Triangles, congruence and similarity

A triangle needs three distinct, noncollinear vertices

We continue in a Euclidean plane. A nondegenerate triangle ABC is formed by three distinct points A, B and C that are not all on one straight line, together with the segments AB, BC and CA. These points are its vertices and these segments its sides. The boundary encloses a triangular region; when we discuss area, we mean the area of that region. A side segment is an object, while its length is a positive number in a chosen length unit.

The interior angles are ∠BAC at A, ∠ABC at B and ∠ACB at C. Each has degree measure strictly between 0° and 180°. When the intended angle is unambiguous, we write ∠A, ∠B or ∠C. A side is opposite the vertex it does not contain: BC is opposite A, CA opposite B and AB opposite C. In the common notation a = BC, b = CA and c = AB, a lowercase letter denotes a side length opposite the corresponding uppercase vertex.

Three collinear points do not give a triangle under this convention, even when their labels and pairwise distances are different. Repeating a vertex also fails the distinct-point requirement. Such boundary configurations can help us understand a limiting case, but we will explicitly call them degenerate. A sketch with three letters or three strokes cannot replace these conditions. Check the stated points, angle measures and lengths before applying a triangle theorem.

Side and angle classifications answer different questions

An equilateral triangle has three equal side lengths. An isosceles triangle has at least two equal side lengths under our inclusive convention, so every equilateral triangle is also isosceles. A scalene triangle has three pairwise different side lengths. Some books reserve 'isosceles' for exactly two equal sides; state that alternative convention when using it. Our categories equilateral and isosceles therefore overlap deliberately, while scalene does not overlap either.

An acute triangle has all three interior angles less than 90°. A right triangle has one interior angle equal to 90°. An obtuse triangle has one interior angle greater than 90°. The side opposite a right angle is the hypotenuse; the two sides meeting at that angle are the legs. These are angle classifications, so an isosceles triangle can also be acute, right or obtuse. Equal-looking strokes or an almost square corner are not exact side or angle data.

Equal sides of a triangle have equal opposite angles, and equal angles have equal opposite sides. We will justify this relationship using congruence below. An equilateral triangle therefore has three equal angles, each 60° by the angle sum proved next. That sum also shows that every nondegenerate triangle has at least two acute angles: two angles of at least 90° would already total at least 180°, leaving no positive third angle. Thus there cannot be two right angles or two obtuse angles in a triangle.

Parallel-line properties explain the 180° sum

Take any nondegenerate triangle ABC. Through A draw the line parallel to BC. This uses the Euclidean parallel-line setting: through a point outside a line there is exactly one parallel line, and alternate interior angles made by a transversal of two parallel lines are equal. Choose points X and Y on the two opposite rays of the parallel through A so that, in the half-plane containing B and C, the successive rays are AX, AB, AC and AY. The three adjoining angles ∠XAB, ∠BAC and ∠CAY then partition a straight angle.

Line AB is a transversal, so m∠XAB = m∠ABC. Line AC is another transversal, so m∠CAY = m∠ACB. Substitute those equal measures into the straight-angle equation to obtain m∠ABC + m∠BAC + m∠ACB = 180°. This argument applies to every permitted Euclidean triangle; measuring several pictures only checks those pictures. The Euclidean parallel-line properties are part of the reasoning, so the result is not a claim about triangles on every possible curved surface.

Now extend side BC through C to D, with B, C and D collinear and C strictly between B and D. The ordinary exterior angle ∠ACD and the interior angle ∠ACB form a proper linear pair, giving m∠ACD = 180° − m∠ACB. The triangle sum makes this m∠BAC + m∠ABC, the sum of the two remote interior angles. Because both remote measures are positive, this exterior measure exceeds either one separately. The extension and the selected ordinary angle matter: a reflex sector or an unrelated angle outside the drawing is not this exterior angle.

