Show two figures are congruent by describing a sequence of rotations, reflections and translations

Learning goal: Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections and translations, and, given two congruent figures with or without coordinates, describe a sequence of rigid motions that exhibits the congruence.

The definition. Two figures are congruent when one can be moved exactly onto the other by a sequence of rigid motions: some rotations, reflections and translations, applied one after another, possibly only one of them. "Exactly onto" means every point of the first figure lands on a point of the second and nothing is left over. Because each rigid motion preserves lengths, angles and parallel lines, a sequence of them does too. That gives the familiar consequence: congruent figures have equal corresponding sides and equal corresponding angles. Turn that around and you get the test used in practice: if any corresponding side or angle differs, no sequence of rigid motions can exist, because none of them changes a length or an angle.

What a description must contain. "Flip it and slide it" is not a description. Each motion needs its full information: a translation needs its direction and distance, written on a grid as a rule (x, y) → (x + a, y + b); a reflection needs its line; a rotation needs its center, angle and direction. The description should also say the order, because the same motions in a different order usually give a different image.

A worked model with coordinates. Triangle PQR has P(1, 1), Q(4, 1) and R(1, 3). Triangle P″Q″R″ has P″(−1, −4), Q″(−4, −4) and R″(−1, −2). First check that congruence is plausible: PQ is horizontal with length 4 − 1 = 3, and P″Q″ is horizontal with length from −4 to −1, also 3. PR is vertical with length 2, and P″R″ is vertical with length from −4 to −2, also 2. The angle at P is a right angle (a horizontal side meets a vertical side) and so is the angle at P″. The matching of vertices, P to P″, Q to Q″, R to R″, is forced by these lengths.

Next check orientation. Walking P → Q → R in the pre-image goes right, then up-and-left: a counterclockwise turn. Walking P″ → Q″ → R″ goes left, then up-and-right: a clockwise turn. The orientation is reversed, so the sequence must contain a reflection (an odd number of reflections). Rotations and translations alone never reverse orientation.

Now build the sequence. Step 1: reflect across the y-axis, rule (x, y) → (−x, y). P(1, 1) → P′(−1, 1), Q(4, 1) → Q′(−4, 1), R(1, 3) → R′(−1, 3). The triangle now has the right orientation and the right shape but sits 5 units too high: P′ is at height 1 and P″ at height −4. Step 2: translate 5 units down, rule (x, y) → (x, y − 5). P′(−1, 1) → (−1, −4) = P″. Q′(−4, 1) → (−4, −4) = Q″. R′(−1, 3) → (−1, −2) = R″. Every vertex lands on its match, so the sequence "reflect across the y-axis, then translate 5 units down" exhibits the congruence. Do not stop after moving one vertex; the check on all three is what shows the sequence works.

Sequences are not unique, and order matters. Translating 5 units down first and then reflecting across the y-axis gives the same result here, because the translation is parallel to the line of reflection: P(1, 1) → (1, −4) → (−1, −4). But change the translation to (x, y) → (x − 2, y − 5). Reflect first, then translate: P(1, 1) → (−1, 1) → (−3, −4). Translate first, then reflect: P(1, 1) → (−1, −4) → (1, −4). Different points, so the order is part of the description. Any correct sequence is acceptable; the requirement is that yours is fully described and checked.

Without coordinates. Two identical triangular tiles laid edge to edge along their longest sides form a parallelogram when the second tile is placed with its vertices swapped end for end along the shared side, so that the two tiles' equal angles sit at opposite ends of it; placed the other way, with equal angles at the same end, they form a kite instead. In the parallelogram placement the second tile is the image of the first under a rotation of 180° about the midpoint of the shared side. A translation cannot do it: a translation keeps every side pointing in its original direction, but in the parallelogram the second tile's sides point the opposite way. To describe rigid motions on paper without a grid, name the center of a rotation as a specific point ("the midpoint of side AB"), the line of a reflection as a specific line ("the perpendicular bisector of AB"), and the translation by a specific segment ("the translation that takes A to D").

A non-example. Triangle with vertices (0, 0), (4, 0), (0, 3) and triangle with vertices (0, 0), (8, 0), (0, 6) are not congruent. The side along the x-axis is 4 units in one and 8 units in the other. No rotation, reflection or translation changes a length, so no sequence of them can take a 4-unit side to an 8-unit side. The second triangle is an enlargement, which is the subject of similarity, not congruence. A figure with the same side lengths but different angles, such as a square and a non-square rhombus of side 3, is also not congruent, because angles are preserved too.

How to find a sequence. Match corresponding vertices using side lengths and angles. Compare the orientation of the two figures; if it differs, include one reflection. Use a translation to carry one vertex onto its match, then a rotation about that vertex if the figure still needs turning. Finally apply every motion to every vertex and confirm each lands where it should. If a vertex misses, the sequence is wrong, however plausible the picture looked.

Worked, translation only: Triangle ABC has A(0, 0), B(3, 0), C(0, 2); triangle DEF has D(4, −3), E(7, −3), F(4, −1). From A to D is 4 right and 3 down, so try the translation (x, y) → (x + 4, y − 3). B(3, 0) → (7, −3) = E and C(0, 2) → (4, −1) = F. All three vertices land, so one translation exhibits the congruence.

Worked, reflection needed: The same triangle ABC and triangle GHK with G(0, 0), H(−3, 0), K(0, 2). Walking A → B → C turns counterclockwise; walking G → H → K turns clockwise, so a reflection is required. Reflect across the y-axis, (x, y) → (−x, y): A → (0, 0) = G, B(3, 0) → (−3, 0) = H, C(0, 2) → (0, 2) = K. One reflection does it.

Worked, rotation: The same triangle ABC and triangle LMN with L(0, 0), M(0, 3), N(−2, 0). Both are counterclockwise, so no reflection. The 3-unit side has turned from pointing right to pointing up, a quarter turn counterclockwise. Rotate 90° counterclockwise about the origin, (x, y) → (−y, x): B(3, 0) → (0, 3) = M and C(0, 2) → (−2, 0) = N. Confirmed.

Worked non-example: Triangle ABC and the triangle with vertices (0, 0), (3, 0), (0, 3). Side AC has length 2 but the corresponding side has length 3. Since every rigid motion keeps lengths, no sequence can change 2 into 3, and the triangles are not congruent. No amount of clever sequencing can fix a length mismatch.

Recording task: Draw triangle PQR with P(1, 1), Q(4, 1), R(1, 3) and triangle P″Q″R″ with P″(−1, −4), Q″(−4, −4), R″(−1, −2) on one grid. Trace PQR on tracing paper, flip the paper over the y-axis and then slide it 5 units down, checking that it covers P″Q″R″ exactly. Record a three-column table: pre-image vertex, vertex after the reflection, vertex after the translation. Then write the same sequence in words and rules: "reflect across the y-axis, (x, y) → (−x, y); then translate 5 down, (x, y) → (x, y − 5)."

Independent performance: Take quadrilateral with vertices (1, 1), (4, 1), (4, 3), (1, 2) and produce three different images: one by a translation of your choice, one by a reflection across an axis followed by a translation, and one by a rotation of 90° about the origin followed by a translation. Give each image to a partner (or set it aside for yourself the next day) as a list of four coordinates, without saying how it was made, and then describe a sequence of rigid motions that exhibits the congruence between the original and each image. Success criteria: each motion is fully described (rule, line, or center with angle and direction); the order is stated; every vertex is checked and lands on its match; the orientation check correctly predicts whether a reflection is present; and one sentence explains why a figure with a different side length could not be reached by any sequence.