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Congruence and Similarity Notes — Sec 3 E-Math

Two figures are congruent if they have identical shape and identical size. They are similar if they have the same shape but not necessarily the same size: every length is multiplied by the same scale factor k, and every angle stays unchanged. Congruence is the special case of similarity with k = 1.

Almost every O-Level geometry question rests on these two ideas, from proving triangles congruent, to reading map scales, to comparing the areas and volumes of models. This topic covers the congruence and similarity tests, the exam proof layout, enlargement and scale drawings, perpendicular and angle bisector constructions, and the ratio of areas and volumes of similar figures.

What these notes cover

  • •Congruent and similar figures
  • •Reading the vertex order in a statement
  • •Conditions for similar polygons
  • •The four congruence tests
  • •The three similarity tests
  • •Exam-standard proof layout
  • •Nested and overlapping triangles
  • •Enlargement and scale factor
  • •Scale drawings and map scales
  • •Perpendicular bisectors and angle bisectors
  • •Ratio of areas and volumes of similar figures
  • •Common mistakes and exam technique

Key definitions

Congruent figures
Figures with identical shape and identical size. One can be placed exactly on top of the other. Translations, rotations and reflections all produce congruent copies. Symbol: ≅.
Similar figures
Figures with exactly the same shape but not necessarily the same size. All corresponding angles are equal and all corresponding sides are in the same ratio k. Symbol: ∼.
Scale factor (k)
The number that every length is multiplied by in an enlargement, measured from a fixed centre of enlargement O. If A maps to A′, then OA′ = k × OA.
Perpendicular bisector of AB
The line that cuts AB at its midpoint at 90°. It is the set of all points P with PA = PB. The property works in both directions.
Angle bisector
The ray that splits an angle into two equal angles. It is the set of all points equidistant from the two arms, where distance means the perpendicular distance to each arm.

Key formulas

Corresponding sides of similar triangles

ABPQ=BCQR=ACPR=k\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k

Applies when ΔABC ∼ ΔPQR. Read the pairs from the letter order: A ↔ P, B ↔ Q, C ↔ R. When k = 1, the triangles are congruent.

Scale factor of an enlargement

k=image lengthobject lengthk = \frac{\text{image length}}{\text{object length}}

Multiply by k to go from object to image, and divide by k to go back. k > 1 gives a larger image, 0 < k < 1 gives a reduction, and k = 1 gives a congruent image.

Scale 1 : n conversions

actual length=n×plan length,plan length=actual lengthn\text{actual length} = n \times \text{plan length}, \quad \text{plan length} = \frac{\text{actual length}}{n}

Both lengths must be in the same unit. On a 1 : 20 000 map, 1 cm represents 20 000 cm = 200 m = 0.2 km.

Ratio of areas of similar figures

A1A2=(l1l2)2=k2\frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 = k^2

l₁ and l₂ can be any pair of corresponding lengths: heights, radii, diagonals or perimeters.

Ratio of volumes of similar solids

V1V2=(l1l2)3=k3\frac{V_1}{V_2} = \left(\frac{l_1}{l_2}\right)^3 = k^3

Given an area ratio, square-root it to get k first, then cube k. Never go straight from an area ratio to a volume ratio.

Ratio of masses of similar solids

m1m2=(l1l2)3\frac{m_1}{m_2} = \left(\frac{l_1}{l_2}\right)^3

Only applies when both solids are made of the same material, because mass is then proportional to volume.

Worked example

In triangle ABC, the line DE is parallel to BC, with D on AB and E on AC, and AD : DB = 2 : 1. The area of triangle ADE is 20 cm². Find the area of the trapezium DBCE.

  1. 1Since DE ∥ BC, triangles ADE and ABC are similar (AA).
  2. 2AD : DB = 2 : 1 compares AD with only part of the side. The full side is AB = AD + DB, so AD : AB = 2 : 3.
  3. 3Square the length ratio for areas: area of ΔADE ÷ area of ΔABC = (2 ÷ 3)² = 4 ÷ 9.
  4. 4Area of ΔABC = 20 × 9 ÷ 4 = 45 cm².
  5. 5Area of trapezium DBCE = 45 − 20 = 25 cm².

Watch out: The trapezium is not similar to anything, so no ratio applies to it directly. Find it as big triangle minus small triangle. Using AD : DB = 2 : 1 as the length ratio is the classic slip.

Common mistakes

Treating AAA as a congruence test

Wrong: All three pairs of angles are equal, so ΔABC ≅ ΔPQR (AAA).

Right: Equal angles fix the shape but not the size, so AA or AAA proves similarity only. Every congruence test (SSS, SAS, ASA/AAS, RHS) needs at least one pair of equal sides.

Using k instead of k² for areas

Wrong: The sides are in the ratio 2 : 3, so the areas are in the ratio 2 : 3.

Right: Areas scale by k², so the area ratio is 4 : 9. In reverse, an area ratio of 25 : 49 gives the length ratio k = √(25 ÷ 49) = 5 ÷ 7.

Wrong correspondence in overlapping triangles

Wrong: For ΔPQR ∼ ΔPAB, writing the ratio PQ ÷ AB.

Right: The correspondence is P ↔ P, Q ↔ A, R ↔ B, so PQ ÷ PA = PR ÷ PB = QR ÷ AB. Before writing any ratio, redraw the two triangles side by side, the same way up.

Questions students ask

What is the difference between congruent and similar figures?

Congruent figures have identical shape and size, so all corresponding sides and angles are equal. Similar figures have the same shape only: corresponding angles are equal and corresponding sides are in the same ratio k. Congruence is similarity with scale factor k = 1.

What are the congruence tests for triangles?

There are four tests: SSS, SAS, ASA/AAS and RHS. In SAS, the equal angle must be the included angle, between the two equal sides. SSA is not a test because two different triangles can fit the same information, and AAA proves similarity, not congruence.

How do you prove two triangles are similar?

Use one of three tests: AA (two pairs of equal angles), three equal ratios of corresponding sides, or two equal ratios of sides with equal included angles. For polygons other than triangles, you must check both conditions: all corresponding angles are equal and all side ratios are equal.

How do you find the volume ratio from the area ratio of similar solids?

Work through the length ratio k. Square-root the area ratio to get k, then cube it. For example, an area ratio of 4 : 9 gives k = 2 ÷ 3 and a volume ratio of 8 : 27.

In the full PDF

  • •Labelled diagrams of congruent and similar figures, nested and crossed triangles, and enlargement from a centre
  • •Step-by-step compass constructions for the perpendicular bisector and the angle bisector, with the SSS proofs of why they work
  • •Further worked examples: exam-layout congruence proofs on a kite and a rectangle, scale drawings and map scales, and the lamp-post shadow problem
  • •More area, volume and mass ratio examples using photo frames, jugs, ornaments and cones
  • •The remaining common slips and the full exam technique checklist
  • •A summary for each section and a one-page summary of the whole chapter