Math 8 · 8.G.A.2 · G.CO.B.6

Rigid motions and congruence

Two shapes look like copies of each other. What exactly does “the same shape and the same size” mean, and how could you ever be sure?

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▴ 6 above Rotating about a point other than the origin What rigid motions preserve

Rigid motions and congruence

Rigid motions defined precisely, and what they preserve Two reflections in a row

Why this exists

“Congruent” is meant to say “the same shape and the same size”. That sounds clear until you try to check it. Two triangles on a page sit in different places and point different ways, and one may be a mirror image of the other. Which features have to match? The sides? The angles as well? Is the same area enough? A definition has to say exactly.

The answer here uses motion. Trace one figure on tracing paper, then slide the paper, turn it, or flip it over. If some run of those moves lays the tracing exactly on the other figure, the two are congruent. Slides, turns and flips are the rigid motions: they move a figure without stretching, shrinking or bending it. So “same shape and size” becomes something you can do and check. You name the moves and follow every point.

The definition pays off twice. Because a rigid motion keeps every length and every angle, congruent figures have matching sides and matching angles. And it explains the familiar shortcuts for triangles, SSS, SAS and ASA, which say that three matching parts are enough. A course can take those three as starting points, and the triangle congruence criteria does. This lesson gives a second route to the same three facts: each one can be proved by building the motion that carries one triangle onto the other.

Definition

A rigid motion is a translation, a rotation, a reflection, or a sequence of them done one after another. Composing rigid motions gives another rigid motion. Rigid motions defined precisely starts from the other end, calling any move that keeps every distance a rigid motion; in the plane the two descriptions pick out the same moves.

Two figures are congruent when some rigid motion carries the first exactly onto the second: every point of the first lands on a point of the second, and every point of the second is landed on. For figures F and G this is written F \cong G, read “F is congruent to G”.

The motion also pairs up the parts. If it sends the point A to the point D, then A and D correspond. The image of A after one motion is written A' (“A prime”), and after a second motion A'' (“A double prime”). For polygons the pairing is built into the name:

\triangle ABC \cong \triangle DEF

says that some rigid motion sends A \mapsto D, B \mapsto E and C \mapsto F, in that order. The order of the letters carries information, which congruence statements works through.

A mirror image counts. A reflection is a rigid motion, so a triangle and its mirror image are congruent, even though no sliding or turning within the page would carry one onto the other.

Intuition

Figure 1After a slide and a quarter turn (dashed), A sits on P and B on Q, but the triangle faces the wrong way. A flip in line PQ finishes the job.

Figure 1 shows the tracing-paper idea for two triangles. Slide the tracing so that one corner lands on its partner: here A onto P. Then turn the tracing about that corner until side \overline{AB} lies along its partner \overline{PQ}. The two sides have the same length, so B lands exactly on Q. Now two corners are in place, and the third is either on its partner already or sitting at the mirror image of it, on the other side of line PQ. In the figure it is on the wrong side, so one flip over line PQ finishes the job.

That is the whole picture: a slide, a turn and a flip are always enough for two congruent triangles. Often fewer are needed. If the turned copy already faces the right way, the flip is skipped; if the slide alone does it, so is the turn. Example 4 works through Figure 1 with coordinates.

The second part of the lesson runs the same picture backwards. Instead of being told the triangles are congruent, you are told that three parts match, and the question is whether those three parts are enough to make the slide, turn and flip land every corner. For SSS, SAS and ASA they are.

Results

Part 1: what a rigid motion keeps, and what congruence gives

A rigid motion preserves:

the distance between any two points, so every segment goes to a segment of the same length;
every angle measure;
lines and betweenness: the image of a line is a line, and a point between two others lands between their images;
parallel lines, which go to parallel lines, and perpendicular lines, which go to perpendicular lines.

So it keeps perimeter and area as well. What rigid motions preserve collects these facts. Rigid motions defined precisely proves the first two, but for a general rotation and for angle measure it leans on SAS and SSS, so Part 2 below, which proves SAS and SSS, takes them as starting assumptions instead.

A rigid motion can change where a figure is, which way its sides point, and its orientation. Read the corners A, B, C in order: they run either counterclockwise or clockwise. A translation or a rotation keeps that sense; a reflection reverses it. So a sequence with an even number of reflections keeps orientation, and a sequence with an odd number reverses it.

Congruent figures have congruent corresponding parts. If \triangle ABC \cong \triangle DEF, then all six pairs match:

\overline{AB} \cong \overline{DE}, \quad \overline{BC} \cong \overline{EF}, \quad \overline{CA} \cong \overline{FD}

\angle A \cong \angle D, \quad \angle B \cong \angle E, \quad \angle C \cong \angle F

Three motions are enough. If two triangles are congruent, a translation, then a rotation, then at most one reflection carries one onto the other.

Part 2: SAS, ASA and SSS from rigid motions

Each statement below is about two triangles whose parts correspond in the order given.

SAS. If two sides and the included angle (the angle between them) of one triangle are congruent to two sides and the included angle of the other, then a rigid motion carries the first triangle onto the second, so the triangles are congruent.

