Prove some results in a cube...

Question

Question

Let $ABCDA'B'C'D'$ be a cube and the points $M, N, Q$ the means of the sides $A'B', A'D', DC$. We denote by $\alpha=(MNQ)$.

a) If the line $D'C'$ intersects the plane $\alpha$ at the point $T$, prove that $A'T\equiv MD'$

b) Let $P$ be the intersection of the plane $\alpha$ with $DD'$ and let $S$ be the intersection of the plane $\alpha$ with $BB'$. Prove that the quadrilateral $PQSM$ is a rectangle

Idea

drawing

The first thought that came into my mind was showing that $A'MD'T$ is a parallelogram.

Because $T\in D'C' => TD'|| A'M$

We also know that $N$ is mid point of $A'D'$

Using the midline theorem we can express $NM=\frac{l\sqrt{2}}{2}$, where $l$ is the side of the cube.

If we can show that $TN$ is also equal to $NM$ the first point will be done, but I don't know where to start and what to do(I'm kinda new to the geometry in space).

answer using @Calvin Lin's idea

Hope one of you can help! Thank you!

Answer 1

I think your proof is fine.

In the following, I'm going to write some comments and another proof.

Some comments : In your proof that $M,N,T$ are collinear, it might be better to explicitly say that you used the fact that given three non collinear points, only one plane goes through them. Also, it might be better to add a few explanations about why $PQSM$ is a rectangle when $PQ=MS,PQ\parallel MS$ and $PS=MQ$.

Another proof that $M,N,T$ are collinear : Using the fact that the intersection of two non-parallel planes in three-dimensional space is a line, we can say that the intersection of $\alpha$ and the plane $A'B'C'D'$ is the line $MN$. It follows that $M,N,T$ are collinear. (Also, it follows that $M,N,T'$ are collinear.)

Answer 2

Steps towards a solution. Fill in the details as needed.
If you're stuck, explain what you've tried, and why it doesn't work.
Geometry in 2-D is often easier than geometry in 3-D, so try to force things to happen on a plane as much as possible.

a) You're on the right track.
Show that $ T = MN \cap D'C'$. -> This demystifies the intersection of the plane and some line.
Hence $TND' \cong MNA'$ by ASA.
Hence $A'T = MD'$.

b) Let $R$ be the midpoint of $BC$.
Show that $MNQR$ is in the same plane, and form a rectangle.
Show that $P$ is the midpoint of $DD'$, and $S$ is the midpoint of $BB'$. -> This is arguably the crux.
Show that $MNPQRS$ is a regular hexagon.
Hence $PQSM$ is also a rectangle.