In the figure,$ABC$ is a triangle in which $\angle ABC < 90^{\circ}$ and $AD \perp BC$. Prove that $AC^{2} = AB^{2} + BC^{2} - 2BC \cdot BD$.

(N/A) Applying the Pythagoras theorem in $\triangle ADB$,we obtain:
$AD^{2} + DB^{2} = AB^{2}$
$\Rightarrow AD^{2} = AB^{2} - DB^{2} \dots(1)$
Applying the Pythagoras theorem in $\triangle ADC$,we obtain:
$AD^{2} + DC^{2} = AC^{2}$
Substituting the value of $AD^{2}$ from equation $(1)$:
$(AB^{2} - BD^{2}) + DC^{2} = AC^{2}$
Since $D$ lies on $BC$,we have $DC = BC - BD$. Substituting this into the equation:
$AB^{2} - BD^{2} + (BC - BD)^{2} = AC^{2}$
$AB^{2} - BD^{2} + (BC^{2} + BD^{2} - 2BC \cdot BD) = AC^{2}$
$AB^{2} + BC^{2} - 2BC \cdot BD = AC^{2}$
Thus,$AC^{2} = AB^{2} + BC^{2} - 2BC \cdot BD$.