Congruency and Similarity

Unit 15 · Compulsory Mathematics · 12 min read

Congruency and Similarity

What the syllabus asks for 4
  • congruent figures and congruent triangles
  • tests of congruence (SSS, SAS, ASA, RHS)
  • similar figures and similar triangles
  • ratio of areas of similar triangles

Two figures are congruent when they are exactly the same shape and the same size, so that one can be placed on the other and fit perfectly. They are similar when they are the same shape but not necessarily the same size, so that one is an enlargement of the other. Both ideas are tests: given a few measurements, they let you say that two triangles must match without measuring everything.

What this chapter asks of you

You should be able to state the five tests of congruence (SSS, SAS, ASA, AAS, RHS), the conditions for two triangles to be similar, use corresponding parts of congruent triangles in a proof, and use the ratio of areas and sides of similar triangles.

15.1 Congruent figures

Two figures are congruent if they have exactly the same shape and size. When triangles ABC and PQR are congruent we write

\triangle ABC\cong\triangle PQR

and the order of the letters matters: it tells you which vertex matches which. So A matches P, B matches Q and C matches R, and therefore AB=PQ, BC=QR, CA=RP, and the corresponding angles are equal.

Corresponding parts of congruent triangles are equal — usually written CPCTC, and used as the reason in every proof of this chapter.

The order of the letters is information

\triangle ABC\cong\triangle PQR says AB corresponds to PQ. If the congruence is really with \triangle QRP, then AB corresponds to QR and the statement is false as written. Examiners deduct for a wrong order even when the figure is right.

15.2 The tests of congruence

To prove two triangles congruent you need only three matched facts, and the three must be in one of these five patterns.

Test What must match Why it works
SSS all three sides only one triangle can be built from three given sides
SAS two sides and the included angle the angle fixes the shape between the sides
ASA two angles and the side between them the side and the two angles at its ends fix the triangle
AAS two angles and a side not between them the third angle is forced (angles add to 180°), so it becomes ASA
RHS a right angle, the hypotenuse and one other side in a right triangle the third side follows by Pythagoras

figure
Figure 1. The five tests of congruence. The thick blue sides and the red arcs are the parts that must be equal in the two triangles. In SAS the angle lies between the two sides; in ASA the side lies between the two angles; in AAS the side does not.

There is no SSA test

Two sides and a non-included angle do not fix a triangle: two different triangles can often be drawn with the same two sides and the same angle somewhere else. Congruence must be proved with SSS, SAS, ASA, AAS or RHS — never with "SSA", which does not exist.

A congruence proof

Problem. In the figure, AB=AC and D is the midpoint of BC. Prove that \triangle ABD\cong\triangle ACD, and hence that AD is perpendicular to BC.

Solution.

Given. AB=AC and BD=DC (D is the midpoint).

To prove. \triangle ABD\cong\triangle ACD and \angle ADB=90°.

Proof.

Statement Reason
AB=AC given
BD=DC D is the midpoint of BC
AD=AD common side
\triangle ABD\cong\triangle ACD SSS
\angle ADB=\angle ADC corresponding parts of congruent triangles

Now \angle ADB+\angle ADC=180°, because BDC is a straight line. Equal angles adding to 180° are each 90°, so \angle ADB=90° and therefore AD\perp BC. ∎

What it shows. In an isosceles triangle the line from the apex to the midpoint of the base is perpendicular to the base — and it also bisects the apex angle, since it makes the two triangles congruent.

A proof using ASA

Problem. Two triangles have \angle B=\angle Q, \angle C=\angle R and BC=QR. Prove the triangles congruent.

Solution.

Proof. In a triangle the angles add to 180°, so

\angle A=180°-(\angle B+\angle C),\qquad \angle P=180°-(\angle Q+\angle R)

Since \angle B=\angle Q and \angle C=\angle R, it follows that \angle A=\angle P. We now have \angle B=\angle Q, BC=QR (the side between them) and \angle C=\angle R. Therefore

\triangle ABC\cong\triangle QRP\quad\text{(ASA)}

Note. We never measured \angle A; it was forced by the angle sum. This is why ASA works with only two angles.

Check. The equal side must be the one lying between the two equal angles. Here BC lies between angles B and C ✓.

Finding an unknown from congruent triangles

Problem. \triangle ABC\cong\triangle XYZ with AB=(1.5x-1.5) cm and XY=(x+0.5) cm. Find x and the length of AB.

