Triangles: Congruency and Inequalities ICSE Class 9 with Solved Proofs

What Triangles: Congruency and Inequalities is about

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Triangles: Congruency and Inequalities is in the 4. Geometry (i) Triangles (a)-(b) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

The four congruence conditions

Two triangles are congruent if any one of these holds:

• SSS: all three sides of one equal the three sides of the other.
• SAS: two sides and the angle between them (the included angle) are equal.
• AAS (also written ASA): two angles and a side are equal, provided the sides correspond (both opposite the same equal angle, or both between the equal angles).
• RHS: in right-angled triangles, the hypotenuse and one other side are equal.

Not tests: AAA (gives same shape, not same size) and SSA where the angle is not included (two different triangles may fit).

Worked example

In $△ABC$ and $△DEF$, $AB=DE$, $\angle A=\angle D$ and $AC=DF$. Name the congruence and the pairs of equal parts that follow.

Try it first, then show the solutionSolution

1. Two sides with the included angle: SAS.
2. $△ABC\cong △DEF$.
3. By CPCT: $BC=EF$, $\angle B=\angle E$, $\angle C=\angle F$.

Answer: SAS; BC = EF, ∠B = ∠E, ∠C = ∠F.

Writing a congruence proof

A good proof has three parts:
1. Name the two triangles you will compare.
2. List three pairs of equal parts, each with a reason in brackets (given, common, vertically opposite angles, alternate angles, radii, and so on).
3. Conclude with the congruence and its test, written with matching vertex order, then state the required result 'by CPCT'.

A common side or common angle counts as one of the three pairs.

Worked example

$AB$ and $CD$ bisect each other at $O$. Prove that $AC=BD$.

Try it first, then show the solutionSolution

1. In $△AOC$ and $△BOD$:
2. $AO=BO$ ($O$ is the midpoint of $AB$); $CO=DO$ ($O$ is the midpoint of $CD$); $\angle AOC=\angle BOD$ (vertically opposite angles).
3. $△AOC\cong △BOD$ (SAS).
4. $AC=BD$ (CPCT).

Answer: AC = BD, proved by SAS.

Isosceles triangles: equal sides and equal angles

Result: angles opposite equal sides of a triangle are equal. Converse: sides opposite equal angles are equal.

So in $△ABC$ with $AB=AC$, the base angles $\angle B$ and $\angle C$ are equal. If $\angle A=50^{\circ }$, then $\angle B=\angle C=\frac{180^{\circ }-50^{\circ }}{2}=65^{\circ }$.

An equilateral triangle has all angles $60^{\circ }$. In an isosceles triangle, the perpendicular from the vertex to the base bisects the base (by RHS).

Worked example

In $△ABC$, $AB=AC$ and side $BC$ is produced to $D$. If $\angle ACD=115^{\circ }$, find $\angle A$.

Try it first, then show the solutionSolution

1. $\angle ACB=180^{\circ }-115^{\circ }=65^{\circ }$ (linear pair).
2. $\angle ABC=\angle ACB=65^{\circ }$ (angles opposite equal sides).
3. $\angle A=180^{\circ }-130^{\circ }=50^{\circ }$.

Answer: ∠A = 50°

Greater side, greater angle

Result: if two sides of a triangle are unequal, the angle opposite the greater side is greater. Converse: the side opposite the greater angle is longer.

So the longest side is opposite the largest angle, and the shortest side opposite the smallest angle. In a right-angled triangle, the hypotenuse is the longest side (it faces the $90^{\circ }$ angle).

To compare sides, first find all three angles; to compare angles, compare the opposite sides.

Worked example

In $△PQR$, $\angle P=70^{\circ }$ and $\angle Q=60^{\circ }$. Name the longest and shortest sides.

Try it first, then show the solutionSolution

1. $\angle R=180^{\circ }-130^{\circ }=50^{\circ }$.
2. Largest angle $\angle P$, so the longest side is $QR$.
3. Smallest angle $\angle R$, so the shortest side is $PQ$.

