Sec 2 Congruence and Similarity: Concepts & Worked Examples

Secondary 2 G3 Mathematics · Study guide

Match corresponding vertices, use equal angles and scale factors, and solve nested-triangle and algebraic-length problems with four worked examples and an explorer.

Concept → worked examples → practice16 min read · practice linked below

Same shape, same size, or both?

Congruent figures have the same shape and size. They can be moved, turned or reflected to fit exactly on top of each other. Similar figures have the same shape: corresponding angles are equal and every corresponding length changes by the same positive scale factor.

Congruent figures are similar with scale factor 1. Similar figures need not be congruent. A photograph enlarged uniformly keeps its shape; stretching only its width changes its shape.

Order matters: $\triangle ABC\sim\triangle PQR$ means $A\leftrightarrow P$ , $B\leftrightarrow Q$ , $C\leftrightarrow R$ . Therefore AB matches PQ, BC matches QR, and AC matches PR. Follow these matches even when one triangle is rotated or reflected.

Name corresponding sides and angles

Reflection changes the orientation, but it preserves lengths, angles and the order of correspondence.

Worked example 1 — read a congruence statement

It is given that $\triangle ABC\cong\triangle PQR$ , with AB = 6 cm, AC = 8 cm and BC = 10 cm as shown. Find PQ, PR and QR, and name the angle equal to $\angle ACB$ .

Write the matches from the order: A ↔ P, B ↔ Q, C ↔ R.
AB matches PQ, so PQ = 6 cm. AC matches PR, so PR = 8 cm. BC matches QR, so QR = 10 cm.
Replace each letter in ACB with its match: P R Q. Thus $\angle ACB=\angle PRQ$ .

Check: the middle letter identifies the vertex. Angle ACB is at C, so the corresponding angle must be at R. Matching the leftmost corners by appearance would give the wrong result after reflection.

Use equal angles before doing arithmetic

Worked example 2 — congruence and a straight line

Triangles ABC and DEF are congruent in that order. In ABC, $\angle A=48^\circ$ and $\angle B=67^\circ$ . The side EF is extended beyond F to G. Find $\angle DFG$ .

Angles in triangle ABC sum to $180^\circ$ , so $\angle C=180^\circ-48^\circ-67^\circ=65^\circ$ .
C corresponds to F, so $\angle DFE=65^\circ$ , using corresponding angles of congruent triangles.
E, F and G lie on a straight line. The adjacent interior and exterior angles sum to $180^\circ$ , giving $\angle DFG=180^\circ-65^\circ=\mathbf{115^\circ}$ .

Check: the exterior angle also equals the sum of the two opposite interior angles: $48^\circ+67^\circ=115^\circ$ . State the angle reason at each step; the picture alone is not a reason.

Use one consistent scale factor

Choose a direction before writing a ratio. From a smaller triangle to a larger triangle, divide a larger side by its corresponding smaller side:

$k=\frac{\text{larger length}}{\text{smaller length}}.$

Every side and the perimeter multiply by k. An enlargement has $k>1$ ; a reduction has $0<k<1$ .

Worked example 3 — nested similar triangles and a missing length

In the diagram, D lies on AB, E lies on AC, and DE is parallel to BC. AD = 4 cm, DB = 6 cm, AE = 6 cm, EC = 9 cm and BC = 15 cm. Explain why ADE and ABC are similar and find DE.

The triangles share angle A. Parallel lines give equal corresponding angles at D and B, and at E and C. The sketch is not to scale.

Angle A is shared. Since DE ∥ BC, $\angle ADE=\angle ABC$ and $\angle AED=\angle ACB$ . Thus $\triangle ADE\sim\triangle ABC$ .
Use whole corresponding sides: $AB=4+6=10$ cm and $AC=6+9=15$ cm.
From ADE to ABC, $k=AB/AD=10/4=2.5$ . Therefore $BC=2.5DE$ , giving $DE=15/2.5=\mathbf{6\text{ cm}}$ .

Check: $AC/AE=15/6=2.5$ , the same factor. Do not use DB instead of AB: DB is only the remaining part of a side.

Turn a side ratio into an equation

Worked example 4 — similarity with algebraic lengths

It is given that $\triangle ABC\sim\triangle PQR$ . AB = 5 cm, BC = 7 cm, AC = 8 cm, PQ = $(x+1)$ cm and QR = $(2x-1)$ cm. Find x, PR and the perimeter of PQR.

AB matches PQ and BC matches QR. Use the same direction in both ratios: $\frac{x+1}{5}=\frac{2x-1}{7}$ .
Multiply by 35: $7(x+1)=5(2x-1)$ . Expanding gives $7x+7=10x-5$ , so $12=3x$ and $x=4$ .
Then PQ = 5 cm and QR = 7 cm, so the scale factor is $k=5/5=1$ . Hence PR = 8 cm, and the perimeter is $5+7+8=\mathbf{20\text{ cm}}$ .

Interpretation: these similar triangles are also congruent because k = 1. Similarity does not require a different size.

Check: both ratios equal 1, all side lengths are positive, and $5+7>8$ , so the lengths can form a triangle. The expressions require $x+1>0$ and $2x-1>0$ ; x = 4 satisfies both.

Explore enlargement, reduction and reflection

Keep the correspondence as the size changes

Triangle ABC has side lengths 3, 4 and 5 units. Change the scale factor to create PQR. Reflecting PQR changes its orientation but keeps each vertex matched to the same letter.

Lengths are in units. A ↔ P, B ↔ Q and C ↔ R for every setting.

Enlargement by factor 2. PQ = 6, PR = 8 and QR = 10 units. Perimeter changes from 12 to 24 units. Corresponding angles remain equal.

Looking ahead: tests and higher-dimensional scale factors

The central skills here are matching vertices, equal angles and one consistent length ratio. Two useful later connections build on them:

How can triangle congruence or similarity be proved?

Congruence tests include SSS (three corresponding sides), SAS (two sides and the included angle), ASA or AAS (two angles and a corresponding side), and RHS (a right angle, hypotenuse and one other corresponding side). AAA gives the shape but does not fix the size. Two sides and a non-included angle do not in general establish congruence.

Similarity tests include AA (two equal corresponding angles), SSS (three proportional pairs of sides), and SAS (two proportional pairs with an equal included angle). State the correspondence alongside the test.

How do areas and volumes scale?

For similar figures with positive length scale factor k, areas scale by $k^2$ . For similar solids, volumes scale by $k^3$ . At k = 2, lengths double, areas become four times as large and volumes become eight times as large. Reverse these with a square root or cube root: an area ratio of 9 gives a length ratio of 3; a volume ratio of 27 also gives a length ratio of 3.

These statements require similarity throughout. Doubling only a rectangle’s width doubles its area but changes its shape. If similar solids have the same material density, their masses also scale with their volumes.

Common mistakes and checks

Matching by position: write the vertex correspondence first; rotated or reflected diagrams can mislead the eye.
Reversing just one ratio: use new/old on both sides, or old/new on both sides.
Matching a part to a whole: for nested triangles, mark the full corresponding sides before adding lengths.
Assuming similarity from appearance: use given similarity or explain the angle and ratio conditions.
Scaling angles: corresponding angles stay equal; lengths and perimeters change by k.
Missing a geometric restriction: reject zero or negative side lengths and check a proposed side factor against another corresponding pair.

Practise with our topic questions

Use the Sec 2 congruence and similarity question bank to apply these methods. Attempt each question before opening its worked solution. Compare your reasoning as well as the final answer.