Congruence from Rigid Motions
Emma is sewing a quilt with 40 triangular patches, all cut from one cardboard template. Her friend Malik asks, "How do you know they're all identical? Did you measure every side and every angle of every patch?" She says no. She only needs to check a few parts, as long as they're the right ones. Which parts are enough, and why? By the end of this page you'll be able to answer, and prove it.
Rigid motions and congruent figures
Imagine sliding a cardboard triangle across a table, flipping it over, or spinning it. The triangle moves, but its size and shape don't change. Mathematicians call these moves rigid motions.
In each picture the blue triangle is the preimage and the orange triangle is the image. Points of the image are named with a prime: $A$ maps to $A'$ ("A prime").
This definition has a big consequence. If $\triangle ABC$ is mapped onto $\triangle DEF$ by rigid motions, then lengths and angles don't change. So all six pairs of corresponding parts are congruent:
$$\overline{AB} \cong \overline{DE},\quad \overline{BC} \cong \overline{EF},\quad \overline{AC} \cong \overline{DF},\quad \angle A \cong \angle D,\quad \angle B \cong \angle E,\quad \angle C \cong \angle F$$The letters in $\triangle ABC \cong \triangle DEF$ are listed in matching order: $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$.
Writing a proof
In geometry you have to justify each claim. A two-column proof lists statements on the left and a reason for each on the right. A paragraph proof says the same thing in sentences.
| Statements | Reasons |
|---|---|
| 1. A reflection maps $\triangle ABC$ onto $\triangle DEF$. | 1. Given |
| 2. $\overline{AB} \cong \overline{DE}$ | 2. Rigid motions preserve distance. |
A reflection across line $\ell$ maps $\triangle ABC$ exactly onto $\triangle DEF$. You are given that segment $AB$ maps to segment $DE$, segment $AC$ maps to segment…
A reflection across line $\ell$ maps $\triangle ABC$ exactly onto $\triangle DEF$. You are given that segment $AB$ maps to segment $DE$, segment $AC$ maps to segment $DF$, and $\angle A$ maps to $\angle D$.
State which triangle congruence criterion (SSS, SAS, ASA, AAS, or HL) can be directly deduced from this rigid motion, and explain why the preservation of distance and angle measure by the reflection guarantees that the two triangles are congruent.
A landscape architect in Seattle is designing a triangular flower bed, $\triangle PQR$. She decides to move the entire design $15 \text{ feet}$ to the east and…
A landscape architect in Seattle is designing a triangular flower bed, $\triangle PQR$. She decides to move the entire design $15 \text{ feet}$ to the east and $5 \text{ feet}$ north to avoid a water pipe, creating a new triangular bed, $\triangle P'Q'R'$.
Since a translation is a rigid motion, all side lengths remain the same. Write a short explanation of how you can use the definition of a translation to prove that $\triangle PQR \cong \triangle P'Q'R'$ using the SSS congruence criterion.
Complete the missing reason in the following two-column proof snippet: Given: A rotation about point $P$ maps $\triangle LMN$ to $\triangle RST$. 1. A rotation is a…
Complete the missing reason in the following two-column proof snippet:
Given: A rotation about point $P$ maps $\triangle LMN$ to $\triangle RST$.
-
A rotation is a rigid motion. (Reason: Given)
-
Rigid motions preserve angle measures. (Reason: Definition of rigid motion)
-
$\angle L \cong \angle R$ and $\angle M \cong \angle S$. (Reason: ___)
-
Rigid motions preserve segment lengths. (Reason: Definition of rigid motion)
-
Segment $LM \cong Segment RS$. (Reason: Corresponding parts of figures mapped by a rigid motion)
-
$\triangle LMN \cong \triangle RST$. (Reason: ASA Congruence Criterion)
SSS: three sides
Emma's cardboard template has three sides. If a new patch has three sides that match the template, must it be the same triangle? Yes, because a triangle can't flex.
