This is a problem that comes with an answer, but I can't understand it. My book says:
"We denote $H_A$ simply by $A$. Then $P{\circ}Q{\circ}R=X$, where $X$ is the fourth vertex of the parallelogram $(P,Q,R)$. Furthermore, $X{\circ}S{\circ}T=A$. Thus we have constructed A".
And the rest is obviously simple as we can reflect our result to find the other points. But I don't understand why $(X,S,T,A)$ forms a parallelogram. Can anyone please explain?
We have Varignon's Theorem, which says that if we take any quadrilateral and connect its midpoints, the resulting shape is a parallelogram. So taking $P$, $Q$, and $R$, we find $X$, the midpoint of $AD$, the fourth side of the quadrilateral $ABCD$.
But $X$ is on a diagonal of the pentagon, and as such it divides it into a quadrilateral and a triangle. And if we take the midpoints of a triangle, we have the medial triangle, which we can spin across its sides (forming more parallelograms) to get the actual corners of the triangle. This particular formulation has found the one between $X$ and $T$, which is also the corner of the pentagon between $P$ and $T$, which we'll call $A$.
Once we have one of the corners (and know which pair of midpoints it's between), we can finish the job by simply reflecting each successive corner across another midpoint.