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YEAR 10 · GEOMETRY AND PROOF · 20 MIN

Connected deductive arguments

Arrange given facts and deductions into a proof that establishes the stated conclusion.

In this lesson

Before you start
  • Congruence and similarity
  • Circle angle theorems
01 / LEARN

Understand the idea.

Plan backwards, write forwards

Plan a deductive argument by identifying the target and asking what would establish it. To prove equal angles, for example, look for congruent triangles or a suitable circle theorem and check the facts each requires.

Write the final proof forwards from the givens. Name the exact segments or angles, state each reason, and put triangle names in corresponding order so the reader can trace every deduction.

Every link needs support

Each link must use a given, a definition or something already proved. A common side supplies equal lengths; equal radii and a midpoint definition can supply other pairs for congruence.

Measurements and sketches help you search for a proof but cannot establish exact equality for every allowed figure. Several successful examples support a conjecture; one valid counterexample can disprove it.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A chord midpoint proof

O is the centre of a circle. M is the midpoint of a chord AB that is not a diameter. Prove OM⊥AB.

Worked solution

  1. 1

    Use the centre and midpoint information.

    O⁢A=O⁢B,A⁢M=B⁢M

    OA and OB are radii of one circle, and M is the midpoint of AB.

  2. 2

    Include the shared side.

    O⁢M=O⁢M

    The same segment belongs to both triangles.

  3. 3

    Prove the triangles congruent.

    △O⁢M⁢A≅△O⁢M⁢B

    All three corresponding side pairs are equal, so SSS applies.

  4. 4

    Relate the adjacent angles at M.

    ∠⁢O⁢M⁢A=∠⁢O⁢M⁢B,∠⁢O⁢M⁢A+∠⁢O⁢M⁢B=180∘

    Congruence gives equal angles; A, M and B are collinear, so those angles form a straight angle.

  5. 5

    Conclude perpendicularity.

    ∠⁢O⁢M⁢A=∠⁢O⁢M⁢B=90∘,O⁢M⟂A⁢B

    Two equal angles summing to 180 degrees are right angles. The non-diameter condition ensures O and M are distinct.

Answer

OM⊥AB

EXAMPLE 02

Cyclic angle proof

Explain why opposite interior angles of a cyclic quadrilateral sum to 180°.

Worked solution

  1. 1

    Name the arcs for an opposite angle pair.

    x+y=360∘

    Name the cyclic vertices A, B, C, D in order. Let x and y be the central angle measures of the arcs intercepted by angles A and C; together they cover the circle.

  2. 2

    Express the circumference angles.

    ∠⁢A=x2,∠⁢C=y2

    Each circumference angle is half the central angle on its intercepted arc.

  3. 3

    Add the opposite angles.

    ∠⁢A+∠⁢C=x+y2=360∘2=180∘

    The intercepted arcs together cover a full circle. The same reasoning applies to the other opposite pair.

Answer

Opposite interior angles are supplementary.

03 / YOUR TURN

Try it yourself.

Check your understanding

Why does one shared side supply equal lengths in a congruence proof?

Show answer & reasoning

A segment has the same length as itself.

A segment has one fixed length, so a side shared by two triangles gives an equality such as A⁢C=A⁢C. No extra measurement or assumption is needed.

Keep practising

6 worksheet questions · 16 marks

Read the questions online
  1. In triangle A⁢B⁢C,A⁢B=A⁢C. D is the midpoint of BC. Prove triangles ABD and ACD congruent.
    3 marks
  2. Use the previous conditions to prove A⁢D⟂B⁢C.
    3 marks
  3. A proof says “A⁢B=C⁢D because the picture looks symmetric.” Explain why this is insufficient.
    2 marks
  4. Prove that an angle in a semicircle is 90∘ using the centre-angle theorem.
    3 marks
  5. Triangles ABC and DEF are similar with correspondence A↔D,B↔E,C↔F. A⁢B=D⁢E. Prove they are congruent.
    3 marks
  6. A student verifies a circle claim for three drawings and calls it proved. What is still needed?
    2 marks
Read the worked solutions

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

  1. In triangle A⁢B⁢C,A⁢B=A⁢C. D is the midpoint of BC. Prove triangles ABD and ACD congruent.
    3 marks

    A⁢B=A⁢C; B⁢D=C⁢D; AD is shared.

    Apply SSS.

    SSS congruence.
  2. Use the previous conditions to prove A⁢D⟂B⁢C.
    3 marks

    SSS gives ∠⁢A⁢D⁢B=∠⁢A⁢D⁢C.

    They total 180∘, so each is 90∘.

    The two adjacent angles at D are 90∘.
  3. A proof says “A⁢B=C⁢D because the picture looks symmetric.” Explain why this is insufficient.
    2 marks

    A sketch may not be to scale.

    Prove symmetry or use a given equality.

    Visual appearance is not a stated mathematical condition.
  4. Prove that an angle in a semicircle is 90∘ using the centre-angle theorem.
    3 marks

    A diameter subtends a 180∘ central angle.

    The circumference angle on the same arc is 90∘.

    The inscribed angle is half 180∘.
  5. Triangles ABC and DEF are similar with correspondence A↔D,B↔E,C↔F. A⁢B=D⁢E. Prove they are congruent.
    3 marks

    A⁢BD⁢E=1.

    All corresponding lengths therefore match, so SSS congruence follows.

    Their similarity scale factor is 1.
  6. A student verifies a circle claim for three drawings and calls it proved. What is still needed?
    2 marks

    Three cases do not cover every possible allowed configuration.

    A general argument from valid definitions and theorems.
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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum

Angle and similarity reasoning; congruent/similar triangles in proofs; chords, radii and tangents; angles at the centre and circumference; circle theorems and their applications; connected deductive arguments; construction/spatial algorithms; testing and justifying solutions.

Official curriculum source ↗

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