NCERT Solutions for Class 9 Maths Chapter 7: Triangles — Free PDF

Chapter 7 is one of the most important chapters in Class 9 geometry, and it forms the foundation for all proof-based geometry you will encounter in Class 10 and beyond. The chapter focuses on triangle congruence — when two triangles are identical in shape and size — and derives several powerful properties from congruence criteria.

You will learn five congruence rules (SAS, ASA, AAS, SSS, RHS), explore the properties of isosceles triangles, and prove the triangle inequality theorem. Each of these concepts is tested heavily in CBSE board exams, typically carrying 6-8 marks in the form of proof-based questions.

The chapter has five exercises (7.1 to 7.5) that progress systematically from basic congruence to inequality theorems. Exercise 7.1 introduces SAS congruence. Exercise 7.2 covers properties of isosceles triangles (angles opposite equal sides are equal, and its converse). Exercise 7.3 deals with SSS and RHS congruence. Exercises 7.4 and 7.5 cover inequalities in triangles. Mastering the proof-writing technique here is crucial — the same structured approach is used in Chapter 8 (Quadrilaterals) and Chapter 9 (Circles).

Congruence of Triangles: Two triangles are congruent if all their corresponding sides and corresponding angles are equal. We write $\triangle ABC \cong \triangle DEF$, where the order of vertices indicates the correspondence: $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$.

CPCT (Corresponding Parts of Congruent Triangles): Once two triangles are proved congruent, we can conclude that all their corresponding parts are equal. This is abbreviated as CPCT and is the primary tool for deriving further results.

Congruence Criteria:

• SAS (Side-Angle-Side): Two sides and the included angle of one triangle are equal to two sides and the included angle of another.
  - ASA (Angle-Side-Angle): Two angles and the included side of one triangle are equal to two angles and the included side of another.
  - AAS (Angle-Angle-Side): Two angles and a non-included side of one triangle are equal to the corresponding parts of another. (This follows from ASA since the third angle is automatically determined.)
  - SSS (Side-Side-Side): All three sides of one triangle are equal to all three sides of another.
  - RHS (Right angle-Hypotenuse-Side): Both triangles have a right angle, their hypotenuses are equal, and one other pair of sides is equal. This applies only to right-angled triangles.

Invalid Criteria:
- AAA does NOT prove congruence — it only proves similarity (same shape, not necessarily same size).
- SSA (Side-Side-Angle, where the angle is not included) is NOT valid in general because it can give two different triangles (the ambiguous case).

Isosceles Triangle Properties:
- Angles opposite to equal sides are equal.
- Sides opposite to equal angles are equal (converse).
- The altitude from the vertex angle of an isosceles triangle bisects the base and the vertex angle.

Triangle Inequality Theorem: The sum of any two sides of a triangle is strictly greater than the third side. If $a$, $b$, $c$ are the sides, then $a + b > c$, $b + c > a$, and $a + c > b$.

Problem 1: In quadrilateral $ABCD$, $AC = AD$ and $AB$ bisects $\angle A$. Show that $\triangle ABC \cong \triangle ABD$. What can you say about $BC$ and $BD$?

Solution:
In $\triangle ABC$ and $\triangle ABD$:
- $AC = AD$ (given)
- $\angle CAB = \angle DAB$ ($AB$ bisects $\angle A$)
- $AB = AB$ (common side)

By SAS congruence rule: $\triangle ABC \cong \triangle ABD$

By CPCT: $BC = BD$

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Problem 2: $ABCD$ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$. Prove that $AC = BD$.

Solution:
In $\triangle DAB$ and $\triangle CBA$:
- $AD = BC$ (given)
- $\angle DAB = \angle CBA$ (given)
- $AB = BA$ (common)

By SAS: $\triangle DAB \cong \triangle CBA$

By CPCT: $DB = CA$, i.e., $AC = BD$ $\blacksquare$

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Problem 3: $AB$ is a line segment and $P$ is the midpoint. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle APD = \angle BPE$. Show that $AD = BE$.

