Question

Prove by vector method that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer 1

 

Let ABCD be a parallelogram such that AC and BD are its two diagonals. Taking A as the origin, let the position vectors of B and D be \[\vec{b}\]  and \[\vec{d}\] respectively.

Then,   \[\vec{AB} = \vec{b}\]  and \[\vec{AD} = \vec{d}\]

Using triangle law of vector addition, we have

\[\vec{AD} + \vec{DB} = \vec{AB} \] 
\[ \Rightarrow \vec{DB} = \vec{b} - \vec{d}\] 

In ∆ABC,  

\[\vec{AC} = \vec{AB} + \vec{BC} = \vec{AB} + \vec{AD} = \vec{b} + \vec{d}\] 

Now, 

\[\left| \vec{AB} \right|^2 + \left| \vec{BC} \right|^2 + \left| \vec{CD} \right|^2 + \left| \vec{DA} \right|^2 \]
\[ = \left| \vec{AB} \right|^2 + \left| \vec{AD} \right|^2 + \left| - \vec{AB} \right|^2 + \left| - \vec{AD} \right|^2 \]
\[ = 2 \left| \vec{AB} \right|^2 + 2 \left| \vec{AD} \right|^2 \]
\[ = 2 \left| \vec{b} \right|^2 + 2 \left| \vec{d} \right|^2 . . . . . \left( 1 \right)\] 

Also, 

\[\left| \vec{DB} \right|^2 + \left| \vec{AC} \right|^2 \]
\[ = \left| \vec{b} - \vec{d} \right|^2 + \left| \vec{b} + \vec{d} \right|^2 \]
\[ = \left( \vec{b} - \vec{d} \right) . \left( \vec{b} - \vec{d} \right) + \left( \vec{b} + \vec{d} \right) . \left( \vec{b} + \vec{d} \right)\]
\[ = \left| \vec{b} \right|^2 - 2 \vec{b} . \vec{d} + \left| \vec{d} \right|^2 + \left| \vec{b} \right|^2 + 2 \vec{b} . \vec{d} + \left| \vec{d} \right|^2 \]
\[ = 2 \left| \vec{b} \right|^2 + 2 \left| \vec{d} \right|^2 . . . . . \left( 2 \right)\] 

From (1) and (2), we have 

\[\left| \vec{AB} \right|^2 + \left| \vec{BC} \right|^2 + \left| \vec{CD} \right|^2 + \left| \vec{DA} \right|^2 = \left| \vec{DB} \right|^2 + \left| \vec{AC} \right|^2\]