Answer: Longest QR, shortest PQ.

Triangle inequality

Result: in every triangle, any two sides added together are longer than the remaining side.

To test three lengths, check that the two smaller add up to more than the largest. 3 cm, 4 cm, 8 cm fail ($3+4=7<8$).

If two sides are $a$ and $b$ with $a>b$, the third side $x$ must satisfy $a-b<x<a+b$.

Worked example

Two sides of a triangle are 5 cm and 9 cm. Between what limits must the third side lie? Can it be 14 cm?

Try it first, then show the solutionSolution

1. $9-5<x<9+5$.
2. $4<x<14$.
3. 14 cm is not allowed: $5+9=14$ is not greater than 14.

Answer: Between 4 cm and 14 cm (exclusive); 14 cm is not possible.

The perpendicular is the shortest

Result: from a point outside a line, the perpendicular to the line is shorter than every other segment joining the point to the line.

This is why the 'distance' from a point to a line always means the perpendicular distance. It follows from the greater-angle result: in the right triangle formed, the slanting line faces the $90^{\circ }$ angle, so it is the hypotenuse and is longer.

The syllabus asks you to use this result; its proof is not required.

Worked example

$P$ is a point 6 cm from a straight road $l$ (perpendicular distance). $Q$ and $R$ are points on $l$ with $PQ=6.5$ cm and $PR=5.5$ cm claimed. Which claim must be wrong?

Try it first, then show the solutionSolution

1. The perpendicular distance 6 cm is the shortest distance from $P$ to $l$.
2. No line from $P$ to $l$ can be shorter than 6 cm.
3. So $PR=5.5$ cm is impossible.

Answer: PR = 5.5 cm is wrong.

Maths in real life: Triangles: Congruency and Inequalities

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Congruence Conditions, Writing a Proof and Isosceles Triangles
• Inequalities in a Triangle: Greater Side, Greater Angle and the Triangle Inequality

Triangles: Congruency and Inequalities formula sheet

Congruence tests

$SSS,SAS,AAS,RHS$

AAA and SSA are not tests.

Isosceles

$AB=AC⟺\angle B=\angle C$

Result and converse.

Base angle

$\angle B=\angle C=\frac{180^{\circ }-\angle A}{2}$

When AB = AC.

Greater side

$AB>AC⟺\angle C>\angle B$

Side and opposite angle.

Triangle inequality

$a-b<c<a+b$

For a ≥ b.

Angle sum and exterior angle

$\angle A+\angle B+\angle C=180^{\circ }$

Exterior angle = sum of interior opposite angles.

Triangles: Congruency and Inequalities: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Triangles: Congruency and Inequalities

The question patterns to expect, and the method that works for each.

Tricks to remember Triangles: Congruency and Inequalities

Short memory aids. Say each one aloud once, then use it on a question the same day.

• SAS needs the angle sandwiched

  In SAS the angle sits between the two sides like the filling in a sandwich.

  Use it when: Avoiding the invalid SSA case.

• Big faces big

  The biggest angle looks across at the biggest side.

  Use it when: Comparing sides and angles.

• Two short sticks must reach

  The two shorter sticks together must be longer than the long one, or they cannot meet.

  Use it when: Triangle inequality.

• Order matters

  Write ΔABC ≅ ΔPQR only if A matches P, B matches Q, C matches R.

  Use it when: Reading off correct CPCT pairs.

Triangles: Congruency and Inequalities: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Triangles: Congruency and Inequalities

The slips examiners see most often, and what to write instead.

• ✗Mistake: Wrong vertex order in the congruence statement

  $△ABC\cong △QPR$ when $A$ actually matches $P$ leads to wrong CPCT pairs.

  ✓Do this: Match vertices by the equal parts you listed, then write both triangles in that order.

  Usually costs: 1 mark, and wrong conclusions.

• ✗Mistake: Using SSA as a test

  Two sides and a non-included angle do not fix a triangle (except in the RHS case).

