Transformational proof
Two figures look like copies of each other. How do you prove it, rather than just say it?
Where you are 18 in Geometric proof
When you reach for it
Congruence has a definition in terms of motion: two figures are congruent exactly when some sequence of rigid motions (translations, rotations and reflections) carries one onto the other. That is congruence as a sequence of rigid motions. Similarity has the same kind of definition with a dilation allowed as well, which is similarity by transformations.
So the most direct way to prove two figures congruent is to produce the motion: say exactly which motion it is, and show that it carries every point of one figure to the matching point of the other. That is a transformational proof. It is usually written as a paragraph, but like any two-column or paragraph proof, every sentence carries its reason.
Reach for it in three situations.
- The figures sit on a coordinate grid. Each motion has a rule, so where each point lands is a calculation, and the proof is short and certain.
- A figure has a symmetry. An isosceles triangle, a kite or a parallelogram can be carried onto itself by a reflection or a half turn that swaps two of its parts. Whatever a motion swaps is congruent, with no pair of triangles and no congruence criterion needed.
- You are proving a criterion itself. SAS, ASA and SSS are theorems only because, for any two triangles that meet them, a motion can be built. Building it is a transformational proof.
The procedure
A transformational proof has three parts: say which motion, show where each point lands and why, and read off the conclusion.
Name the motion completely. “Flip it over” is not yet a motion, because there are infinitely many lines to flip in. A motion is fixed only when its data are given.
| Motion | What must be named |
|---|---|
| translation | the vector, such as \langle 3, -2 \rangle, or “the vector from A to D” |
| rotation | the center, the angle, and the direction of turn |
| reflection | the mirror line |
| dilation | the center and the scale factor |
A sequence of motions is named by naming each one in order, since the order can change the result.
Land each point, with a reason. For each vertex, say where it goes and why it has to go there. On a coordinate grid the reason is the rule, and the landing is computed. These are the rules the examples use; each comes from its own lesson, such as rotating about the origin and reflecting over the axes.
| Motion | Rule |
|---|---|
| translation by \langle a, b \rangle | (x, y) \mapsto (x + a,\ y + b) |
| rotation of 90^\circ counterclockwise about the origin | (x, y) \mapsto (-y,\ x) |
| rotation of 90^\circ clockwise about the origin | (x, y) \mapsto (y,\ -x) |
| rotation of 180^\circ about the origin | (x, y) \mapsto (-x,\ -y) |
| reflection in the x-axis | (x, y) \mapsto (x,\ -y) |
| reflection in the y-axis | (x, y) \mapsto (-x,\ y) |
| reflection in the line y = x | (x, y) \mapsto (y,\ x) |
| dilation from the origin by the factor k | (x, y) \mapsto (kx,\ ky) |
Without coordinates, the reasons come from a short list of facts, and only from these.
- A rigid motion keeps every distance and every angle measure, sends lines to lines, and sends a segment to the segment joining the images of its endpoints. That is what rigid motions preserve.
- A reflection leaves every point of its mirror where it is, sends every other point to the opposite side of the mirror, and the mirror is the perpendicular bisector of the segment from a point to its image. So a reflection in the perpendicular bisector of \overline{PQ} swaps P and Q.
- A rotation leaves its center where it is. A half turn, a rotation of 180^\circ, about M sends each point X to the point X' for which M is the midpoint of \overline{XX'}.
- From a point on a line, on a chosen side of that line, there is only one ray that makes a given angle with it. On a ray there is only one point at a given distance from its endpoint.
The last pair is what turns “lands roughly in the right place” into “lands exactly on it”. Most landings without coordinates go the same way: the image ray makes the right angle on the right side, so it is the right ray; the image point is the right distance along that ray, so it is the right point.
One rule keeps the argument honest. A point may never be said to land on its partner “because the figures are congruent”. That is the thing being proved.
Conclude. Once every vertex of a polygon lands on its partner, every side does too, because a segment goes to the segment between the images of its ends. The motion carries one figure onto the other, so they are congruent by definition. Better still, every part the motion carries onto another part is congruent to it, since lengths and angles are kept. That is where the useful results come from.
Here is the whole procedure on one claim, the Base Angles Theorem.
Given. In \triangle ABC, \overline{AB} \cong \overline{AC}.
