If AD is the median of ∆ABC, using vectors, prove that ${AB}^{2} + {AC}^{2} = 2 ({AD}^{2} + {CD}^{2})$ .

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Solution

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,

${AB}^{2} + {AC}^{2} = {|\vec{AB}|}^{2} + {|\vec{AC}|}^{2} = {|\vec{b}|}^{2} + {|\vec{c}|}^{2}$ .....(1)

Also,

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

From (1) and (2), we have

${AB}^{2} + {AC}^{2} = 2 ({AD}^{2} + {CD}^{2})$

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Δ A B C,

A B^{2} + A C^{2} = 2 A D^{2} + 2 B D^{2} ?

Δ A B C

A D

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{A B}^{2} + {A C}^{2} = 2 {A D}^{2} + \frac{1}{2} {B C}^{2}

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