ASA. If two angles and the included side (the side between them) of one triangle are congruent to two angles and the included side of the other, then a rigid motion carries the first onto the second, so the triangles are congruent.

SSS. If the three sides of one triangle are congruent to the three sides of the other, then a rigid motion carries the first onto the second, so the triangles are congruent.

Six pairs, or a motion. Two triangles are congruent exactly when all six pairs of corresponding sides and angles are congruent. If the triangles are congruent, the motion gives all six pairs. If all six pairs are congruent, then in particular two sides and the angle between them are, and SAS gives the motion.

These are the same three statements that the triangle congruence criteria takes as postulates. Here they are theorems, proved from the definition of congruence and two starting assumptions: every rigid motion keeps distance, and every rigid motion keeps angle measure. Either way, once they are known they are used the same way.

Why it is true

Congruent figures have congruent parts. Suppose a rigid motion sends A \mapsto D, B \mapsto E and C \mapsto F. A rigid motion sends a segment to the segment joining the images of its ends, so \overline{AB} goes to \overline{DE}, and lengths are kept, so AB = DE. The same holds for each side, and since angle measures are kept, \angle ABC goes to \angle DEF with the same measure, and so on for each angle.

What the second route assumes. A proof of SAS cannot rest on SAS. So the proofs below use only the definition of congruence, the starting assumption that rigid motions keep distance and angle measure, and two facts about the plane that are part of what a line and an angle are. From a point on a line, on a chosen side of the line, there is exactly one ray making a given angle with a given ray along the line. And on a ray there is exactly one point at a given distance from its endpoint.

The common start. Each of SAS, ASA and SSS includes a matching pair of sides. Name the triangles so that this pair is \overline{AB} and \overline{DE}, with AB = DE.

Stage 1, a slide. Translate by the vector from A to D. Then A lands on D. Call the new places of B and C B_1 and C_1.

Stage 2, a turn. Rotate about D through the angle from ray DB_1 to ray DE. The center D stays put, and ray DB_1 lands along ray DE. Distance has been kept, so the new place of B is at distance AB = DE from D along ray DE. Only one point of that ray is that far from D, namely E. So B is now on E. Call the new place of C C_2. Skip this stage if ray DB_1 is already ray DE.

Figure 2After the slide and the turn, A is on D and B is on E. If the third corner lands at C_2, on the far side of line DE, a reflection in that line carries it across.

Now \triangle DEC_2 is a rigid-motion image of \triangle ABC, so it has all the lengths and angles of \triangle ABC, and it shares side \overline{DE} with \triangle DEF. Everything comes down to whether C_2 is F.

Stage 3, a flip if needed. If C_2 is on the other side of line DE from F, reflect in line DE. Points D and E lie on the mirror and stay put, C_2 crosses to F’s side, and every length and angle is kept. So from here on C_2 and F are on the same side of line DE, and each criterion finishes in its own way.

SAS finishes (given AB = DE, \angle A \cong \angle D and AC = DF). The angle at D is kept, so m\angle EDC_2 = m\angle BAC = m\angle EDF, and C_2 and F are on the same side. Only one ray from D on that side makes that angle with ray DE, so ray DC_2 is ray DF. Distance is kept, so DC_2 = AC = DF, and only one point of the ray is that far from D. So C_2 = F.

ASA finishes (given \angle A \cong \angle D, AB = DE and \angle B \cong \angle E). The angle at D puts C_2 on ray DF, exactly as for SAS. The same argument at the other end, m\angle DEC_2 = m\angle ABC = m\angle DEF, puts C_2 on ray EF. The rays DF and EF lie on two different lines, and two different lines cross at most once; these cross at F. So C_2 = F. This is where “included” matters: the two angles sit at the two ends of the side that was placed.

SSS finishes (given AB = DE, BC = EF and CA = FD). No angle is given, so the argument uses distances. DC_2 = AC = DF, so D is the same distance from C_2 as from F, and likewise EC_2 = BC = EF. Could C_2 and F be different points? A point the same distance from two points lies on the perpendicular bisector of the segment joining them, so both D and E would lie on the perpendicular bisector of \overline{C_2F}. Two points fix a line, so line DE would be that perpendicular bisector, and then C_2 and F would sit on opposite sides of it. They are on the same side. So they cannot be different points, and C_2 = F.

That fact about the perpendicular bisector needs no congruence criterion either. Suppose DC_2 = DF with C_2 \ne F, and reflect in the line that bisects \angle C_2DF. The reflection keeps the angle each ray makes with the mirror and carries each side of the mirror to the other, so it sends ray DC_2 onto ray DF; the distance from D is kept, so it sends C_2 to F. The mirror of a reflection is the perpendicular bisector of the segment from a point to its image, and D is on the mirror. (If D lies on \overline{C_2F}, it is the midpoint, which is on the bisector anyway.)

In every case the sequence sends A \mapsto D, B \mapsto E and C \mapsto F. Each side goes to the segment joining the images of its ends, so the whole of \triangle ABC lands on \triangle DEF, and the triangles are congruent by the definition. No congruence criterion was used anywhere, not even hidden inside “rigid motions keep distance and angle”, since that is the starting assumption of this route rather than something proved from SAS or SSS. That is why this is a genuine second route to SSS, SAS and ASA, and not a circle. It also proves the Part 1 result that a slide, a turn and at most one flip are always enough.