Prove. \angle B \cong \angle C.
The triangle looks as if it folds onto itself down the middle. The proof turns that fold into a reflection and checks every point.
The motion. Let \ell be the line along the bisector of \angle BAC, which exists and is unique, and let D be where it meets \overline{BC}. Reflect in \ell.
Where A goes. A lies on the mirror, so A' = A.
Where B goes. Ray AD is part of the mirror, so it stays put. The reflection keeps angle measure, so the image of ray AB makes the angle m\angle BAD with ray AD, and it lies on the other side of \ell from B, which is the side C is on. Ray AC makes that same angle with ray AD on that side, because \ell bisects \angle BAC. There is only one such ray, so ray AB lands on ray AC. Then B' lies on ray AC, and since distance is kept, AB' = AB = AC. The only point of ray AC at that distance from A is C. So B' = C.
Where C goes. The same argument with B and C exchanged gives C' = B.
The conclusion. The reflection sends A \mapsto A, B \mapsto C and C \mapsto B. It carries \angle ABC onto \angle ACB, and a reflection keeps angle measure, so \angle B \cong \angle C.
The same proof in two columns:
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \cong \overline{AC} | Given |
| 2. \ell bisects \angle BAC; reflect in \ell | Every angle has exactly one bisector |
| 3. A' = A | A is on the mirror |
| 4. Ray AB lands on ray AC | The angle with ray AD is kept, the side switches, and only one ray fits |
| 5. B' = C | AB' = AB = AC, and only one point of ray AC is at that distance |
| 6. C' = B | Steps 4 and 5 with B and C exchanged |
| 7. \angle ABC \mapsto \angle ACB | Steps 3, 5 and 6 |
| 8. \angle B \cong \angle C | A reflection keeps angle measure |
Nothing in it needs a second triangle. It also gives more than was asked: the reflection swaps B and C, so its mirror is the perpendicular bisector of \overline{BC}, which means D is the midpoint of the base and \overline{AD} \perp \overline{BC}. The lesson on proving the base angles theorem reaches the same facts through congruent halves.
Worked examples
A single rotation on a grid
Triangle ABC has vertices A(1, 1), B(5, 2) and C(2, 4). Triangle DEF has vertices D(-1, 1), E(-2, 5) and F(-4, 2). Prove that \triangle ABC \cong \triangle DEF by giving a rigid motion that sends A \mapsto D, B \mapsto E and C \mapsto F.
First, a guess from the picture. A is up and to the right of the origin, D is up and to the left, the dashed segments \overline{OA} and \overline{OD} are the same length and meet at a right angle, and the whole triangle seems to have turned with them. So try the rotation of 90^\circ counterclockwise about the origin. A guess proves nothing; the rule decides it.
The rule is (x, y) \mapsto (-y, x). Apply it to each vertex:
Every vertex lands on its partner. A rotation sends each side to the segment joining the images of its ends, so \overline{AB} \mapsto \overline{DE}, \overline{BC} \mapsto \overline{EF} and \overline{CA} \mapsto \overline{FD}. The rotation carries \triangle ABC onto \triangle DEF, so \triangle ABC \cong \triangle DEF.
Check. A rotation keeps length, so matching sides must agree. \displaystyle AB = \sqrt{4^2 + 1^2} = \sqrt{17} and \displaystyle DE = \sqrt{1^2 + 4^2} = \sqrt{17}. They do. This is a check, not the proof: two equal lengths alone would prove nothing about the triangles.
A reflection and a slide
Triangle PQR has vertices P(1, 2), Q(5, 2) and R(2, 5). Triangle JKL has vertices J(-1, -4), K(-5, -4) and L(-2, -1). Find a sequence of rigid motions that sends P \mapsto J, Q \mapsto K and R \mapsto L.
Start with the order of the letters. Going P \to Q \to R you move right, then up and to the left: the triangle is lettered counterclockwise. Going J \to K \to L you move left, then up and to the right: clockwise. A translation or a rotation keeps the direction in which a figure is lettered, and a reflection reverses it. So no sequence of translations and rotations alone can work, and the sequence must contain a reflection.