Why it works. Suppose $AB = A'B'$, $BC = B'C'$, and $AC = A'C'$.
-
Translate $A$ onto $A'$. Then rotate about $A'$ until ray $AB$ lies along ray $A'B'$. Since $AB = A'B'$, point $B$ lands exactly on $B'$.
-
Where is $C$ now? It is $AC = A'C'$ away from $A'$ and $BC = B'C'$ away from $B'$. Two circles with different centers meet in at most two points, and those two points are mirror images across the line through $A'$ and $B'$.
-
The point $C'$ is one of those two points. If $C$ landed on $C'$, you're done. If it landed on the mirror image, reflect across line $A'B'$. That reflection fixes $A'$ and $B'$ and sends $C$ to $C'$.
A sequence of rigid motions has mapped $\triangle ABC$ onto $\triangle A'B'C'$, so the triangles are congruent.
A carpenter cuts a triangular piece of wood, $\triangle ABC$, and then uses it as a template to trace a second triangle, $\triangle DEF$, on another wooden board. This…
A carpenter cuts a triangular piece of wood, $\triangle ABC$, and then uses it as a template to trace a second triangle, $\triangle DEF$, on another wooden board. This tracing process acts as a rigid motion (a translation and a rotation).
If the carpenter measures and confirms that $AB = DE$, $BC = EF$, and $AC = DF$, write a two-step logical argument showing that SSS congruence is a direct consequence of the template being moved as a rigid object.
Copy and complete the following two-column proof. Given: A sequence of rigid motions maps $\triangle ABC$ onto $\triangle WXY$ such that $AB$ maps to $WX$, $BC$ maps to…
Copy and complete the following two-column proof.
Given: A sequence of rigid motions maps $\triangle ABC$ onto $\triangle WXY$ such that $AB$ maps to $WX$, $BC$ maps to $XY$, and $AC$ maps to $WY$.
Prove: $\triangle ABC \cong \triangle WXY$
| Statements | Reasons |
|---|---|
| 1. A sequence of rigid motions maps $\triangle ABC$ onto $\triangle WXY$. | 1. ______ |
| 2. Rigid motions preserve distance. | 2. Definition of rigid motion |
| 3. $AB \cong WX$, $BC \cong XY$, $AC \cong WY$ | 3. ______ |
| 4. $\triangle ABC \cong \triangle WXY$ | 4. ______ |
In a mechanical linkage, two triangles, $\triangle GHI$ and $\triangle GJI$, share a common side $GI$. You are given that $GH \cong GJ$ and $HI \cong JI$. If you were to…
In a mechanical linkage, two triangles, $\triangle GHI$ and $\triangle GJI$, share a common side $GI$. You are given that $GH \cong GJ$ and $HI \cong JI$.
If you were to reflect $\triangle GHI$ across the line containing segment $GI$, describe how the properties of rigid motions and circles (considering $G$ and $I$ as centers of rotation for the segments) prove that point $H$ must map exactly onto point $J$. Conclude which congruence theorem this illustrates.
SAS: two sides and the angle between them
What if you only know two sides? You also need the angle between them (the included angle). Together those three parts pin the triangle down.
Why it works. Suppose $AB = DE$, $\angle B \cong \angle E$, and $BC = EF$.
-
Translate $B$ onto $E$. Rotate about $E$ so ray $BA$ lies along ray $ED$. Since $BA = ED$, point $A$ lands on $D$.
-
The image of ray $BC$ is a ray from $E$ that makes an angle of $m\angle B = m\angle E$ with $\overrightarrow{ED}$. There are only two such rays, one on each side of line $ED$.
-
If the image is on the wrong side of line $ED$, reflect across line $ED$. This fixes $E$ and $D$. Now the image of ray $BC$ lies along ray $EF$.
-
Since $BC = EF$, point $C$ lands on $F$.
All three vertices match, so $\triangle ABC \cong \triangle DEF$.