Solution:
In $\triangle APD$ and $\triangle BPE$:
- $AP = BP$ ($P$ is midpoint of $AB$)
- $\angle APD = \angle BPE$ (given)
- $\angle DAP = \angle EBP$ (given: $\angle BAD = \angle ABE$)

By ASA: $\triangle APD \cong \triangle BPE$

By CPCT: $AD = BE$ $\blacksquare$

Theorem: Angles opposite to equal sides of an isosceles triangle are equal.

Theorem (Converse): Sides opposite to equal angles of a triangle are equal.

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Problem 1: In an isosceles triangle $ABC$ with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Show that $OB = OC$.

Solution:
Since $AB = AC$, we have $\angle B = \angle C$ (angles opposite equal sides).

$\therefore \dfrac{1}{2}\angle B = \dfrac{1}{2}\angle C$

$\implies \angle OBC = \angle OCB$ (since $BO$ and $CO$ are bisectors)

In $\triangle BOC$, since $\angle OBC = \angle OCB$:

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Problem 2: In $\triangle ABC$, $AB = AC$ and $D$ is the midpoint of $BC$. Show that $AD \perp BC$.

Solution:
In $\triangle ABD$ and $\triangle ACD$:
- $AB = AC$ (given)
- $BD = CD$ ($D$ is midpoint of $BC$)
- $AD = AD$ (common)

By SSS: $\triangle ABD \cong \triangle ACD$

By CPCT: $\angle ADB = \angle ADC$

But $\angle ADB + \angle ADC = 180^\circ$ (linear pair)

$\therefore 2\angle ADB = 180^\circ \implies \angle ADB = 90^\circ$

Hence $AD \perp BC$ $\blacksquare$

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Problem 3: $\triangle ABC$ is isosceles with $AB = AC$. Show that $\angle B = \angle C$. (Proof of the base angles theorem.)

Solution:
Draw $AD$, the bisector of $\angle A$, meeting $BC$ at $D$.

In $\triangle ABD$ and $\triangle ACD$:
- $AB = AC$ (given)
- $\angle BAD = \angle CAD$ ($AD$ bisects $\angle A$)
- $AD = AD$ (common)

By SAS: $\triangle ABD \cong \triangle ACD$

By CPCT: $\angle ABD = \angle ACD$, i.e., $\angle B = \angle C$ $\blacksquare$

Problem 1: $\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and on the same side of $BC$. Show that $\angle ABD = \angle ACD$.

Solution:
Since $\triangle ABC$ is isosceles with $AB = AC$, and $\triangle DBC$ is isosceles with $DB = DC$.

In $\triangle ABD$ and $\triangle ACD$:
- $AB = AC$ (given)
- $DB = DC$ (given)
- $AD = AD$ (common)

By SSS: $\triangle ABD \cong \triangle ACD$

By CPCT: $\angle ABD = \angle ACD$ $\blacksquare$

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Problem 2: $ABC$ is a right triangle in which $\angle B = 90^\circ$ and $D$ is the midpoint of $AC$. Prove that $BD = \dfrac{1}{2}AC$.

Solution:
Construct point $E$ on $BD$ produced such that $DE = BD$. Join $EC$.

In $\triangle ADB$ and $\triangle CDE$:
- $AD = DC$ ($D$ is midpoint of $AC$)
- $BD = DE$ (by construction)
- $\angle ADB = \angle CDE$ (vertically opposite angles)

By SAS: $\triangle ADB \cong \triangle CDE$

By CPCT: $AB = CE$ and $\angle ABD = \angle DCE$

Since $\angle ABD + \angle ABC = 180^\circ$ (we need to check the angle at $B$):
$\angle ABC = 90^\circ$, and since $\angle ABD = \angle DCE$ (CPCT), we get $\angle BCE = 90^\circ$.