Worked examples

The lesson is taught in two parts. Examples 1 to 6 belong to the first: what a rigid motion keeps, and finding the motions that carry one figure onto another. Examples 7 to 9 belong to the second: ASA and SSS built from motions, and a case where no motion can be built. The SAS construction runs on a grid exactly as Example 4 does, and Exercises 18 and 19 carry it out.

Example 1

Same area, not congruent

\triangle ABC is a right triangle with legs 3 and 4. \triangle PQR is a right triangle with legs 2 and 6. Show that the two triangles have the same area, and decide whether they are congruent.

Solution

The legs of a right triangle are a base and a height, so the areas are

\tfrac{1}{2}(3)(4) = 6 \qquad\text{and}\qquad \tfrac{1}{2}(2)(6) = 6

The same. Now the sides. By the Pythagorean theorem the hypotenuses are

\begin{aligned} \sqrt{3^2 + 4^2} &= 5 \\ \sqrt{2^2 + 6^2} &= \sqrt{40} = 2\sqrt{10} \end{aligned}

So \triangle ABC has sides 3, 4, 5 and \triangle PQR has sides 2, 6, \displaystyle 2\sqrt{10}. A rigid motion keeps every length, so any image of \triangle ABC has a side of length 3. \triangle PQR has none. No rigid motion carries one onto the other: the triangles are not congruent.

Congruent figures always have equal areas, because area is kept. The reverse fails, as here: equal area is a consequence of congruence, not a test for it.

Example 2

What a flip and a slide keep, and what they change

\triangle ABC has vertices A(1, 1), B(5, 1) and C(1, 4). Reflect it in the y-axis, then translate by \langle 2, -6 \rangle. Find \triangle A''B''C'', and compare its side lengths, its right angle and its orientation with those of \triangle ABC.

Solution

Reflection in the y-axis changes the sign of the first coordinate, (x, y) \mapsto (-x, y):

A'(-1, 1), \quad B'(-5, 1), \quad C'(-1, 4)

The translation adds 2 to each first coordinate and -6 to each second:

A''(1, -5), \quad B''(-3, -5), \quad C''(1, -2)

Lengths. \overline{AB} is level with length 5 - 1 = 4, and \overline{A''B''} is level with length 1 - (-3) = 4. \overline{AC} is upright with length 3, and \overline{A''C''} is upright with length -2 - (-5) = 3. The third sides are \displaystyle \sqrt{4^2 + 3^2} = 5 in both. Every length is kept.

The right angle. At A a level side meets an upright side, and at A'' the same is true, so \angle A'' is still a right angle.

Orientation. Walk from A to B: you head east, and C is north of A, on your left. So A, B, C run counterclockwise. Walk from A'' to B'': you head west, and C'' is north of A'', which is now on your right. So A'', B'', C'' run clockwise. The one reflection reversed the orientation; the translation did nothing to it.

Example 3

Finding a slide and a turn

\triangle ABC is the triangle of Example 2, with A(1, 1), B(5, 1), C(1, 4). \triangle DEF has D(7, 1), E(7, 5), F(4, 1), with A matching D, B matching E and C matching F. Find rigid motions that carry \triangle ABC onto \triangle DEF.

Solution

First check that a motion is possible. DE = 5 - 1 = 4 = AB, DF = 7 - 4 = 3 = AC, and \displaystyle EF = \sqrt{3^2 + 4^2} = 5 = BC. Next, orientation. Heading from D north to E, the point F is to the west, on the left, so D, E, F run counterclockwise, the same as A, B, C. No reflection is needed.

Slide A onto D. The vector from A(1, 1) to D(7, 1) is \langle 6, 0 \rangle:

A \mapsto (7, 1), \quad B \mapsto (11, 1), \quad C \mapsto (7, 4)

Turn about D. The ray from D to the new B points east, and ray DE points north, so the turn is 90^\circ counterclockwise about D(7, 1). To turn about a point other than the origin, take each point’s offset from the center, turn the offset, and add it back. A quarter turn counterclockwise sends an offset (a, b) to (-b, a).

(11, 1) has offset (4, 0), which turns to (0, 4), landing at (7, 5) = E.
(7, 4) has offset (0, 3), which turns to (-3, 0), landing at (4, 1) = F.

The translation by \langle 6, 0 \rangle, then the rotation of 90^\circ counterclockwise about D, carries \triangle ABC onto \triangle DEF, so \triangle ABC \cong \triangle DEF.

Example 4

When a flip is needed

\triangle ABC is again A(1, 1), B(5, 1), C(1, 4). \triangle PQR has P(8, 2), Q(8, 6), R(11, 2), with A matching P, B matching Q and C matching R. Find a sequence of rigid motions that carries \triangle ABC onto \triangle PQR.

Solution

The lengths match: PQ = 4, PR = 3 and \displaystyle QR = \sqrt{3^2 + 4^2} = 5. Orientation: heading from P north to Q, the point R is to the east, on the right, so P, Q, R run clockwise. The corners A, B, C run counterclockwise, so the sequence must contain a reflection.

