HSG.CO.B.8: Explaining the ASA, SAS and SSS Criteria with Rigid Motions
In plain English: HSG.CO.B.8 is the Common Core geometry standard that asks students to explain why the ASA, SAS and SSS criteria prove two triangles congruent. Students start from the definition of congruence in terms of rigid motions and show that the three given parts force a rigid motion to carry one triangle onto the other. It is usually taught in high school Geometry.
Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions.
Common Core State Standards for Mathematics · Domain: Congruence (CO) · Cluster: Understand congruence in terms of rigid motions Also written as HSG-CO.B.8 or G-CO.8 · Official standard
Students already know the definition: two triangles are congruent when a sequence of rigid motions maps one onto the other. This lesson explains why only three well-chosen parts are needed to guarantee such a motion. For SAS, ASA and SSS, students follow the same opening moves: translate so one vertex lands on its partner, rotate so a side lines up, and reflect across that side if the third vertex is on the wrong side. Then each criterion needs its own final argument for why the third vertex lands exactly where it should.
The finishing arguments are the heart of the standard. With SAS, the matching angle puts the third vertex on the right ray and the matching side puts it at the right point. With ASA, two rays meet in only one point. With SSS, two circles meet in at most two points, which are mirror images across the matched side. Students also see where the same argument breaks down for AAA and SSA, which explains why those are not criteria.
Learning Objectives
By the end of this lesson, students will be able to:
Describe the translate, rotate and reflect steps that map two vertices of one triangle onto the matching vertices of another
Explain why, in SAS, the included angle and the second side force the third vertex onto its partner
Explain why, in ASA, the two angles force the third vertex onto its partner because two rays meet in at most one point
Explain why, in SSS, the three sides force the third vertex onto its partner or its reflection across the matched side
Identify which criterion, if any, applies to given information, and explain why AAA and SSA do not guarantee congruence
Prior Knowledge Required
Students should already be comfortable with:
Deciding when given sides and angles determine a unique triangle 7.G.A.2
Describing a sequence of rigid motions that maps one figure onto another HSG.CO.A.5
Congruent triangles have congruent corresponding sides and angles, and the converse HSG.CO.B.7
A circle as the set of points at a fixed distance from its center HSG.CO.A.1
The distance formula on the coordinate plane 8.G.B.8
Before class, write three part sets on the board: (1) sides 6 cm and 9 cm with a 50° angle between them, (2) sides 5 cm, 7 cm and 10 cm, (3) angles 30°, 60° and 90°.
Warm-Up Prompt
"Sketch a triangle for each set. Then compare with your neighbor. For which sets are your two triangles sure to fit on top of each other? For which could they be different?"
Students usually find that sets 1 and 2 give matching triangles and set 3 does not: one partner draws a small 30-60-90 triangle and the other a large one. Record the question the lesson answers: why do some sets of three parts lock the triangle in place, using only what we know about rigid motions?
Direct Instruction20 minutes
The shared opening. In all three proofs, the goal is a rigid motion that maps △ABC onto △DEF. Present the opening moves once, since SAS, ASA and SSS all reuse them:
Match one side. The side AB is congruent to DE in every criterion (SAS and ASA give it directly; SSS gives it as one of the three sides). Translate so A lands on D, then rotate about D so ray AB lies along ray DE. Rigid motions keep length, so AB = DE puts B exactly on E.
Choose the side of line DE. If the image of C is on the other side of line DE from F, reflect across line DE. This reflection keeps D and E where they are. Now C and F are on the same side.
SAS finish. Given ∠A ≅ ∠D and AC = DF. Ray AC makes the same angle with ray DE as ray DF does, on the same side, so ray AC lies along ray DF. Since AC = DF, C lands on F.
ASA finish. Given ∠A ≅ ∠D and ∠B ≅ ∠E. Ray AC lies along ray DF and ray BC lies along ray EF, for the same reason as in SAS. C is on both rays, and so is F. Two rays that are not on the same line meet in at most one point, so C lands on F.
