SSS, SAS, ASA, AAS
Two triangles look like copies of each other. How few matching parts does it take to be sure they are?
Where you are 18 in Triangles
Why this exists
Two triangles are congruent when all six pairs of corresponding parts match: three pairs of sides and three pairs of angles. Checking six facts is slow, in a proof it is six lines, and in most problems you are handed only two or three of them.
A triangle, though, is rigid. Three rods pinned at their ends make a frame that cannot be pushed out of shape, which is why bridges and roofs are built from triangles. Once a few parts of a triangle are fixed, the rest have no freedom left. The four criteria in this lesson say exactly which three parts are enough, so that three matching pairs prove all six.
Definition
Two triangles are congruent when their vertices can be paired so that all six pairs of parts are congruent; \triangle ABC \cong \triangle DEF pairs them in letter order, A with D, B with E, C with F (congruence statements and correspondence). A congruence criterion is a list of three pairs of corresponding parts which, once they match, force the other three pairs to match as well.
Two words do most of the work. In a triangle, the included angle of two sides is the angle between them, at the vertex they share. The included side of two angles is the side that joins their vertices. In Figure 1, \angle Q is included between \overline{PQ} and \overline{QR}, and \overline{PR} is included between \angle P and \angle R.
Each criterion is named by its parts, S for a side and A for an angle, read in order around the triangle. SSS is three sides. SAS is two sides and their included angle. ASA is two angles and their included side. AAS is two angles and a side that is not between them. Figure 2 marks the three parts of each; as in congruent segments, matching ticks show congruent sides and matching arcs show congruent angles.
Intuition
Try to build a triangle from the parts each criterion gives, and see how much choice is left.
SSS. Three sticks of fixed lengths. Lay one down and swing the other two about its ends until their free ends meet. They meet in one place above the first stick (and in its mirror image below, the same triangle flipped over).
SAS. Two sticks hinged at a fixed angle. Their free ends are now a fixed distance apart, so the third side is whatever length closes the gap.
ASA. One stick, with the angle fixed at each end. Two rays leave the ends of the side, and two rays that are not parallel cross in exactly one point, the third vertex (Figure 3).
AAS. The angles of a triangle add to 180^\circ, so two angles fix the third. AAS therefore gives all three angles and a side, and that side lies between two of the known angles: it is ASA again.
Each time, the three parts leave no freedom, so any two triangles built from them are copies.
Results
SSS postulate. If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.
SAS postulate. If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent.
ASA postulate. If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.
AAS theorem. If two angles and a non-included side of one triangle are congruent to the corresponding two angles and side of another triangle, then the triangles are congruent.
In every case the pairs must correspond: each part is matched with the part in the same position in the other triangle, and the congruence statement lists the vertices in that order. Then the three pairs never given are congruent too, because corresponding parts of congruent triangles are congruent (its own lesson).
Why it is true
SSS, SAS and ASA are postulates, accepted without proof as starting points, like the segment addition postulate. Euclid did prove SAS, as Proposition 4 of Book I, by laying one triangle on top of the other, but his axioms never say that a figure may be moved, and in 1899 David Hilbert made a form of SAS one of the axioms of geometry instead. Euclid proves SSS as Proposition 8, and ASA and AAS together as Proposition 26. Proclus, commenting on Proposition 26, reports that Eudemus credited it to Thales, on the grounds that Thales’s way of finding how far ships were from the shore must have used it. Books built on transformations derive SSS, SAS and ASA from rigid motions instead. This course takes them as starting points, and rigid motions and congruence shows how each one also follows from slides, turns and flips.
AAS is a theorem, and it follows from ASA with one extra step.
Given. \angle A \cong \angle D; \angle B \cong \angle E; \overline{BC} \cong \overline{EF}.
Prove. \triangle ABC \cong \triangle DEF.
Plan. \overline{BC} is not between \angle A and \angle B, so ASA cannot be cited yet. But \overline{BC} is between \angle B and \angle C. Two pairs of angles match, so the third pair matches, and that gives an angle at each end of \overline{BC}.
| Statements | Reasons |
|---|---|
| 1. \angle A \cong \angle D | given |
| 2. \angle B \cong \angle E | given |
| 3. \overline{BC} \cong \overline{EF} | given |
| 4. \angle C \cong \angle F | Third Angle Theorem: lines 1 and 2 |
| 5. \triangle ABC \cong \triangle DEF | ASA postulate: lines 2, 3 and 4 |
Line 5 can cite ASA because \overline{BC} joins the vertices of \angle B and \angle C, and \overline{EF} joins those of \angle E and \angle F. Line 4 is the triangle angle sum at work: each third angle is 180^\circ minus the other two, and those two are equal.