In $\triangle BCE$: $BC^2 + CE^2 = BE^2$

But $CE = AB$, so $BC^2 + AB^2 = BE^2 = (2BD)^2 = 4BD^2$.

Also $BC^2 + AB^2 = AC^2$ (Pythagoras in $\triangle ABC$).

$\therefore 4BD^2 = AC^2 \implies BD = \dfrac{1}{2}AC$ $\blacksquare$

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Problem 3: $\triangle ABC$ is an isosceles triangle with $AB = AC$. $AP \perp BC$. Show that $\angle BAP = \angle CAP$.

Solution:
In $\triangle ABP$ and $\triangle ACP$:
- $AB = AC$ (given)
- $AP = AP$ (common)
- $\angle APB = \angle APC = 90^\circ$ ($AP \perp BC$)

By RHS: $\triangle ABP \cong \triangle ACP$

By CPCT: $\angle BAP = \angle CAP$ $\blacksquare$

Key Theorems:
- In a triangle, the side opposite to the larger angle is longer.
- In a triangle, the angle opposite to the longer side is larger.
- The sum of any two sides of a triangle is greater than the third side: $a + b > c$.

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Problem 1: Show that in a right-angled triangle, the hypotenuse is the longest side.

Solution:
In $\triangle ABC$ with $\angle B = 90^\circ$.

Since $\angle B = 90^\circ$ and $\angle A + \angle C = 90^\circ$:

Since the side opposite to the larger angle is longer:

So $AC$ (the hypotenuse) is the longest side. $\blacksquare$

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Problem 2: $D$ is a point on side $BC$ of $\triangle ABC$ such that $AD = AC$. Show that $AB > AD$.

Solution:
In $\triangle ACD$: $AD = AC$ (given)
$\therefore \angle ACD = \angle ADC$ (angles opposite equal sides)

$\angle ADC$ is an exterior angle of $\triangle ABD$:

$\therefore \angle ACD > \angle ABD$, i.e., $\angle ACB > \angle ABC$

In $\triangle ABC$, the side opposite to the larger angle is longer:

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Problem 3: In the figure, $PR > PQ$ and $PS$ bisects $\angle QPR$. Prove that $\angle PSR > \angle PSQ$.

Solution:
Since $PR > PQ$, the angle opposite $PR$ is larger than the angle opposite $PQ$:

Since $PS$ bisects $\angle QPR$: $\angle QPS = \angle RPS \quad \cdots (2)$

In $\triangle PQS$: $\angle PSR = \angle PQS + \angle QPS$ (exterior angle of $\triangle PQS$)

In $\triangle PRS$: $\angle PSQ = \angle PRS + \angle RPS$ (exterior angle of $\triangle PRS$)

From (1): $\angle PQS > \angle PRS$ (since $\angle PQR = \angle PQS$ and $\angle PRQ = \angle PRS$)

From (2): $\angle QPS = \angle RPS$

$\therefore \angle PQS + \angle QPS > \angle PRS + \angle RPS$

$\therefore \angle PSR > \angle PSQ$ $\blacksquare$

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Problem 4: Is it possible to construct a triangle with sides $5$ cm, $3$ cm, and $9$ cm?

Solution:
Check the triangle inequality: $5 + 3 = 8 < 9$.

Since $5 + 3$ is NOT greater than $9$, a triangle with these sides is not possible.

Example 1: In $\triangle ABC$, $\angle A = 50^\circ$ and $\angle B = 60^\circ$. Determine the longest and shortest sides.

Solution:
$\angle C = 180^\circ - 50^\circ - 60^\circ = 70^\circ$

Since $\angle C > \angle B > \angle A$, the sides opposite to these angles follow the same order:

$AB$ (opposite the largest angle $C$) is the longest. $BC$ (opposite the smallest angle $A$) is the shortest.

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Example 2: $ABCD$ is a parallelogram. The bisector of $\angle A$ meets $DC$ at $E$ and $BC$ produced at $F$. Prove that $\triangle AEF$ is isosceles.