\overline{PQ} points to the right and \overline{JK} points to the left, both level. A reflection in a vertical line reverses left and right and leaves level segments level, so try the y-axis, (x, y) \mapsto (-x, y):
These are the dashed triangle. Each of its vertices is exactly 6 above its target: (-1, 2) against J(-1, -4), and the same for the other two. So finish with the translation by \langle 0, -6 \rangle:
The reflection in the y-axis followed by the translation by \langle 0, -6 \rangle carries \triangle PQR onto \triangle JKL, so the two are congruent.
Here the two motions could have been done in either order, because the slide runs along the direction of the mirror. That is special. A reflection followed by a slide along its mirror is a glide reflection, and no single reflection does the same job here.
Proving that no motion exists
Triangle ABC has vertices A(0, 0), B(3, 0) and C(0, 4). Triangle DEF has vertices D(2, 1), E(2, 4) and F(6, 2). Are the two triangles congruent?
SolutionThey look alike. \overline{AB} and \overline{DE} both have length 3, and both triangles have area 6: for DEF, the base \overline{DE} lies on the line x = 2 and F is 4 across from it, so the area is \displaystyle \frac{1}{2}(3)(4) = 6.
To prove “yes” you produce a motion. To prove “no”, trying a few motions and failing is not enough, because there are infinitely many to try. Instead, find something that every rigid motion keeps and on which the two triangles differ. Distance will do.
A rigid motion carrying \triangle ABC onto \triangle DEF would carry its three sides onto the three sides of \triangle DEF with every length unchanged, so the two lists of lengths would be the same. They are 3, 4, 5 and \displaystyle 3, \sqrt{17}, \sqrt{20}, which are not. No rigid motion carries one triangle onto the other, and the triangles are not congruent.
The equal areas were a trap. Area is kept by every rigid motion too, so unequal areas would have settled the question at once, but equal areas settle nothing.
A dilation for similarity
Triangle ABC has vertices A(0, 0), B(2, 0) and C(0, 1). Triangle DEF has vertices D(5, 1), E(5, 5) and F(3, 1). Show that a dilation followed by rigid motions carries A \mapsto D, B \mapsto E, C \mapsto F, and give the scale factor.
SolutionCompare matching sides to choose the factor. AB = 2 and DE = 4; AC = 1 and DF = 2. Every length doubles, so dilate from the origin by the factor 2, (x, y) \mapsto (2x, 2y):
The triangle now has the right size but the wrong direction: its side from the image of A to the image of B points right, while \overline{DE} points up. A quarter turn counterclockwise about the origin, (x, y) \mapsto (-y, x), fixes that:
Last, translate by \langle 5, 1 \rangle to move the origin onto D:
The dilation by 2, then the rotation, then the translation carry \triangle ABC onto \triangle DEF. So \triangle ABC \sim \triangle DEF with scale factor 2.
Check. The third sides should be in the same ratio: \displaystyle BC = \sqrt{2^2 + 1^2} = \sqrt{5} and \displaystyle EF = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}. They are.
A kite folds along its diagonal
Kite ABCD has \overline{AB} \cong \overline{AD} and \overline{CB} \cong \overline{CD}, with B \ne D. Prove that \angle ABC \cong \angle ADC, and that the diagonal \overline{AC} bisects \angle BAD.
The kite looks as if it folds along \overline{AC}, so the motion to try is the reflection in line AC. The work is in showing that this reflection really does swap B and D, and the look of the figure is not a reason.
The motion. A is the same distance from B as from D, so A lies on the perpendicular bisector of \overline{BD}, by the converse part of the perpendicular bisector theorem. For the same reason C lies on it. A and C are two different points, and through two points there is exactly one line, so line AC is the perpendicular bisector of \overline{BD}. Reflect in line AC.
The landings. A and C are on the mirror, so A' = A and C' = C. The mirror is the perpendicular bisector of \overline{BD}, so the reflection swaps B and D: B' = D.
The conclusion. The reflection carries \triangle ABC onto \triangle ADC. It carries \angle ABC onto \angle ADC, so they are congruent. It also carries \angle BAC onto \angle DAC, so those two are congruent, and \overline{AC} splits \angle BAD into two equal parts: it bisects \angle BAD. The same reflection shows it bisects \angle BCD as well. These are the angle facts listed under kite properties.
