45-45 right triangle and Pythagoras theorem

ABCD is a square and E is the intersection point of the diagonals . If N is any point on AE , show that $\displaystyle A{{B}^{2}}-B{{N}^{2}}=AN.NC$ .

Solution

$\displaystyle \text{Let}\ AB=BC=x\ \ \text{and}\ \ NE=y$

$\displaystyle \Delta \ BCE\ \ \text{is}\ \ {{45}^{\circ }}\text{-}\ {{45}^{\circ }}\ rt\ \Delta .$ 

$\displaystyle \therefore \ BE=CE=\frac{x}{{\sqrt{2}}}$ 

$\displaystyle A{{B}^{2}}-B{{N}^{2}}={{x}^{2}}-\left( {B{{E}^{2}}+N{{E}^{2}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}-\left( {{{{\left( {\frac{x}{{\sqrt{2}}}} \right)}}^{2}}+{{y}^{2}}} \right)$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}-\frac{{{{x}^{2}}}}{2}-{{y}^{2}}$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}{{x}^{2}}-{{y}^{2}}$ 

$\displaystyle AN.NC=\left( {AE-NE} \right)\left( {CE+NE} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\left( {\frac{x}{{\sqrt{2}}}-y} \right)\left( {\frac{x}{{\sqrt{2}}}+y} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}{{x}^{2}}-{{y}^{2}}$ 

$\displaystyle \therefore \ A{{B}^{2}}-B{{N}^{2}}=AN.NC$