Slide A onto P by \langle 7, 1 \rangle:

A \mapsto (8, 2), \quad B \mapsto (12, 2), \quad C \mapsto (8, 5)

Turn about P. The ray to the new B points east and ray PQ points north: a quarter turn counterclockwise about P(8, 2). The offset (4, 0) turns to (0, 4), so (12, 2) lands on (8, 6) = Q. The offset (0, 3) turns to (-3, 0), so (8, 5) lands on (5, 2).

Flip over line PQ. Two corners are in place. The third is at (5, 2) but R is at (11, 2), on the other side of line PQ, which is the line x = 8. Reflecting in x = 8 sends (x, y) \mapsto (16 - x, y), because a point 3 units left of the mirror lands 3 units right of it. So (5, 2) \mapsto (11, 2) = R, while P and Q, on the mirror, stay put.

The translation by \langle 7, 1 \rangle, the rotation of 90^\circ counterclockwise about P, and the reflection in x = 8 carry \triangle ABC onto \triangle PQR. This is Figure 1. Other sequences work too, but every one of them contains an odd number of reflections, since the orientation is reversed.

Example 5

Reading off what the motion keeps

A rotation carries quadrilateral ABCD onto quadrilateral PQRS, with A \mapsto P, B \mapsto Q, C \mapsto R and D \mapsto S. You know that AB = 7 cm, BC = (2x + 3) cm, QR = 15 cm, m\angle B = 110^\circ, m\angle Q = (4y - 10)^\circ, and \overline{AD} \parallel \overline{BC}. Find x, y and PQ, and say what is true of \overline{PS} and \overline{QR}.

Solution

The rotation sends \overline{BC} to \overline{QR}, and lengths are kept:

\begin{aligned} 2x + 3 &= 15 \\ x &= 6 \end{aligned}

It sends \angle ABC to \angle PQR, and angle measures are kept:

\begin{aligned} 4y - 10 &= 110 \\ y &= 30 \end{aligned}

\overline{AB} goes to \overline{PQ}, so PQ = 7 cm.

Parallel lines go to parallel lines. \overline{AD} goes to \overline{PS} and \overline{BC} goes to \overline{QR}, so \overline{PS} \parallel \overline{QR}. A rotation also keeps orientation, so if A, B, C, D run counterclockwise, then so do P, Q, R, S.

Example 6

A slanted pair, matched by lengths

\triangle STU has S(1, 1), T(4, 2), U(2, 5). A second triangle has X(-2, -5), Y(-1, -1), Z(-4, -2). Decide whether the triangles are congruent. If they are, write a congruence statement with the corners in matching order, and find one rigid motion that does the job.

Figure 3Each corner of \triangle STU has its partner straight through O and the same distance beyond it, which is what a half turn about O does.

Solution

Neither triangle has a level or upright side, so measure with the distance formula:

\begin{aligned} ST &= \sqrt{3^2 + 1^2} = \sqrt{10} \\ TU &= \sqrt{2^2 + 3^2} = \sqrt{13} \\ SU &= \sqrt{1^2 + 4^2} = \sqrt{17} \end{aligned}

\begin{aligned} YZ &= \sqrt{3^2 + 1^2} = \sqrt{10} \\ ZX &= \sqrt{2^2 + 3^2} = \sqrt{13} \\ XY &= \sqrt{1^2 + 4^2} = \sqrt{17} \end{aligned}

The same three lengths, all different. A motion keeps lengths, so sides must pair up by length: \overline{ST} with \overline{YZ}, \overline{TU} with \overline{ZX}, \overline{SU} with \overline{XY}. Then the corners pair up too. S is where the \displaystyle \sqrt{10} side meets the \displaystyle \sqrt{17} side, and in the other triangle that corner is Y. In the same way T matches Z and U matches X.

Now look for the motion. \overrightarrow{ST} runs 3 right and 1 up, while \overrightarrow{YZ} runs 3 left and 1 down. \overrightarrow{SU} runs 1 right and 4 up, while \overrightarrow{YX} runs 1 left and 4 down. Every direction is exactly reversed, which is what a half turn does. Try the half turn about the origin, (x, y) \mapsto (-x, -y):

\begin{aligned} S(1, 1) &\mapsto (-1, -1) = Y \\ T(4, 2) &\mapsto (-4, -2) = Z \\ U(2, 5) &\mapsto (-2, -5) = X \end{aligned}

So \triangle STU \cong \triangle YZX, by a rotation of 180^\circ about the origin. Writing \triangle STU \cong \triangle XYZ would be false: it would pair S with X, and the side \overline{ST} of length \displaystyle \sqrt{10} with \overline{XY}, of length \displaystyle \sqrt{17}.

Example 7

ASA by one reflection

Points C and D lie on opposite sides of line AB, with \angle CAB \cong \angle DAB and \angle CBA \cong \angle DBA. Use a rigid motion to prove that AC = AD and BC = BD.

Figure 4Equal angles at both ends of \overline{AB}, on opposite sides of it. The reflection in line AB carries C to D.