Three positive lengths must satisfy strict triangle inequalities

If a, b and c are the side lengths of a nondegenerate Euclidean triangle, then a + b > c, b + c > a and c + a > b. Geometrically, the broken route through the third vertex is longer than the direct side joining the other two vertices; equality in the Euclidean distance inequality occurs exactly when the intermediate point lies on that direct segment. Noncollinearity excludes that equality here. Positive lengths and all three strict inequalities are also sufficient for such a triangle to exist. The ruler-and-compasses construction is a later subject.

For positive a and b, these conditions are equivalent to |a − b| < c < a + b. The upper bound comes from a + b > c; the other two inequalities give c > a − b and c > b − a, together giving the strict lower bound. For example, sides 4 cm and 7 cm allow a third side x cm exactly when 3 < x < 11. This describes every allowed real length, not just a few trial values. If lengths are sorted as 0 < a ≤ b ≤ c, checking a + b > c is enough: the other two inequalities then follow from positivity and the order.

Equality is a boundary, not a valid triangle. Lengths 4, 7 and 11 can lie on one line with the 11-unit segment equal to the other two put end to end; lengths 4, 7 and 3 give the corresponding lower-bound collinear case. Either has zero enclosed area. If the longest proposed side exceeds the sum of the other two, even that broken route cannot reach its endpoint, so no triangle has those lengths. A zero or negative proposed side already fails the positive-length condition. A calculation that solves an equation must still pass these checks.

Congruence preserves shape and size under an ordered matching

Two triangles are congruent when a distance-preserving movement maps one onto the other. In the plane, translations, rotations and reflections, or their compositions, supply such movements. A reflected triangle may have the opposite orientation and still be congruent. Under the matching, corresponding side lengths and interior angle measures agree. The two triangular regions consequently have the same area and their boundaries the same perimeter.

Write △ABC ≅ △PQR only when A corresponds to P, B to Q and C to R. That order gives AB ↔ PQ, BC ↔ QR and CA ↔ RP, together with ∠A ↔ ∠P, ∠B ↔ ∠Q and ∠C ↔ ∠R. If A instead matches R, B matches P and C matches Q, write △ABC ≅ △RPQ. Renaming the same triangle in a different order does not change its point set, but it changes the correspondence asserted by a two-triangle statement. Establish one consistent matching before collecting equalities.

Congruence tests let us establish the complete matching from a smaller collection of exact data. After a valid test proves congruence, the remaining corresponding parts are equal. Do not reverse that workflow by taking an unproved angle equality from the picture and using it to establish the congruence that was meant to justify it. Likewise, equal perimeter or equal area alone does not establish congruence: each is only one number and does not specify the side and angle arrangement.

Each congruence test specifies which data must match

For two existing nondegenerate Euclidean triangles with a fixed vertex matching, the tests in the table are sufficient for congruence. SSS uses all three side pairs. SAS uses two side pairs and their included angle: if the sides are AB and AC, the required angle is ∠BAC at their shared endpoint A. The corresponding angle in the second triangle must lie between the corresponding two sides. A side length is positive, and an included interior measure lies strictly between 0° and 180°.

ASA uses two angle pairs and the side between those angle vertices; AAS uses two angle pairs and a specified corresponding side that is not between them. Two matching angle measures force the third pair to agree by the 180° sum, so the nonincluded-side version can be reduced to ASA by using the appropriate known angles. RHS, also called hypotenuse–leg or HL, requires both triangles to be right triangles, equal hypotenuses and an equal corresponding leg. The hypotenuse must be identified opposite the given right angle. We use these established Euclidean criteria here; a full development of every criterion belongs to a later axiomatic treatment.

Congruence now explains the equal-side relationship used in classification. If AB = AC in triangle ABC, compare △ABC with △ACB: all three corresponding sides agree, so SSS makes them congruent and gives ∠ABC = ∠ACB. Conversely, if those two angles agree, compare the same ordered triangles. Their angles at B and C match in reverse order and the included side BC equals CB, so ASA gives AB = AC. This is a general argument with the hypotheses stated, rather than an appeal to an apparently symmetric sketch.