Worked examples
Included or not
Use triangle PQR in Figure 1.
(a) Which angle is included between \overline{QR} and \overline{RP}?
(b) Which side is included between \angle P and \angle Q?
(c) In \triangle PQR and \triangle STU, \overline{PQ} \cong \overline{ST}, \overline{QR} \cong \overline{TU} and \angle R \cong \angle U. Does SAS prove the triangles congruent?
Solution(a) \overline{QR} and \overline{RP} share the vertex R, so the angle between them is \angle R.
(b) The side joining the vertices P and Q is \overline{PQ}.
(c) No. \overline{PQ} and \overline{QR} meet at Q, so the angle they include is \angle Q. The given \angle R sits at the far end of \overline{QR}, facing \overline{PQ}. Read around the triangle, the parts are side, side, angle: SSA, which is not a criterion. The triangles might still be congruent, but these three facts do not force it.
Naming the criterion
In Figure 5, which criterion proves the triangles congruent? Write the congruence statement.
The second triangle has been turned over, so match the marks, not the positions on the page. One arc: \angle J and \angle R. Two arcs: \angle K and \angle P. One tick: \overline{KL} and \overline{PQ}.
That is two angles and a side. The side between \angle J and \angle K is \overline{JK}, and the ticked side is \overline{KL}, so the side is not included: this is AAS.
AAS needs the sides to correspond, not just to be ticked. \overline{KL} faces \angle J, and \overline{PQ} faces \angle R, the angle matched with \angle J, so they correspond. The pairing is J \leftrightarrow R, K \leftrightarrow P, L \leftrightarrow Q, and
A shared side
In Figure 6, \overline{AB} \cong \overline{AD} and \overline{CB} \cong \overline{CD}. Prove \triangle ABC \cong \triangle ADC.
Plan. Two pairs of sides are given. The diagonal \overline{AC} is a side of both triangles, and the reflexive property makes it the third pair.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \cong \overline{AD} | given |
| 2. \overline{CB} \cong \overline{CD} | given |
| 3. \overline{AC} \cong \overline{AC} | reflexive property of congruence |
| 4. \triangle ABC \cong \triangle ADC | SSS postulate: lines 1, 2 and 3 |
The pairing A \leftrightarrow A, B \leftrightarrow D, C \leftrightarrow C matches exactly lines 1, 2 and 3. As a bonus, \angle BAC \cong \angle DAC, since they correspond: the diagonal of a kite cuts the angle at A in half.
Angles from parallel lines
In Figure 7, \overline{AB} \parallel \overline{DC} and \overline{AB} \cong \overline{DC}, and \overline{AC} and \overline{BD} cross at E. Prove \triangle ABE \cong \triangle CDE.
Plan. Only one pair of sides is given, so the other two pairs must be angles. Each segment through E is a transversal of the parallel segments, and alternate interior angles along it are congruent. That puts an angle at each end of \overline{AB} and of \overline{CD}.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \parallel \overline{DC} | given |
| 2. \overline{AB} \cong \overline{DC} | given |
| 3. \angle BAE \cong \angle DCE | Alternate Interior Angles Theorem: line 1, transversal \overline{AC} |
| 4. \angle ABE \cong \angle CDE | Alternate Interior Angles Theorem: line 1, transversal \overline{BD} |
| 5. \triangle ABE \cong \triangle CDE | ASA postulate: lines 3, 2 and 4 |
\overline{AB} joins the vertices of \angle BAE and \angle ABE, and \overline{CD} joins those of \angle DCE and \angle CDE, so line 5 is ASA with A \leftrightarrow C, B \leftrightarrow D, E \leftrightarrow E. (Line 3, the vertical angles at E and line 2 would also do it, as AAS.) Now corresponding parts give \overline{AE} \cong \overline{CE} and \overline{BE} \cong \overline{DE}: E is the midpoint of both segments.
When no criterion applies
In each half of Figure 8, the marked parts of the two triangles are congruent. Does a criterion prove the triangles congruent? If not, what one extra pair would?
Top. The marks give \overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF} and \angle A \cong \angle D. The angle between \overline{AB} and \overline{BC} is \angle B, not \angle A, so this is SSA, and no criterion applies. The figure shows why none can: both triangles fit the marks, and they are different triangles. Building such pairs is why side side angle fails. One more pair would settle it: \angle B \cong \angle E gives SAS, and \overline{AC} \cong \overline{DF} gives SSS.
