Why AAA and SSA fail
Three matching parts make two triangles congruent in SSS, SAS, ASA and AAS. Why not three matching angles, or two sides and an angle that is not between them?
Where you are 3 in Congruent triangles
Why this exists
A triangle has six parts, three sides and three angles. SSS, SAS, ASA and AAS each name three matching parts that force two triangles to be congruent. SSS, SAS and ASA are taken as postulates on this map, and AAS is proved from ASA. (The three postulates can also be derived from rigid motions.)
Three parts of a triangle can be arranged in six ways: SSS, SAS, ASA, AAS, AAA and SSA. Four are criteria. This lesson is about the other two.
AAA, the three angles and no side at all, sounds like the most natural test of all. It fails for a reason you can see at once: a photograph and its enlargement have the same angles.
SSA, two sides and an angle that is not between them, is the one that fools people. It uses the same ingredients as SAS, and AAS shows that a part off to one side can still do its job, so it seems it ought to work.
Neither does. This lesson builds the triangles that break each one, shows the one extra fact that turns AAA into a criterion, and says exactly when SSA data does pin a triangle down after all.
Definition
A criterion for congruence is a list of matching parts that forces two triangles to be congruent every time, whatever the triangles look like. A general rule is broken by a single case where it fails: one pair of triangles that matches the list and still is not congruent shows the list is not a criterion. Such a pair is a Counterexample.
Angle angle angle (AAA)
Two triangles satisfy AAA under a correspondence when each angle of one is congruent to the matching angle of the other. For \triangle ABC and \triangle DEF matched A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F, that means
To say that AAA is not a congruence criterion is to say that these three facts do not force \triangle ABC \cong \triangle DEF. All it takes is two triangles with the same three angles that are not congruent.
Side side angle (SSA)
SSA (side, side, angle) describes two triangles in which two sides of one and an angle that is not included between them are congruent to the corresponding parts of the other. In \triangle ABC with the angle at A, the two sides are \overline{AB}, one arm of the angle, and \overline{BC}, the side opposite the angle. The angle touches one given side and faces the other.
Compare SAS: there the angle is at B, the corner where \overline{AB} and \overline{BC} meet. The ingredients are the same; only the position of the angle has moved.
Intuition
Why angle angle angle fails
Enlarge a photograph of a triangle. Every side gets longer, but no corner opens or closes: the angles of the enlargement are the angles of the original. The two triangles have the same shape and different sizes.
In Figure 1 the arcs say the angles match, one arc with one arc, two with two, three with three. But every side of \triangle DEF is about twice the matching side of \triangle ABC, and a rigid motion never changes a length, so no sliding or turning will lay one exactly on the other. The angles have said nothing about length.
There is a second way to see it, by counting. The angles of a triangle add to 180^\circ (the triangle angle sum), so once two angles match, the third matches automatically. AAA looks like three facts but is really only two, and every criterion on the list needs three.
Why side side angle fails
Build a triangle from SSA data and watch where the freedom is.
Draw the angle at A and measure \overline{AB} along one ray, so B is fixed. In SAS the second side would now be measured along the other ray of the same angle, finishing the triangle.
In SSA the second side, \overline{BC}, starts at B. You know its length but not its direction. Think of it as a stick hinged at B whose free end must land on the other ray, because that is where C has to be. Swing it down.
The free end traces the circle around B with radius BC, gray in Figure 2. If that circle crosses the ray twice, the stick can land at C_1 or at C_2, and both landings finish a triangle with the same angle at A, the same \overline{AB} and the same length BC. One is narrow and one is wide.
The hinge is the whole story. In SAS the angle fixes the direction of the second side. In SSA nothing does.
Results
Angle angle angle
AAA is not a congruence criterion. Two triangles can have all three pairs of corresponding angles congruent and still not be congruent.
The third angle comes free (the Third Angle Theorem). If two angles of one triangle are congruent to two angles of another, the third angles are congruent too, because each third angle is 180^\circ minus the other two.
One side settles it. If all three angles match and one pair of corresponding sides is congruent, the triangles are congruent, by ASA. Every side of a triangle runs between two of its angles, so the matching side is always the included side of two matching angles.
Side side angle
SSA is not a congruence criterion. Two triangles can have two pairs of congruent sides and a pair of congruent angles not included between those sides, and still not be congruent.
One pair of triangles settles it. In Figure 3, AB = DE = 20, BC = EF = 13 and \angle A \cong \angle D, but AC = 11 while DF = 21:
When two triangles appear. Fix the angle at A and the length AB, and let h be the distance from B to the line of the other ray. SSA data builds two different triangles exactly when the angle at A is acute and
In every other case it builds one triangle or none (see Boundaries), but one trouble case is enough to lose “every time”.