SSS finish. Given AC = DF and BC = EF. The image of C is at distance DF from D and distance EF from E, so it lies on the circle with center D and radius DF and on the circle with center E and radius EF. Those circles meet in at most two points, which are reflections of each other across line DE (Diagram 1). Since C and F are on the same side, C lands on F.
Where the argument breaks. With AAA, nothing fixes the length of the first side, so B cannot be placed on E (a dilation can change size without changing angles). With SSA, the circle around B can cross ray DF twice, so C can land in two places.
SAS with coordinates (Diagram 2)
△ABC has A(0, 0), B(4, 0), C(3, 2) and △DEF has D(6, 1), E(6, 5), F(4, 4). AB = DE = 4, AC = DF = √13, and ∠A ≅ ∠D. Carry out the SAS argument.
Equation: Translate by (6, 1): A′(6, 1), B′(10, 1), C′(9, 3). Rotate 90° counterclockwise about D: B″(6, 5) = E and C″(4, 4) = F. C and F were already on the same side, so no reflection is needed
ASA: the rays can meet only once
△PQR and △STU have m∠P = m∠S = 50°, PQ = ST = 6 cm and m∠Q = m∠T = 70°. After P is mapped to S and Q to T, where can R land?
Equation: After a reflection across line ST if needed, R is on the same side as U. R must be on the ray from S that makes 50° with ray ST and on the ray from T that makes 70° with ray TS. Those rays meet at exactly one point, which is U, so R lands on U. (Both third angles are 180° - 50° - 70° = 60°.)
SSS: two circles, two candidates (Diagram 1)
D is at (0, 0) and E is at (5, 0). The third vertex must be 4 units from D and 3 units from E. Find every possible position.
Equation: x² + y² = 16 and (x - 5)² + y² = 9 give x = 3.2 and y = ±2.4. The two points (3.2, 2.4) and (3.2, -2.4) are reflections across line DE, so at most one reflection is needed
Choosing the criterion
In △ABC and △DEF, AB = DE, ∠B ≅ ∠E, BC = 2x + 3 and EF = 11. For what value of x does a criterion prove △ABC ≅ △DEF, and which one?
Equation: 2x + 3 = 11, so x = 4. Then AB = DE, ∠B ≅ ∠E and BC = EF, with ∠B between the two sides, so SAS applies
After the third example, ask: in the SSS argument, which fact plays the role that the angle played in SAS? (The second circle: it pins C down instead of the ray.) Stress that the criteria are not new axioms here; each one is a short argument built from the definition of congruence.
Guided Practice15 minutes
Pairs complete a proof frame for ASA with △GHK and △MNP, given ∠G ≅ ∠M, GH = MN and ∠H ≅ ∠N. Each blank needs a motion or a reason.
Translate △GHK so that G lands on ____. (M)
Rotate about M so that ray GH lies along ray ____. H lands on N because ____. (MN; rigid motions preserve length and GH = MN.)
If the image of K is on the opposite side of line MN from P, ____. This does not move M or N because ____. (Reflect across line MN; points on the line of reflection stay fixed.)
Ray GK now lies along ray MP because ____, and ray HK lies along ray NP because ____. (∠G ≅ ∠M; ∠H ≅ ∠N.)
K lands on P because ____. So a rigid motion maps △GHK onto △MNP, and the triangles are congruent by the definition. (K and P are both the single point where those two rays meet.)
As pairs finish, ask them to point to the step where ASA and SAS stop being the same proof. Then ask one pair to explain why the third step is needed at all.
Independent Practice15 minutes
Students work alone. (1) D is at (0, 0) and E is at (7, 0). Find every point that is 5 units from D and √32 units from E, and explain what the answer shows about the SSS argument. (x² + y² = 25 and (x - 7)² + y² = 32 give x = 3 and y = ±4: the points (3, 4) and (3, -4), mirror images across DE.) (2) For each set, name the criterion or write "not enough", and point to the step of the proof that needs the missing information: (a) AC = DF, ∠C ≅ ∠F, CB = FE; (b) ∠A ≅ ∠D, ∠B ≅ ∠E, ∠C ≅ ∠F; (c) ∠B ≅ ∠E, BC = EF, ∠C ≅ ∠F. (SAS; not enough, because no side fixes where B lands; ASA.)
