Transformations and Symmetry

Call \(O\) the midpoint of the diagonal \(\Segment{AC}\): the half-turn about \(O\) sends \(A\) to \(C\) and \(C\) to \(A\). By the conservation properties of a point reflection, the image of the line \(\Line{AB}\) is a line parallel to \(\Line{AB}\) through the image \(C\) of \(A\): it is the line \(\Line{CD}\), since \(ABCD\) is a parallelogram. In the same way, the image of the line \(\Line{CB}\) is the parallel to \(\Line{CB}\) through \(A\): it is \(\Line{AD}\). The point \(B\) lies on both \(\Line{AB}\) and \(\Line{CB}\), so its image lies on both \(\Line{CD}\) and \(\Line{AD}\): it is \(D\).
The half-turn about \(O\) therefore sends \(A\) to \(C\) and \(B\) to \(D\): the parallelogram falls back exactly onto itself, \(O\) is a centre of symmetry, and the properties follow at once:

• the half-turn maps side \(\Segment{AB}\) onto side \(\Segment{CD}\), and side \(\Segment{AD}\) onto side \(\Segment{CB}\), so the opposite sides are equal and the opposite angles are equal;
• the half-turn sends \(B\) onto \(D\) while leaving \(O\) fixed, so \(O\) is also the midpoint of \(\Segment{BD}\): the diagonals cross at \(O\) and bisect each other.

The last property follows from these: the four angles of a quadrilateral add up to \(360^\circ\) (cut it along a diagonal: two triangles, \(2 \times 180^\circ\)), and since the opposite angles are equal, two adjacent angles make half of that, \(180^\circ\) — the adjacent angles are supplementary.

Two rotations about the same centre \(O\), one after the other, make one rotation about \(O\): the angles add when both turn the same way, and they subtract when the second one turns back. Let \(a^\circ\) be the smallest angle, with \(0 < a \le 360\), whose rotation about \(O\) maps the figure onto itself (such angles exist, since the full turn works, and there are only \(n\) of them).

• Every multiple of \(a\) works. The rotation through \(a^\circ\) maps the figure onto itself; applying it again maps the result, which is the figure, onto itself again: the rotation through \(2a^\circ\) works, then \(3a^\circ\), and so on.
• No other angle works. Suppose that the rotation through \(b^\circ\), with \(0 < b \le 360\), maps the figure onto itself, and that \(b\) is not a multiple of \(a\). Then \(b\) lies between two consecutive multiples of \(a\): \(k \times a < b < (k+1) \times a\) for a whole number \(k\). Turn the figure through \(b^\circ\), then turn it back through \(k \times a^\circ\): each move maps the figure onto itself, so the two together do too, and together they make the rotation through \((b - k\times a)^\circ\). But \(0 < b - k \times a < a\): this rotation would work with an angle smaller than \(a\), which is impossible, since \(a\) is the smallest.
• Counting the angles. The full turn works, so \(360\) is a multiple of \(a\): \(360 = m \times a\) for a whole number \(m\). The angles that work, between \(0\) and \(360\), are then \(a, 2a, \dots, m \times a = 360\): there are \(m\) of them. The order is the number of these angles, so \(m = n\) and \(a = \dfrac{360}{n}\).

Hence the rotations that map the figure onto itself are exactly those through the multiples of \(\dfrac{360^\circ}{n}\), up to \(360^\circ\).

The first sentence is the definition of a centre of symmetry, since the point reflection in \(O\) is the half-turn about \(O\), the rotation about \(O\) through \(180^\circ\).
By the Proposition “Order and Angle”, the rotations about \(O\) that map the figure onto itself are those through the multiples of \(\dfrac{360^\circ}{n}\). So the half-turn works exactly when there is a whole number \(k\) such that $$\begin{aligned}[t] 180 &= k \times \frac{360}{n}\\ 180 \times n &= 360 \times k && \text{(both sides multiplied by } n\text{)}\\ n &= 2 \times k. && \text{(both sides divided by } 180\text{)} \end{aligned}$$ Each of these three equalities says the same thing as the others. A whole number \(k\) with \(n = 2 \times k\) exists exactly when \(n\) is even. Hence \(O\) is a centre of symmetry exactly when \(n\) is even.