As shown in the adjoining diagram,$\angle B$ is an obtuse angle and $\overline{AM}$ is an altitude in $\Delta ABC$. Prove that $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.
(N/A) In $\Delta AMC$,$\angle AMC = 90^{\circ}$. By Pythagoras theorem,$AC^{2} = AM^{2} + MC^{2}$.
Since $MC = MB + BC$,we have $AC^{2} = AM^{2} + (MB + BC)^{2}$.
Expanding the square,$AC^{2} = AM^{2} + MB^{2} + BC^{2} + 2 \cdot MB \cdot BC$.
In $\Delta AMB$,$\angle AMB = 90^{\circ}$. By Pythagoras theorem,$AB^{2} = AM^{2} + MB^{2}$.
Substituting $AB^{2}$ for $AM^{2} + MB^{2}$ in the equation,we get $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.
Since $MC = MB + BC$,we have $AC^{2} = AM^{2} + (MB + BC)^{2}$.
Expanding the square,$AC^{2} = AM^{2} + MB^{2} + BC^{2} + 2 \cdot MB \cdot BC$.
In $\Delta AMB$,$\angle AMB = 90^{\circ}$. By Pythagoras theorem,$AB^{2} = AM^{2} + MB^{2}$.
Substituting $AB^{2}$ for $AM^{2} + MB^{2}$ in the equation,we get $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.
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