Question

In a quadrilateral ABCD, prove that \[{AB}^2 + {BC}^2 + {CD}^2 + {DA}^2 = {AC}^2 + {BD}^2 + 4 {PQ}^2\] where P and Q are middle points of diagonals AC and BD. 

Answer 1

 

Let ABCD be the quadrilateral. Taking A as the origin, let the position vectors of B, C and D be \[\vec{b} , \vec{c}\] and \[\vec{d}\] respectively. 

 Then, 

Position vector of P =\[\frac{\vec{c}}{2}\]........(Mid-point formula) 

Position vector of Q = \[\frac{\vec{b} + \vec{d}}{2}\]..............(Mid-point formula) 

Now, 

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

Also, 

\[{AC}^2 + {BD}^2 + 4 {PQ}^2 \]
\[ = \left| \vec{AC} \right|^2 + \left| \vec{BD} \right|^2 + 4 \left| \vec{PQ} \right|^2 \]
\[ = \left| \vec{c} \right|^2 + \left| \vec{d} - \vec{b} \right|^2 + 4 \left| \frac{\vec{b} + \vec{d}}{2} - \frac{\vec{c}}{2} \right|^2 \]
\[ = \left| \vec{c} \right|^2 + \left| \vec{d} - \vec{b} \right|^2 + \left| \vec{b} + \vec{d} \right|^2 - 2\left( \vec{b} + \vec{d} \right) . \vec{c} + \left| \vec{c} \right|^2 \]
\[ = 2 \left| \vec{c} \right|^2 + 2 \left| \vec{d} \right|^2 + 2 \left| \vec{b} \right|^2 - 2 \vec{b} . \vec{c} - 2 \vec{d} . \vec{c} \]
\[ = 2 \left| \vec{b} \right|^2 + 2 \left| \vec{c} \right|^2 + 2 \left| \vec{d} \right|^2 - 2 \vec{b} . \vec{c} - 2 \vec{c} . \vec{d} . . . . . \left( 2 \right)\] 

From (1) and (2), we have 

\[{AB}^2 + {BC}^2 + {CD}^2 + {DA}^2 = {AC}^2 + {BD}^2 + 4 {PQ}^2\]