Three equal angles and general SSA do not guarantee congruence

AAA means three corresponding angle pairs agree. It fixes a triangle's shape, but supplies no length to fix its size. An equilateral triangle of side 2 cm and another of side 5 cm each have angles 60°, 60° and 60°, yet they are not congruent because their side lengths differ. Two corresponding angle pairs are already enough to force the third pair equal; they will provide a similarity test below. They become a congruence test only when an appropriate corresponding side length is fixed as well.

On a first reading, focus on why AAA does not fix size and why SAS needs the included angle. The coordinate check below is optional: return to it when you are comfortable with squared coordinate distances and fractions. You can use SSS, SAS, ASA, AAS and RHS with their stated conditions without this calculation.

SSA gives two side pairs and an angle that is not included between those sides. In general it can permit two noncongruent triangles. Here is an exact coordinate counterexample. Set A = (0, 0), B = (8, 0), C = (4, 3) and C′ = (156/25, 117/25), in one common coordinate length unit. Since C′ = (39/25)C with a positive multiplier, rays AC and AC′ coincide. Thus △ABC and △ABC′ have the same angle at A. Both have AB = 8, and BC = 5 while BC′² = (−44/25)² + (117/25)² = 25, so BC′ = 5.

However, AC = 5 and AC′ = (39/25) × 5 = 39/5. Both third vertices have positive second coordinates, so the two triangles are distinct valid noncollinear configurations, and their third side lengths differ. The known angle at A is between AB and AC, whereas the supplied side pair is AB and BC: it is not the included angle needed for SAS. This single verified counterexample disproves a universal SSA congruence rule. It does not say every SSA specification has two solutions. RHS is a valid additional right-triangle case, so it must retain its right-angle and hypotenuse conditions.

Similarity preserves shape with one positive length factor

Two triangles are similar when one can be mapped onto the other by uniform scaling combined with distance-preserving movements. Corresponding interior angles agree, and every corresponding length changes by the same positive factor. Translations, rotations and reflections are permitted, so a mirror image can be similar. The scaling must be uniform in every direction; stretching a drawing only horizontally generally changes its angles and is not a similarity transformation.

For △ABC ∼ △PQR, keep the order A ↔ P, B ↔ Q, C ↔ R and define k from the first triangle to the second by k = PQ/AB = QR/BC = RP/CA > 0. Each denominator is positive because the triangles are nondegenerate. If k > 1 the second triangle is larger; if 0 < k < 1 it is smaller; if k = 1 the triangles are congruent. In the reverse direction the factor is 1/k. A factor of zero collapses vertices and is excluded; a negative sign is not a length ratio under this convention. Reflection is handled by the permitted movement, while k remains positive.

Use matching sides, the same direction and common units in every ratio. A length of 2 m divided by 50 cm is 200 cm / 50 cm = 4, not 2/50. The resulting scale factor is dimensionless. Similarity tells us which quantities share that factor; it does not make every visible segment in a complex diagram corresponding. Identify each endpoint pair and, for internal heights or other constructed lengths, explain why the chosen construction is carried to its counterpart by the similarity.

Similarity tests combine matching shape with consistent ratios

AA is sufficient: two pairs of corresponding interior angles are equal. The third pair then agrees by the angle sum. To see why size is the only remaining freedom, choose one pair of corresponding sides and scale the first triangle by their positive length ratio. Scaling preserves its angles and makes that side equal to its counterpart. ASA or AAS then makes the scaled triangle congruent to the second one. Therefore the original triangles are similar. AAA is the same sufficient information with the third equal-angle pair repeated.

SSS similarity requires all three ratios of corresponding sides to equal one positive k. Scale the first triangle by k; now all three side pairs agree, so SSS congruence applies. SAS similarity requires two corresponding side ratios equal to one positive k and equal included angles. After scaling, the two side pairs and their included angles agree, so SAS congruence applies. These arguments explain the tests for every valid specification and clarify why similarity compares ratios while congruence compares lengths.

