So I tried solving this problem and ended up with the solution as:
I showed this by showing that two diagonals intersect at midpoints.
And made the vertices as ABCD which would then be, AC and BD.
Let M1 be the midpoint of AC and M2 be the midpoint of BD.
Where we need to show that M1 = M2 -> AM1 = AM2.
Now, using vectors we have
AM1 = 1/2AC
We need to use the fact that ABCD is a parallelogram. From this we get
AB = DC -> AB = 1/2 AB + 1/2AB = 1/2AB + 1/2DC
AM2 = AB + 1/2BD = 1/2AB + 1/2 DC + 1/2 BD = 1/2 (AB + BD + DC) = 1/2 AC
Where we have shown that AM1 = AM2.
Would this be the right approach to answer the question below?
What don't I understand is that there are no vertices to name and only w and v are given?
Thanks for your help in advance!
Best Answer
I have a simpler solution:
Draw a parallelogram, one diagonal coincident to x-axis and the intersect of two diagonals on origin. One diagonal is divided to be $\vec{a}$ and m$\vec{a}$, the other is $\vec{b}$ and n$\vec{b}$.
click here to see the parallelogram
$\vec{a}$ + $\vec{l}$ = $\vec{b}$ and $\vec{l}$ + m$\vec{a}$ = n$\vec{b}$
(where m and n are scalars)
$\vec{a}$ - $\vec{b}$ = m$\vec{a}$ - n$\vec{b}$
(m-1)$\vec{a}$ = (n-1)$\vec{b}$
$\vec{a}$=$\begin{pmatrix} |a| \\ 0\end{pmatrix}$ $\vec{b}$=$\begin{pmatrix} |b|cos\theta \\ |b|sin\theta\end{pmatrix}$
y direction:
$0=(n-1)|b|sin\theta$
In the parallelogram, $\theta>0$ and $|b|>0$
$\therefore n=1$ and $\vec{b}=n\vec{b}$
diagonal b is bisected.
$\vec{a}=m\vec{a}$
diagonal a is bisected.