  Wrong working

  $AB=PQ$, $BC=QR$, $\angle A=\angle P$, so congruent (SAS)

  Right working

  The equal angle must be $\angle B=\angle Q$ (between the sides) for SAS

  ✓Do this: Check the angle lies between the two sides.

  Usually costs: The proof.

• ✗Mistake: Giving no reasons

  Each equal pair needs a reason (given, common, vertically opposite, alternate angles…); bare statements lose method marks.

  ✓Do this: Write the reason in brackets after every line.

  Usually costs: 1 mark per missing reason.

• ✗Mistake: Matching the greater side with the wrong angle

  Students compare a side with an adjacent angle instead of the opposite one.

  ✓Do this: Draw the triangle and shade the angle facing each side before comparing.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Writing proofs of results that are not required

  The syllabus says proofs of the inequality results are not required; long proofs waste time without extra marks.

  ✓Do this: Quote the result by name and apply it.

  Usually costs: Time in the exam.

Practice questions on Triangles: Congruency and Inequalities

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   In $△ABC$, $\angle A=50^{\circ }$ and $\angle B=60^{\circ }$. The longest side is:
   (a) $AB$
   (b) $BC$
   (c) $CA$
   (d) cannot be found

   Hint

   Find ∠C.

   Answer

   (a): $\angle C=70^{\circ }$ is largest, and $AB$ is opposite it.

2. Q2 · 2 marks · Short answerExam level

   Two sides of a triangle are 6 cm and 10 cm. Which of these cannot be the third side: 5 cm, 4 cm, 15 cm, 9 cm?

   Hint

   4 < x < 16.

   Answer

   4 cm (it must be more than 4 cm). 5, 9 and 15 cm are all possible.

3. Q3 · 2 marks · Short answerExam level

   In $△PQR$, $PQ=PR$ and $\angle P$ is twice $\angle Q$. Find all the angles.

   Hint

   Let ∠Q = ∠R = x.

   Answer

   $2x+x+x=180$, $x=45$. $\angle P=90^{\circ }$, $\angle Q=\angle R=45^{\circ }$.

4. Q4 · 1 mark · Assertion-reasonHarder exam level

   Assertion (A): If two triangles have their three angles equal in pairs, they are congruent.
   Reason (R): Equal angles fix the shape of a triangle but not its size.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Think of two equilateral triangles of different sizes.

   Answer

   (b)

5. Q5 · 3 marks · Short answerExam level

   $ABCD$ is a quadrilateral with $AB=AD$ and $CB=CD$. Prove that $AC$ bisects $\angle A$ and $\angle C$.

   Hint

   Compare ΔABC and ΔADC.

   Answer

   $AB=AD$, $CB=CD$, $AC$ common: $△ABC\cong △ADC$ (SSS). So $\angle BAC=\angle DAC$ and $\angle BCA=\angle DCA$ (CPCT).

6. Q6 · 3 marks · Short answerHarder exam level

   In $△ABC$, $AB=AC$. $D$ and $E$ are points on $BC$ such that $BD=CE$. Prove that $AD=AE$.

   Hint

   Use ∠B = ∠C.

   Answer

   In $△ABD$ and $△ACE$: $AB=AC$; $\angle B=\angle C$ (opposite equal sides); $BD=CE$. SAS, so $AD=AE$ (CPCT).

7. Q7 · 3 marks · Short answerHarder exam level

   In $△ABC$, $AB=AC$. $BC$ is produced to $D$ and $AC$ is produced to $E$ so that $\angle DCE=40^{\circ }$. Find $\angle BAC$.

   Hint

   ∠ACB and ∠DCE are vertically opposite.

   Answer

   $\angle ACB=40^{\circ }$ (vertically opposite), $\angle ABC=40^{\circ }$, so $\angle BAC=100^{\circ }$.

