Spin it a half-turn and it lands on itself — then run every property backward into a test.
Flip a parallelogram a half-circle about the point where its two diagonals cross, and it drops back onto itself, corner for corner — it has point symmetry. That single picture re-explains, in one stroke, why its opposite sides and angles come in matched pairs. Then we turn the whole question around. Instead of starting from a parallelogram and reading off its properties, we ask: what is the least you must check to prove a four-sided figure really is one? The answer is a tidy set of five tests — from the sides, from the angles, or from the diagonals — each one the converse of a property you already know. We will also meet the classic trap that catches almost everyone.
Take a figure and rotate it 180° — a half-turn — about some fixed point. If the figure lands exactly on top of where it started, we say it has point symmetry (also called central symmetry), and the fixed point is the center of symmetry. Every point of the figure swaps places with the point directly opposite it through the center, like a pinwheel turned half a revolution.
This is a different kind of symmetry from the mirror symmetry you met with isosceles triangles. Line symmetry folds a figure across a line so the two halves match. Point symmetry spins it a half-turn about a point. Some figures have one, some the other, some both, some neither.
A figure has point symmetry if a 180° rotation about some point — the center of symmetry — maps it exactly onto itself. The letters S, N, Z, a standard playing card, and the parallelogram all have it; an equilateral triangle does not.
Point symmetry behaves a lot like the translation you met in translating figures — both slide a whole figure onto a congruent copy — but instead of equal, parallel arrows, a half-turn pivots everything through the center. The defining fact is simple and worth memorizing.
If a half-turn about O sends point P to its image P′, then O is the midpoint of segment PP′: the segment PP′ passes straight through O, and OP = OP′.
Three consequences follow at once, and we will use all three in the next section:
Now the payoff. Where is a parallelogram's center of symmetry? Exactly at O, the intersection of its diagonals. Recall from the previous lesson that the diagonals of a parallelogram bisect each other — so O is the midpoint of both AC and BD.
That is all we need. Because O is the midpoint of AC, the half-turn about O sends A→C. Because O is the midpoint of BD, it sends B→D. The four vertices simply trade places in pairs, so the whole figure lands back on itself.
The half-turn carries side AB onto side CD, so they are equal and parallel; likewise AD onto CB. And it carries each half-diagonal onto the other, so the diagonals bisect each other. The properties of 16.2 fall out of one spin. ∴ ▱ABCD is point-symmetric about O.
Here is the turn of thought that makes this lesson. Every property we proved in 16.2 ran forward: "if it is a parallelogram, then the opposite sides are equal." A test runs the same statement backward — it is the converse: "if the opposite sides are equal, then it is a parallelogram." Read each property in reverse and you get a way to prove a quadrilateral is one.
Four of the five tests come from the sides and angles. A quadrilateral is a parallelogram if any one of these holds:
1. Both pairs of opposite sides are parallel (this is the definition).
2. Both pairs of opposite sides are equal.
3. One pair of opposite sides is both parallel and equal.
4. Both pairs of opposite angles are equal.
Each is a genuine converse of a 16.2 fact, and each can be proved by drawing one diagonal and finding a pair of congruent triangles. Test 3 is the most useful in practice — you only have to check one pair of sides, as long as that pair is both parallel and equal.
Test 3 needs the same pair of sides to be parallel and equal. If instead one pair is parallel and the other pair is equal, that is not enough — the figure could be an isosceles trapezoid. Watch for this; it is the single most common mistake on this topic.
The fifth test comes from the diagonals, and it is the converse of 16.2.3.
5. If the diagonals bisect each other — that is, they cross at their common midpoint, with AO = OC and BO = OD — then the quadrilateral is a parallelogram.
The proof is quick. If AO = OC, BO = OD, and ∠AOB = ∠COD (vertical angles), then △AOB ≅ △COD (SAS). Matching parts give AB = CD and ∠OAB = ∠OCD — but those equal alternate angles mean AB ∥ CD. So one pair of sides is parallel and equal, which is Test 3. ∴ the figure is a parallelogram.