A half turn between parallel lines
Parallel lines m and n are cut by a transversal t, at P on m and at Q on n. Let M be the midpoint of \overline{PQ}. Use the half turn about M to prove that alternate interior angles are congruent.
The lesson on proving alternate interior angles congruent gets this from the corresponding angles postulate. Here it comes from a motion and the Parallel Postulate.
The motion. The half turn about M sends each point X to the point X' for which M is the midpoint of \overline{XX'}. Since M is the midpoint of \overline{PQ}, it sends P \mapsto Q and Q \mapsto P. Line t passes through M, so each of its points goes to a point of t, and t maps onto itself.
Where m goes. A rigid motion sends a line to a line; call the image of m the line m'. It passes through P' = Q. The claim is that m' is parallel to m. Suppose instead that m and m' had a point Z in common. Since Z is on m', it is the image of some point W of m, and doing the half turn twice brings every point back, so Z' = W, which is on m. Since Z is on m, its image Z' is on m'. So Z and Z' both lie on both lines. If Z \ne Z', the two lines share two points, so they are one line, and it contains M, the midpoint of \overline{ZZ'}. If Z = Z', then Z = M, since the center is the only point a half turn leaves in place. Either way m would pass through M. But m meets t only at P, and M \ne P. So m and m' have no point in common, and m' \parallel m.
So m' is a line through Q parallel to m. By the Parallel Postulate there is only one such line, and n is one, so m' = n. The half turn carries m onto n.
The angles. Take a point A of m on one side of t, so \angle APQ is an interior angle at P. Its image is \angle A'P'Q' = \angle A'QP, with A' on n because m goes onto n. The midpoint M of \overline{AA'} lies on t and A does not, so \overline{AA'} crosses t and A' is on the other side of t from A. So \angle A'QP is the interior angle at Q on the opposite side of the transversal: the alternate interior angle to \angle APQ. A rotation keeps angle measure, so
Proving SAS
Triangles ABC and DEF have \overline{AB} \cong \overline{DE}, \angle A \cong \angle D and \overline{AC} \cong \overline{DF}. Prove that \triangle ABC \cong \triangle DEF by building a rigid motion that carries one onto the other.
SAS is the thing being proved, so no congruence criterion may be used. Only the definition of congruence and the list of facts about motions are allowed. The motion is built in up to three stages, each one keeping what the one before achieved.
Stage 1: a translation puts A on D. Translate by the vector from A to D. Then A lands on D. Call the images of B and C at this stage B_1 and C_1.
Stage 2: a rotation about D puts B on E. Rotate about D through the angle from ray DB_1 to ray DE. D is the center, so it stays put, and by the choice of angle ray DB_1 lands on ray DE. Call the new images B_2 and C_2. Both motions so far keep distance, so DB_2 = AB = DE. The only point of ray DE at that distance from D is E, so B_2 = E. (If ray DB_1 already is ray DE, this stage is skipped.)
Stage 3: if needed, a reflection puts C on F. Now D and E are in place. Angle measure and distance have been kept, so m\angle EDC_2 = m\angle BAC = m\angle EDF and DC_2 = AC = DF.
If C_2 is on the same side of line DE as F, then rays DC_2 and DF make the same angle with ray DE on the same side, so they are the same ray. C_2 and F are the same distance along it from D, so C_2 = F, and no reflection is needed.
If C_2 is on the other side, reflect in line DE. D and E are on the mirror, so they stay put, and the image of C_2 is on F’s side, with the angle at D and the distance from D unchanged. The argument of the last paragraph now puts it exactly on F.
The conclusion. The sequence sends A \mapsto D, B \mapsto E and C \mapsto F, so it carries \triangle ABC onto \triangle DEF, and the triangles are congruent. In the figure DEF is a mirror image of ABC, so all three stages are needed.
This is why SAS may be quoted as a shortcut afterwards: its proof is this construction, done once for every pair of triangles.
Boundaries
A motion has to be named, not pictured. “It fits if you turn it” names no center and no angle, and “it lands there” with no reason shows nothing. A figure drawn slightly wrong can look as if it fits and not fit.
A failed search is not a “no”. To prove that two figures are not congruent, find a quantity every rigid motion keeps, such as a length, an angle measure or an area, on which they differ. Trying three motions and finding that none works proves only that those three do not.
