Congruence and similarity in geometric proofs | Year 11

01 / LEARN

The idea, step by step

Match triangles before matching lengths

Congruent triangles have equal corresponding lengths and angles. Valid tests include SSS, SAS, ASA/AAS and RHS for right triangles. The included angle matters in SAS. Three equal angles establish similarity, not congruence, because they do not fix scale.

Write the vertices in matching order. If $△ A B C \sim △ D E F$ , A matches D, B matches E and C matches F. Then $A B / D E = B C / E F = A C / D F$ . Reversing just one ratio destroys the correspondence.

Turn the triangle fact into the target result

A proof has a sequence: identify triangles, state enough equal angles/sides with reasons, establish a test, then use corresponding parts. In circle proofs, equal radii, vertical angles and angles on one chord often supply the needed facts.

For similarity, say why the triangles are similar before writing a proportional equation. Check that a length equation uses the correct corresponding sides and that both triangles are non-degenerate. The aim is to explain why the result holds for every permitted diagram.

A common trap

Treating AAA as a congruence test.

Instead: AAA fixes shape but not size; use similarity unless a suitable length condition fixes scale.

02 / WORKED EXAMPLES

See the thinking

Example 1

A median in an isosceles triangle

In triangle ABC, AB $= A C$ and D is the midpoint of BC. Prove AD is perpendicular to BC.

Triangles ADB and ADC have AB $= A C$ , BD $= D C$ and the shared side AD.
They are congruent by SSS, so angles ADB and ADC are equal.
Those angles form a straight line and total $180^{\circ}$ ; each is $90^{\circ}$ .

AD is perpendicular to BC.

Example 2

An altitude in a right triangle

A right triangle ABC has right angle A. The altitude AH meets hypotenuse BC at H. Prove $A H^{2} = B H \times H C$ .

Triangles ABH and CAH are right-angled at H.
Their other acute angles correspond because each is complementary to angle C or angle B in ABC. Thus they are similar by AA.
With correspondence $A \leftrightarrow C$ , $B \leftrightarrow A$ , $H \leftrightarrow H$ , obtain $B H / A H = A H / C H$ .
Cross-multiply.

A H^{2} = B H \times H C

Check your understanding

Similar triangles have matching side lengths 4 and 6. What is their large-to-small area ratio?

Show answer

$9 : 4$ .

The length scale is $6 / 4 = 3 / 2$ , so the area scale is ${(3 / 2)}^{2} = 9 / 4$ .

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Triangles ABC and DEF have AB $= D E$ , BC $= E F$ and AC $= D F$ . State a valid conclusion with a test.
2 marks
A student proves all three corresponding angles equal and claims congruence. Give a concrete counterexample.
2 marks
Triangles ABC and DEF are similar in that order. AB $= 6$ , BC $= 9$ and DE $= 10$ . Find EF.
2 marks
AB and AC are tangents from A to a circle with centre O. Prove AO bisects angle BAC.
4 marks
An altitude to the hypotenuse divides it into lengths 4 and 9. Find the altitude and explain the theorem used.
3 marks
In the same altitude construction, the hypotenuse pieces are p and q. Explain why $p + q$ is not the altitude in general.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Triangles ABC and DEF have AB $= D E$ , BC $= E F$ and AC $= D F$ . State a valid conclusion with a test.
2 marks
All three corresponding sides are equal.
$△ A B C ≅ △ D E F$ by SSS.
A student proves all three corresponding angles equal and claims congruence. Give a concrete counterexample.
2 marks
All corresponding angles are $60^{\circ}$ , but their side lengths differ.
Equilateral triangles with side lengths 2 and 5.
Triangles ABC and DEF are similar in that order. AB $= 6$ , BC $= 9$ and DE $= 10$ . Find EF.
2 marks
Scale from ABC to DEF is $10 / 6 = 5 / 3$ ; $E F = 9 (5 / 3) = 15$ .
15 units.
AB and AC are tangents from A to a circle with centre O. Prove AO bisects angle BAC.
4 marks
OB $= O C$ as radii; AO is a shared hypotenuse; tangent-radius angles B and C are right.
Corresponding angles BAO and OAC are equal.
Triangles ABO and ACO are congruent by RHS.
An altitude to the hypotenuse divides it into lengths 4 and 9. Find the altitude and explain the theorem used.
3 marks
By similarity, $h^{2} = 4 \times 9 = 36$ . Take $h = 6 > 0$ .
6 units.
In the same altitude construction, the hypotenuse pieces are p and q. Explain why $p + q$ is not the altitude in general.
3 marks
The geometric-mean theorem gives $h^{2} = p q$ .
For example, p $= 4$ , q $= 9$ give hypotenuse 13 and altitude 6, not 13.
It is the full hypotenuse, whereas the altitude is $\sqrt{p q}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: geometric reasoning and real-number/proof material, regrouped into a school chapter; all exercises are original.

Official curriculum source ↗

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