Why it is true
Angle angle angle: the parallel cut
The counterexample can be built inside any triangle. Start with \triangle ABC, choose a point D strictly between A and B, and draw the line through D parallel to \overline{BC}. It meets \overline{AC} at a point E.
Line AB is a transversal of the parallel lines DE and BC, so \angle ADE \cong \angle ABC by the Corresponding Angles Theorem. Line AC is another transversal, so \angle AED \cong \angle ACB for the same reason. And \angle A \cong \angle A by the reflexive property. Matched A \leftrightarrow A, D \leftrightarrow B, E \leftrightarrow C, the two triangles satisfy AAA.
They are not congruent under that match. The side \overline{AD} would have to be congruent to \overline{AB}, but D lies between A and B, so AD + DB = AB with DB > 0, and AD is shorter than AB.
Could some other matching of vertices make them congruent? Not if \triangle ABC has three different angles, such as 58^\circ, 72^\circ and 50^\circ in Figure 4. Congruent triangles have congruent corresponding angles, and a 58^\circ angle can only be matched with the other 58^\circ angle, so the match above is the only one to try. It fails, so \triangle ADE and \triangle ABC are the counterexample.
Side side angle: two triangles from any three lengths
One counterexample is enough, but here is how to build one from any three lengths h < BC < AB.
Draw a line and mark a point H on it. Put B at height h straight above H, so that \overline{BH} is perpendicular to the line. Put A on the line to the left of H with \triangle ABH a right triangle of hypotenuse AB; by the Pythagorean theorem,
Mark C_1 and C_2 on the line, one on each side of H, each at the distance
By the Pythagorean theorem in the right triangles \triangle BHC_1 and \triangle BHC_2, BC_1 = BC_2 = BC. Because BC < AB, we get HC_1 < AH, so both points lie on the ray \overrightarrow{AH}.
\triangle ABC_1 and \triangle ABC_2 share \overline{AB}, have BC_1 = BC_2, and share the same angle at A, made of the same two rays. That is SSA. But AC_1 = AH - HC_1 and AC_2 = AH + HC_2 differ by 2\,HC_1 \neq 0, so the triangles are not congruent.
Worked examples
Examples 1 to 6 are about angle angle angle; Examples 7 to 11 are about side side angle.
Checking whether the angles match
In \triangle ABC, m\angle A = 35^\circ and m\angle B = 65^\circ. In \triangle DEF, m\angle D = 35^\circ and m\angle F = 80^\circ. Do the triangles have the same three angles? Are they congruent?
SolutionEach triangle is missing one angle, and the triangle angle sum gives it:
Both triangles have angles 35^\circ, 65^\circ and 80^\circ. Matching equal angles gives A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F, and under that match all three pairs of angles are congruent.
That is only AAA. With no length given at all, the triangles might be congruent or might not.
An AAA counterexample with numbers
In Figure 5, m\angle A = 50^\circ, m\angle B = 70^\circ, AD = 4, DB = 6 and \overline{DE} \parallel \overline{BC}. Find the three angles of \triangle ADE and of \triangle ABC, and show the triangles are not congruent.
In \triangle ABC the third angle is m\angle C = 180^\circ - 50^\circ - 70^\circ = 60^\circ.
In \triangle ADE, \angle A is the same angle, 50^\circ. With \overleftrightarrow{AB} as the transversal, \angle ADE and \angle ABC are corresponding angles, so m\angle ADE = 70^\circ. With \overleftrightarrow{AC} as the transversal, \angle AED and \angle ACB are corresponding angles, so m\angle AED = 60^\circ. As a check, 50 + 70 + 60 = 180.
The angles are all different, so the only match that pairs equal angles is A \leftrightarrow A, D \leftrightarrow B, E \leftrightarrow C. Under it, \overline{AD} corresponds to \overline{AB}. But AD = 4 and AB = AD + DB = 4 + 6 = 10. The triangles share all three angles and are not congruent.
Angles in terms of x, then a side
In \triangle PQR, m\angle P = (2x)^\circ, m\angle Q = (3x + 10)^\circ and m\angle R = (x + 20)^\circ. In \triangle XYZ, m\angle X = 50^\circ and m\angle Y = 85^\circ. Also QR = 12 and YZ = 12. Find x, and decide whether the triangles are congruent.
SolutionThe angles of \triangle PQR add to 180^\circ:
So m\angle P = 50^\circ, m\angle Q = 85^\circ and m\angle R = 45^\circ. In \triangle XYZ, m\angle Z = 180^\circ - 50^\circ - 85^\circ = 45^\circ. The triangles satisfy AAA with P \leftrightarrow X, Q \leftrightarrow Y, R \leftrightarrow Z.
