$\Delta ABC$ is an acute-angled triangle and $\overline{AM}$ is an altitude. Prove that $AC^{2} = AB^{2} + BC^{2} - 2 \cdot BC \cdot BM$.

(N/A) In $\Delta ABC$,$\overline{AM}$ is the altitude to side $\overline{BC}$,so $\angle AMB = 90^{\circ}$ and $\angle AMC = 90^{\circ}$.
In $\Delta AMC$,by Pythagoras theorem: $AC^{2} = AM^{2} + MC^{2}$.
Since $M$ lies on $BC$,we have $MC = BC - BM$.
Substituting this into the equation: $AC^{2} = AM^{2} + (BC - BM)^{2}$.
Expanding the square: $AC^{2} = AM^{2} + BC^{2} + BM^{2} - 2 \cdot BC \cdot BM$.
In $\Delta AMB$,by Pythagoras theorem: $AB^{2} = AM^{2} + BM^{2}$.
Substituting $AM^{2} + BM^{2} = AB^{2}$ into the equation: $AC^{2} = AB^{2} + BC^{2} - 2 \cdot BC \cdot BM$.
Hence,the result is proved.