Show that the line joining any vertex of a parallelogram to the mid points of an opposite side divides the opposite diagonal in the ratio 2:1.
I have seen many questions related to this. But no question describes how to solve it using vectors.
Any help is appreciated.
Let $A,B,C,D$ be the vertices of the parallelogram with direct orientation, meaning that
$$\tag{0}\vec{AB}=\vec{DC}, \ \ \ \ \ \ \vec{AD}=\vec{BC}$$
(draw a figure!)
The line joining point $A$ to the midpoint $I$ of $BC$ has
$$\tag{1}\vec{AI}=\vec{AB}+\frac12 \vec{BC}$$
as its directing vector. We have to show that point $H$ defined by
$$\tag{2} \vec{AH}=\frac23 \vec{AI}$$
is such that
$$\tag{3}\vec{AH}=a\vec{AB} + (1-a) \vec{AD}$$
for a certain positive value $a$ ($H$, being on segment $BD$, is a weighted average of $B$ and $D$)
It shouldn't be difficult now, using (1),(2),(3) AND the second equation in (0), to get such an $a$.
Without loss of generality, let $M$ be a midpoint of side $AB$ in parallelogram $ABCD$.
Let $P$ is a point on diagonal $AC$, with $AP:PC = 1:2$.
Observe that $\vec{DM} = \frac12\vec{AB} - \vec{AD}$
Also, observe that $\frac13(\vec{AB}+\vec{AD})=\vec{AD}+\vec{DP}$
Rearranging, $\vec{DP} = \frac13\vec{AB}-\frac23\vec{AD}$
This is equal to $\frac23\vec{DM}$. Since $\vec{DP}$ is a scalar multiple of $\vec{DM}$, $DP$ and $DM$ must be collinear, and therefore $DM$ is indeed the line that divides $AC$ in $1:2$