Given $\triangle LMN$ and $\triangle XYZ$ where $LM \cong XY$, $\angle M \cong \angle Y$, and $MN \cong YZ$. A student performs a translation mapping point $M$ to point…
Given $\triangle LMN$ and $\triangle XYZ$ where $LM \cong XY$, $\angle M \cong \angle Y$, and $MN \cong YZ$.
A student performs a translation mapping point $M$ to point $Y$, followed by a rotation mapping ray $ML$ onto ray $YX$. Since $LM \cong XY$, point $L$ maps to point $X$.
Write the next logical steps in this paragraph proof to show that point $N$ must map to point $Z$, and conclude why this sequence of rigid motions proves the SAS congruence criterion.
You are given two triangles on a coordinate plane, $\triangle ABC$ and $\triangle DEF$. It is known that $AB = DE = 5 \text{ inches}$, $BC = EF = 7 \text{ inches}$, and…
You are given two triangles on a coordinate plane, $\triangle ABC$ and $\triangle DEF$.
It is known that $AB = DE = 5 \text{ inches}$, $BC = EF = 7 \text{ inches}$, and $\angle B \cong \angle E$.
Construct a formal two-column proof to show that $\triangle ABC \cong \triangle DEF$. Your proof must explicitly use rigid motions: define a translation that maps $B$ to $E$, followed by a rotation about $E$ that maps ray $BA$ to ray $ED$. Justify why point $C$ must map to point $F$ after these motions, establishing the SAS criterion fundamentally.
ASA: two angles and the side between them
Instead of two sides and the angle between them, you can use two angles and the side between them (the included side).
Why it works. Suppose $\angle A \cong \angle D$, $AB = DE$, and $\angle B \cong \angle E$.
-
Translate $A$ onto $D$ and rotate about $D$ until $B$ lands on $E$. This is possible because $AB = DE$.
-
If the triangle is on the wrong side of line $DE$, reflect across line $DE$.
-
Equal angle measures at $A$ and $B$ mean the images of rays $AC$ and $BC$ lie along rays $DF$ and $EF$. Two rays that start at different points on the same side of a line meet in exactly one point. So $C$ lands on $F$.
In a graphic design software, a triangular logo $\triangle JKL$ is rotated $90^\circ$ clockwise around the origin to create $\triangle MNO$. The designer knows that…
In a graphic design software, a triangular logo $\triangle JKL$ is rotated $90^\circ$ clockwise around the origin to create $\triangle MNO$. The designer knows that $\angle J \cong \angle M$, side $JK \cong MN$, and $\angle K \cong \angle N$.
Identify the congruence criterion that applies to this situation. Briefly explain how the properties of rotations support the use of this specific criterion to conclude that the two logos are identical in size and shape.
You are given two triangles, $\triangle ABC$ and $\triangle XYZ$. A sequence of rigid motions (a translation followed by a reflection) maps $A$ to $X$, $B$ to $Y$, and…
You are given two triangles, $\triangle ABC$ and $\triangle XYZ$. A sequence of rigid motions (a translation followed by a reflection) maps $A$ to $X$, $B$ to $Y$, and $C$ to $Z$.
If you know that the rigid motions mapped $\angle B$ to $\angle Y$, $\angle C$ to $\angle Z$, and the side $BC$ to the side $YZ$, which congruence criterion is demonstrated here? State the corresponding parts that align to form this criterion.
Two triangular sails, $\triangle DEF$ and $\triangle RST$, are designed for a yacht. The manufacturer states that $\angle D \cong \angle R$, $\angle E \cong \angle S$,…
Two triangular sails, $\triangle DEF$ and $\triangle RST$, are designed for a yacht. The manufacturer states that $\angle D \cong \angle R$, $\angle E \cong \angle S$, and the included side $DE \cong RS$.