AAA alone would stop there. But under this match the partner of \overline{QR} is \overline{YZ}, and both are 12. So \angle Q, \overline{QR}, \angle R are congruent to \angle Y, \overline{YZ}, \angle Z: two angles and the side between them. By ASA, \triangle PQR \cong \triangle XYZ.
A side of the same length, in the wrong place
Both triangles in Figure 6 have angles 40^\circ at A and D, 60^\circ at B and E, and 80^\circ at C and F. In \triangle ABC, BC = 5. In \triangle DEF, DE = 5. Are the triangles congruent?
Three matching angles and a side of 5 each look like enough, but the sides of 5 are not partners. With the angles all different, the only match is A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F, and under it \overline{BC} corresponds to \overline{EF}, while \overline{DE} corresponds to \overline{AB}.
Suppose the triangles were congruent. Corresponding parts of congruent triangles are congruent, so AB = DE = 5, and BC = 5 as given, so AB = BC. Now compare \triangle ABC with itself read the other way round, as \triangle CBA. Side \overline{AB} matches \overline{CB}, the angle \angle B matches itself, and \overline{BC} matches \overline{BA}, so by SAS, \triangle ABC \cong \triangle CBA. Corresponding angles would then give \angle A \cong \angle C, which says 40^\circ = 80^\circ. That is false, so the supposition was false: the triangles are not congruent. This is an indirect proof.
Had the second triangle had EF = 5 instead, \overline{BC} and \overline{EF} would be partners, and ASA would make the triangles congruent. One side settles AAA only when it is the corresponding side.
A proof where all three angles come for free
Given. \angle ABC and \angle EDC are right angles; \overline{AE} and \overline{BD} cross at C; \overline{AC} \cong \overline{EC}.
Prove. \triangle ABC \cong \triangle EDC.
Plan. The right angles match, the crossing gives vertical angles at C, and then the third angles match too, so the two triangles share every angle. That alone is AAA and proves nothing. The given sides \overline{AC} and \overline{EC} are a corresponding pair, each facing the right angle, and \overline{AC} lies between the angles at A and C, which is ASA.
| Statements | Reasons |
|---|---|
| 1. \angle ABC and \angle EDC are right angles | given |
| 2. \angle ABC \cong \angle EDC | Right Angle Congruence Theorem: line 1 |
| 3. \overline{AE} and \overline{BD} cross at C | given |
| 4. \angle ACB \cong \angle ECD | Vertical Angles Theorem: line 3 |
| 5. \angle BAC \cong \angle DEC | Third Angle Theorem: lines 2 and 4 |
| 6. \overline{AC} \cong \overline{EC} | given |
| 7. \triangle ABC \cong \triangle EDC | ASA postulate: lines 5, 6 and 4 |
After line 5 all three pairs of angles match, and line 6 is the one pair of corresponding sides that settles it: \overline{AC} joins the vertices of \angle BAC and \angle ACB. (The AAS theorem with lines 2, 4 and 6 would also do, without line 5.) Without line 6 every angle still matches, but if the crossing point sat nearer B, the left triangle would be the smaller one. The angles fix the shape of both triangles; the side fixes the size.
Cutting matching braces
A carpenter cuts triangular corner braces for a set of shelves. Each brace has corners of 35^\circ and 55^\circ, cut with the saw at two fixed settings. Will every brace come out the same? If not, what single measurement should the carpenter repeat on each one to make them identical?
SolutionThe third corner is 180^\circ - 35^\circ - 55^\circ = 90^\circ, so every brace has the angles 35^\circ, 55^\circ and 90^\circ. That is AAA, and it does not make the braces identical: a larger brace with the same saw settings has the same corners and longer edges.
One matching side fixes it, provided it is the same side on every brace. The long edge, opposite the right angle, runs between the 35^\circ corner and the 55^\circ corner. Cut that edge to the same length on every brace, say 60 cm, and any two braces agree in two angles and the side between them, so ASA makes them congruent. “Some edge is 60 cm” would not do: as in Example 4, it might be the long edge on one brace and a short edge on another.
SAS or SSA?
In \triangle PQR and \triangle XYZ, \overline{PQ} \cong \overline{XY} and \overline{QR} \cong \overline{YZ}. For each extra fact, say whether the triangles must be congruent.
(a) \angle Q \cong \angle Y (b) \angle P \cong \angle X (c) \angle R \cong \angle Z
SolutionThe two given sides meet at Q in the first triangle and at Y in the second, so those are the included angles.
(a) \angle Q is included between \overline{PQ} and \overline{QR}, and \angle Y between the matching sides. This is SAS, so \triangle PQR \cong \triangle XYZ.
(b) \angle P sits at the far end of \overline{PQ} and faces \overline{QR}. This is SSA. The triangles might be congruent, but these facts do not force it.
