Names you may know: triangle congruency; SSS, SAS, ASA and AAS; RHS or HL congruence; CPCTC.

Geometry path · G02

Before this lesson: Angle chasing, Geometric language and ratios

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You will learn: Match congruent triangles, choose SSS, SAS, ASA, AAS or RHS (HL), and write a proof using corresponding parts.

Before you start: Side and angle notation; angles in a triangle total 180° and angles on a straight line total 180°. Review angle chasing first if these facts are new.

Same shape, same size

Imagine cutting out a triangle and placing it on another. If it fits exactly after sliding, turning or flipping, the triangles are congruent. Every matching side and every matching angle is equal. Enlarging a triangle keeps its shape but changes its size, so an enlargement is generally similar, not congruent.

Congruence is also called congruency. This lesson is about triangles in plane geometry. For congruences such as 17 ≡ 2 (mod 5), use the remainders and modular arithmetic lesson.

The order of the letters matters

Write △ABC ≅ △PQR to say that A matches P, B matches Q and C matches R. The symbol △ means triangle; ≅ means congruent. The matching sides are AB = PQ, BC = QR and AC = PR. The angle ∠ABC is the angle at B: the middle letter names the vertex. Its matching angle is ∠PQR.

Matching vertices of congruent trianglesTriangle ABC has A at the top, B at the lower left and C at the lower right. Its reflected copy PQR has P at the top, Q at the lower right and R at the lower left. A matches P, B matches Q, C matches R.ABCPQR
The second triangle is a reflected copy. Match A ↔ P, B ↔ Q and C ↔ R; position on the page does not decide the correspondence.

Write the vertex matches before choosing a side or an angle. A turned or reflected triangle may look different on the page, but the correspondence stays the same.

Which facts are enough?

You do not have to check all six parts separately. The following tests prove congruence for non-degenerate triangles: three distinct vertices that do not lie on one line. In each test, the facts must match under the same vertex correspondence.

SSS — side, side, side
All three pairs of matching sides are equal. Once a base and the distances from its ends to the third vertex are fixed, the two possible positions on opposite sides of the base are reflections, so they give congruent triangles.
SAS — side, angle, side
Two pairs of sides and the angle between them are equal. Think of two rods with a fixed angle between them: their free ends determine the third side.
ASA — angle, side, angle
Two pairs of angles and the side joining those angle vertices are equal. The two rays from the ends of that fixed side meet at the third vertex.
AAS — angle, angle, side
Two pairs of angles and a matching side not between them are equal. The third pair of angles is also equal because a triangle’s angles total 180°. You can then apply ASA using the given side.
RHS — right angle, hypotenuse, side
Both triangles are right angled, their hypotenuses are equal, and one pair of corresponding legs is equal. A leg is a side next to the right angle; the hypotenuse is opposite it. This test is also called HL (hypotenuse–leg).

These are congruence tests, not a rule that any three equal measurements will do. For RHS, Pythagoras’ theorem explains why the remaining legs agree: their squares are each “hypotenuse squared minus known leg squared”. Since lengths are positive, those legs are equal too, giving SSS. You can use the RHS test here without carrying out that calculation.

Worked example 1: choosing the right test

Suppose AB = DE = 5, AC = DF = 7 and ∠BAC = ∠EDF = 60°. Can we prove △ABC ≅ △DEF?

Two sides and their included angleIn triangle ABC, AB is 5, AC is 7 and angle BAC is 60 degrees. The angle is between the two known sides.ABC5760°
The included angle is ∠BAC: its two arms are AB and AC.

The side pairs give the matches A ↔ D, B ↔ E and C ↔ F. The angle at A is between AB and AC; the angle at D is between DE and DF. Therefore the given facts fit SAS, and △ABC ≅ △DEF. We may now conclude BC = EF even though neither length was given.

Worked example 2: use a shared side

In triangle ABC, AB = AC and D is the midpoint of BC. Prove that AD bisects ∠BAC. A midpoint lies on the segment and divides it into two equal parts. An angle bisector divides an angle into two equal angles.

An isosceles triangle split at the midpoint of its baseTriangle ABC has A at the top, B and C on a horizontal base, and D its midpoint. AB and AC have one tick each; BD and DC have two ticks each. AD is drawn, with no right angle assumed.ABCD
Given AB = AC and BD = DC. The segment AD is shared. The drawing is not a reason to assume a right angle.
  1. Compare △ABD and △ACD. AB = AC is given.
  2. BD = CD because D is the midpoint of BC.
  3. AD = AD because the two triangles share this side. A common side counts as a pair of equal sides.
  4. The three pairs of sides are equal, so △ABD ≅ △ACD by SSS. The correspondence is A ↔ A, B ↔ C, D ↔ D.
  5. The matching angles ∠BAD and ∠CAD are equal. Thus AD bisects ∠BAC, as required.

