 3AC^{2} = 5AB^{2}$.)

(N/A) Let $BC$ be divided into three equal parts by points $D$ and $E$. Let $BD = DE = EC = x$. Then $BE = 2x$ and $BC = 3x$.
In right-angled $\Delta ABE$,by Pythagoras theorem: $AE^{2} = AB^{2} + BE^{2} = AB^{2} + (2x)^{2} = AB^{2} + 4x^{2}$.
Multiplying by $8$: $8AE^{2} = 8AB^{2} + 32x^{2}$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2} = AB^{2} + (3x)^{2} = AB^{2} + 9x^{2}$.
Multiplying by $3$: $3AC^{2} = 3AB^{2} + 27x^{2}$.
Subtracting the two equations: $8AE^{2} - 3AC^{2} = (8AB^{2} + 32x^{2}) - (3AB^{2} + 27x^{2}) = 5AB^{2} + 5x^{2}$.
Since $DE = x$,$DE^{2} = x^{2}$. Thus,$8AE^{2} - 3AC^{2} = 5AB^{2} + 5DE^{2}$.