Bottom. All three pairs of angles match, yet one triangle is a smaller copy of the other. Angles fix the shape and not the size (why angle angle angle fails). Any one pair of corresponding sides fixes the size: a side between two of the marked angles gives ASA, and any other gives AAS.
“No criterion applies” means the marks do not force congruence. It does not mean the triangles are not congruent: other triangles with the same marks may well be copies of each other.
Solving for a value that makes SAS work
In \triangle ABC and \triangle DEF, AB = DE = 11, m\angle B = 58^\circ, m\angle E = (5x - 7)^\circ, BC = 9 and EF = 2x - 17. Find the value of x that makes SAS prove \triangle ABC \cong \triangle DEF.
Solution\angle B is included between \overline{AB} and \overline{BC}, and \angle E between \overline{DE} and \overline{EF}, so SAS needs two more pairs to match: the angles and the second sides. Each gives an equation.
Both give x = 13, so one value makes both pairs match, and SAS applies.
Check. 5(13) - 7 = 58 and 2(13) - 17 = 9.
Had EF been 2x - 16, the side would need x = 12.5 and the angle x = 13, and no value of x would make SAS apply.
Measuring across a pond
A surveyor needs the distance AB across a pond. From a point C on dry land she walks from A through C and on the same distance again, to D, so that CD = CA; likewise from B through C to E, so that CE = CB. She measures DE = 84 m (Figure 9). Explain why AB = 84 m.
Compare \triangle ACB, across the water, with \triangle DCE, on dry land.
- \overline{CA} \cong \overline{CD}, because she paced CD to equal CA.
- \angle ACB \cong \angle DCE, because A, C, D lie on one line and so do B, C, E: the two angles are vertical angles.
- \overline{CB} \cong \overline{CE}, because she paced CE to equal CB.
The angle at C is included between the two sides in each triangle, so SAS gives \triangle ACB \cong \triangle DCE, with A \leftrightarrow D, C \leftrightarrow C, B \leftrightarrow E. Under that pairing \overline{AB} corresponds to \overline{DE}, and corresponding parts of congruent triangles are congruent. So AB = DE = 84 m, and nothing was measured across the water.
Boundaries
Included means between. SSA, two sides and an angle not between them, is not a criterion: two different triangles can share those parts (Example 5, why side side angle fails).
Three angles are never enough. AAA fixes the shape but not the size (why angle angle angle fails).
Right triangles have one more. A congruent hypotenuse and leg do force two right triangles to be congruent, though that is side, side and a non-included angle (HL).
The parts must correspond. In AAS the side must face the matching angle in both triangles. Two triangles that each have a 50^\circ angle, a 60^\circ angle and a side of 8 need not be congruent, because the 8 may sit in a different place in each (Exercise 23).
The parts must be known, not seen. Each pair must be given, marked, or proved on an earlier line. Parts that look equal in a drawing count for nothing.
Only triangles. Four matching sides do not fix a quadrilateral: a square and a squashed rhombus can have the same four sides and different angles (Exercise 24).
Where this goes next
The criteria are the engine of most proofs in plane geometry: find two triangles, match three pairs of parts, cite a criterion, then use the other three pairs, which have just become congruent for free. Writing those proofs is proving triangles congruent, which starts from planning which triangles to use and goes on to triangles that overlap.
The first theorem the method proves is the Base Angles Theorem. SSS is also why building a triangle from three sides gives just one triangle. Later, solving a triangle given SAS finds the rest of the triangle in numbers, and SAS similarity keeps the letters but lets the sides be in proportion.
Exercises
Included parts
- In \triangle XYZ, name the angle included between \overline{XY} and \overline{YZ}.
- In \triangle XYZ, name the side included between \angle X and \angle Z.
- In \triangle XYZ, name the two sides that include \angle Z, and the two angles that include \overline{YZ}.
Reading marked figures
Use Figure 10. For each panel, name the criterion that proves the two triangles congruent and write the congruence statement, or say that no criterion applies.
- Panel (a).
- Panel (b).
- Panel (c).
- Panel (d).
From listed parts
In Exercises 8 to 13 the parts listed are congruent. Name the criterion that proves \triangle ABC \cong \triangle DEF, or say that none of the four applies.
