Question

If AD is the median of ∆ABC, using vectors, prove that \[{AB}^2 + {AC}^2 = 2\left( {AD}^2 + {CD}^2 \right)\] 

Answer 1

 

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

It is given that AD is the median of ∆ABC. 
∴ Position vector of mid-point of BC = \[\vec{AD} = \frac{\vec{b} + \vec{c}}{2}\]................(Mid-point formula) 

Now, 

\[{AB}^2 + {AC}^2 = \left| \vec{AB} \right|^2 + \left| \vec{AC} \right|^2 = \left| \vec{b} \right|^2 + \left| \vec{c} \right|^2\] 

Also, 

\[2\left( {AD}^2 + {CD}^2 \right)\]
\[ = 2\left( \left| \vec{AD} \right|^2 + \left| \vec{CD} \right|^2 \right)\]
\[ = 2\left[ \left( \frac{\vec{b} + \vec{c}}{2} \right) . \left( \frac{\vec{b} + \vec{c}}{2} \right) + \left( \frac{\vec{b} + \vec{c}}{2} - \vec{c} \right) . \left( \frac{\vec{b} + \vec{c}}{2} - \vec{c} \right) \right]\]
\[ = 2\left[ \left( \frac{\vec{b} + \vec{c}}{2} \right) . \left( \frac{\vec{b} + \vec{c}}{2} \right) + \left( \frac{\vec{b} - \vec{c}}{2} \right) . \left( \frac{\vec{b} - \vec{c}}{2} \right) \right]\]
\[ = \frac{\left| \vec{b} \right|^2 + 2 \vec{b} . \vec{c} + \left| \vec{c} \right|^2}{2} + \frac{\left| \vec{b} \right|^2 - 2 \vec{b} . \vec{c} + \left| \vec{c} \right|^2}{2}\]
\[ = \frac{2 \left| \vec{b} \right|^2 + 2 \left| \vec{c} \right|^2}{2}\]
\[ = \left| \vec{b} \right|^2 + \left| \vec{c} \right|^2 . . . . . \left( 2 \right)\] 

From (1) and (2), we have 

\[{AB}^2 + {AC}^2 = 2\left( {AD}^2 + {CD}^2 \right)\]