Solution. Corresponding sides of congruent triangles are equal, so 1.5x-1.5=x+0.5 → 0.5x=2 → x=4. AB=1.5\times4-1.5=4.5 cm (and XY=4+0.5=4.5 cm ✓).

A proof with a perpendicular bisector

Problem. In the figure, D is the midpoint of AB and CD\perp AB. Prove that \triangle ACD\cong\triangle BCD.

Solution.

Statement Reason
\angle ADC=\angle BDC=90° CD\perp AB (given)
AD=BD D is the midpoint of AB
CD=CD common side
\triangle ACD\cong\triangle BCD SAS (side AD, included right angle, side CD)

So AC=BC (corresponding sides of congruent triangles).

15.3 Similar figures

Two figures are similar if they have the same shape, even if they differ in size. One is an enlargement of the other by a fixed scale factor.

For two similar triangles:

  • the three angles of one equal the three angles of the other, and
  • the three pairs of corresponding sides are in the same ratio.

\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}=k

where k is the scale factor.

Conditions for two triangles to be similar.

Condition Meaning
AA (or AAA) two angles of one equal two angles of the other; the third follows
SSS similarity the three pairs of sides are in the same ratio
SAS similarity two pairs of sides are in the same ratio and the included angles are equal

figure
Figure 2. Similar triangles: the same angles, and every side of the larger is 2 times the matching side of the smaller (3 → 6, 2 → 4). The ratio of corresponding sides is 1 : 2.

Similar triangles and a missing side

Problem. Triangles ABC and PQR are similar, with AB=6 cm, BC=9 cm and PQ=4 cm. Find QR.

Solution. Corresponding sides are in the same ratio:

\frac{AB}{PQ}=\frac{BC}{QR}\;\Longrightarrow\;\frac64=\frac{9}{QR}

Cross-multiplying: 6\times QR=36, so QR=6 cm.

Answer. 6 cm.

Check. The scale factor is \frac{6}{4}=1.5, and 9\div1.5=6 ✓ — the same answer from the other direction.

The area ratio

Problem. Two similar triangles have corresponding sides in the ratio 2:5. Find the ratio of their areas.

Solution. For similar figures, the ratio of areas is the square of the ratio of corresponding sides:

\frac{\text{area}_1}{\text{area}_2}=\left(\frac25\right)^2=\frac{4}{25}

Answer. 4:25.

Note. The enlarged triangle has sides 2.5 times as long but an area 6.25 times as large. This is why a map of twice the scale needs four times the paper, and why a model house has far less surface than the house it copies.

Similar is not congruent

Two triangles with sides 3, 4, 5 and 6, 8, 10 are similar but not congruent. Every pair of congruent triangles is similar, but the reverse is false. Read the question: "find the length" usually means similar, "prove equal" means congruent.

Using the angle sum to prove similarity

Problem. In triangles ABC and DEF, \angle A=50°, \angle B=70°, \angle D=50° and \angle E=70°. Prove the triangles similar and write the ratio of corresponding sides.

Solution. Two angles of the first equal two angles of the second: \angle A=\angle D and \angle B=\angle E. So

\triangle ABC\sim\triangle DEF\quad\text{(AA)}

The third angles are equal too: \angle C=\angle F=180-120=60°. The ratio of corresponding sides is

\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}

Check. The matched order is A with D, B with E, C with F — because those are the equal angles. Writing the correspondence in the wrong order gives wrong ratios.

Chapter summary

  • Congruent figures have the same shape and size; similar figures have the same shape at different sizes.
  • The tests of congruence are SSS, SAS, ASA, AAS and RHS; there is no SSA test.
  • The order of the letters in \triangle ABC\cong\triangle PQR gives the correspondence, and corresponding parts of congruent triangles are equal.
  • Two triangles are similar if two angles match (AA), or if the three pairs of sides are in the same ratio (SSS similarity), or if two pairs of sides are in the same ratio with equal included angles (SAS similarity).
  • In similar triangles, corresponding sides are in the same ratio k and areas are in the ratio k^2.

Practice questions

Group A — Objective (1 mark each)

  1. Two triangles are congruent if [1]atwo angles are equalball three sides are equalcthey have the same areadthey look alike
  2. Which of these is not a test of congruence? [1]aSSSbSAScSSAdRHS
  3. In SAS the given angle must be [1]aany anglebthe largest anglecbetween the two given sidesda right angle
  4. If two triangles are similar with sides in the ratio 3:4, the ratio of their areas is [1]a3:4b9:16c6:8d4:3
  5. Two triangles with all three angles equal are [1]aalways congruentbalways similarcnever similardequal in area
  6. Congruent figures have [1]athe same shape onlybthe same size onlycthe same shape and sizedthe same area only
Answers to Group A

1. (b) — that is the SSS test.

