Please help me understand this geometry problem ABCD is a quadrilateral; P,Q,R and S are the points of trisection of sides AB,BC,CD and DA respectively and are adjacent to A and C; prove that PQRS is a parallelogram.
In the above question it is mentioned that "P,Q,R,S" are the points of trisection.What does this statement mean?
As far as I know points of trisections are the points which divides the line segment into three equal parts and for this, there must exist two points on that segment.But in the above question there exists only one point on a  given line segment
For example in question it is provided that P is the point of trisection of segment AB.
Where am I wrong?
 A: As stated in several comments, e.g., achille hui's comment, the problem is indicating that when choosing among any of the mentioned $2$ possible internal trisection points, always choose the ones closest (which would also be the ones most "adjacent") to the points $A$ and $C$. This is shown in the diagram below. Also, the green $w$, $x$, $y$ and $z$ represent the lengths between each set of respective trisection points, and I've added the quadrilateral diagonals in red.

Next, note
$$\frac{|DR|}{|DC|} = \frac{2y}{3y} = \frac{2}{3} = \frac{2z}{3z} = \frac{|DS|}{|DA|} \tag{1}\label{eq1A}$$
Since $\angle RDS$ is a common angle, we have by SAS (i.e., side-angle-side, with this being the second bullet point in the triangle similarity section) that $\triangle RDS \sim \triangle CDA$ (i.e., the triangles are similar). Thus, $\angle DRS = \angle DCA$ and $\angle DSR = \angle DAC$. This means $SR \parallel AC$ (due to the corresponding angles of the transversal being congruent, which is point $3$ in the Condition for parallelism section). Repeating these arguments with the corresponding triangles on the other side of $AC$ shows $\triangle PBQ \sim \triangle ABC$ and $PQ \parallel AC$. Since $SR$ and $PQ$ are each parallel to $AC$, this means $SR \parallel PQ$.
Repeating what was done above with the angles and triangles, but with using the diagonal $BD$ instead, results in $PS \parallel QR$. This means both sets of opposite sides of $PQRS$ are parallel to each other, thus making it a parallelogram (this is the first bullet point in the Characterizations section, but note I could have instead (or in addition to) used the similar triangle properties to get that $\left| SR \right| = \frac{2}{3}\left| AC \right| = \left| PQ \right|$ and $\left| PS \right| = \frac{1}{3}\left| BD \right| = \left| QR \right|$, meaning the opposite sides' lengths are the same, thus matching the second bullet point).