- \overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}, \overline{CA} \cong \overline{FD}
- \angle A \cong \angle D, \overline{AB} \cong \overline{DE}, \angle B \cong \angle E
- \overline{AB} \cong \overline{DE}, \angle B \cong \angle E, \overline{BC} \cong \overline{EF}
- \angle A \cong \angle D, \angle B \cong \angle E, \angle C \cong \angle F
- \angle A \cong \angle D, \angle C \cong \angle F, \overline{BC} \cong \overline{EF}
- \overline{AB} \cong \overline{DE}, \overline{AC} \cong \overline{DF}, \angle C \cong \angle F
- \overline{AB} \cong \overline{DE} and \angle B \cong \angle E. Name one more pair that would complete SAS, one that would complete ASA, and one that would complete AAS.
Using algebra
- In \triangle ABC and \triangle DEF, AB = DE = 10, \angle B \cong \angle E, BC = 3x + 2 and EF = 5x - 8. Find the value of x for which SAS proves the triangles congruent, and the length of \overline{BC}.
- In \triangle ABC and \triangle DEF, m\angle A = (4x + 6)^\circ, m\angle D = (6x - 20)^\circ, m\angle B = m\angle E = 50^\circ and AB = DE = 9. Find x so that ASA applies, then find m\angle C.
- In \triangle ABC and \triangle DEF, m\angle A = m\angle D = 40^\circ, m\angle B = (3x + 1)^\circ, m\angle E = (x + 45)^\circ and BC = EF = 6. Find x so that AAS applies, then find m\angle F.
- One triangle has sides 7, 9 and 12. Another has sides 7, 2y + 1 and y + 8. Find every value of y for which SSS makes the triangles congruent.
- In the pond of Example 7, suppose instead that DE = 70 m, m\angle CDE = 38^\circ and m\angle DCE = 105^\circ. Find AB, m\angle CAB and m\angle CBA.
Short proofs
Write each as a two-column proof.
- Given. In Figure 10(a), without its arcs, \overline{AB} \parallel \overline{CD} and \overline{AB} \cong \overline{CD}. Prove. \triangle ABD \cong \triangle CDB.
- Given. \overline{AD} and \overline{BC} cross at E, as in Figure 10(b); E is the midpoint of \overline{AD}; \angle B \cong \angle C. Prove. \triangle AEB \cong \triangle DEC.
Reasoning
- The proof of the AAS theorem cites the Third Angle Theorem before it cites ASA. What does that line supply, and why could ASA not be cited without it?
- Two triangles each have a 50^\circ angle, a 60^\circ angle and a side of length 8. Must they be congruent? Explain.
- A triangle made of three rods pinned at the corners cannot be pushed out of shape, but a square made of four rods can be pushed into a rhombus. Explain what the SSS postulate says about the triangle, and why there is no “SSSS” criterion for four-sided figures.
Answers
- \angle Y. The two sides share the vertex Y.
- \overline{XZ}. It joins the vertices X and Z.
- \overline{XZ} and \overline{YZ} include \angle Z, since they meet at Z. \angle Y and \angle Z include \overline{YZ}, since it joins their vertices.
- SAS: \triangle ABD \cong \triangle CDB. \overline{AB} \cong \overline{CD} (ticks), \angle ABD \cong \angle CDB (arcs), and \overline{BD} \cong \overline{DB} by the reflexive property. Each marked angle is included between its ticked side and \overline{BD}.
- AAS: \triangle AEB \cong \triangle DEC. \angle B \cong \angle C (arcs), \angle AEB \cong \angle DEC (vertical angles), and \overline{AE} \cong \overline{DE} (ticks). \overline{AE} faces \angle B and \overline{DE} faces \angle C, so the side is not between the two angles, and it corresponds.
- No criterion applies. The marks are \overline{XY} \cong \overline{UV}, \overline{YZ} \cong \overline{VW} and \angle X \cong \angle U, and \angle X is not between \overline{XY} and \overline{YZ}: SSA. The triangles are drawn alike and may be congruent, but the marks do not force it. \angle Y \cong \angle V would give SAS, or \overline{XZ} \cong \overline{UW} SSS.
- ASA: \triangle GHK \cong \triangle NLM. \angle G \cong \angle N (one arc), \overline{GH} \cong \overline{NL} (tick) and \angle H \cong \angle L (two arcs), and each ticked side joins the two marked angles.
- SSS.
- ASA. \overline{AB} joins the vertices of \angle A and \angle B.
- SAS. \angle B is included between \overline{AB} and \overline{BC}.