(c) \angle R touches \overline{QR} and faces \overline{PQ}. Again SSA, and again no conclusion.
The test is always the same: find the corner where the two given sides meet. Only the angle at that corner makes SAS.
Building the two SSA triangles
In Figure 8, B is 12 units above the ray from A, with \overline{BH} perpendicular to the ray, and AB = 20. Points C_1 and C_2 on the ray satisfy BC_1 = BC_2 = 13. Find AH, HC_1, AC_1 and AC_2.
\triangle ABH has a right angle at H and hypotenuse \overline{AB}, so by the Pythagorean theorem
\triangle BHC_1 also has a right angle at H, with hypotenuse \overline{BC_1}:
The same working in \triangle BHC_2 gives HC_2 = 5. C_1 lies between A and H, and C_2 lies beyond H, so by segment addition
The triangles share \overline{AB} and the angle at A, and BC_1 = BC_2 = 13, but their third sides are 11 and 21: the non-congruent pair of Figure 3.
How many triangles close?
The angle at A is fixed, AB = 10, and B is 6 units from the line of the other ray, so the foot H of the perpendicular is 8 units from A. How many triangles ABC, with C on the ray, have BC = 5? BC = 6? BC = 8? BC = 12?
A point of the line at distance BC from B sits \displaystyle \sqrt{BC^2 - 36} from H, by the Pythagorean theorem in the right triangle with leg 6 and hypotenuse BC.
- BC = 5: 5 < 6, so even the nearest point of the line, H, is out of reach. No triangle.
- BC = 6: the circle just touches the line at H, so C = H and the angle at C is a right angle. One triangle.
- BC = 8: the landings are \displaystyle \sqrt{64 - 36} = \sqrt{28} = 2\sqrt{7} on either side of H. Since \displaystyle 2\sqrt{7} \approx 5.29 is less than 8, both lie on the ray, at \displaystyle AC = 8 - 2\sqrt{7} and \displaystyle AC = 8 + 2\sqrt{7}. Two triangles.
- BC = 12: the landings are \displaystyle \sqrt{144 - 36} = \sqrt{108} = 6\sqrt{3} \approx 10.39 from H. That is more than 8, so the landing on the A side falls behind A, off the ray. Only \displaystyle AC = 8 + 6\sqrt{3} counts. One triangle.
Here h = 6 and \displaystyle AB = \sqrt{8^2 + 6^2} = 10, so, as the Results say, only 6 < BC < 10 gives two triangles.
A proof that cannot be finished
In Figure 10, C and D lie on a ray from A, with C between A and D, and \overline{BC} \cong \overline{BD}. Try to prove \triangle ABC \cong \triangle ABD.
Three facts come at once:
| Statements | Reasons |
|---|---|
| 1. \overline{BC} \cong \overline{BD} | given |
| 2. \overline{AB} \cong \overline{AB} | reflexive property of congruence |
| 3. \angle BAC \cong \angle BAD | reflexive property of congruence: \overrightarrow{AC} and \overrightarrow{AD} are the same ray |
The sides in lines 1 and 2 meet at B, but the angle in line 3 is at A. The angle is not included between them, so this is SSA, and no criterion supplies a line 4.
No other route can finish the proof either. If the triangles were congruent, corresponding parts of congruent triangles would be congruent, so \overline{AC} \cong \overline{AD}. But C is between A and D, so by segment addition AC + CD = AD with CD > 0, and AC < AD. The statement to be proved is false, and a chain of valid reasons cannot end on a false statement. The figure is itself a counterexample: lines 1 to 3 hold, and the triangles are not congruent.
Same measurements, different areas
A triangle ABC has AB = 25 and BC = 17, and the angle at A is acute, sized so that B sits exactly 15 units from the line AC. Find every possible length AC, and the area of each triangle.
SolutionDrop the perpendicular from B to the line, with foot H. In the right triangle ABH,
and in each right triangle BHC,
C can be 8 units either side of H, and 8 < 20, so both positions are on the ray from A:
Both triangles have base \overline{AC} on the line and height 15, so by the area of a triangle:
Three measurements match, yet one triangle has more than twice the area of the other, and congruent triangles have equal areas.
Boundaries
Angle angle angle
AAA fails for congruence, not for shape. Two triangles with the same three angles always have the same shape: one is an enlargement of the other. That relationship is similarity, written \triangle ADE \sim \triangle ABC, and for triangles matching angles are exactly enough to force it. Congruence is similarity with the scale factor 1.
The side added must correspond. AAA with one pair of congruent corresponding sides gives congruence by ASA. Equal lengths on sides that are not partners do not, as Example 4 shows.
