1. Skew Lines

Let L1 and L2 be skew lines; let P1, and P2 be any two distinct points on L1 and let Q1 and Q2 be any two distinct points on L2; let M be a plane parallel to both L1 and L2. Show that the intersections of M with the four lines determined by the line-segments P1Q1, P1Q2,P2Q1, and P2Q2 form the vertices of a parallelogram.

I tried to solve this for awhile. I started writing things that I knew about the situation down, and I found that the (shortest) distance between the plane and either line was constant (though a different constant for both lines). I figured out a few more things but I am not sure if there are necessary for the argument.

2. Very hard to see, but let me give it a try. But beware: how you draw the figure can make it difficult to see.

I'll make more definitions:
Let distance of L1 from M be p.
Let distance of L1 from M be q.
Let normal from P1 intersects M at A1.
Let normal from P2 intersects M at A2.
Let normal from Q1 intersects M at B1.
Let normal from Q2 intersects M at B2.

Let P1Q1 intersects M at X.
Let P1Q2 intersects M at Y.
Let P2Q1 intersects M at Z.
Let P2Q2 intersects M at T.
Clear? Are you still here?

Now (for point P2)
|ZB1|/|ZA2| = q/p (similar triangles)
|TB2|/|TA2| = q/p (similar triangles)
=> triangles: A2B1B2 & A2ZT are similar.
=> ZT is parallel to B1B2. (1)

Do the same thing for point A1:
|XB1|/|XA1| = q/p (similar triangles)
|YB2|/|YA1| = q/p (similar triangles)
=> triangles: A1B1B2 & A1XY are similar.
=> XY is parallel to B1B2. (2)

(1) and (2) => ZT is parallel to XY.

Now I will stop here. You try to see:
ZX || A1A2 and TY || A1A2

Beware: how you draw the figure can make it difficult to see.

Do you need a figure? I won't promise if I can do it. But I may try.

-O believes - deeply.

3. I get the definitions you made(even tho I think you switched forgot to use L2 for q), and I understand what you are trying to show, but I don't get why q/p and |ZB1|/|ZA2| are parts of similar triangles. I am still working on it, but if you happen to read this before I figure it out then maybe you could help. Thanks for the post

EDIT:

Alright I think I see why they are similar triangles - you were right about drawing the figure correctly
When you draw the in a segment from B1 to L1 normal to M and a segment from A2 to L2 that is also normal to M, and you draw in ZA1 and ZA2, there are two right triangles with a vertical angle. I think...

Still working lol...

4. Ok I think I understand what you wrote enough to do it myself now - but I have one last question

isn't it
|ZB1|/|ZA2| = p/q (similar triangles)
|TB2|/|TA2| = p/q (similar triangles)

because p is part of the triangle that contains ZB1 and q is part of the triangle that contains ZA2?

And also, when I finish showing that all the sides are parallel, I think I still need to show that the distances of at least 1 pair of opposite sides are equal - I have to stop working on it tonight tho so I will try again tomorrow.

Thanks very much again

5. Depends on how you draw the figure. Please see that attached picture. One line is thicker to give you the feeling of perspective.

|ZB1|/|ZA2| = q/p (similar triangles)
|TB2|/|TA2| = q/p (similar triangles)

Can you see: TZ || B1B2 ?

Please note the figure show only 25% of the solution. Then you have to look at point P1.
Then Do the above 2 steps for line A1A2.

Good luck.

-O sleeps if need be.