Question
If $AD$ is the median of $\triangle\text{ABC},$ using vectors, prove that $\text{AB}^2+\text{AC}^2=2\big(\text{AD}^2+\text{CD}^2\big).$
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Answer
Take A as origin, let the position vectors of B and C are $\vec{\text{b}}$ and $\vec{\text{c}}$ respectively. position vector of $\text{D} = \frac{\vec{\text{b}} + \vec{\text{c}}}{2},\vec{\text{AB}} = \vec{\text{b}}$ and $\vec{\text{AC}} = \vec{\text{c}.}$$\vec{\text{AD}} = \frac{\vec{\text{b}} + \vec{\text{c}}}{2} - 0 = \frac{\vec{\text{b}} +\vec{\text{c}}}{2}$
consider,$ 2 (AD^2 + CD^2)$
$= 2\Big[\Big(\frac{\vec{\text{b}}+ \vec{\text{c}}}{2}\Big)^{2} + \Big(\frac{\vec{\text{b}} + \vec{\text{c}}}{2}-\vec{\text{c}}\Big)^{2}\Big]$
$=2\Big[\Big(\frac{\vec{\text{b}} + \vec{\text{c}}}{2}\Big)^{2}\Big(\frac{\vec{\text{b}} - \vec{\text{c}}}{2}\Big)^{2}\Big]$
$=\frac{1}{2}\big[\big(\vec{\text{b}} + \vec{\text{c}}\big)^{2} + \big(\vec{\text{b}} - \vec{\text{c}}\big)^{2}\big]$
$= (\vec{\text{b}})^{2} + (\vec{\text{c}})^{2}$
$= \big(\vec{\text{AB}}\big)^{2} + \big(\vec{\text{AC}}\big)^{2}$
$= \text{AB}^2 + \text{AC}^2$
Hence proved.
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