Beyond triangles, matching angles do not even fix the shape. A square and a long thin rectangle both have four right angles, yet one is not an enlargement of the other. It is special to triangles that the angles decide the shape, and even there they never decide the size.
Side side angle
Wherever SSA data builds at most one triangle, two triangles with that data must be congruent. The count depends on the angle and on how BC compares with h and with AB.
Too short, or just long enough. With an acute angle at A: if BC < h the stick cannot reach the line, and no triangle exists. If BC = h it just touches the line at the foot H, and exactly one triangle closes, with a right angle at C.
When the angle faces the longer given side. Still with an acute angle, suppose BC \geq AB. The landings are \displaystyle \sqrt{BC^2 - h^2} either side of H, and H is \displaystyle \sqrt{AB^2 - h^2} from A. Since BC \geq AB, the first distance is at least the second, so the near landing is A itself or falls behind A, off the ray. Exactly one triangle closes.
In Figure 11, AB = 13, B is 12 above the line and BC = 20. The foot is \displaystyle \sqrt{169 - 144} = 5 from A, and the landings are \displaystyle \sqrt{400 - 144} = 16 either side of it: 21 to the right of A and 11 to the left. Only C, on the ray, makes a triangle.
A right or obtuse angle. If the angle at A is right, the foot is A itself; if obtuse, the foot lies behind A. Either way the near landing can never be on the ray, and the far one is on the ray only when BC > AB. So a triangle exists exactly when the given angle faces the longer given side, and then there is exactly one. The right-angle case is a criterion of its own, hypotenuse leg congruence, proved there from the Pythagorean theorem and SSS.
So SSA fails in one situation only: the angle is acute, it faces the shorter of the two given sides, and that side is longer than the distance h from B to the ray. That is the situation of Examples 8, 10 and 11, and of BC = 8 in Example 9.
“Not a criterion” does not mean “never congruent”. This holds for AAA and SSA alike. Two triangles with matching AAA or SSA data may well be congruent. What fails is the promise: the data alone does not guarantee it, so in a proof it is never a reason.
Where this goes next
The shape that AAA does pin down is the subject of angle angle similarity, where two pairs of matching angles make two triangles similar, and similar triangles have their sides in a constant ratio. The parallel cut in Figure 4 returns as the triangle proportionality theorem, which measures how the small triangle’s sides compare with the large one’s.
Similarity also explains a surprise: two triangles can agree in five of their six measurements, all three angles and two side lengths, and still not be congruent. The triangles with sides 8, 12, 18 and 12, 18, 27 are one such pair; the similarity lessons show why their angles agree. As in Example 4, the equal lengths are not partners: 12 is the middle side of the first triangle but the shortest side of the second.
The fact that a triangle’s angles depend only on its shape, never its size, is what makes the trigonometric ratios possible.
The most useful safe case of SSA, the right angle, becomes hypotenuse leg congruence, the one extra tool for right triangles.
In trigonometry the same two SSA triangles return as the ambiguous case. The law of sines turns SSA data into a value for the sine of the angle at C, and two supplementary angles share that sine: exactly the angles at C_1 and C_2. The distance h, which decided everything here, is computed there as AB times the sine of the angle at A, in comparing the given side to the altitude.
Knowing exactly why these two arrangements fail is what makes the working criteria trustworthy when you prove triangles congruent, and it is why a construction from two sides and the included angle always gives one triangle.
Exercises
Do the angles match?
Use the triangle angle sum to find each missing angle, then say whether the two triangles have the same three angles. If they do, give the match of vertices.
- In \triangle ABC, m\angle A = 45^\circ and m\angle B = 65^\circ. In \triangle DEF, m\angle D = 45^\circ and m\angle F = 70^\circ.
- In \triangle JKL, m\angle J = 30^\circ and m\angle K = 100^\circ. In \triangle MNP, m\angle M = 30^\circ and m\angle N = 50^\circ.
- In \triangle ABC, m\angle A = 72^\circ and m\angle B = 54^\circ. In \triangle XYZ, m\angle X = 72^\circ and m\angle Y = 64^\circ.
- In \triangle RST, m\angle R = (x + 10)^\circ, m\angle S = (2x)^\circ and m\angle T = (3x - 10)^\circ. In \triangle UVW, m\angle U = 40^\circ and m\angle V = 80^\circ.
- One triangle has angles of 35^\circ and 55^\circ. Another has angles of 55^\circ and 90^\circ.
Adding one side to AAA
In Exercises 6 to 8, \angle A \cong \angle D, \angle B \cong \angle E and \angle C \cong \angle F. Say whether the extra fact makes \triangle ABC \cong \triangle DEF, and if it does, name the criterion and the parts it uses.