Rigid motions give congruence, and nothing more. A dilation by the factor k multiplies every length by k. With dilations in the sequence the result is a similarity, whose scale factor is the product of all the factors, and the figures are congruent only when that product is 1.
Letter order is kept or reversed, never scrambled. Translations and rotations keep the direction in which a figure is lettered, and each reflection reverses it. If the two figures are lettered in opposite directions, the sequence must contain an odd number of reflections.
A coordinate rule belongs to one center or one mirror. (x, y) \mapsto (-y, x) is the quarter turn about the origin and no other point. About any other center, or in any other mirror, the rule is different and has to be worked out, as in Exercises 4 and 5.
The correspondence is part of the answer. A motion that sends A \mapsto E, B \mapsto D and C \mapsto F proves \triangle ABC \cong \triangle EDF, not \triangle ABC \cong \triangle DEF. The order of the letters in a congruence statement records where each vertex went, as in congruence statements.
Vertices are enough only for figures made of segments. A polygon goes wherever its vertices go. A circle goes wherever its center goes, with its radius unchanged. A curved shape given only by a picture has no short list of points that settles where it goes.
Where this goes next
Every triangle congruence criterion is a transformational proof done once and for all. SAS is Example 7 and ASA is Exercise 16. SSS needs one more fact, that two circles meet in at most two points, one on each side of the line joining their centers, and then goes the same way. After that, the criteria can be quoted and the motion left unsaid.
With a dilation in front, the same method proves figures similar, which is what similarity by transformations builds on.
A motion that carries a figure onto itself is a symmetry of it, and every symmetry hands over congruences for free. The reflection in Example 5 is the line symmetry of a kite. The half turn of Exercise 14, which rests on Example 6, is the point symmetry of a parallelogram, and it is the quickest route to the facts in proofs about parallelograms.
And the motions themselves can be reduced. Two reflections in a row make a rotation or a translation (two reflections in a row), and every rigid motion of the plane turns out to be at most three reflections. So in the end every transformational proof of congruence is a proof by mirrors.
Exercises
Images under a motion
- Rotate \triangle ABC, with A(2, 1), B(6, 1) and C(3, 4), by 90^\circ clockwise about the origin. Give the vertices of the image.
- Reflect \triangle PQR, with P(-1, 3), Q(4, 2) and R(0, -2), in the line y = x. Give the vertices of the image.
- Translate \triangle DEF, with D(4, -2), E(7, 0) and F(5, -4), by the vector \langle -3, 5 \rangle. Give the vertices of the image.
- Reflect \triangle KLM, with K(5, 1), L(3, 4) and M(0, -1), in the vertical line x = 2. Give the vertices of the image.
- Rotate \triangle STU, with S(3, 5), T(6, 2) and U(1, 0), by 180^\circ about the point (1, 2). Give the vertices of the image.
Naming the motion
- \triangle ABC has A(1, 2), B(4, 2), C(1, 6) and \triangle A'B'C' has A'(1, -2), B'(4, -2), C'(1, -6). Name one rigid motion that sends A \mapsto A', B \mapsto B' and C \mapsto C', and show that it does.
- \triangle ABC has A(1, 1), B(3, 1), C(1, 4) and \triangle DEF has D(2, 3), E(2, 5), F(-1, 3). Find a rotation about the origin followed by a translation that sends A \mapsto D, B \mapsto E and C \mapsto F.
- \triangle ABC has A(0, 0), B(4, 0), C(1, 3) and \triangle DEF has D(5, 5), E(5, 1), F(2, 4). Explain why any motion sending A \mapsto D, B \mapsto E, C \mapsto F must include a reflection, then find a single reflection that does it.
- \triangle ABC has A(0, 0), B(4, 0), C(1, 3) and \triangle GHI has G(5, 5), H(5, 1), I(2, 3). Are the triangles congruent? Prove your answer.
- \triangle ABC has A(1, 0), B(2, 0), C(1, 2) and \triangle A'B'C' has A'(3, 1), B'(6, 1), C'(3, 7). Find a dilation centered at the origin followed by a translation that sends each vertex to its primed partner, and state the scale factor of the similarity.
