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Transformations that preserve distance, and the criteria that force two triangles to be congruent.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you meet the rigid motions — translations, rotations and reflections — the transformations that leave every distance and every angle unchanged, and you work out their effect on coordinates. Congruence is then defined by them: two figures are congruent exactly when some sequence of rigid motions carries one onto the other. You then learn the four criteria that force two triangles to be congruent from only three matching parts, SSS, SAS, ASA and AAS, why SSA is not among them, and how congruence lets you claim that every remaining pair of parts is equal.
You can plot a point and read one off a grid, you can handle negative numbers, and you have already slid, flipped and turned figures on a coordinate plane. What is new is the single word for what those three moves have in common, and the definition of congruence that is built on it.
Rigid motion: a transformation that leaves every distance and every angle unchanged. Translations, rotations and reflections are rigid; a dilation is not.
Pre-image and image: the figure before and after the motion.
Congruent: two figures are congruent, written $\cong$, exactly when some sequence of rigid motions carries one onto the other.
Composition: one motion followed by another, which is itself a rigid motion.
Orientation: which way round the figure is read. A reflection reverses it; a translation and a rotation do not.
Rotational symmetry: a turn about a centre that carries a figure onto itself.
A rigid motion moves a figure without changing any length or angle: a translation, a rotation or a reflection. In coordinates the rules are short. A half turn about the origin sends $(x, y)$ to $(-x, -y)$, so $(3, -4)$ becomes $(-3, 4)$. A $90^\circ$ anticlockwise turn about the origin sends $(x, y)$ to $(-y, x)$. Reflecting in the $x$-axis gives $(x, -y)$ and in the $y$-axis gives $(-x, y)$; reflecting in the slanted mirror $y = x$ swaps the coordinates, so $(2, 5)$ becomes $(5, 2)$. Motions compose: two reflections in perpendicular mirrors give a half turn, since $(x, y) \to (x, -y) \to (-x, -y)$. Two figures are congruent exactly when some composition of rigid motions carries one onto the other — that is the definition, not a test. A figure's rotational symmetries are the turns that map it onto itself, and for a regular polygon with $n$ sides the smallest is $360 \div n$, so a hexagon turns by $60^\circ$.
Another way: picture
A triangle and its image after a half turn about the origin: each vertex, the origin and the matching image vertex lie on one straight line, with the origin exactly halfway along it.
Another way: story
Trace a figure onto tracing paper. Sliding, turning and flipping the paper are the three rigid motions, and the figures you can land on are exactly the congruent ones. Stretching the paper is not allowed.
Counting a dilation as a rigid motion. Enlarging keeps the shape but not the size, so the image is similar and not congruent. Only the three distance-preserving moves belong here.
Reflecting in $y = x$ by changing a sign. That mirror is slanted, so it trades the two axes: $(x, y)$ becomes $(y, x)$, with no sign change at all. The sign changes belong to the two axis mirrors.
Refusing to call a flipped figure congruent. A reflection is a rigid motion, so a mirror image is congruent to the original even though you cannot slide one onto the other in the page. Lift the tracing paper and turn it over.
The $x$-axis mirror keeps $x$ and flips $y$: $(x, y) \to (x, -y)$.
Do one motion at a time and write the point down in between.
The $y$-axis mirror then flips $x$: $(x, -y) \to (-x, -y)$.
$(x, y) \to (-x, -y)$ is the half-turn rule, so two perpendicular mirrors make a rotation.
The composition of two rigid motions is again a rigid motion.
Its six vertices are evenly spaced round the centre, so the smallest turn that works moves each vertex to the next.
$360 \div 6 = 60^\circ$.
The other symmetries are multiples of it.
That mirror swaps the coordinates and changes no signs.
$(-3, 7) \to (7, -3)$.
Rotate the point $(6, 2)$ by $180^\circ$ about the origin.
(x, y)
Reflect the point $(-4, 3)$ in the line $y = x$.
(x, y)
A regular polygon has $5$ equal sides. What is the smallest turn, in degrees, that carries it onto itself?
answer
The point $P = (5, -2)$ is moved by each rigid motion below. Match each motion to the image of $P$.
| $(5, 2)$ | $(-5, -2)$ | $(-5, 2)$ | $(-2, 5)$ | |
|---|---|---|---|---|
| Reflect $P$ in the $x$-axis | ||||
| Reflect $P$ in the $y$-axis | ||||
| Rotate $P$ by $180^\circ$ about the origin | ||||
| Reflect $P$ in the line $y = x$ |
You have just defined congruence by rigid motions, you know the angles of a triangle add to $180^\circ$, and you know that vertical angles at a crossing are equal. Beyond the criteria themselves, those are the only facts a congruence argument here will use.
Included angle: the angle between two named sides. Included side: the side between two named angles. The middle letter of SAS and of ASA is the included part.
Correspondence: the pairing the letters record. Writing $ABC \cong DEF$ says $A$ matches $D$, $B$ matches $E$ and $C$ matches $F$.
Corresponding parts: every side and angle matched by that pairing. Once the triangles are congruent, all six pairs are equal.
