The question
Let $ABCDA'B'C'D'$ be a cube, and the points $O$ and $O'$ are the centres of the faces $ABCD$ and $A'B'C'D'$ respectively. We consider the points $P\in (AD')$ and $Q \in (B'C)$ and denote by $M$ the midpoint of the segment $PQ$. Prove that $PQ \parallel (ABC)$ if and only if $M \in (OO')$.
My drawing
My idea
So we have only $2$ cases:
case $1$) We know that $PQ \parallel (ABC)$, we show that $M \in (OO')$.
case $2$) We know that $M \in (OO')$, we show that $PQ \parallel (ABC)$.
Even though I wasn't able to do any of these, I think that for the first case we have to show that any of $OM$ or $O'M$ are parallel to one of the heights of the cube, because $OO'$ is already parallel to them and this will mean by Euclid's theorem that $O, O', M$ are colinear and that $M\in OO'$. I think of this parallelism because $M$ and $O$ or $O'$ are already mid-points of $PQ$,$AC$ or $A'C'$.
A similar rationament would be for case $2$.
I tried showing this but it got me to nowhere.
I hope one of you can help me! Thank you!
Best Answer
We may suppose that the length of a side of the cube is $1$.
Let $E$ be a point on $AD$ such that $PE\perp AD$.
Let $F$ be a point on $BC$ such that $QF\perp BC$.
Let $G$ be the midpoint of $AB$.
Let $H$ be a point on the plane $ABCD$ such that $MH\perp (ABCD)$.
Then, we can see that $M$ is on the plane $OO'G$ which is parallel to the plane $AA'D'D$.
Let $s:=BF$ and $t:=AE$.
Then, we can write $QF=1-s$ and $PE=t$.
If $PQ\parallel (ABC)$, then $QF=PE$, so we have $1-s=t$. This can be written as $\frac{s+t}{2}=\frac 12$, so $\frac{BF+AE}{2}=\frac 12$, i.e. $HG=\frac 12$ (since we already know that $M$ is on the plane $OO'G$). So, we get $H=O$, and $M\in (OO')$ follows.
If $M\in (OO')$, then we have $H=O$ and $HG=\frac 12$, so $\frac{BF+AE}{2}=\frac 12$, i.e. $\frac{s+t}{2}=\frac 12$. This can be written as $1-s=t$, so $QF=PE$. Therefore, $PQ\parallel (ABC)$ follows.