- A half turn sends A(1, 4) to C(7, 0) and B(5, 2) to D(3, 2). Find its center, and explain how you know a single half turn does both.
- Find an equation of the mirror line of the reflection that sends A(0, 4) to A'(6, 2).
Proofs
- \ell is the perpendicular bisector of \overline{AB} and P is any point on \ell. Use the reflection in \ell to prove that PA = PB.
- ABCD is a parallelogram and M is the midpoint of the diagonal \overline{AC}. Use the half turn about M to prove that \overline{AB} \cong \overline{CD} and \overline{AD} \cong \overline{CB}. You may use what Example 6 proved: a half turn carries a line not through its center onto a parallel line.
- Two lines cross at O. Use the half turn about O to prove that vertical angles are congruent.
- Triangles ABC and DEF have \angle A \cong \angle D, \overline{AB} \cong \overline{DE} and \angle B \cong \angle E. Prove that \triangle ABC \cong \triangle DEF by building a rigid motion, as in Example 7.
- In \triangle ABC, \overline{AB} \cong \overline{AC}. K is the midpoint of \overline{AB} and N is the midpoint of \overline{AC}. Use the reflection from the model proof to show that \overline{CK} \cong \overline{BN}.
Reasoning
- Someone writes: “Slide \triangle ABC over until it sits on top of \triangle DEF, so they are congruent.” Explain why this is not yet a proof, and what would make it one.
- Someone writes: “Reflect \triangle ABC in line DE. Then C lands on F, because the triangles are congruent.” What is wrong with the reason?
- \triangle ABC is lettered counterclockwise, and its image under some sequence of rigid motions is lettered clockwise. What can you say about the reflections in the sequence? Explain.
- A sequence of transformations includes a dilation by the factor 3. Can the sequence still carry a figure onto a congruent figure? Explain.
Answers
- A'(1, -2), B'(1, -6), C'(4, -3). The rule is (x, y) \mapsto (y, -x), so for example A(2, 1) \mapsto (1, -2).
- P'(3, -1), Q'(2, 4), R'(-2, 0). The rule is (x, y) \mapsto (y, x): swap the coordinates.
- D'(1, 3), E'(4, 5), F'(2, 1). The rule is (x, y) \mapsto (x - 3, y + 5).
- K'(-1, 1), L'(1, 4), M'(4, -1). The mirror is the perpendicular bisector of the segment joining each point to its image, so a point x - 2 to the right of the line lands x - 2 to the left of it, at 2 - (x - 2) = 4 - x. The rule is (x, y) \mapsto (4 - x, y).
- S'(-1, -1), T'(-4, 2), U'(1, 4). The center (1, 2) is the midpoint of each point and its image, so \displaystyle \frac{x + x'}{2} = 1 and \displaystyle \frac{y + y'}{2} = 2. The rule is (x, y) \mapsto (2 - x, 4 - y).
- The reflection in the x-axis, (x, y) \mapsto (x, -y). It sends A(1, 2) \mapsto (1, -2), B(4, 2) \mapsto (4, -2) and C(1, 6) \mapsto (1, -6), which are A', B' and C'.
- Rotate 90^\circ counterclockwise about the origin, then translate by \langle 3, 2 \rangle. \overline{AB} points right and \overline{DE} points up, which picks the quarter turn. The rotation gives (-1, 1), (-1, 3), (-4, 1), and the translation then gives (2, 3), (2, 5), (-1, 3). (Together the two make a single quarter turn about \displaystyle \left(\frac{1}{2}, \frac{5}{2}\right).)
- The reflection in the line x + y = 5. A \to B \to C turns counterclockwise and D \to E \to F turns clockwise, so an odd number of reflections is needed. The mirror must be the perpendicular bisector of \overline{AD}: its midpoint is \displaystyle \left(\frac{5}{2}, \frac{5}{2}\right) and it has slope 1, so the mirror has slope -1 through that point. The reflection in it is (x, y) \mapsto (5 - y, 5 - x), which sends A(0, 0) \mapsto (5, 5), B(4, 0) \mapsto (5, 1) and C(1, 3) \mapsto (2, 4).