Criterion: a short list of matching parts that forces congruence.
Counterexample: one pair of triangles that fits the list but is not congruent — enough to sink a would-be criterion.
Two triangles have six parts each, but three of them, chosen well, already force congruence. SSS: three matching sides. Draw one side, swing an arc of each of the other two lengths from its ends, and they cross in exactly one place on each side of the line, so there is only one triangle and a rigid motion carries either copy onto the other. SAS: two sides and the angle between them. ASA: two angles and the side between them. AAS: two angles and a side not between them, which works because the third angle is $180^\circ$ minus the other two, turning it into ASA. SSA is not a criterion: with a fixed angle and a fixed adjacent side, an arc of the second side's length can cut the far ray twice, giving two different triangles. Once congruence is established, every corresponding part is equal, so $ABC \cong DEF$ with $\angle A = 40^\circ$ and $\angle B = 65^\circ$ gives $\angle F = \angle C = 75^\circ$.
Another way: picture
The SSA failure drawn: one fixed angle, one fixed side along it, and a compass arc of the second side's length cutting the far ray at two different points — two triangles from the same three facts.
Another way: steps
List the matching parts. Mark each as a side or an angle in the order they go round the triangle. Read the word that spells out: SSS, SAS, ASA or AAS proves it; SSA does not.
Treating SSA as a criterion. Two sides and an angle not between them do not fix a triangle: the far side swings like a compass arc and can meet the other ray in two places. That is a counterexample, so SSA never appears in a proof.
Reading the correspondence off the picture. The letters decide, not the drawing. In $ABC \cong DEF$ the side $EF$ matches $BC$ whatever the two triangles look like on the page.
Believing three equal angles are enough. AAA gives the same shape at any size, which is similarity, not congruence. Every criterion contains at least one side, because something has to fix the size.
$\angle A = 40^\circ$ and $\angle B = 65^\circ$, so $\angle C = 180 - 40 - 65 = 75^\circ$.
The angle sum, inside the first triangle.
The letters pair $F$ with $C$, so $\angle F = \angle C$.
Corresponding parts of congruent triangles are equal.
$\angle F = 75^\circ$.
Given $AB = DE$, $BC = EF$ and $\angle B = \angle E$: the angle is between the two sides, so this is SAS and the triangles are congruent.
Check where the angle sits before naming anything.
Given $AB = DE$, $BC = EF$ and $\angle A = \angle D$ instead: $\angle A$ is not between $AB$ and $BC$, so this is SSA and proves nothing.
Same three quantities, a different position, a different verdict.
$\angle A$ lies between the sides $AB$ and $AC$, so it is included.
Two sides and the included angle: SAS, so the triangles are congruent.
Triangles $ABC$ and $DEF$ are congruent, with $\angle A = 38^\circ$ and $\angle B = 71^\circ$. What is $\angle F$, in degrees?
answer
Triangles $ABC$ and $DEF$ are congruent, with $AB = 8$, $BC = 14$ and $DF = 13$. What is the perimeter of triangle $DEF$?
answer
Triangles $ABC$ and $DEF$ are compared four times, and different parts are measured each time. Match each set of measurements to what it proves.
| SAS: two sides and the angle between them | ASA: two angles and the side between them | AAS: two angles and a side not between them | Nothing: two sides and an angle not between them is SSA | |
|---|---|---|---|---|
| $AB = DE = 6$, $BC = EF = 13$, $\angle B = \angle E = 46^\circ$ | ||||
| $\angle A = \angle D = 46^\circ$, $AB = DE = 6$, $\angle B = \angle E = 68^\circ$ | ||||
| $\angle A = \angle D = 46^\circ$, $\angle B = \angle E = 68^\circ$, $BC = EF = 13$ | ||||
| $AB = DE = 6$, $BC = EF = 13$, $\angle A = \angle D = 46^\circ$ |
Two triangles both have sides $11$, $13$ and $6$, and nothing else is measured. They are congruent. Why do the three sides settle it on their own?
A surveyor wants the distance across a pond from a rock at $A$ to a post at $B$, and she cannot walk on the water. She picks a point $C$ on dry land, measures $CA = 50$ m and $CB = 33$ m, then walks from $A$ straight through $C$ and on for a further $50$ m to a peg $A'$, and from $B$ straight through $C$ and on for a further $34$ m to a peg $B'$. She measures $A'B'$ over dry ground and calls it the width of the pond. Is she right?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Rotate the point $(5, 5)$ by $180^\circ$ about the origin.
(x, y)
Triangles $ABC$ and $DEF$ are congruent, with $\angle A = 64^\circ$ and $\angle B = 56^\circ$. What is $\angle F$, in degrees?
answer
You can move a figure by a rigid motion, write the coordinates of its image, and decide which criterion forces two triangles to be congruent. From memory, name the four criteria, and explain why two sides and an angle that is not between them prove nothing.
8. Your turn: reflect $(-3, 7)$ in the line $y = x$, step 2
19. Your turn: $AB = DE$, $\angle A = \angle D$ and $AC = DF$. Which criterion?, step 2