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YEAR 11 · PROOF AND CIRCLE GEOMETRY · 22 MIN

Congruence and similarity in geometric proofs

Build geometric proofs by establishing congruence or similarity before claiming equal or proportional lengths.

In this lesson

Before you start
  • Triangle tests, angle relationships and circle theorems
01 / LEARN

Understand the idea.

Match triangles before matching lengths

Congruent triangles have matching lengths and angles; similar triangles have matching angles and proportional lengths. Use a valid test before taking corresponding parts: SSS, SAS, ASA/AAS or RHS for congruence, and AA or proportional-side tests for similarity.

Write triangle names in corresponding order. From △A⁢B⁢C∼△D⁢E⁢F, match A to D, B to E and C to F. Keep every ratio in the same direction; three equal angles establish shape but do not fix size.

Turn the triangle fact into the target result

Build a geometric proof in four stages: name the triangles, justify enough equal sides or angles, state a congruence or similarity test, then extract the target fact. Equal radii, shared sides, vertical angles and angles on one chord often provide the links.

For a proportional equation, check which endpoints correspond before selecting lengths. Do not assume a right angle or midpoint from appearance. State the given facts and derived relations so that the argument applies to every non-degenerate configuration meeting the hypotheses.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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.

Worked solution

  1. 1

    List the three matched side equalities.

    A⁢B=A⁢C,B⁢D=D⁢C,A⁢D=A⁢D

    The first equality is given, the second follows from D being the midpoint, and AD is shared.

  2. 2

    Establish congruence and equal angles.

    △A⁢D⁢B≅△A⁢D⁢C⇒∠⁢A⁢D⁢B=∠⁢A⁢D⁢C

    SSS proves the triangles congruent, so corresponding angles are equal.

  3. 3

    Use the straight line through B, D and C.

    ∠⁢A⁢D⁢B+∠⁢A⁢D⁢C=180∘∠⁢A⁢D⁢B=∠⁢A⁢D⁢C=90∘

    Equal adjacent angles on a straight line must both be right angles.

  4. 4

    State the perpendicularity.

    A⁢D⟂B⁢C

    AD meets BC at a right angle at D.

Answer

AD is perpendicular to BC.

EXAMPLE 02

An altitude in a right triangle

A right triangle ABC has right angle A. The altitude AH meets hypotenuse BC at H. Prove AH2=B⁢H×H⁢C.

Worked solution

  1. 1

    Identify the right-angle pair.

    ∠⁢A⁢H⁢B=∠⁢C⁢H⁢A=90∘

    AH is an altitude to BC, and B, H and C are collinear.

  2. 2

    Match another angle pair.

    ∠⁢A⁢B⁢H=90∘−∠⁢A⁢C⁢B=∠⁢C⁢A⁢H

    Both angles are complementary to angle C: one in triangle ABC and the other in triangle CAH.

  3. 3

    Establish similarity in matching order.

    △A⁢B⁢H∼△C⁢A⁢H

    AA gives the correspondence A to C, B to A and H to H.

  4. 4

    Compare corresponding sides and multiply.

    B⁢HA⁢H=A⁢HC⁢HAH2=B⁢H×C⁢H

    The positive altitude permits cross-multiplication; CH and HC denote the same segment.

Answer

AH2=B⁢H×H⁢C.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

9:4.

The large-to-small length ratio is 6:4=3:2. Areas scale as the square of the length factor, so the area ratio is 9:4.

Keep practising

6 worksheet questions · 16 marks

Read the questions online
  1. Triangles A⁢B⁢C and D⁢E⁢F have A⁢B=D⁢E, B⁢C=E⁢F and A⁢C=D⁢F. State a valid conclusion with a test.
    2 marks
  2. A student proves all three corresponding angles equal and claims congruence. Give a concrete counterexample.
    2 marks
  3. Triangles A⁢B⁢C and D⁢E⁢F are similar in that order. A⁢B=6, B⁢C=9 and D⁢E=10. Find E⁢F.
    2 marks
  4. A⁢B and A⁢C are tangents from A to a circle with centre O. Prove A⁢O bisects angle B⁢A⁢C.
    4 marks
  5. An altitude to the hypotenuse divides it into lengths 4 and 9. Find the altitude and explain the theorem used.
    3 marks
  6. 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.

  1. Triangles A⁢B⁢C and D⁢E⁢F have A⁢B=D⁢E, B⁢C=E⁢F and A⁢C=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.
  2. A student proves all three corresponding angles equal and claims congruence. Give a concrete counterexample.
    2 marks

    All corresponding angles are 60∘, but their side lengths differ.

    Equilateral triangles with side lengths 2 and 5.
  3. Triangles A⁢B⁢C and D⁢E⁢F are similar in that order. A⁢B=6, B⁢C=9 and D⁢E=10. Find E⁢F.
    2 marks

    Scale from A⁢B⁢C to D⁢E⁢F is 106=53; E⁢F=9⁢(53)=15.

    15 units.
  4. A⁢B and A⁢C are tangents from A to a circle with centre O. Prove A⁢O bisects angle B⁢A⁢C.
    4 marks

    O⁢B=O⁢C as radii; A⁢O is a shared hypotenuse; tangent-radius angles B and C are right.

    Corresponding angles B⁢A⁢O and O⁢A⁢C are equal.

    Triangles A⁢B⁢O and A⁢C⁢O are congruent by RHS.
  5. An altitude to the hypotenuse divides it into lengths 4 and 9. Find the altitude and explain the theorem used.
    3 marks

    By similarity, h2=4×9=36. Take h=6>0.

    6 units.
  6. 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 h2=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 p⁢q.
Checking saved progress…
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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