- No. Rigid motions keep distance, but the side lengths differ: \triangle ABC has 4, \displaystyle \sqrt{10} and \displaystyle \sqrt{18}, while \triangle GHI has GH = 4, \displaystyle GI = \sqrt{13} and \displaystyle HI = \sqrt{13}.
- Dilate by the factor 3, then translate by \langle 0, 1 \rangle; the scale factor is 3. AB = 1 and A'B' = 3 suggest the factor. The dilation gives (3, 0), (6, 0), (3, 6), and moving each up 1 gives A', B' and C'.
- The center is (4, 2). The center of a half turn is the midpoint of each point and its image. The midpoint of \overline{AC} is \displaystyle \left(\frac{1 + 7}{2}, \frac{4 + 0}{2}\right) = (4, 2), and the midpoint of \overline{BD} is \displaystyle \left(\frac{5 + 3}{2}, \frac{2 + 2}{2}\right) = (4, 2). The two midpoints agree, so one half turn does both.
- y = 3x - 6. The mirror is the perpendicular bisector of \overline{AA'}. The midpoint is (3, 3), and \overline{AA'} has slope \displaystyle \frac{2 - 4}{6 - 0} = -\frac{1}{3}, so the mirror has slope 3: y - 3 = 3(x - 3).
- P is on the mirror, so P' = P. The mirror is the perpendicular bisector of \overline{AB}, so the reflection swaps A and B. It therefore carries \overline{PA} onto \overline{PB}, and a reflection keeps length, so PA = PB.
- M is the midpoint of \overline{AC}, so the half turn swaps A and C. Line AB does not pass through M (if it did, it would be line AC, and A, B, C would be in a line), so it goes to the line through C parallel to AB, which by the Parallel Postulate is line CD. In the same way line CB goes to line AD. B is where lines AB and CB meet, so its image is where lines CD and AD meet, which is D. So A \mapsto C and B \mapsto D, the half turn carries \overline{AB} onto \overline{CD}, and \overline{AB} \cong \overline{CD}. Since the half turn also sends D \mapsto B (doing it twice brings every point back), it carries \overline{AD} onto \overline{CB}, so \overline{AD} \cong \overline{CB}.
- The half turn about O sends each point X to the point X' with O the midpoint of \overline{XX'}, so X' is on line XO, on the far side of O. Each line through O therefore maps onto itself, and each ray from O onto the opposite ray. An angle formed by rays OA and OB goes to the angle formed by the two opposite rays, which is its vertical angle. A rotation keeps angle measure, so vertical angles are congruent.
- Translate by the vector from A to D, then rotate about D so that ray AB lands on ray DE; since AB = DE, B lands on E. If C is now on the other side of line DE from F, reflect in line DE, which leaves D and E in place. Now the image of C is on F’s side. The ray from D to it makes the angle m\angle A = m\angle D with ray DE, so it is ray DF. The ray from E to it makes the angle m\angle B = m\angle E with ray ED, so it is ray EF. The image of C lies on both rays, and the lines DF and EF meet only at F, so it is F. The motion carries \triangle ABC onto \triangle DEF.
- The reflection in the bisector of \angle BAC sends A \mapsto A, B \mapsto C and C \mapsto B. It carries \overline{AB} onto \overline{AC}, and K, the point of \overline{AB} halfway from A, goes to the point of \overline{AC} halfway from A, since distance is kept: that is N. So \overline{CK} goes to \overline{BN}, and \overline{CK} \cong \overline{BN}.
- It names no motion and gives no reason for any landing. A proof names the vector, such as the vector from A to D, and then shows why each of B and C lands on its partner, by a coordinate rule or by the angle and distance facts. It may also turn out that no slide works at all, because the second triangle is turned or flipped.
- The reason assumes the conclusion. Congruence is what is being proved, so it cannot be why C lands on F. The reason has to come from what the reflection keeps: for example, the angle at D and the distance from D, which put C' on ray DF at the distance DF.
- The sequence contains an odd number of reflections, so at least one. Each reflection reverses the direction of the lettering and translations and rotations keep it, so the lettering ends reversed exactly when the number of reflections is odd.
- Yes, if the other dilations bring the total factor back to 1. Every length is multiplied by the product of all the factors, so the image is congruent exactly when that product is 1. For example, dilating by 3 and then by \displaystyle \frac{1}{3} from the same center puts every point back where it started.