Imagine a rigid motion that maps segment $DE$ perfectly onto segment $RS$. Because rigid motions preserve angle measure, explain what must happen to the rays $DF$ and $EF$ during this mapping, and how this logically proves the ASA congruence criterion.
AAS: two angles and a side not between them
What if the side you know is not between the two angles? You can still prove the triangles are congruent. This time you don't need a new rigid-motion argument. You reduce AAS to ASA.
Why it works. The angles of a triangle add to $180^\circ$. If two angles of one triangle match two angles of another, the third angles match too. Once all three angles are known, the side you were given turns out to be between two known angles, and that is ASA.
Two triangular metal plates, $\triangle VWX$ and $\triangle YZA$, are produced in a factory in Detroit. A robotic arm picks up $\triangle VWX$, translates it, and places…
Two triangular metal plates, $\triangle VWX$ and $\triangle YZA$, are produced in a factory in Detroit. A robotic arm picks up $\triangle VWX$, translates it, and places it perfectly on top of $\triangle YZA$.
If the robot's programming ensures that $\angle V$ maps to $\angle Y$, $\angle X$ maps to $\angle A$, and the non-included side $VW$ maps to $YZ$, identify which triangle congruence theorem justifies that the plates are congruent.
A surveyor in Texas is mapping two triangular plots of land, $\triangle KLM$ and $\triangle NOP$. A translation maps point $K$ to point $N$. After the translation, a…
A surveyor in Texas is mapping two triangular plots of land, $\triangle KLM$ and $\triangle NOP$. A translation maps point $K$ to point $N$. After the translation, a rotation about point $N$ maps ray $KL$ onto ray $NO$.
You are given that $\angle K \cong \angle N$, $\angle M \cong \angle P$, and the side $KL$ is congruent to side $NO$.
Write a two-column proof demonstrating that $\triangle KLM \cong \triangle NOP$. Be sure to state which congruence criterion (ASA or AAS) applies based on the given information.
The AAS (Angle-Angle-Side) congruence theorem can be proven using the ASA theorem and the Triangle Angle Sum Theorem. Given $\triangle JKL$ and $\triangle MNO$ with…
The AAS (Angle-Angle-Side) congruence theorem can be proven using the ASA theorem and the Triangle Angle Sum Theorem.
Given $\triangle JKL$ and $\triangle MNO$ with $\angle J \cong \angle M$, $\angle K \cong \angle N$, and non-included side $JL \cong MO$.
Write a multi-step paragraph proof that first uses the Triangle Angle Sum Theorem to show $\angle L \cong \angle O$. Then, describe the specific sequence of rigid motions that will map $\triangle JKL$ onto $\triangle MNO$ using the newly established ASA condition.
HL: right triangles
Right triangles have a bonus. Because one angle is already known to be $90^\circ$, you can get away with just two other parts, the hypotenuse and one leg.
Why it works. Take a utility pole that stands straight up from flat ground. It is held up by two wires of equal length, anchored on the ground.
-
Reflection route. Reflect the left triangle across the pole. The pole stays put. The ground makes a right angle with the pole, so the reflected ground leg lies along the ground on the other side. The reflected wire still has length $45$ ft and starts at the top of the pole. Only two points on the ground are $45$ ft from the top, and they are mirror images of each other across the pole. So the reflected wire lands exactly on the other wire.
-
Pythagorean route. Let the hypotenuse be $c$ and a known leg be $a$. The other leg is $b = \sqrt{c^2 - a^2}$, so it is the same in both triangles. Now all three sides match, and SSS finishes the proof.
Consider a right triangle $\triangle GHI$ with a right angle at $H$. A reflection across the $y$-axis maps $\triangle GHI$ to $\triangle STU$. If you are given that the…
Consider a right triangle $\triangle GHI$ with a right angle at $H$. A reflection across the $y$-axis maps $\triangle GHI$ to $\triangle STU$.