2. (c) — SSA does not determine a triangle uniquely.

3. (c) — the angle must be the included one.

4. (b) — the area ratio is the square of the side ratio.

5. (b) — equal angles give the same shape, hence similarity; sizes may differ.

6. (c) — same shape and same size.

Group B — Short answer (2 marks each)

  1. State the five tests of congruence, naming the parts each uses. [2]
  2. \triangle ABC\cong\triangle PQR with AB=7 cm, BC=9 cm and \angle B=55°. Write down the other measurements you can. [2]
  3. Two similar triangles have corresponding sides 8 cm and 12 cm. Find the scale factor and the ratio of their areas. [2]
  4. In two triangles \angle A=\angle D=65° and \angle B=\angle E=45°. Prove the triangles similar and state the third pair of equal angles. [2]
Answers to Group B

1. SSS — all three sides. SAS — two sides and the included angle. ASA — two angles and the side between them. AAS — two angles and a side not between them. RHS — the right angle, the hypotenuse and one other side.

2. Because the order of the letters gives the correspondence, P matches A, Q matches B and R matches C. So PQ=AB=7 cm, QR=BC=9 cm, and \angle Q=\angle B=55°. We also know RP=CA, \angle P=\angle A and \angle R=\angle C (with \angle P+\angle R=180°-55°=125°), although their values are not given.

3. Scale factor =\frac{12}{8}=1.5; ratio of areas =\left(\frac{8}{12}\right)^2=\frac{64}{144}=\frac49, that is 4:9.

4. Two angles of one equal two angles of the other, so the triangles are similar by AA. The third pair is \angle C=\angle F=180-(65+45)=70°.

Group C — Long answer (4 marks each)

  1. In a quadrilateral ABCD, the diagonal AC bisects both \angle A and \angle C. Prove that \triangle ABC\cong\triangle ADC and hence that AB=AD and CB=CD. [4]
  2. A vertical pole 2 m high casts a shadow 3 m long at the same time as a tower casts a shadow 21 m long. Using similar triangles, find the height of the tower, and state which angles you used. [4]
Answers to Group C

1. Given. \angle BAC=\angle DAC and \angle BCA=\angle DCA (the diagonal bisects both angles). To prove. \triangle ABC\cong\triangle ADC, so AB=AD and CB=CD.

Statement Reason
\angle BAC=\angle DAC given, AC bisects \angle A
AC=AC common side
\angle BCA=\angle DCA given, AC bisects \angle C
\triangle ABC\cong\triangle ADC ASA
AB=AD and CB=CD corresponding parts of congruent triangles

The side AC used in the ASA test is the one lying between the two pairs of equal angles ✓. 2. Let the sun's rays make the same angle with the ground for both objects, so \angle(\text{ray, ground}) is equal; and both the pole and the tower stand vertically, so each makes a right angle with the ground. The two triangles therefore have two equal angles and are similar by AA.

\frac{\text{height of tower}}{\text{height of pole}} =\frac{\text{shadow of tower}}{\text{shadow of pole}}

\frac{h}{2}=\frac{21}{3}=7\;\Longrightarrow\;h=14\ \text{m}

Answer. The tower is 14 m high. Angles used. The right angle at the base of each object and the angle between the sun's ray and the ground — the AA condition. Check. The scale factor of the shadows is 21\div3=7, so the heights must also be in the ratio 7:1, and 14=7\times2 ✓.

Objective practice

Choose the correct answer. [5×1=5]

1. Which test does not prove two triangles congruent? (a) SSS (b) SAS (c) AAA (d) ASA

2. In two similar triangles the corresponding sides are (a) equal (b) proportional (c) perpendicular (d) unrelated

3. Triangles with sides 5, 12, 13 cm and 10, 24, 26 cm are (a) congruent (b) similar only (c) neither (d) equal in area

4. The RHS test applies to (a) all triangles (b) right-angled triangles (c) isosceles triangles (d) equilateral triangles

5. If two figures are congruent then they are (a) also similar (b) never similar (c) similar only if squares (d) equal in perimeter but not area

Objective practice — answers

1. (c) — equal angles fix the shape, not the size.

2. (b) — proportional.

3. (b) — the ratio is 1:2, so similar but not congruent.

4. (b) — right angle, hypotenuse and one side.

5. (a) — congruent figures are similar with ratio 1.

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