- AB = DE = 8
- AC = 6 and EF = 6
- BC = EF = 11
- \triangle PQR and \triangle XYZ have \angle P \cong \angle X, \angle Q \cong \angle Y and \angle R \cong \angle Z. Also PR = 3x - 2 and XZ = x + 10. Find the value of x that makes the triangles congruent, and the length of \overline{PR} then.
- Two triangles have the same three angles, matched K \leftrightarrow S, L \leftrightarrow T, M \leftrightarrow U. KL = 2y + 5 and ST = 4y - 7. For what value of y are the triangles congruent? How long are those sides then?
- In Example 4, change the second triangle so that EF = 5 instead of DE = 5. Are the triangles congruent now? By which criterion?
Building AAA counterexamples
- In \triangle ABC, m\angle A = 35^\circ and m\angle B = 75^\circ. Point D is on \overline{AB} with AD = 3 and DB = 9, and \overline{DE} \parallel \overline{BC} with E on \overline{AC}, as in Figure 5. Find the three angles of \triangle ADE, and the lengths AD and AB that show \triangle ADE is not congruent to \triangle ABC.
- Describe two triangles, each with angles 30^\circ, 60^\circ and 90^\circ, that are not congruent. Explain how you know they are not.
- In a figure like Figure 7, \angle ABC and \angle EDC are right angles and \overline{AE} crosses \overline{BD} at C, but now AC = 4 and CE = 10. Which pairs of angles are congruent? Explain why \triangle ABC \cong \triangle EDC is false.
- In Figure 4, is there any position of D strictly between A and B that makes \triangle ADE \cong \triangle ABC under the match A \leftrightarrow A, D \leftrightarrow B, E \leftrightarrow C? Explain.
Naming the arrangement
In Exercises 16 to 20, \triangle ABC and \triangle DEF correspond as the letters are written. For each set of facts, name the arrangement (SSS, SAS, ASA, AAS or SSA) and say whether the triangles must be congruent.
- \overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}, \angle B \cong \angle E
- \overline{AB} \cong \overline{DE}, \overline{BC} \cong \overline{EF}, \angle C \cong \angle F
- \overline{AB} \cong \overline{DE}, \overline{AC} \cong \overline{DF}, \angle A \cong \angle D
- \angle A \cong \angle D, \angle B \cong \angle E, \overline{BC} \cong \overline{EF}
- \overline{AC} \cong \overline{DF}, \overline{BC} \cong \overline{EF}, \angle A \cong \angle D
Counting SSA triangles
In Exercises 21 to 26 the angle at A is acute, AB = 13, and B is 5 units from the line of the other ray. Say how many triangles ABC, with C on that ray, have the given length BC.
- BC = 4
- BC = 5
- BC = 12
- BC = 13
- BC = 20
- Find every length BC that gives exactly two triangles.
Building both SSA triangles
In Exercises 27 to 31 the angle at A is acute and B is at height h above the line of its other ray. Find every possible length AC.
- AB = 20, h = 12, BC = 15
- AB = 17, h = 8, BC = 10
- AB = 26, h = 24, BC = 25
- Find the areas of the two triangles in Exercise 28.
- AB = 15, h = 9, BC = 41
Reasoning
- AAA names three parts, but it carries only two independent facts. Explain why.
- Explain why AAA together with any pair of congruent corresponding sides gives congruence by ASA, so AAS is never needed once all three angles match.
- Given. \angle ABC and \angle EDC are right angles; \overline{AE} and \overline{BD} cross at C, as in Figure 7 but without the ticks; C is the midpoint of \overline{BD}. Prove. \triangle ABC \cong \triangle EDC. Write a two-column proof.
- Give two quadrilaterals whose four angles match pair by pair, but which do not have the same shape. What does this say about angles alone for figures with more than three sides?
- Kai says: “Two triangles each have angles of 20^\circ, 70^\circ and 90^\circ. That is three matching parts, and three matching parts make triangles congruent.” Explain what is wrong with Kai’s reasoning.
- In Figure 2, explain why \angle BAC_1 and \angle BAC_2 are the same angle, not merely two congruent angles.
- Two triangles have AB = DE = 9, BC = EF = 12 and \angle A \cong \angle D. Must they be congruent? Explain.
- In Figure 8, show that H is the midpoint of \overline{C_1C_2}.
- Someone says: “SSA is not a criterion, so two triangles with SSA data are never congruent.” What is wrong with this?
- AAS also uses a part that is not between the other two, yet AAS is a criterion. Explain the difference: why does knowing a second angle stop the side from swinging?
Answers
- Yes. m\angle C = 180^\circ - 45^\circ - 65^\circ = 70^\circ and m\angle E = 180^\circ - 45^\circ - 70^\circ = 65^\circ. Both triangles have 45^\circ, 65^\circ, 70^\circ, matched A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F.
