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  PQRS is a parallelogram with right-angle corners

Step-by-step explanation:

We know that the midsegment of a triangle is parallel to the base.

  QR is the midsegment of triangle BCD, so is parallel to BD.

  SP is the midsegment of triangle DAB, so is parallel to BD.

QR and SP are both parallel to BD, so are parallel to each other.

  RS is the midsegment of triangle CAD, so is parallel to AC.

  PQ is the midsegment of triangle ABC, so is parallel to AC.

RS and PQ are both parallel to AC, so are parallel to each other.

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We have shown that opposite sides of PQRS are parallel to each other, so the figure is at least a parallelogram.

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By virtue of the congruence of corresponding angles where a transversal crosses parallel lines, each of the so-far named lines can be shown to be perpendicular to any of the lines it meets.* Hence the figure PQRS must be a parallelogram with right angles, a rectangle.

* Transversal BD crosses PQ, AC, and RS at right angles. Hence, transversals RS and PQ cross QR, BD, and SP at right angles. That is, the angles at corners P, Q, R, and S of the parallelogram are right angles.

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}