# \ #
# \mbox{One can do this conveniently using the dot product,} \ \ \mbox{recalling its relationship to the magnitude of a vector, } \ \ \mbox{its commutativity, and its bilinearity}. #
# \mbox{In particular, recall that [ all multiplications here are} \ \ \mbox{dot product]:} #
# \mbox{i)} \qquad \ \ |a|^2=a\cdot a #
# \mbox{ii)} \qquad \ \ a \cdot
b = b \cdot a #
# \mbox{iii)} \quad \mbox{[bilinearity]} #
# \qquad \qquad \qquad a \cdot (b\pm c)=(a \cdot b) \pm (a \cdot c) #
# \qquad \qquad \qquad (a \pm b) \cdot c=(a \cdot c) \pm (b \cdot c). #
# \ #
# \mbox{[Each of these fundamental properties can be easily} \ \ \mbox{established by writing the vectors in coordinate form,} \ \ \mbox{and then computing each item in the equations.} #
# \mbox{Allow me to defer these simpler proofs to an Addendum,} \ \ \mbox{which follows the main argument here.]} #
# \ #
# \underline{ \mbox{Main Argument} } #
# \mbox{Using the above properties, we simply compute:} #
# \mbox{1)} \qquad |a+b|^2 = (a+b) \cdot \underbrace{(a+b)}_{c} #
# \qquad \qquad \qquad \qquad \qquad = [ a \cdot \underbrace{(a+b)}_{c} ] + #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad [ b \cdot \underbrace{(a+b)}_{c} ] #
# = (a \cdot a) + (a \cdot b) + (b \cdot a) + (b \cdot b) #
# = (a \cdot a) + (a \cdot b) + (a \cdot b) + (b \cdot b) #
# = (a \cdot a) + 2 (a \cdot b) + (b \cdot b) #
# = |a|^2 + 2 (a \cdot b) + |b|^2. #
# \mbox{2) Similarly:} #
# \qquad \qquad |a-b|^2 = |a|^2 - 2 (a \cdot b) + |b|^2. #
# \mbox{3) Combining (1) and (2) above:} #
# \qquad \qquad |a+b|^2 + |a-b|^2 = 2 |a|^2 + 2 |b|^2. \quad \mbox{QED} #
# \ #
# \ #
# \ #
# \underline{\mbox{Addendum:}} \qquad
\mbox{Proofs of Properties (i)--(iii) above.} #
# \mbox{i)} \qquad ( x \hat{i} + y \hat{j} ) \cdot ( x \hat{i} + y \hat{j} ) = #
# \qquad \qquad \qquad x^2 + y^2 = |x \hat{i} + y \hat{j} |^2. #
# \mbox{ii)} \qquad ( p \hat{i} + q \hat{j} ) \cdot ( x \hat{i} + y \hat{j} ) = #
# \qquad \qquad \qquad px + qy = xp + yq = #
# \qquad \qquad \qquad \qquad ( x \hat{i} + y \hat{j} ) \cdot ( p \hat{i} + q \hat{j} ). #
# \mbox{iii)} \qquad \ underbrace{ ( p \hat{i} + q \hat{j} ) }_{a} \cdot ( ( s \hat{i} + t \hat{j} ) \pm ( x \hat{i} + y \hat{j} ) ) #
# \qquad \qquad \qquad \quad = ( p \hat{i} + q \hat{j} ) \cdot ( ( s \pm x) \hat{i} + ( t \pm y ) \hat{j} ) #
# \qquad \qquad \qquad \quad = p(s \pm x) + q(t \pm y) #
# \qquad \qquad \qquad \quad = ( ps + qt ) \pm (px + qy) #
# \qquad \qquad \quad = ( \underbrace{ ( p \hat{i} + q \hat{j} ) }_{a} \cdot ( s \hat{i} + t \hat{j} ) ) \pm #
# \qquad \qquad \qquad \qquad \qquad \qquad ( \underbrace{ ( p \hat{i} + q \hat{j} ) }_{a} \cdot ( x \hat{i} + y \hat{j} ) ). #
# \mbox{And similarly for the case:} #
# \qquad \quad \quad ( ( p \hat{i} + q \hat{j} ) \pm ( s \hat{i} + t \hat{j} ) ) \cdot \underbrace{ ( x \hat{i} + y \hat{j} ) }_{c}. #
