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

Explanation

Let ABC be a triangle and N be the midpoints of the sides AB and AC.
then we have to show that MN is parallel to BC and
`l(MN)=(1)/(2)*l(BC)`.
Let `bar(a),bar(b),bar(c),bar(m)andbar(n)` be the position
vectors of A,B,C, M and N respectively. Since M abd N are the midpoints of AB and AC respectively,
`bar(m)=(bar(a)+bar(b))/(2)andbar(n)=(bar(a)+bar(c))/(2)`
`:.bar(MN)=bar(n)-bar(m)`
`=((bar(a)+bar(c))/(2))-((bar(a)+bar(b))/(2))`
`=(1)/(2)(bar(a)+bar(c)-bar(a)-bar(b))`
`=(1)/(2)(bar(c)-bar(b))=(1)/(2)bar(BC)`
Thus `bar(MN)` is non-zero scalar multiple of `bar(BC)`.
`:.bar(MN)` is parallel to `bar(BC)`.
`:.` seg MN is parallel to sed BC.
Also, `|bar(MN)|" ":.l(MN)=(1)/(2)*l(BC)`.