Congruence and similarity of triangles

Learning outcome

Use valid triangle-matching criteria and scale corresponding lengths and areas correctly.

Concepts and assumptions

Congruent triangles coincide after moving, turning or reflecting; shape and size both match. In triangle ABC ≅ triangle DEF, the order fixes A↔D, B↔E, C↔F. Thus AB↔DE, BC↔EF and AC↔DF.

Congruence tests require equal corresponding measurements: SSS (three sides), SAS (two sides and included angle), ASA (two angles and included side), AAS (two angles and a nonincluded side), and RHS (right angles, hypotenuses and a leg). Two angles determine the third, explaining AAS.

The included angle lies between the two known sides. SSA—two sides and a nonincluded angle—is not a general congruence test. The third vertex can sometimes occupy two positions preserving those data. RHS is a special right-triangle case.

Similar triangles have equal corresponding angles and proportional corresponding sides. Tests are AA, three proportional sides (SSS), and two proportional sides with equal included angles (SAS). AAA gives similarity, not necessarily congruence.

Scale factor k is target length divided by corresponding original length. Then k > 0; perimeters scale by k, but areas by k². Both base and perpendicular height scale by k, explaining the square. Do not apply this area rule without similarity.

Equal area alone proves neither congruence nor similarity. Right triangles with perpendicular legs 3, 8 cm and 4, 6 cm both have area 12 cm², but their leg pairs neither match nor have a common ratio.

Worked examples

Example 1 — Correspondence matters. In triangle ABC, AB = 5 cm, BC = 7 cm, AC = 8 cm. Triangle PQR has PR = 5 cm, PQ = 7 cm, QR = 8 cm. Match AB↔RP, BC↔PQ and AC↔RQ. SSS gives triangle ABC ≅ triangle RPQ, not triangle PQR.

Example 2 — Length and area scaling. Triangles ABC and DEF have ∠B = ∠E = 90° and ∠A = ∠D. Given AB = 6 cm, BC = 8 cm and DE = 9 cm, AA gives ABC ∼ DEF. Scale factor = 9/6 = 3/2, so EF = 8 × 3/2 = 12 cm. Original area = 6 × 8/2 = 24 cm²; target area = 24 × (3/2)² = 54 cm².

Example 3 — Reverse area scaling. Similar triangles have smaller:larger area ratio 9:25. Their length ratio is √9:√25 = 3:5. A 12 cm side in the smaller triangle therefore corresponds to 12 × 5/3 = 20 cm in the larger.

Common mistakes

Keep correspondence order throughout. Area ratios are squared length ratios, not ordinary length ratios.

Practice questions

1. Triangles ABC and DEF are right at A and D. BC = EF = 13 cm and AB = DE = 5 cm. State the congruence order and criterion.
2. PQR ∼ XYZ, with PQ = 10 cm, XY = 15 cm and QR = 14 cm. Find YZ.
3. Similar triangles have corresponding sides 5 cm and 8 cm. The smaller area is 25 cm². Find the larger area.
4. Two triangles each have sides 7 cm and 10 cm, with a 30° angle opposite the 7 cm side. Do these data guarantee congruence?

Worked solutions

1. Right angles correspond A↔D, hypotenuses BC↔EF and legs AB↔DE. Hence ABC ≅ DEF by RHS.
2. Correspondence gives QR↔YZ. Scale factor = 15/10 = 3/2, so YZ = 14 × 3/2 = 21 cm.
3. Area factor = (8/5)² = 64/25. Larger area = 25 × 64/25 = 64 cm².
4. No. This is SSA, with a nonincluded angle. Two different third-vertex positions are possible for these data, so congruence is not guaranteed.