Prove that the line segment joining the mid point of the two sides of a triangle is parallel to the third side and equal to half the third side.

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Solution

Let A be origin and $B (\to b), C (\to 0)$
$\Rightarrow$ D and E as mid-points would have p.vs $\frac{\to b}{2}$ and $\frac{\to c}{2}$
$\Rightarrow D E$ is parallel to $\frac{\to b - \to c}{2}$ or $\to b - \to c$
Also BC is parallel to $\to b - \to c$
$\Rightarrow D E | | B C; | \to D E | = \frac{| \to b - \to c |}{2}$ ;
$| \to B C | = | \to b - \to c |$
$\Rightarrow | \to D E | = \frac{| \to B C |}{2}$

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Prove that the line segment joining the mid point of the two sides of a triangle is parallel to the third side and equal to half the third side.

Let A be origin and B(→b),C(→0)⇒ D and E as mid-points would have p.vs →b2 and →c2⇒DE is parallel to →b−→c2 or →b−→cAlso BC is parallel to →b−→c⇒DE||BC;|→DE|=|→b−→c|2;|→BC|=|→b−→c|⇒|→DE|=|→BC|2