In the cube ABCDA1B1C1D1, through an arbitrary point M of segment AA1, a plane is drawn parallel to the intersection

In the cube ABCDA1B1C1D1, through an arbitrary point M of segment AA1, a plane is drawn parallel to the intersection of segments BB1; CC1; DD1, respectively, at points N; P; Q. Prove that MP is perpendicular to NQ

Since ABCDA1B1C11 is a cube, then all its opposite faces are parallel, and the neighboring ones are perpendicular, then the plane АА1В1В is perpendicular to BB1С1В. By condition, MN is perpendicular to B1B, and therefore MN is parallel to A1B1 and AB. Then ABMN is a rectangle and MN = AB.

Similarly, we prove that NP = BC, HQ = DC, MQ = AD.

And since AB = AB = CD = AD, then MN = NP = HQ = MQ, which means MNPQ is a square.

NQ and МР are the diagonals of the square, which means they intersect at an angle of 90, which was required to be proved.