Solution

The two triangles, \triangle ABC and \triangle ABD, share side \overline{AB}, and the angles at both ends of it match. That is ASA with the side already in place, so the slide and the turn are not needed: only the flip. Reflect in line AB.

Where A and B go. Both are on the mirror, so they stay put.

Where ray AC goes. A reflection keeps angle measure and sends each side of the mirror to the other side. So the image of ray AC makes an angle of m\angle CAB with ray AB, on D’s side. Ray AD does exactly that, and only one ray from A on that side does. So ray AC lands on ray AD.

Where ray BC goes. The same argument at B, with \angle CBA \cong \angle DBA, sends ray BC onto ray BD.

Where C goes. C lies on both ray AC and ray BC, so its image lies on both ray AD and ray BD. Those two rays cross at one point only, D. So the reflection sends C to D.

The reflection keeps lengths, and it sends \overline{AC} to \overline{AD} and \overline{BC} to \overline{BD}. So AC = AD and BC = BD, and \triangle ABC \cong \triangle ABD.

Example 8

The last stage of SSS, in coordinates

After the slide and the turn of the SSS proof, the placed copy has corners D(0, 0), E(4, 4) and C_2(1, 5), and the target is \triangle DEF with F(5, 1). Check that DC_2 = DF and EC_2 = EF, show that line DE is the perpendicular bisector of \overline{C_2F}, and find the motion that finishes.

Solution

By the distance formula,

\begin{aligned} DC_2 &= \sqrt{1^2 + 5^2} = \sqrt{26} \\ DF &= \sqrt{5^2 + 1^2} = \sqrt{26} \\ EC_2 &= \sqrt{3^2 + 1^2} = \sqrt{10} \\ EF &= \sqrt{1^2 + 3^2} = \sqrt{10} \end{aligned}

The midpoint of \overline{C_2F} is

\left(\frac{1 + 5}{2}, \frac{5 + 1}{2}\right) = (3, 3)

which lies on line DE, the line y = x. The slope of \overline{C_2F} is \displaystyle \frac{1 - 5}{5 - 1} = -1 and the slope of line DE is 1. Their product is -1, so the two are perpendicular. Line DE crosses \overline{C_2F} at its midpoint at a right angle: it is the perpendicular bisector, as the SSS proof said it must be whenever C_2 \ne F.

So C_2 and F are on opposite sides of line DE, and the finishing motion is the reflection in y = x, which swaps the coordinates, (x, y) \mapsto (y, x). It sends C_2(1, 5) \mapsto (5, 1) = F, and it leaves D(0, 0) and E(4, 4), which are on the mirror, where they are.

Example 9

Two sides and a non-included angle are not enough

A is at (0, 0) and B at (4, 3). The point C is on the positive x-axis, with \displaystyle BC = \sqrt{10}. Find every possible position of C, and explain why this shows that two sides and an angle not between them (SSA) do not force two triangles to be congruent.

Figure 5Triangles ABC_1 and ABC_2 share \overline{AB} and the angle at A, and BC_1 = BC_2, yet AC_1 = 3 and AC_2 = 5.

Solution

Write C = (c, 0) with c > 0. The distance from B to C gives

\begin{aligned} (c - 4)^2 + 3^2 &= 10 \\ (c - 4)^2 &= 1 \\ c - 4 = 1 \quad &\text{or} \quad c - 4 = -1 \end{aligned}

so c = 5 or c = 3. Both are positive, so there are two positions: C_1(3, 0) and C_2(5, 0).

The triangles ABC_1 and ABC_2 have \displaystyle AB = \sqrt{4^2 + 3^2} = 5 in common, the same angle at A (both use ray AB and the positive x-axis), and \displaystyle BC_1 = BC_2 = \sqrt{10}. That is two sides and an angle, matching. But AC_1 = 3 and AC_2 = 5. A rigid motion keeps lengths, so no motion carries one triangle onto the other: they are not congruent.

In the language of the proofs: after \overline{AB} is placed, the angle at A fixes the ray that C lies on, exactly as in SAS. But the given length is BC, measured from B, not from A, and the points at that distance from B form a circle, which can cross the ray twice. The construction cannot decide between them.

Boundaries

Only rigid motions. A dilation changes lengths, so the image of a figure under a dilation is similar to it, not congruent, unless the scale factor is 1. A stretch in one direction, such as (x, y) \mapsto (2x, y), is not rigid either.

Some matches are not enough. Equal area or equal perimeter does not make figures congruent (Example 1). Three matching angles are not enough: equilateral triangles of side 2 and side 5 both have three 60^\circ angles. Two sides and a non-included angle are not enough in general (Example 9), though in some cases they are, for instance when the angle is a right angle, which is the hypotenuse leg criterion.

The included part has to be the included part. SAS needs the angle between the two given sides. ASA needs the side between the two given angles.

The criteria are about triangles. Four matching sides do not make two quadrilaterals congruent: a square with sides of 2 and a rhombus with sides of 2 and a 60^\circ angle have the same four side lengths, but a rigid motion keeps angles and their angles differ.