The final step uses CPCTC: corresponding parts of congruent triangles are congruent. Some books shorten this to CPCT. First prove that the triangles are congruent; only then use matching parts to reach your target.

Two traps: AAA and SSA

AAA does not prove congruence. An equilateral triangle of side 3 and an equilateral triangle of side 6 both have three 60° angles. They have the same shape but different sizes.

SSA does not prove congruence in general. Two sides and an angle that is not between them can allow two different triangles. RHS is a valid special case with a right angle; it is not permission to use SSA for any angle.

Two possible triangles for the same SSA dataA is at the left end of a horizontal ray. AB has length 10 and makes a 30 degree angle with the ray. Two different points C1 and C2 on the ray each lie at distance 6 from B.ABC₁C₂106630°
Both triangles have the same 30° angle at A, AB = 10 and BC = 6, but AC₁ and AC₂ are different. This is SSA, with the angle opposite the side of length 6.

A proof you can check line by line

  1. Name the two triangles in matching order.
  2. List each side or angle equality and explain its reason: given, shared side, midpoint or a fact already proved.
  3. Name the congruence test supported by those facts.
  4. State the matching part that answers the question. Never use the equality you are trying to prove as one of your starting facts.

Practise at your next step

A six-question session chooses from nine questions. Two correct answers in a row without hints move you up a level; an incorrect answer brings a simpler next question where one is available. Hints keep you at the same level. This is a practice suggestion, not an exam score or proof of mastery.

Interactive practice loads here. You can also use the complete question set below.

Write a proof of your own

In a convex quadrilateral ABCD (all interior angles less than 180°), AB = AD and BC = DC. Prove that AC bisects ∠BCD. Use the diagram below; begin by naming the two triangles in matching order.

Two triangles sharing a diagonalA quadrilateral ABCD has A at left, B above, C at right and D below. AB equals AD, BC equals DC. Diagonal AC is drawn.ABCD
Given AB = AD and BC = DC. Compare the triangles on opposite sides of AC.

Write your reasoning on paper before comparing. The practice checker does not grade a written proof.

Compare your proof with a full solution

Compare △ABC and △ADC. We know AB = AD and BC = DC, and the side AC is common to both triangles. Hence △ABC ≅ △ADC by SSS, with A matching A, B matching D and C matching C. The corresponding angles at C are therefore equal: ∠BCA = ∠DCA. Since AC lies inside ∠BCD in this quadrilateral, it divides that angle into two equal parts. Thus AC bisects ∠BCD.

Check: did you state the assumptions, explain the key step, and reach the requested conclusion?

All nine practice questions, hints and solutions

This complete set works without the interactive practice. Hide each solution until you have made an attempt.

1. What does it mean for two triangles to be congruent?

Foundation

  1. They have the same area only
  2. They have the same shape and size
  3. They have equal angles but may have different sizes
  4. They point in the same direction
Hint

Think about fitting one cut-out exactly over the other.

Answer and explanation

They have the same shape and size. Congruent triangles fit exactly after sliding, turning or flipping. Their matching sides and angles are equal, so they have the same shape and size. Equal area alone is not enough.

2. Given △ABC ≅ △QPR, which side equals BC?

Foundation

  1. QP
  2. QR
  3. PR
  4. AB
Hint

Read the vertices in order: A matches Q, B matches P and C matches R.

Answer and explanation

PR. B matches P and C matches R. Therefore the side joining B to C matches the side joining P to R, so BC = PR. The matching order, not the position in a drawing, determines the answer.

3. Each of two triangles has side lengths 4, 5 and 6. Which test proves they are congruent?

Foundation

  1. AAA
  2. SSA
  3. SSS
  4. RHS
Hint

Pair the sides of length 4, then 5, then 6.

Answer and explanation

SSS. All three pairs of matching sides have equal lengths, so SSS applies. These lengths form a triangle because 4 + 5 > 6. No right angle is stated or needed.

4. AB = DE = 5, AC = DF = 7 and ∠BAC = ∠EDF = 60°. Which test uses these given facts directly?

Core

Two sides and their included angleIn triangle ABC, AB is 5, AC is 7 and angle BAC is 60 degrees. The angle is between the two known sides.ABC5760°
The included angle is ∠BAC: its two arms are AB and AC.
  1. SSS
  2. SAS
  3. ASA
  4. RHS
Hint

Check whether the given angle is between the two given sides.

Answer and explanation

SAS. The angle at A lies between AB and AC, and the angle at D lies between DE and DF. We have two matching sides and their included angle, so △ABC ≅ △DEF by SAS.

5. ∠A = ∠D = 50°, ∠B = ∠E = 70° and BC = EF = 8. Which test uses these given facts directly for △ABC and △DEF?

Core

  1. SSS
  2. SAS
  3. AAS
  4. RHS
Hint

The side between the given angles A and B would be AB. Which side is actually given?