Two proportional side pairs without the included-angle equality are insufficient. For example, an equilateral triangle of sides 4, 4, 4 and an isosceles triangle of sides 8, 8, 12 each exist, but matching the first two sides gives factor 2 while the third gives factor 3. They are not similar. Likewise, replacing the included angle in SAS similarity by a general nonincluded angle does not give a universal test; the SSA counterexample above already has two side ratios equal to 1 and the nonincluded angle equal, but unequal third side ratios.

Perimeters scale by k and areas by k²

Let a, b and c be the positive side lengths of the first triangle, and let the second similar triangle have sides ka, kb and kc. Its perimeter is ka + kb + kc = k(a + b + c), so P₂ = kP₁. The equation applies to enlargement and reduction, and each perimeter keeps a length unit. It follows from adding all three scaled sides, not from measuring a selected pair of pictures.

For area, choose a base of length b and the positive perpendicular height h from the opposite vertex to the supporting line of that base. The foot of the perpendicular can lie beyond a side in an obtuse triangle; the height is still the perpendicular distance to the line. A similarity maps the supporting line and perpendicular to their counterparts, preserves right angles and multiplies both lengths by k. Thus S₂ = ½(kb)(kh) = k²(½bh) = k²S₁. Area uses square units. A factor 3 gives nine times the area; a factor 1/2 gives one quarter of the area.

For similar nondegenerate triangles, k = P₂/P₁ and k = √(S₂/S₁), using the positive square root. Equal areas imply congruence only with similarity already established: then k² = 1 and k > 0 give k = 1. Without similarity, equal areas can accompany different shapes. In an application such as a shadow comparison, AA needs geometric conditions: vertical objects, level ground and parallel sunlight rays give matching right angles and matching light-ray angles. Compare measurements made under those conditions, keep corresponding height-to-shadow ratios consistent, and distinguish estimated field measurements from exact mathematical givens.

Inspect exact side data and distinguish the chosen matching from the pair

The comparison mode starts with A = (0, 0), B = (4, 0) and C = (0, 3). Its six target scenes show an unchanged triangle, a translation, a reflection, a uniform scaling by 3/2, a right isosceles triangle with different proportions, and a collinear configuration. Choose one of the six possible vertex matchings. The model reports whether that chosen order meets SSS congruence or SSS similarity, and separately whether any vertex matching establishes the relationship for the pair. A wrong chosen order can fail even when the two triangles are congruent. All six bijections exhaust the possible vertex matchings for this particular pair.

Comparisons use exact squared side lengths, so a similarity requires every corresponding squared-length quotient to equal one positive number λ. The length factor is then k = √λ, since λ = k²; it is reported only after the nondegenerate triangles pass that similarity test. Reflection preserves congruence, while the 3/2 scaling gives λ = 9/4, perimeter factor 3/2 and area factor 9/4. Coordinate calculations determine lengths and unsigned areas; the equal-scale graph illustrates them without measuring pixels. These scenes inspect side-based tests and do not individually demonstrate all the angle-based criteria taught above.

The side-length mode offers eight ordered triplets, with a = BC, b = CA and c = AB, covering valid triangles, a positive equality boundary, an impossible specification and nonpositive lengths. It checks positivity and all three strict inequalities and shows the full permitted interval |b − c| < a < b + c for side a when b and c are fixed and positive. For valid data it places A = (0, 0), B = (c, 0) and C = (x, h), with x = (b² + c² − a²)/(2c), h² = b² − x² and h > 0. Positive equality data give a collinear zero-area boundary. Impossible or nonpositive data do not receive an invented triangle vertex. The selected cases support inspection; the definitions and general arguments establish the wider results. The next lesson develops the side-length relation for a right triangle.