The parts must correspond. In Example 6, \triangle STU \cong \triangle YZX is true and \triangle STU \cong \triangle XYZ is false. A triangle with two equal sides can be matched with a copy of itself in more than one way, and then more than one statement is true.

Mirror images count. In the plane a flip is allowed, so a figure and its mirror image are congruent. That is a choice the definition makes, and it differs from everyday speech, where a left shoe is not a copy of a right shoe.

Where this goes next

Allow a dilation as well as the rigid motions and the same definition gives similarity: two figures are similar when a dilation followed by rigid motions carries one onto the other. That is similarity transformations, and the proofs there follow the pattern of this lesson, with lengths scaled instead of kept.

Producing the motion is itself a way to prove things. Transformational proof writes such arguments out in full, for symmetric figures such as isosceles triangles and kites as well as for the criteria themselves.

Once SSS, SAS and ASA are known, by this route or as postulates, most of classical geometry runs on them: two triangles are shown congruent from three parts, and then the other three parts follow, which is proving parts congruent. The right triangle shortcut, hypotenuse leg, is one more criterion of the same kind.

The motions themselves can be pared down further. Two reflections in a row make a rotation or a translation, so the slide, the turn and the flip can each be rebuilt from reflections alone.

Exercises

Exercises 1 to 14 go with the first part of the lesson (Examples 1 to 6), Exercises 15 to 21 with the second (Examples 7 to 9), and the reasoning group draws on both.

Part 1: what a rigid motion keeps

A rigid motion carries \triangle JKL onto \triangle MNP with J \mapsto M, K \mapsto N and L \mapsto P. JK = 9, KL = 12, m\angle K = 64^\circ and m\angle J = 51^\circ. Find MN, NP, m\angle N and m\angle P.
A pentagon has sides of 4, 5, 6, 7 and 9 cm and an area of 52 cm². A reflection followed by a translation carries it onto a second pentagon. Find the perimeter and the area of the second pentagon.
A sequence of rigid motions carries the segment with endpoints A(2, 3) and B(8, 11) onto \overline{CD}. How long is \overline{CD}?
A translation carries \triangle ABC onto \triangle DEF with A \mapsto D, B \mapsto E and C \mapsto F. AB = 3x - 4, DE = x + 10, m\angle C = (5y + 2)^\circ and m\angle F = 72^\circ. Find x, y and AB.
A rotation carries quadrilateral ABCD onto quadrilateral WXYZ, with the corners matching in that order. \overline{AB} \parallel \overline{CD} and \angle A is a right angle. What can you say about \overline{WX}, \overline{YZ} and \angle W?

Part 1: following a sequence

\triangle PQR has P(1, 2), Q(4, 2) and R(4, 6).

Rotate \triangle PQR by 180^\circ about the origin, then translate by \langle 5, 3 \rangle. Find P'', Q'' and R'', and check that P''Q'' = PQ and Q''R'' = QR.
Reflect \triangle PQR in the x-axis, then in the y-axis. Find P'', Q'' and R''. Which single rigid motion has the same effect?
For each sequence, say whether a figure keeps its orientation or has it reversed: (a) two reflections, then a rotation; (b) a translation, then a reflection; (c) three reflections.

Part 1: finding a motion

In Exercises 9 to 12, \triangle ABC has A(1, 1), B(3, 1) and C(1, 4), and A, B, C match D, E, F. Find a rigid motion, or a sequence of them, that carries \triangle ABC onto \triangle DEF.

D(6, 3), E(8, 3), F(6, 6)
D(-1, 1), E(-3, 1), F(-1, 4)
D(-1, 1), E(-1, 3), F(-4, 1)
D(5, 6), E(5, 4), F(2, 6)
\triangle GHI has G(0, 0), H(4, 0), I(0, 3), and \triangle JKL has J(5, 5), K(5, 1), L(8, 4). Can any rigid motion carry one onto the other? Explain.
\triangle RST has R(1, 0), S(4, 1), T(2, 4), and a second triangle has U(1, -4), V(4, -2), W(0, -1). Write a congruence statement with the corners in matching order, and name a single rigid motion that carries \triangle RST onto the second triangle.

Part 2: corresponding parts

\triangle ABC \cong \triangle XYZ. List all six pairs of congruent corresponding parts.
\triangle PQR \cong \triangle STU, with m\angle P = 40^\circ, m\angle Q = 75^\circ and QR = 11. Find m\angle U and TU.
\triangle KLM \cong \triangle XYZ, with KL = 2a + 5, XY = 3a - 4, m\angle M = (b + 30)^\circ and m\angle Z = (2b - 15)^\circ. Find a, b, KL and m\angle M.

Part 2: the criteria from rigid motions

\triangle ABC has A(0, 0), B(4, 0) and C(1, 3) in Exercises 18 and 19.