- Yes. m\angle L = 180^\circ - 30^\circ - 100^\circ = 50^\circ and m\angle P = 180^\circ - 30^\circ - 50^\circ = 100^\circ. The match is J \leftrightarrow M, K \leftrightarrow P, L \leftrightarrow N. The letters in alphabetical order do not always pair up.
- No. m\angle C = 180^\circ - 72^\circ - 54^\circ = 54^\circ and m\angle Z = 180^\circ - 72^\circ - 64^\circ = 44^\circ. The angles are 72^\circ, 54^\circ, 54^\circ against 72^\circ, 64^\circ, 44^\circ.
- Yes. (x + 10) + 2x + (3x - 10) = 180 gives 6x = 180, so x = 30 and the angles of \triangle RST are 40^\circ, 60^\circ, 80^\circ. In \triangle UVW, m\angle W = 180^\circ - 40^\circ - 80^\circ = 60^\circ. The match is R \leftrightarrow U, S \leftrightarrow W, T \leftrightarrow V.
- Yes. The first triangle’s third angle is 180^\circ - 35^\circ - 55^\circ = 90^\circ and the second’s is 180^\circ - 55^\circ - 90^\circ = 35^\circ. Both have 35^\circ, 55^\circ, 90^\circ.
- Congruent, by ASA: \angle A, \overline{AB}, \angle B match \angle D, \overline{DE}, \angle E, and \overline{AB} lies between the two angles.
- Not decided. \overline{AC} corresponds to \overline{DF}, not to \overline{EF}, so these are not partner sides and the facts given are still only AAA.
- Congruent, by ASA: \angle B, \overline{BC}, \angle C match \angle E, \overline{EF}, \angle F.
- x = 6, and PR = 16. The sides are partners, so they must be equal: 3x - 2 = x + 10 gives 2x = 12. Then 3(6) - 2 = 16 = 6 + 10, and ASA with \angle P, \overline{PR}, \angle R gives the congruence.
- y = 6, and both sides are 17. From 2y + 5 = 4y - 7, 12 = 2y. Check: 2(6) + 5 = 17 and 4(6) - 7 = 17. With KL = ST, ASA applies to \angle K, \overline{KL}, \angle L.
- Yes, by ASA. \overline{BC} and \overline{EF} are partners under the only match A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F, and each lies between the 60^\circ and 80^\circ angles.
- m\angle A = 35^\circ, m\angle ADE = 75^\circ, m\angle AED = 70^\circ; AD = 3 and AB = 12. The third angle of \triangle ABC is 180^\circ - 35^\circ - 75^\circ = 70^\circ, and the corresponding angles at D and E copy \angle B and \angle C. The angles are all different, so the only match is A \leftrightarrow A, D \leftrightarrow B, E \leftrightarrow C, which pairs AD = 3 with AB = 3 + 9 = 12.
- One way: draw a 30^\circ, 60^\circ, 90^\circ triangle ABC, then cut it through the midpoint D of \overline{AB}, parallel to \overline{BC}, as in Figure 4. The small triangle has the same angles by the Corresponding Angles Theorem. The angles are all different, so the only match pairs equal angles, and under it \overline{AD}, half of \overline{AB}, is paired with \overline{AB}. So they are not congruent.
- \angle ABC \cong \angle EDC (right angles), \angle ACB \cong \angle ECD (vertical angles), and so \angle BAC \cong \angle DEC (Third Angle Theorem). The statement pairs \overline{AC} with \overline{EC}, and 4 \ne 10, so the corresponding sides are not congruent and the statement is false. The triangles have the same angles and different sizes.
- No. The match pairs \overline{AD} with \overline{AB}, and AD + DB = AB with DB > 0, so AD < AB wherever D is placed.
- SAS, congruent. \angle B is where \overline{AB} and \overline{BC} meet.
- SSA, not necessarily congruent. \angle C touches \overline{BC} but faces \overline{AB}; the given sides meet at B.
- SAS, congruent. \overline{AB} and \overline{AC} meet at A, the given angle.
- AAS, congruent. \overline{BC} is not between \angle A and \angle B, but two angles and a non-included side is a criterion.
- SSA, not necessarily congruent. \overline{AC} and \overline{BC} meet at C, but the angle is at A, facing \overline{BC}.
- None. 4 < 5, so \overline{BC} cannot reach the line.
- One. BC equals the distance to the line, so C is the foot of the perpendicular and the angle at C is right.
- Two. The foot is \displaystyle \sqrt{169 - 25} = 12 from A, and the landings are \displaystyle \sqrt{144 - 25} = \sqrt{119} \approx 10.91 either side of it. \displaystyle \sqrt{119} < 12, so both are on the ray, at \displaystyle AC = 12 \pm \sqrt{119}.