If you are given that the hypotenuse $GI$ maps to $SU$, and the leg $GH$ maps to $ST$, state the specific congruence criterion for right triangles that proves $\triangle GHI \cong \triangle STU$. Explain how the rigid motion ensures the necessary conditions for this criterion are met.
A utility pole in Florida is supported by two guy wires anchored to the flat ground. The pole forms a right angle with the ground. The two guy wires, representing the…
A utility pole in Florida is supported by two guy wires anchored to the flat ground. The pole forms a right angle with the ground. The two guy wires, representing the hypotenuses of two right triangles, are of equal length ($45 \text{ feet}$). The pole acts as the shared leg for both triangles.
Using the concept of rigid motions (specifically, reflecting one triangle across the utility pole), explain how you can deduce the Hypotenuse-Leg (HL) congruence theorem to prove the two triangles are congruent.
Symmetry in proofs
A line of symmetry is a reflection that maps a figure onto itself. A rotational symmetry is a rotation that does the same. They are rigid motions, so they hand you congruent parts for free.
The Isosceles Triangle Theorem
Proof with a reflection. Let $PQ \cong PR$ and let $\ell$ be the bisector of $\angle P$.
-
Reflect across $\ell$. Point $P$ is on $\ell$, so it stays fixed.
-
$\ell$ bisects $\angle P$, so the reflection swaps ray $PQ$ and ray $PR$.
-
Reflections preserve distance from $P$. Since $PQ = PR$, point $Q$ lands exactly on $R$. Reflecting again undoes the move, so $R$ lands on $Q$.
-
So $\angle PQR$ maps onto $\angle PRQ$. Rigid motions preserve angle measure, so $\angle Q \cong \angle R$.
Rotational symmetry
Suppose an equilateral triangle $ABC$ is inscribed in a circle with center $O$. The vertices are equally spaced around the circle, so each central angle is $360^\circ \div 3 = 120^\circ$. A $120^\circ$ rotation about $O$ sends $A \to B$, $B \to C$, and $C \to A$. Rigid motions map segments to segments of the same length, so $\overline{AB}$ maps to $\overline{BC}$, and $\overline{BC}$ maps to $\overline{CA}$. That gives $AB = BC = CA$.
Symmetry in quadrilaterals
-
Kite. In kite $ABCD$ with $AB \cong AD$ and $CB \cong CD$, diagonal $\overline{AC}$ is shared. By SSS, $\triangle ABC \cong \triangle ADC$. This is exactly the reflection across line $AC$, which swaps $B$ and $D$.
-
Parallelogram. A $180^\circ$ rotation about the point where the diagonals meet sends each vertex to the opposite vertex, so $\overline{AB}$ maps to $\overline{CD}$ and $\overline{BC}$ maps to $\overline{DA}$.
The Isosceles Triangle Theorem states that if two sides of a triangle are congruent, then the angles opposite those sides are congruent. Consider an isosceles triangle…
The Isosceles Triangle Theorem states that if two sides of a triangle are congruent, then the angles opposite those sides are congruent.
Consider an isosceles triangle $\triangle PQR$ with $PQ \cong PR$. Let $\ell$ be the angle bisector of $\angle P$.
Write a paragraph proof using rigid motions to prove this theorem. Specifically, analyze a reflection of $\triangle PQR$ across the line $\ell$. Explain where each vertex maps and why, and how this rigid motion proves that $\angle Q \cong \angle R$.
Consider an equilateral triangle $\triangle ABC$ inscribed in a circle with center $O$. A rigid motion consisting of a $120^\circ$ clockwise rotation about the center…
Consider an equilateral triangle $\triangle ABC$ inscribed in a circle with center $O$.
A rigid motion consisting of a $120^\circ$ clockwise rotation about the center $O$ is performed.
Write an analytical argument explaining how this specific rotation maps the vertices of the triangle onto each other (e.g., $A$ to $B$, $B$ to $C$, $C$ to $A$). Use the properties of this rigid motion to prove that all three sides of the triangle must be congruent ($AB \cong BC \cong CA$), thereby establishing that the triangle is equilateral based on the SSS framework derived from rotational symmetry.