Closure5 minutes
Exit ticket: (1) The three proofs share their first two steps. Write those steps. (2) For SSS, what replaces the angle that SAS uses to place the third vertex? (3) Why does "all three angles congruent" fail to guarantee congruence?
Differentiation Strategies
For Struggling Students
Give a three-column organizer (SAS, ASA, SSS) with the shared opening already written in each column, so students only write the finish
Let students act out each proof with a patty-paper tracing of △ABC on top of a printed △DEF before writing any sentences
Use color: mark the given parts in one color on both triangles and ask students to point to where each given part is used in the proof
For Advanced Students
Ask students to explain why AAS also guarantees congruence by turning it into ASA with the triangle angle sum (a challenge beyond the three criteria in this standard)
Ask students to write the SSS finish without circles, using the fact that a point equidistant from C and F lies on the perpendicular bisector of CF
Ask students to find conditions under which SSA does work, for example when the given angle is a right angle
Assessment Guidance
What to Look For
Look for explanations that name a rigid motion at each step and give the reason each vertex lands on its partner. Weak answers say "the triangles line up" or restate the criterion as its own reason. For SAS, check that students use the angle to place the ray and the side to place the point. For ASA, listen for "two rays meet in only one point". For SSS, listen for the two circles, or the perpendicular bisector, and for why a reflection may be needed. Students should also be able to name the missing step when a set of parts is not enough.
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Classroom Activities
3 Activities
1
Three Parts, One Triangle?
20 minPairs
Partners build triangles from the same three parts without looking at each other's work, then use patty paper to test whether one triangle can be moved onto the other. The results show which part sets lock a triangle in place.
The 4 Part Sets
Set 1: sides 7 cm and 9 cm with a 50° angle between them
Set 2: angles 40° and 65° with an 8 cm side between them
Set 3: sides 6 cm, 7 cm and 9 cm
Set 4: sides 8 cm and 6 cm with a 40° angle that is next to the 8 cm side but not the 6 cm side
Procedure
Each partner builds a triangle for each set with a ruler, protractor and compass
Trace your triangle on patty paper and try to place the tracing on your partner's triangle. Record the motions you used (translation, rotation, reflection)
For every set that matched, write the name of the criterion (SAS, ASA or SSS). For Set 4, compare with other pairs: in some pairs the triangles will not match, because the 6 cm side can swing to two positions
Discussion Questions
For each matching set, which given part decided where the last vertex went?
In Set 4, why could the 6 cm side meet the other ray in two places?
Modification for Distance Learning
Students build each set in dynamic geometry software and drag the free vertices. They screenshot any set that allows two different triangles.
2
Proof Relay: Same Start, Different Finish
20 minGroups of 3
Each group writes all three rigid-motion proofs on one poster. Every proof is split into three stages, and each student writes a different stage for each criterion.
Stages
Stage 1: translate and rotate so the matched side lines up and two vertices land on their partners, with the reason
Stage 2: decide whether a reflection is needed and explain why it does not move the two matched vertices
Stage 3: the criterion-specific reason the third vertex lands on its partner
Procedure
Round 1 is SAS, round 2 is ASA and round 3 is SSS. Roles rotate each round, so every student writes Stage 3 once
After each round, the group reads the whole proof aloud and circles the sentence in Stage 3 that uses each given part
Groups post their posters and do a gallery walk, leaving a sticky note wherever a reason is missing
Discussion Questions
Stages 1 and 2 were almost identical in all three rounds. Why?
Which Stage 3 was hardest to write, and what fact did it depend on?
3
Criterion Sort
15 minPairs
Pairs sort 8 cards, each listing three congruent pairs of parts of △ABC and △DEF, into SAS, ASA, SSS and "not enough". For every "not enough" card they name the step of the proof that fails.