Answer and explanation

AAS. The given side BC is not the side joining the given angle vertices A and B. Two angle pairs and this matching non-included side give AAS. Equivalently, ∠C = ∠F = 180° − 50° − 70° = 60°, after which ASA can be applied using BC and EF.

6. △ABC and △DEF are right angled at B and E. AC = DF = 13 and AB = DE = 5. Which test applies directly?

Core

A right triangle with hypotenuse 13 and leg 5Triangle ABC is right angled at B. AB is 5 and AC, opposite the right angle, is the hypotenuse of length 13.ABC513
AC is the hypotenuse because it is opposite the right angle at B.
  1. AAA
  2. RHS (HL)
  3. ASA
  4. No test is possible
Hint

Identify the side opposite each right angle.

Answer and explanation

RHS (HL). AC and DF are opposite the right angles, so they are the hypotenuses. They are equal, and the corresponding legs AB and DE are equal. Both triangles are right angled, so RHS, also called HL, proves congruence.

7. AB = AC and D is the midpoint of BC. SSS gives △ABD ≅ △ACD. Why is ∠ADB = 90°?

Stretch

An isosceles triangle split at the midpoint of its baseTriangle ABC has A at the top, B and C on a horizontal base, and D its midpoint. AB and AC have one tick each; BD and DC have two ticks each. AD is drawn, with no right angle assumed.ABCD
Given AB = AC and BD = DC. The segment AD is shared. The drawing is not a reason to assume a right angle.
  1. Every midpoint makes a right angle
  2. The picture looks perpendicular
  3. Matching angles ∠ADB and ∠ADC are equal and their sum is 180°
  4. Congruent triangles always have right angles
Hint

D lies on BC, so DB and DC are opposite rays.

Answer and explanation

Matching angles ∠ADB and ∠ADC are equal and their sum is 180°. The congruence gives ∠ADB = ∠ADC. Since B, D and C lie on a straight line with D between B and C, these two angles sum to 180°. Two equal angles with sum 180° must each be 90°. It is the congruence argument, not the midpoint alone, that proves perpendicularity.

8. In quadrilateral ABCD, AB = AD and BC = DC. Which argument proves ∠BAC = ∠DAC?

Stretch

Two triangles sharing a diagonalA quadrilateral ABCD has A at left, B above, C at right and D below. AB equals AD, BC equals DC. Diagonal AC is drawn.ABCD
Given AB = AD and BC = DC. Compare the triangles on opposite sides of AC.
  1. △ABC ≅ △ADC by SSS, then use matching angles
  2. The angles are equal because the drawing is symmetric
  3. Equal perimeter always implies congruence
  4. Use AAA before proving any angles equal
Hint

The two triangles share AC. List all three pairs of sides.

Answer and explanation

△ABC ≅ △ADC by SSS, then use matching angles. Compare △ABC and △ADC: AB = AD, BC = DC and AC = AC. Therefore they are congruent by SSS, with B matching D. The angles at A, ∠BAC and ∠DAC, are corresponding parts, so they are equal.

9. Two triangles each have ∠A = 30°, AB = 10 and BC = 6. Must they be congruent?

Stretch

Two possible triangles for the same SSA dataA is at the left end of a horizontal ray. AB has length 10 and makes a 30 degree angle with the ray. Two different points C1 and C2 on the ray each lie at distance 6 from B.ABC₁C₂106630°
Both triangles have the same 30° angle at A, AB = 10 and BC = 6, but AC₁ and AC₂ are different. This is SSA, with the angle opposite the side of length 6.
  1. Yes, by SAS
  2. Yes, because any three matching measurements are enough
  3. No; the given angle is not included, and two different positions of C are possible
  4. Yes, by RHS
Hint

The angle at A lies between AB and AC. Is AC one of the given sides?

Answer and explanation

No; the given angle is not included, and two different positions of C are possible. No. The known sides are AB and BC, but their included angle is at B, not A. In the diagram, C₁ and C₂ are different points on the same ray from A, each at distance 6 from B. Both triangles meet the given facts while AC₁ ≠ AC₂, so they are not congruent. This is the ambiguity in SSA.

Where to go next

Move on to the mixed proof-practice plan · Explore the wider Olympiad topic map

Teaching examples prepared for this guide, not attributed to a contest paper. Familiar elementary problems may appear in other learning materials. Report an unclear step or an error.

Extend the method

Use congruence to justify an auxiliary construction

When adding a point, specify the ray, distance and side of a line so the construction is unambiguous. Then name the three pieces of congruence evidence and their matching vertices. Only after congruence may you infer the remaining corresponding sides and angles. A correct SSS/SAS/ASA/AAS/RHS argument cannot be replaced by “the triangles look the same”.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

Choose your next step

Continue to Triangle inequalities. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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