A kite $PQRS$ is defined such that $PQ \cong PS$ and $RQ \cong RS$. The main diagonal $PR$ is drawn. Write a rigorous two-column proof demonstrating that…
A kite $PQRS$ is defined such that $PQ \cong PS$ and $RQ \cong RS$. The main diagonal $PR$ is drawn.
Write a rigorous two-column proof demonstrating that $\triangle PQR \cong \triangle PSR$. Your proof must utilize the definition of a reflection across the line containing $PR$. Explain why points $P$ and $R$ remain fixed, why point $S$ must map to point $Q$, and conclude with the appropriate triangle congruence criterion (SSS) deduced from this transformation.
Why AAA and SSA don't work
Knowing that a criterion works is only half the story. You also need to know which patterns fail.
AAA only gives similar triangles
A phone screen and a TV screen can have the same $16:9$ shape. Cut each screen along a diagonal and you get two right triangles with the same three angles, but the second is much bigger. The picture shows the idea with two small triangles: same angles, one with a $4$ in. base and the other with an $8$ in. base.
Rigid motions preserve distance. A segment of length $4$ can never be moved onto a segment of length $8$, so no sequence of translations, reflections, and rotations can map one triangle onto the other. To enlarge or shrink you need a dilation (here with scale factor $2$). AAA, then, only establishes similarity. You'll study similar triangles in the next chapter.
SSA is not enough
Take $AB = 8$, $\angle A = 30^\circ$, and $BC = 5$. The angle $A$ is not between the two known sides. Point $B$ is $8 \sin 30^\circ = 4$ units above the horizontal ray from $A$, and $4 < 5 < 8$. So a circle of radius $5$ about $B$ meets the ray at two points. That gives two different triangles.
While SSS, SAS, ASA, and AAS are valid congruence criteria deduced from rigid motions, AAA is not. Provide a detailed counterexample using a real-world scenario (e.g., a…
While SSS, SAS, ASA, and AAS are valid congruence criteria deduced from rigid motions, AAA is not.
Provide a detailed counterexample using a real-world scenario (e.g., a smartphone screen and a flat-screen TV with the same aspect ratio). Explain mathematically why rigid motions (translations, rotations, reflections) cannot map one triangle onto the other in the AAA scenario, and identify which type of transformation would be required, proving that AAA only establishes similarity, not congruence.
A student claims that if two triangles have two congruent sides and a congruent non-included angle (SSA), the triangles must be congruent. Using the provided diagram…
A student claims that if two triangles have two congruent sides and a congruent non-included angle (SSA), the triangles must be congruent.
Using the provided diagram showing a rigid segment $QR$ swinging along an arc to position $QR'$, explain why SSA is a counterexample to triangle congruence. Discuss how the rigid motion of rotation (swinging the segment) creates two distinct triangles, $\triangle PQR$ and $\triangle PQR'$, that share the same SSA measurements.
Summary
| Criterion | What you need | Rigid-motion idea |
|---|---|---|
| SSS | 3 pairs of sides | Match one side, then the third vertex is one of two mirror-image points. |
| SAS | 2 sides and the included angle | Match one side, then match the other side with the angle. |
| ASA | 2 angles and the included side | Match the side, then two rays meet at one point. |
| AAS | 2 angles and a non-included side | Third angles are equal, so it becomes ASA. |
| HL | Hypotenuse and a leg (right triangles) | Reflect across the shared leg, or use $a^2 + b^2 = c^2$ and SSS. |
Back to Emma's quilt. She does not need to measure all six parts of every patch. If each patch matches the template in three sides (SSS), or in two sides and the included angle (SAS), or in a pattern of two angles and a side (ASA or AAS), a sequence of rigid motions can lay the patch exactly on the template. All of her patches are identical.