The 8 Cards
AB = DE, BC = EF, CA = FD (SSS)
AB = DE, ∠B ≅ ∠E, BC = EF (SAS)
∠A ≅ ∠D, AB = DE, ∠B ≅ ∠E (ASA)
∠A ≅ ∠D, ∠B ≅ ∠E, ∠C ≅ ∠F (not enough: no side places B on E)
AB = DE, BC = EF, ∠A ≅ ∠D (not enough: the angle is not between the two sides, so C can land in two places)
AC = DF, ∠C ≅ ∠F, CB = FE (SAS)
∠B ≅ ∠E, BC = EF, ∠C ≅ ∠F (ASA)
AB = DE, AC = DF, ∠B ≅ ∠E (not enough: ∠B is not between sides AB and AC)
Procedure
Sketch both triangles on each card and mark the given parts with ticks and arcs
Place the card in a pile and write the name of the included angle or included side if there is one
For "not enough" cards, sketch two triangles that fit the card but are not congruent
Challenge Variation
Add a ninth card: ∠A ≅ ∠D, ∠B ≅ ∠E, BC = EF. Ask pairs to explain why the angle sum turns this card into ASA. This goes beyond the three criteria named in the standard.
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Diagrams & Visual Aids
2 diagrams
Diagram 1: The SSS Argument with Two Circles
Drawn to scale with D(0, 0) and E(5, 0). Every point 4 units from D lies on the blue circle, and every point 3 units from E lies on the navy circle. They meet only at F(3.2, 2.4) and its reflection (3.2, -2.4) across line DE, so SSS pins the third vertex down.
Diagram 2: The SAS Argument on the Coordinate Plane
The SAS proof for △ABC with A(0, 0), B(4, 0), C(3, 2) and △DEF with D(6, 1), E(6, 5), F(4, 4). A translation sends A to D, and a 90° rotation about D sends B to E and C to F. The dashed triangle is the position before each step.
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Homework Assignment
~30 min
HSG.CO.B.8 Homework: Why the Congruence Criteria Work
Directions: In every explanation, name each rigid motion you use and give the reason each vertex lands on its partner. Show your algebra and coordinates.
Part 1: SAS and ASA (Problems 1-3)
In △GHI and △JKL, GH = JK, ∠H ≅ ∠K and HI = KL. Write a paragraph that explains, using rigid motions, why △GHI ≅ △JKL. Your paragraph must say where the given angle is used and where the second given side is used.
△GHI has G(-2, 0), H(3, 0), I(0, 4) and △JKL has J(5, -1), K(5, 4), L(1, 1). (a) Show that GH = JK and GI = JL. (b) You are told that ∠G ≅ ∠J. Carry out the SAS argument: translate G to J, then rotate about J so H lands on K. Give the image of every vertex after each step. (c) Is a reflection needed? Explain.
In △MNO and △QRS, m∠M = m∠Q = 35°, MN = QR = 9 cm and m∠N = m∠R = 80°. (a) Explain, using rigid motions, why O must land on S. (b) Find m∠O and m∠S. (c) Why would the argument fail if you only knew the two angles and not the side?
Part 2: SSS and Choosing a Criterion (Problems 4-6)
A is at (0, 0) and B is at (10, 0). (a) Find every point C with AC = 6 and BC = 8. (b) How are your answers related to line AB? (c) Explain how this computation illustrates the SSS argument.
For each set of given parts of △PQR and △XYZ, name the criterion or write "not enough", and explain which step of the rigid-motion proof decides it: (a) QR = YZ, ∠R ≅ ∠Z, RP = ZX; (b) PQ = XY, QR = YZ, PR = XZ; (c) ∠R ≅ ∠Z, RP = ZX, ∠P ≅ ∠X; (d) PR = XZ, QR = YZ, ∠P ≅ ∠X.