8. Q8 · 4 marks · Long answerHarder exam level

   $P$ is any point inside $△ABC$. Prove that $AB+AC>PB+PC$.

   Hint

   Produce BP to meet AC at Q; use the triangle inequality twice.

   Answer

   Produce $BP$ to meet $AC$ at $Q$. In $△ABQ$: $AB+AQ>BQ=BP+PQ$. In $△PQC$: $PQ+QC>PC$. Add: $AB+AQ+QC+PQ>BP+PQ+PC$, i.e. $AB+AC>PB+PC$.

9. Q9 · 3 marks · Short answerHarder exam level

   In $△ABC$, $AD$ is perpendicular to $BC$. Using the result that the perpendicular is the shortest, prove that $AB>AD$ and $AC>AD$, and hence that $AB+AC>2AD$.

   Hint

   AD is the perpendicular from A to BC.

   Answer

   $AD$ is the perpendicular from $A$ to line $BC$, so it is shorter than any other segment from $A$ to $BC$: $AB>AD$ and $AC>AD$. Adding, $AB+AC>2AD$.

10. Q10 · 4 marks · Long answerHarder exam level

    In $△ABC$, $\angle B=2\angle C$ and the bisector of $\angle B$ meets $AC$ at $D$. Prove that $BD=DC$ and that $AC>AB$.

    Hint

    ∠DBC = ∠C; then compare ∠B with ∠C.

    Answer

    $\angle DBC=\frac{1}{2}\angle B=\angle C$, so $BD=DC$ (sides opposite equal angles). Since $\angle B=2\angle C>\angle C$, the side opposite $\angle B$ (that is $AC$) is longer than the side opposite $\angle C$ (that is $AB$).

11. Q11 · 4 marks · Long answerHOTS / competency

    The sides of a triangle are $x$, $x+3$ and $2x-1$ cm, where $x$ is a whole number. Find the smallest possible value of $x$ and the perimeter then.

    Hint

    Write all three inequalities.

    Answer

    $x+(x+3)>2x-1$ always holds ($3>-1$). $x+(2x-1)>x+3\Rightarrow 2x>4\Rightarrow x>2$. $(x+3)+(2x-1)>x$ always holds. Also $2x-1>0$. Smallest whole number $x=3$: sides 3, 6, 5 cm, perimeter 14 cm.

12. Q12 · 4 marks · Long answerHOTS / competency

    $O$ is a point inside $△ABC$. Prove that $OA+OB+OC>\frac{1}{2}(AB+BC+CA)$.

    Hint

    Triangle inequality in ΔOAB, ΔOBC, ΔOCA.

    Answer

    $OA+OB>AB$, $OB+OC>BC$, $OC+OA>CA$. Add: $2(OA+OB+OC)>AB+BC+CA$; divide by 2.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Name the four congruence conditions.Show answer

   SSS, SAS, AAS, RHS.

2. 2.What does CPCT stand for?Show answer

   Corresponding parts of congruent triangles (are equal).

3. 3.Why is AAA not a congruence test?Show answer

   Triangles with equal angles can be of different sizes.

4. 4.In ΔABC, AB = AC. Which angles are equal?Show answer

   ∠B and ∠C.

5. 5.Which side is opposite the largest angle?Show answer

   The longest side.

6. 6.Can 2 cm, 3 cm and 6 cm form a triangle?Show answer

   No; 2 + 3 < 6.

7. 7.What is the shortest distance from a point to a line?Show answer

   The perpendicular distance.

8. 8.What does the H in RHS stand for?Show answer

   Hypotenuse.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you name the test from the marked equal parts?
• ☐Do you write vertex order correctly in ΔABC ≅ ΔPQR?
• ☐Do you give a reason for every equal pair?
• ☐Can you find base angles of an isosceles triangle quickly?
• ☐Can you name the longest and shortest sides from the angles?
• ☐Can you test three lengths with the triangle inequality?
• ☐Do you know which results do not need proofs?

Triangles: Congruency and Inequalities: frequently asked questions