Prove the diagonals of a parallelpiped bisect each other

Question

I am stuck on how to Prove the diagonals of a parallelpiped bisect each other I have been given the hint to make one of the corners O. If possible I would just like a push in the right direction. Thankyou in advance.

Answer 1

A parallelepiped can be defined by three vectors and its sums. Consider one of its vertices at $O$, then the position vector of the other vertices $A,B,C,D,E,F,G$ are

$\vec{OA}=\vec a\;;\vec{OB}=\vec b\;;\vec{OC}=\vec a+\vec b$

$\vec{OD}=\vec c\;;\vec{OE}=\vec c+\vec a\;;\vec{OF}=\vec c+\vec b\;;\vec{OG}=\vec c+\vec a+\vec b$

Now consider the diagonals $OG$ and $AF$. They bisect. To prove it we can use the formula to calculate the midpoint of a segment knowing the position vector of its endpoints.

Can you follow from here?

Added

Consider the position vector of the midpoint, $M_0$, of $OG$, $\vec{OM_0}=\dfrac12(\vec{OO}+\vec{OG})=\dfrac12(\vec a+\vec b+\vec c)$, it has to be the position vector of the midpoint, $M_1$, of $AF$, $\vec{OM_1}=\dfrac12(\vec{OF}+\vec{OA})=\dfrac12(\vec a+\vec b+\vec c)$ and it is proven.