In △ABC and △DEF, AB = 3x - 1, DE = 2x + 4, m∠B = (5y + 5)°, m∠E = (7y - 15)° and BC = EF = 12. (a) Find x and y so that AB = DE and ∠B ≅ ∠E. (b) Give AB and m∠B. (c) Name the criterion that now proves △ABC ≅ △DEF and explain why it applies.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Rigid Motions
Each motion named and correctly applied
Motions named but one step unclear or wrong
No motions described
Placing the Third Vertex
Uses the correct criterion-specific reason (ray and length, two rays, two circles)
Reason given but incomplete
Restates the criterion as the reason
Computation
Coordinates and algebra all correct
One error
Several errors or missing work
Criterion Choice
Correct criterion or "not enough" with the failing step named
Correct choice without the step
Incorrect choice
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Quiz: 20 Questions
Interactive, with answers
Instructions
Answer each multiple-choice question to see whether you are right, with an explanation. For short-answer questions, write a full explanation first, then compare with the model answer. Reset quiz clears everything.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
In △JKL and △MNO, JK = MN, ∠K ≅ ∠N and KL = NO. Which criterion applies?
Answer: B
∠K is formed by sides JK and KL, so it is the included angle: two sides and the angle between them is SAS. Choice A needs three pairs of sides. Choice D is wrong because the angle is between the two given sides.
Question 2 of 20 · Multiple Choice
In the SAS proof, A has been mapped to D and B to E, and C is on the same side of line DE as F. Why does ray AC now lie along ray DF?
Answer: A
Only one ray from D, on F's side of line DE, makes an angle equal to m∠D with ray DE. The length AC = DF is used in the next step, to put C at F on that ray. Choice B mixes up the two steps.
Question 3 of 20 · Multiple Choice
In the ASA proof, why must the third vertex C land on F?
Answer: D
The two matching angles place C on ray DF and on ray EF. Those rays are on different lines, so they share at most one point, which is F. Choice B is not given in ASA. Choice C describes the SSS argument.
Question 4 of 20 · Multiple Choice
In the SSS proof, after A → D and B → E, why can the image of C be in at most two places?
Answer: C
AC = DF puts C on the circle with center D and radius DF; BC = EF puts it on the circle with center E and radius EF. Those two points are reflections across line DE, which is why SSS may need one reflection. Choice B assumes an angle that SSS does not give.
Question 5 of 20 · Multiple Choice
D is at (0, 0) and E is at (8, 0). Which pair lists every point that is 5 units from D and 5 units from E?
Answer: B
x² + y² = 25 and (x - 8)² + y² = 25 give x = 4, then y² = 9, so y = 3 or y = -3. Choice A forgets the second intersection of the circles, the mirror image across DE. Choice C swaps the coordinates: (3, 4) is 5 from D but √41 from E.
Question 6 of 20 · Multiple Choice
Why is AAA (three pairs of congruent angles) not a congruence criterion?
Answer: C
The proofs start by placing B on E, which needs a known side. A dilation keeps all angles but changes lengths, so a small and a large 30-60-90 triangle have all angles congruent and are not congruent. Choice A is false: rigid motions move angles and preserve their measures.
Question 7 of 20 · Multiple Choice
In all three proofs, why might a reflection across line DE be needed after the translation and rotation?
Answer: A
Reflecting across line DE keeps D and E fixed and flips the side C is on. It is needed only when the triangles have opposite orientation. Choice B is false: many proofs need no reflection. Choice C is false because reflections preserve length.
Question 8 of 20 · Multiple Choice
In △ABC and △DEF, ∠A ≅ ∠D, AC = DF, AB = 4x - 3 and DE = 2x + 9. For SAS to apply, what must AB be?
Answer: D
Set 4x - 3 = 2x + 9, so 2x = 12 and x = 6. Then AB = 4(6) - 3 = 21 and DE = 2(6) + 9 = 21. Choice A gives x instead of the length. Choice B comes from the sign error 2x = 6, which gives x = 3 and DE = 15.
Question 9 of 20 · Multiple Choice
In △ABC and △DEF, AB = DE, ∠B ≅ ∠E, m∠A = (3x + 5)° and m∠D = (5x - 25)°. For ASA to apply, what must m∠A be?
Answer: B
3x + 5 = 5x - 25 gives 30 = 2x and x = 15, so m∠A = 3(15) + 5 = 50°. Choice A is x itself. Choice C is the supplement of 50°.
Question 10 of 20 · Multiple Choice
In △PQR and △XYZ, PQ = XY, QR = YZ and ∠Q ≅ ∠Y. Which statement is justified?
