The midpoints of the adjacent sides of a quadrilateral are joined. Show that the resulting figure is a parallelogram.
I know that you suppose to somehow use vectors to prove this, but I dont know how you are suppose to model this?
The midpoints of the adjacent sides of a quadrilateral are joined. Show that the resulting figure is a parallelogram.
I know that you suppose to somehow use vectors to prove this, but I dont know how you are suppose to model this?
Suppose that $\displaystyle A,~B,~C,~D$ are the consecutive vertices of the quadrilateral.
Let midpoints be $\displaystyle P \in \overrightarrow {AB},~ Q \in \overrightarrow {BC},~ R \in \overrightarrow {CD},~ S \in \overrightarrow {DA} $.
We know that $\displaystyle \overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CD}+\overrightarrow {DA}=0 $.
Moreover, $\displaystyle 0.5(\overrightarrow {AB})=\overrightarrow {PB}~\&~ 0.5(\overrightarrow {BC})=\overrightarrow {BQ} $
So $\displaystyle 0.5(\overrightarrow {AB}+\overrightarrow {BC})=\overrightarrow {PQ} $
Do the same thing for $\displaystyle \overrightarrow {RS} $
Prove that $\displaystyle \overrightarrow {PQ}=-\overrightarrow {RS} $
You will be done.
What if the quadrilateral you were doing this proof in was a trapezium then how would this be true? (since all the sides are unequal). How would the vectors of the sides add up 0?
Sorry but I dont understand why the inner side of a parallelogram = 2 x 1/2 of the sides of a quadrilateral.
This is a standard question. It is done in any geometry course.
I have given you the standard vector proof.
BTW. This theorem is true for any quadrilateral, even non-convex quadrilaterals.
And the proof is the same.
I suspect that you do not understand vector addition.
If that is the case, you no chance proving this.
As I said, the proof I gave is the standard.