Worked examples

Try it yourself

1. Do A = (0, 0), B = (2, 0), C = (5, 0) form a triangle under this lesson's convention? What if C = A? Can 0°, 90° and 90° be the interior measures of a nondegenerate triangle? Finally, classify a triangle with all three sides equal, including its angles and its status under our isosceles convention.

   Show the answer and reasoning

   The collinear triple does not form a nondegenerate triangle, and C = A fails distinctness. The measures 0°, 90° and 90° fail the strictly positive interior-angle condition. An equilateral triangle is also isosceles under our convention and is acute, with three 60° angles.

   All three points of the first triple lie on y = 0, so joining them encloses no triangular region. Repeating A also removes a distinct vertex. A sum of 180° alone is insufficient when one measure is zero. Equal sides give equal opposite angles; the sum then gives 3θ = 180° and θ = 60°. Isosceles means at least two equal sides here, so it includes the three-equal-side case.

2. The interior measures of a nondegenerate Euclidean triangle ABC are (3t + 2)°, (4t + 5)° and (5t + 17)° at A, B and C respectively. Find t, classify the triangle by angles, and find the ordinary exterior angle at C formed by extending BC through C. Verify the exterior-angle theorem.

   Show the answer and reasoning

   t = 13, and the interior measures are 41°, 57° and 82°, so the triangle is acute. The stated exterior measure at C is 98°, equal to 41° + 57°.

   The sum equation is 12t + 24 = 180, hence 12t = 156 and t = 13. Substitution gives 41, 57 and 82; all are positive, below 90 and sum to 180. The extension creates a linear pair with the 82° interior angle, so its ordinary exterior measure is 180° − 82° = 98°. The remote interior sum is also 41° + 57° = 98°. The position of the extension supplies the linear-pair condition.

3. Two sides of a proposed triangle measure 4 cm and 7 cm. Describe every real third-side length x cm that gives a nondegenerate triangle. List every integer x allowed, and classify the proposals x = 3, x = 11, x = 2 and x = 0.

   Show the answer and reasoning

   The full real range is 3 < x < 11. The integer values are 4, 5, 6, 7, 8, 9 and 10. Values 3 and 11 are degenerate collinear boundaries; 2 is impossible and 0 fails positive side length.

   The strict bound is |7 − 4| < x < 7 + 4, giving 3 < x < 11. At x = 3, 4 + 3 = 7; at x = 11, 4 + 7 = 11. These equalities cannot describe three noncollinear vertices. At x = 2 the longest side 7 exceeds 4 + 2, violating the distance condition. An integer answer must still lie strictly inside the real interval; rounding or including either endpoint would change the permitted set.

4. For two nondegenerate triangles with A ↔ D, B ↔ E and C ↔ F, AB = DE and AC = DF are given. Which equal-angle pair makes SAS applicable: ∠A = ∠D or ∠B = ∠E? If instead ∠A = ∠D and ∠B = ∠E are given, which side pair makes ASA applicable? Would AB = EF provide that corresponding side?

   Show the answer and reasoning

   ∠A = ∠D supplies the included angles for SAS; ∠B = ∠E with the same two side pairs is general SSA and is insufficient. With the two angle pairs, ASA uses AB = DE. The equality AB = EF is between noncorresponding sides and does not establish that criterion.

   AB and AC meet at A, and DE and DF meet at D, so their included angles are at those vertices. The angle at B is between BA and BC, while BC was not one of the two supplied sides. For the angle pairs at A, B and D, E, the joining sides are AB and DE. The fixed correspondence sends A to D and B to E, not to E and F. Even accurate equal lengths need the correct endpoint matching to serve as a congruence criterion.

5. Nondegenerate triangle ABC is right-angled at B and triangle PQR is right-angled at R. Their lengths satisfy AC = PQ = 13 cm and AB = PR = 5 cm. Give the ordered congruence statement and criterion, and identify the side corresponding to BC. Would the two side equalities alone justify RHS if the right-angle facts were omitted?