Answer: C
Q is between the given sides PQ and QR, and Y is between XY and YZ, so SAS applies with P↔X, Q↔Y, R↔Z. Choice A pairs Q with X, so it claims QR = XZ, which is not given. Choice B needs a third pair of sides that is not given.
Question 11 of 20 · Multiple Choice
Which angle is included between sides GK and KH of △GHK?
Answer: A
The included angle is at the vertex the two sides share, and GK and KH share K. Choosing ∠G or ∠H would make an SSA arrangement, which is not a criterion.
Question 12 of 20 · Multiple Choice
Which rigid motion comes first in the proofs of SAS, ASA and SSS as taught in this lesson?
Answer: D
The proofs start by moving one vertex onto its partner with a translation, then rotate about that point. Choice B is not a rigid motion. Choice A may come last, only if it is needed.
Question 13 of 20 · Multiple Choice
A student says: "SSS works because if the sides match, the triangles look the same." Which response best explains SSS with rigid motions?
Answer: B
A proof needs a reason each vertex must land on its partner, and the two circles give that reason for C. Choice A relies on appearance. Choice C is false, since SSS works for every triangle. Choice D does not fix any side length.
Question 14 of 20 · Multiple Choice
In each proof, after the translation maps A to D and the rotation lines up ray AB with ray DE, why does B land exactly on E?
Answer: C
B ends up on ray DE at distance AB from D, and the only point of that ray at distance DE = AB from D is E. Choice D uses an angle, which cannot fix a distance. Choice B is false: the rotation is chosen to line up the ray, not the point.
Question 15 of 20 · Short Answer
Explain SAS: given AB = DE, ∠A ≅ ∠D and AC = DF, describe a rigid motion that maps △ABC onto △DEF and give the reason each vertex lands on its partner.
Translate A to D, rotate about D so ray AB lies along ray DE, and reflect across line DE if C is on the wrong side. A lands on D by the translation. B lands on E because AB = DE and rigid motions preserve length. After the possible reflection, C is on F's side, and ∠A ≅ ∠D puts ray AC along ray DF. Since AC = DF, C lands on F. The motion maps the triangle onto △DEF, so they are congruent.
Question 16 of 20 · Short Answer
Explain ASA: given ∠A ≅ ∠D, AB = DE and ∠B ≅ ∠E, why does the rigid motion that maps A to D and B to E also map C to F?
C is forced to the single point where ray DF and ray EF meet, which is F. After A → D and B → E (and a reflection across DE if needed, so C is on F's side), ∠A ≅ ∠D puts ray AC along ray DF, and ∠B ≅ ∠E puts ray BC along ray EF. C lies on both rays. Two rays on different lines meet in at most one point, and F is on both, so C = F.
Question 17 of 20 · Short Answer
Explain SSS: given AB = DE, BC = EF and AC = DF, why does the rigid motion that maps A to D and B to E also map C to F, possibly after a reflection?
C must lie on two circles that meet only at F and its reflection across DE. The image of C is AC = DF from D and BC = EF from E, so it lies on the circle with center D and radius DF and on the circle with center E and radius EF. Two circles meet in at most two points, symmetric across the line DE through their centers. If C lands on the reflection of F, a reflection across line DE, which fixes D and E, moves it onto F.
Question 18 of 20 · Short Answer
A is at (0, 0) and B is at (13, 0). Find every point C with AC = 5 and BC = 12, and explain what the answer shows.
C = (25/13, 60/13) or C = (25/13, -60/13). Subtracting x² + y² = 25 from (x - 13)² + y² = 144 gives -26x + 169 = 119, so x = 25/13, and y² = 25 - 625/169 = 3600/169, so y = ±60/13. The two points are mirror images across line AB, which is exactly why the SSS proof sometimes needs one reflection.
Question 19 of 20 · Short Answer
A is at (0, 0), B is at (4, 3), and C is on the positive x-axis with BC = √13. Find every possible C, and use your answer to explain why SSA is not a congruence criterion.