- One. The landings are \displaystyle \sqrt{169 - 25} = 12 either side of the foot, which is 12 from A, so the near one is A itself. Only AC = 24 makes a triangle.
- One. The landings are \displaystyle \sqrt{400 - 25} = \sqrt{375} \approx 19.36 from the foot, more than 12, so the near one falls behind A. Only \displaystyle AC = 12 + 5\sqrt{15} counts.
- 5 < BC < 13. There are two landings when BC > 5, and the near one stays on the ray while \displaystyle \sqrt{BC^2 - 25} < 12, that is BC < 13.
- AC = 7 or AC = 25. The foot is \displaystyle \sqrt{400 - 144} = 16 from A, and \displaystyle HC = \sqrt{225 - 144} = 9, so AC = 16 \mp 9.
- AC = 9 or AC = 21. The foot is \displaystyle \sqrt{289 - 64} = 15 from A, and \displaystyle HC = \sqrt{100 - 64} = 6, so AC = 15 \mp 6.
- AC = 3 or AC = 17. The foot is \displaystyle \sqrt{676 - 576} = 10 from A, and \displaystyle HC = \sqrt{625 - 576} = 7, so AC = 10 \mp 7. Both are positive, so both triangles exist.
- 36 and 84. Each triangle has height 8: \tfrac{1}{2}(9)(8) = 36 and \tfrac{1}{2}(21)(8) = 84.
- Only AC = 52. The foot is \displaystyle \sqrt{225 - 81} = 12 from A and \displaystyle HC = \sqrt{1681 - 81} = 40. The landing 12 + 40 = 52 is on the ray; the other, 12 - 40 = -28, is 28 behind A. Here BC is the longer given side, so only one triangle closes.
- The three angles of a triangle add to 180^\circ. Once two pairs of angles match, the third angles are each 180^\circ minus the same two numbers, so they match automatically. The third fact adds nothing, and every congruence criterion needs three.
- Each side of a triangle has an angle at each end, and those two angles are among the three that match. Say the side is \overline{AB}, matched with \overline{DE}: then \angle A, \overline{AB}, \angle B and \angle D, \overline{DE}, \angle E are two angles and the side between them, which is ASA.
- Line (1): \angle ABC and \angle EDC are right angles (given). (2): \angle ABC \cong \angle EDC (Right Angle Congruence Theorem: line 1). (3): C is the midpoint of \overline{BD} (given). (4): \overline{BC} \cong \overline{DC} (definition of midpoint: line 3). (5): \overline{AE} and \overline{BD} cross at C (given). (6): \angle ACB \cong \angle ECD (Vertical Angles Theorem: line 5). (7): \triangle ABC \cong \triangle EDC (ASA postulate: lines 2, 4 and 6). ASA fits because \overline{BC} lies between the angles at B and C, and \overline{DC} between the angles at D and C. The third angles, at A and E, match as well, but this time they are not needed.
- A 1 by 1 square and a 1 by 3 rectangle. Each has four 90^\circ angles, but one is not an enlargement of the other, since the square’s sides are equal and the rectangle’s are not. With more than three sides, matching angles do not even fix the shape, let alone the size.
- Three matching parts are enough only for the listed combinations, SSS, SAS, ASA and AAS, each with at least one side. Three angles are not on the list. A 20^\circ, 70^\circ, 90^\circ triangle can be enlarged to any size and keep its angles, so Kai’s two triangles might have different sizes.
- C_1 and C_2 both lie on the same ray from A, so \overrightarrow{AC_1} and \overrightarrow{AC_2} are one ray. Each angle is formed by that ray and \overrightarrow{AB}, so they are one angle with two names.
- Yes. The angle at A faces \overline{BC}, the longer given side (12 > 9). If the angle is acute, the landings are \displaystyle \sqrt{144 - h^2} either side of the foot, which is only \displaystyle \sqrt{81 - h^2} from A, so the near landing falls behind A. If it is right or obtuse, the near landing is never on the ray. Either way only one triangle can be built, so both triangles are that one triangle, and they are congruent.
- \triangle BHC_1 and \triangle BHC_2 are right triangles with the same leg BH = 12 and the same hypotenuse 13. By the Pythagorean theorem \displaystyle HC_1 = \sqrt{13^2 - 12^2} = HC_2, and H lies between C_1 and C_2, so H is the midpoint of \overline{C_1C_2}.
- It confuses “not guaranteed” with “impossible”. SSA fails because some pairs are not congruent; other pairs, such as any pair with the angle facing the longer side, are.
- With two angles known, the triangle angle sum fixes the third, so the angle at the other end of the side is known too and fixes the direction of the next side from that corner: the data becomes ASA. In SSA only one angle is known, and nothing fixes the direction of \overline{BC} at B.