   Show the answer and reasoning

   △ABC ≅ △PRQ by RHS, with A ↔ P, B ↔ R and C ↔ Q. BC corresponds to RQ. The two side equalities alone do not justify RHS without the right-angle facts.

   The sides opposite the given right angles are AC and PQ, so the 13 cm pair is the hypotenuse pair. The 5 cm pair AB and PR consists of corresponding legs. That fixes B ↔ R and A ↔ P, leaving C ↔ Q. Both right angles, the hypotenuse equality and the leg equality meet the RHS conditions. Omitting the right-angle hypotheses removes both the right-triangle classification and the justification for assigning those sides the hypotenuse role.

6. Two nondegenerate triangles satisfy ∠A = ∠P and ∠B = ∠R. In ABC, AB = 6 cm, BC = 8 cm and CA = 7 cm; in PQR, PR = 9 cm. State the ordered similarity justified by AA, find the remaining corresponding lengths and the second perimeter, and decide whether reflecting the second triangle would invalidate the similarity.

   Show the answer and reasoning

   △ABC ∼ △PRQ. The factor from ABC to PRQ is 3/2, so RQ = 12 cm, QP = 21/2 cm and the second perimeter is 63/2 cm. A reflection preserves similarity.

   The two angle pairs give A ↔ P and B ↔ R; the 180° sum then gives C ↔ Q. Thus AB ↔ PR, BC ↔ RQ and CA ↔ QP. Use k = 9/6 = 3/2 consistently: (3/2) × 8 = 12 and (3/2) × 7 = 21/2. The first perimeter is 21 cm, so the second is (3/2) × 21 = 63/2 cm, also equal to 9 + 12 + 21/2. Reflection changes orientation while preserving angles and corresponding distance ratios.

7. Two nondegenerate triangles are known to be similar. The first has area 12 cm² and perimeter 16 cm; the second has area 27 cm². Find the factor from the first to the second and the second perimeter. Explain why the area ratio itself is not the length factor, and whether equal area would prove congruence without similarity.

   Show the answer and reasoning

   The area ratio is 9/4, so the positive length factor is k = 3/2 and the second perimeter is 24 cm. Area ratio is k². Equal area alone does not prove congruence when similarity is not established.

   Divide like area units: 27/12 = 9/4. Then k² = 9/4 and k > 0 give k = 3/2, so P₂ = (3/2) × 16 = 24 cm. There is no negative length factor under our convention. For a counterexample to equal-area congruence, the right triangles with vertices (0, 0), (4, 0), (0, 3) and (0, 0), (6, 0), (0, 2) both have area 6 square coordinate units. The first has side lengths 3, 4 and 5; the second includes a side of length 2 and has no matching 3–4–5 side set, so they are not congruent. Similarity supplies the additional shape condition needed to infer congruence from equal positive area.

What to carry forward

• A nondegenerate Euclidean triangle has three distinct noncollinear vertices, three positive side lengths and three interior measures strictly between 0° and 180°.
• Classify sides and angles separately. With our inclusive convention, equilateral triangles are also isosceles; equal sides correspond to equal opposite angles and conversely.
• The 180° interior-angle sum follows from Euclidean parallel-line properties. An ordinary exterior angle formed by extending a side equals the sum of the two remote interior angles.
• Positive a, b and c form a nondegenerate triangle exactly when all three triangle inequalities are strict. For fixed positive a and b, the third length satisfies |a − b| < c < a + b; equality is degenerate.
• Write triangle vertices in corresponding order. SSS, SAS with its included angle, ASA, AAS with a corresponding side, and RHS for right triangles are sufficient congruence tests; AAA and general SSA are not.
• Similarity allows reflection and one positive, uniform length factor k. AA, three equal corresponding side ratios, or two equal ratios with the included angle establish similarity; k = 1 gives congruence.
• Corresponding lengths and perimeters scale by k, while areas scale by k². Preserve ratio direction, convert units before dividing, and recover k from a positive area ratio using its positive square root.