- None. AAA fixes the shape but not the size, so congruence is not forced.
- AAS. \overline{BC} is not between \angle A and \angle C (that side is \overline{AC}); it faces \angle A, and \overline{EF} faces \angle D, so it corresponds.
- None: SSA. The two sides meet at A, so their included angle is \angle A, not \angle C. The triangles may or may not be congruent; these parts do not force it.
- SAS: \overline{BC} \cong \overline{EF}, so that \angle B is included. ASA: \angle A \cong \angle D, so that \overline{AB} joins the two angles. AAS: \angle C \cong \angle F, since \overline{AB} faces \angle C and is not between \angle B and \angle C.
- x = 5 and BC = 17. \angle B is included between \overline{AB} and \overline{BC}, so SAS needs BC = EF: 3x + 2 = 5x - 8, so 10 = 2x. Then BC = 3(5) + 2 = 17 = 5(5) - 8.
- x = 13 and m\angle C = 72^\circ. \overline{AB} joins \angle A and \angle B, so ASA needs m\angle A = m\angle D: 4x + 6 = 6x - 20, so 26 = 2x. Then m\angle A = 58^\circ and m\angle C = 180^\circ - 58^\circ - 50^\circ = 72^\circ.
- x = 22 and m\angle F = 73^\circ. \overline{BC} faces \angle A and is not between \angle A and \angle B, so AAS needs m\angle B = m\angle E: 3x + 1 = x + 45, so 2x = 44. Then m\angle E = 67^\circ and m\angle F = 180^\circ - 40^\circ - 67^\circ = 73^\circ.
- y = 4 only. 2y + 1 must be one of 7, 9, 12. If 2y + 1 = 9, then y = 4 and y + 8 = 12, which works. If 2y + 1 = 12, then y = 5.5 and y + 8 = 13.5, not 9. If 2y + 1 = 7, then y = 3 and the sides are 7, 7, 11.
- AB = 70 m, m\angle CAB = 38^\circ and m\angle CBA = 37^\circ. The pairing is A \leftrightarrow D, C \leftrightarrow C, B \leftrightarrow E, so \overline{AB} matches \overline{DE} and \angle CAB matches \angle CDE. In \triangle DCE, m\angle CED = 180^\circ - 105^\circ - 38^\circ = 37^\circ, and \angle CBA matches \angle CED.
- Line (1): \overline{AB} \parallel \overline{CD} (given). (2): \overline{AB} \cong \overline{CD} (given). (3): \angle ABD \cong \angle CDB (Alternate Interior Angles Theorem: line 1, transversal \overline{BD}). (4): \overline{BD} \cong \overline{DB} (reflexive property of congruence). (5): \triangle ABD \cong \triangle CDB (SAS postulate: lines 2, 3 and 4). \angle ABD is included between \overline{AB} and \overline{BD}, and \angle CDB between \overline{CD} and \overline{DB}.
- Line (1): E is the midpoint of \overline{AD} (given). (2): \overline{AE} \cong \overline{DE} (definition of midpoint: line 1). (3): \angle B \cong \angle C (given). (4): \overline{AD} and \overline{BC} cross at E (given). (5): \angle AEB \cong \angle DEC (Vertical Angles Theorem: line 4). (6): \triangle AEB \cong \triangle DEC (AAS theorem: lines 3, 5 and 2). \overline{AE} faces \angle B and \overline{DE} faces \angle C.
- It supplies \angle C \cong \angle F. ASA needs an angle at each end of the given side. The given side \overline{BC} has \angle B at one end, but the other given angle, \angle A, is at the far corner, facing it. The angle at the other end of \overline{BC} is the third one, and only the Third Angle Theorem provides it.
- No. Both have angles 50^\circ, 60^\circ and 70^\circ, but the 8 may face a different angle in each. In the same place in both, ASA or AAS makes them congruent; otherwise the parts do not correspond and no criterion applies. To see two that differ, construct an 8 cm side with 50^\circ and 60^\circ at its ends, and another with 60^\circ and 70^\circ at its ends (each a single triangle, by ASA). Measured, the first has other sides of about 6.5 cm and 7.4 cm, the second about 9.0 cm and 9.8 cm. The law of sines, much later, gives these lengths exactly.
- Any shape the rod triangle could be pushed into would have the same three sides, so by SSS it would be congruent to the original: the same angles, no change of shape. Four sides carry no such guarantee: a square and a rhombus with the same side length match in all four sides but not in their angles.