Here are all five tests gathered in one place. Each says: "given this ⇒ it's a parallelogram."
| # | What you are given | Conclusion |
|---|---|---|
| 1 | Both pairs of opposite sides ∥ | ▱ (definition) |
| 2 | Both pairs of opposite sides equal | ▱ |
| 3 | One pair ∥ and equal | ▱ |
| 4 | Both pairs of opposite angles equal | ▱ |
| 5 | Diagonals bisect each other | ▱ |
| ✗ | One pair ∥, the other pair equal | not enough — could be a trapezoid |
Point symmetry: a figure has it if a 180° turn about some center O lands it on itself. Then O is the midpoint of every point-to-image segment, and the image is congruent to the original.
The parallelogram is point-symmetric about the crossing of its diagonals, which re-proves opposite sides equal & parallel and diagonals bisecting in one spin.
The five tests (each a converse of a property): both pairs of sides ∥; both pairs of sides equal; one pair ∥ and equal; both pairs of opposite angles equal; diagonals bisect each other. Trap: one pair ∥ + the other pair equal is not enough.
State the test that is the converse of "the opposite sides of a parallelogram are equal."
If both pairs of opposite sides of a quadrilateral are equal, then it is a parallelogram. (Test 2.)
In quadrilateral ABCD, side AB is parallel to DC, and the other pair AD = BC. Must ABCD be a parallelogram?
No. This is the classic trap: one pair parallel, the other pair equal. The figure could be an isosceles trapezoid. To use Test 3 you need the same pair to be both parallel and equal.
The diagonals of quadrilateral PQRS bisect each other. What kind of figure is PQRS, and which test tells you?
A parallelogram, by Test 5 (diagonals bisect each other). The proof pairs the vertical-angle triangles △POQ ≅ △ROS (SAS).
A quadrilateral has ∠A = ∠C and ∠B = ∠D. Is it a parallelogram? Briefly say why.
Yes — Test 4 (both pairs of opposite angles equal). With the four angles summing to 360°, ∠A + ∠B = 180°, so each pair of consecutive co-interior angles forces a pair of parallel sides.
In ▱ABCD the diagonals cross at O. Name the center of symmetry, and state where the half-turn about it sends A and B.
The center of symmetry is O, the intersection of the diagonals. The half-turn sends A → C and B → D (because O is the midpoint of each diagonal).
Which of these has point symmetry: the letter N, an equilateral triangle, a parallelogram, the letter A?
The letter N and the parallelogram have point symmetry (a half-turn lands each on itself). The equilateral triangle and the letter A do not — though A has line (mirror) symmetry.
Six questions to lock it in. Tap the answer you think is right.
This lesson does two jobs. First it names point (half-turn) symmetry and uses it to re-explain the parallelogram's properties in a single, vivid picture — a 180° rotation about the diagonals' crossing swaps each vertex with its opposite. Second, it pivots from properties to tests: every property of 16.2 is read backward as its converse, giving five conditions that each prove a quadrilateral is a parallelogram. Encourage students to say out loud which converse they are using.
The misconception to watch is the parallel-one-pair / equal-other-pair trap. Students who know "one pair parallel and equal ⇒ parallelogram" (Test 3) often misremember it as "one pair parallel and another pair equal," which is false — an isosceles trapezoid satisfies it. The "Name the test" widget shows the red non-example deliberately; have students explain why the trapezoid slips through. A second, smaller confusion is mixing up point symmetry with line symmetry: a parallelogram has point symmetry but, in general, no line of symmetry at all.
Common Core alignment: HS G-CO.A.3 and G-CO.A.5 (rotational / point symmetry and carrying figures onto themselves), HS G-CO.C.11 (prove and apply the criteria for parallelograms), building on 8.G.A (rotations and congruence).