\triangle DEF has D(6, 2), E(6, 6), F(3, 3), and \angle A \cong \angle D. Check that AB = DE and AC = DF, then follow the SAS construction: give the translation and the rotation, and say whether a reflection is needed.
\triangle DEG has D(6, 2), E(6, 6), G(9, 3), and \angle A \cong \angle D. Follow the SAS construction to carry \triangle ABC onto \triangle DEG with C \mapsto G.
After the slide and the turn of the SSS proof, the placed copy has corners D(0, 1), E(4, 5) and C_2(5, 2), and the target is \triangle DEF with F(1, 6). Check that DC_2 = DF and EC_2 = EF, show that line DE is the perpendicular bisector of \overline{C_2F}, and give the rule of the reflection that finishes.
A is at (0, 0) and B at (3, 4). The point C is on the positive x-axis with \displaystyle BC = \sqrt{17}. Find every position of C and the length AC in each case. What does this say about SSA?

Reasoning

Two rectangles each have an area of 12 square units. Must they be congruent? Explain with an example.
\triangle ABC has three different side lengths. It is reflected in three different lines, one after another, and the final images are A''', B''', C'''. Explain why no single translation or rotation sends A \mapsto A''', B \mapsto B''' and C \mapsto C'''.
The ASA proof places the given side first and then uses an angle at each end of it. Two angles and a side that is not between them also force congruence (AAS). Explain how to turn AAS information into ASA information, so that the same motion argument applies.
Prove from the definition of congruence that if \triangle ABC \cong \triangle DEF and \triangle DEF \cong \triangle GHI, then \triangle ABC \cong \triangle GHI.
Explain why the SSS argument cannot be copied for quadrilaterals, using two quadrilaterals with four matching sides that are not congruent.
A proof of SAS by rigid motions must never say “C lands on F because the triangles are congruent.” Explain why not.