C = (2, 0) or C = (6, 0). BC² = (x - 4)² + 9 = 13 gives (x - 4)² = 4, so x = 2 or x = 6. Both triangles have the same ∠A, AB = 5 and BC = √13, but AC is 2 in one and 6 in the other, so they are not congruent. In the proof, the circle around B crosses ray AC twice, so the third vertex is not pinned down. This is why SSA is not on the list of criteria.
Question 20 of 20 · Short Answer
△PQR has P(0, 0), Q(3, 0), R(0, 2) and △STU has S(2, 5), T(-1, 5), U(2, 3). PQ = ST, PR = SU and both ∠P and ∠S are right angles. Describe the SAS rigid motion and check every vertex.
Translate by (2, 5), then rotate 180° about S. The translation gives (2, 5), (5, 5) and (2, 7). Rotating 180° about S(2, 5) sends (5, 5) to (-1, 5) = T and (2, 7) to (2, 3) = U, while S stays fixed. PQ = ST = 3 and PR = SU = 2, and ∠P and ∠S are both right angles, so each vertex lands where the SAS argument says it should.
0 of 20 answered · 0 correct
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Frequently Asked Questions
10 Questions
What does HSG.CO.B.8 mean?
HSG.CO.B.8 means students explain why ASA, SAS and SSS are enough to prove two triangles congruent, starting from the rigid-motion definition of congruence. Instead of accepting the criteria as rules, students show that three given parts force a translation, a rotation and possibly a reflection to carry one triangle exactly onto the other.
Is HSG.CO.B.8 taught in Geometry?
Yes, it is a high school Geometry standard, usually taught in the congruence unit right after HSG.CO.B.7. Students then use the criteria in proofs about triangles and quadrilaterals (HSG.CO.C.10 and HSG.CO.C.11) and to solve problems (HSG.SRT.B.5).
Do students have to write formal two-column proofs for HSG.CO.B.8?
No particular format is required. The standard asks students to explain, so a clear paragraph that names each rigid motion and the reason each vertex lands on its partner meets it. Flow charts or two-column proofs work as well, as long as the reasons come from rigid motions.
Why are AAA and SSA not congruence criteria?
Because the rigid-motion argument cannot pin the triangle down. With AAA there is no given side to place the second vertex, so triangles of different sizes fit. With SSA, the unmatched side can meet the other ray in two places, giving two different triangles.
What is the difference between HSG.CO.B.7 and HSG.CO.B.8?
HSG.CO.B.7 shows that two triangles are congruent exactly when all six pairs of corresponding parts are congruent. HSG.CO.B.8 shows that three of those pairs, in the arrangements ASA, SAS or SSS, already force the other three. The same translate, rotate, reflect construction is used in both.
Why does the SSS proof talk about circles?
Because a known distance from a known point describes a circle. Once two vertices are matched, the third vertex must be a fixed distance from each of them, so it is on two circles at once. Two circles meet in at most two points, and those points are reflections of each other across the line through the centers.
Is AAS a valid criterion, and is it part of HSG.CO.B.8?
AAS does prove congruence, but the standard names only ASA, SAS and SSS. Many teachers show AAS as a follow-up: if two angles match, the third angles match by the angle sum, and the given side becomes the included side of an ASA pair.
What mistakes do students make when explaining the criteria?
A common error is using the criterion as its own reason ("the triangles are congruent by SAS because of SAS"). Students also forget the reflection step, use an angle that is not included, or say two vertices "line up" without saying which length or angle forces it.
How can I check whether a student really understands why SAS works?
Ask where each given part is used. A student who understands the proof can say that the first side places the second vertex, the angle places the ray, and the second side places the third vertex on that ray. Then ask what would go wrong with the angle in a different position.
How do these ideas show up in later math?
The criteria are the main tools for proving triangle and quadrilateral theorems in Geometry. The same style of argument, with dilations added, proves the AA criterion for similar triangles (HSG.SRT.A.3), which later supports right-triangle trigonometry.
07
Related Standards
5 standards
These standards connect to HSG.CO.B.8: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.G.A.2Prerequisite
Draw shapes from given conditions and notice when they determine a unique triangle