Answers

MN = 9, NP = 12, m\angle N = 64^\circ and m\angle P = 65^\circ. The motion keeps every length and angle, so each part equals its partner: MN = JK, NP = KL, m\angle N = m\angle K. \angle P matches \angle L, and m\angle L = 180^\circ - 64^\circ - 51^\circ = 65^\circ by the triangle angle sum.
Perimeter 31 cm, area 52 cm². The perimeter is 4 + 5 + 6 + 7 + 9 = 31, and both lengths and area are kept by rigid motions.
CD = 10. \displaystyle AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10, and rigid motions keep length.
x = 7, y = 14 and AB = 17. Lengths are kept, so 3x - 4 = x + 10 and 2x = 14. Angles are kept, so 5y + 2 = 72 and 5y = 70. Then AB = 3(7) - 4 = 17.
\overline{WX} \parallel \overline{YZ}, because \overline{AB} goes to \overline{WX}, \overline{CD} goes to \overline{YZ}, and parallel lines go to parallel lines. \angle W is a right angle, because angle measure is kept.
P''(4, 1), Q''(1, 1), R''(1, -3). The half turn gives (-1, -2), (-4, -2), (-4, -6), and adding \langle 5, 3 \rangle gives the images. PQ = 3 and P''Q'' = 4 - 1 = 3; QR = 4 and Q''R'' = 1 - (-3) = 4.
P''(-1, -2), Q''(-4, -2), R''(-4, -6). The first reflection gives (1, -2), (4, -2), (4, -6), and the second changes the sign of each first coordinate. Both signs have changed, so the single motion is the rotation of 180^\circ about the origin, (x, y) \mapsto (-x, -y).
(a) Kept: two reflections, an even number, and a rotation keeps it. (b) Reversed: one reflection. (c) Reversed: three reflections, an odd number.
The translation by \langle 5, 2 \rangle. Each corner moves 5 right and 2 up: (1, 1) \mapsto (6, 3), (3, 1) \mapsto (8, 3), (1, 4) \mapsto (6, 6).
The reflection in the y-axis, (x, y) \mapsto (-x, y): (1, 1) \mapsto (-1, 1), (3, 1) \mapsto (-3, 1), (1, 4) \mapsto (-1, 4).
The rotation of 90^\circ counterclockwise about the origin, (x, y) \mapsto (-y, x): (1, 1) \mapsto (-1, 1), (3, 1) \mapsto (-1, 3), (1, 4) \mapsto (-4, 1).
One answer: translate by \langle 4, 5 \rangle, rotate 90^\circ clockwise about D(5, 6), then reflect in the line x = 5. The slide gives (5, 6), (7, 6), (5, 9). A quarter turn clockwise sends an offset (a, b) to (b, -a): (2, 0) becomes (0, -2), landing on (5, 4) = E, and (0, 3) becomes (3, 0), landing on (8, 6). That is on the wrong side of line DE, which is x = 5, and the reflection sends (8, 6) \mapsto (2, 6) = F. A reflection is unavoidable: A, B, C run counterclockwise, and D, E, F run clockwise.
No. \triangle GHI has sides 4, 3 and 5. \triangle JKL has JK = 4, \displaystyle JL = \sqrt{3^2 + 1^2} = \sqrt{10} and \displaystyle KL = \sqrt{3^2 + 3^2} = 3\sqrt{2}. A rigid motion keeps lengths, and the side lengths differ.
\triangle RST \cong \triangle WUV, by the rotation of 90^\circ clockwise about the origin, (x, y) \mapsto (y, -x): R(1, 0) \mapsto (0, -1) = W, S(4, 1) \mapsto (1, -4) = U, T(2, 4) \mapsto (4, -2) = V. The lengths agree: \displaystyle RS = WU = \sqrt{10}, \displaystyle ST = UV = \sqrt{13}, \displaystyle RT = WV = \sqrt{17}.
\overline{AB} \cong \overline{XY}, \overline{BC} \cong \overline{YZ}, \overline{CA} \cong \overline{ZX}, \angle A \cong \angle X, \angle B \cong \angle Y and \angle C \cong \angle Z. The order of the letters gives the pairs.
m\angle U = 65^\circ and TU = 11. \angle U matches \angle R, and m\angle R = 180^\circ - 40^\circ - 75^\circ = 65^\circ. \overline{TU} matches \overline{QR}.
a = 9, b = 45, KL = 23 and m\angle M = 75^\circ. \overline{KL} matches \overline{XY}, so 2a + 5 = 3a - 4. \angle M matches \angle Z, so b + 30 = 2b - 15. Then KL = 2(9) + 5 = 23 and m\angle M = 45 + 30 = 75 degrees.
AB = 4 = DE, and \displaystyle AC = \sqrt{1^2 + 3^2} = \sqrt{10} = DF, since F is 3 left and 1 up from D. Translate by \langle 6, 2 \rangle: B \mapsto (10, 2), C \mapsto (7, 5). Rotate 90^\circ counterclockwise about D(6, 2): the offset (4, 0) becomes (0, 4), landing on (6, 6) = E, and the offset (1, 3) becomes (-3, 1), landing on (3, 3) = F. No reflection is needed.
Translate by \langle 6, 2 \rangle, rotate 90^\circ counterclockwise about D, then reflect in x = 6. The first two stages are those of Exercise 18 and put C at (3, 3), which is on the other side of line DE (the line x = 6) from G. The reflection sends (3, 3) \mapsto (12 - 3, 3) = (9, 3) = G. As a check, \displaystyle DG = \sqrt{3^2 + 1^2} = \sqrt{10} = AC.
\displaystyle DC_2 = \sqrt{5^2 + 1^2} = \sqrt{26} and \displaystyle DF = \sqrt{1^2 + 5^2} = \sqrt{26}; \displaystyle EC_2 = \sqrt{1^2 + 3^2} = \sqrt{10} and \displaystyle EF = \sqrt{3^2 + 1^2} = \sqrt{10}. The midpoint of \overline{C_2F} is (3, 4), which is on line DE, the line y = x + 1. The slope of \overline{C_2F} is \displaystyle \frac{6 - 2}{1 - 5} = -1 and the slope of DE is 1, so they are perpendicular. The reflection in y = x + 1 is (x, y) \mapsto (y - 1, x + 1), which sends C_2(5, 2) \mapsto (1, 6) = F and leaves D and E where they are.
C(2, 0) with AC = 2, or C(4, 0) with AC = 4. From (c - 3)^2 + 4^2 = 17, (c - 3)^2 = 1, so c = 2 or c = 4, both positive. The two triangles share AB = 5, the angle at A and \displaystyle BC = \sqrt{17}, but their third sides differ, so they are not congruent: SSA is not a congruence criterion.
No. A 3 by 4 rectangle and a 2 by 6 rectangle both have area 12. A rigid motion keeps lengths, so it would carry a side of length 3 to a side of length 3, and the 2 by 6 rectangle has none.
Three reflections, an odd number, reverse orientation, so if A, B, C run counterclockwise, then A''', B''', C''' run clockwise (or the other way round). A translation or a rotation keeps orientation, so it would send A, B, C to three points running the same way as A, B, C. It cannot send them to A''', B''', C''' in that order.
Use the triangle angle sum. If \angle A \cong \angle D, \angle B \cong \angle E and \overline{BC} \cong \overline{EF}, then m\angle C = 180^\circ - m\angle A - m\angle B and m\angle F = 180^\circ - m\angle D - m\angle E, so \angle C \cong \angle F. Now \overline{BC} lies between \angle B and \angle C, and \overline{EF} between \angle E and \angle F: that is ASA, and the motion argument applies with \overline{BC} placed on \overline{EF}.
Some rigid motion m carries \triangle ABC onto \triangle DEF with A \mapsto D, B \mapsto E and C \mapsto F, and some rigid motion n carries \triangle DEF onto \triangle GHI with D \mapsto G, E \mapsto H and F \mapsto I. Doing m and then n is a sequence of rigid motions, so it is a rigid motion, and it sends A \mapsto G, B \mapsto H and C \mapsto I, carrying \triangle ABC onto \triangle GHI. So \triangle ABC \cong \triangle GHI by the definition.
A 2 by 3 rectangle and a parallelogram with sides 2 and 3 and a 60^\circ angle have the same four side lengths, but a rigid motion keeps angles, and the rectangle’s are all 90^\circ. The argument breaks because after two corners are placed, the other two corners of a quadrilateral are not fixed by distances to the placed ones: a quadrilateral frame can lean.
That the triangles are congruent is what the proof is trying to show. Using it as a reason assumes the conclusion, so the argument would prove nothing. Each landing must come from the definition of the motion and what it keeps.