Solution:
Since $ABCD$ is a parallelogram, $AD \parallel BC$.

$\angle DAF = \angle AFB$ (alternate interior angles with $AF$ as transversal).

Since $AF$ bisects $\angle A$: $\angle DAF = \angle FAB$.

$\therefore \angle FAB = \angle AFB$

In $\triangle ABF$: sides opposite equal angles are equal, so $AB = BF$... Actually, let us reconsider. We want to show $\triangle AEF$ is isosceles.

Since $AD \parallel BC$: $\angle DAE = \angle AEF$ (alternate interior angles) ... Hmm, $E$ is on $DC$, which is parallel to $AB$.

Actually, $\angle AEF = \angle DAE$ (alternate angles, $AD \parallel EF$). And $\angle DAE = \angle EAF$ (since $AF$ bisects $\angle A$ and $\angle DAE$ is half of $\angle A$... this needs $E$ to be on $DC$).

Let us use: $\angle DAE = \angle AEF$ (alternate angles, $AD \parallel BC$ extended). Also $\angle DAE = \angle EAF$ (bisector). Therefore $\angle AEF = \angle AEF$... We need to be more careful.

$\angle DAF = \angle AFE$ (alternate interior angles, since $AD \parallel BF$). And $\angle DAE = \angle EAF$ (bisector). Now $\angle AEF = \angle DAE$ (alternate angles, $DE \parallel AB$, since $ABCD$ is a parallelogram so $DC \parallel AB$). So $\angle AEF = \angle EAF$, making $\triangle AEF$ isosceles with $AF = EF$. $\blacksquare$

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Example 3: Show that the sum of the three altitudes of a triangle is less than the sum of the three sides.

Solution:
In $\triangle ABC$, let the altitudes from $A$, $B$, $C$ be $h_a$, $h_b$, $h_c$ respectively.

The altitude from $A$ to $BC$ is the shortest distance from $A$ to line $BC$. Therefore $h_a < AB$ and $h_a < AC$.

Similarly: $h_b < AB$ and $h_b < BC$; $h_c < AC$ and $h_c < BC$.

Adding appropriate inequalities: $h_a < AB$, $h_b < BC$, $h_c < AC$.

$\therefore h_a + h_b + h_c < AB + BC + AC$ $\blacksquare$

Mistake 1: Writing the congruence in the wrong vertex order.
$\triangle ABC \cong \triangle DEF$ means $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. Writing $\triangle ABC \cong \triangle FDE$ (for example) implies a completely different correspondence and is incorrect.

Mistake 2: Not stating the congruence rule.
After listing the three matching elements, you must explicitly write "By SAS congruence rule" (or ASA, SSS, RHS). Examiners deduct marks for missing this.

Mistake 3: Using AAA or SSA for congruence.
AAA only proves similarity, not congruence. SSA is ambiguous and not a valid congruence criterion. Always ensure you have one of the five valid criteria.

Mistake 4: Forgetting to state CPCT.
When extracting equal sides or angles from congruent triangles, you must write "By CPCT" to justify the step.

Mistake 5: Confusing the included angle in SAS.
In SAS, the angle must be between the two sides. If $AB = DE$, $BC = EF$, and $\angle A = \angle D$, this is NOT SAS (because $\angle A$ is not between $AB$ and $BC$). You would need $\angle B = \angle E$ for SAS.

Q1. In $\triangle ABC$, $\angle B = 35^\circ$ and $\angle C = 65^\circ$. Which is the longest side?

Answer: $\angle A = 180^\circ - 35^\circ - 65^\circ = 80^\circ$. Since $\angle A$ is the largest angle, the side opposite to it ($BC$) is the longest side.

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Q2. Can a triangle have sides of lengths $7$ cm, $3$ cm, and $2$ cm?

Answer: No. $3 + 2 = 5 < 7$. The triangle inequality is violated.

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Q3. In $\triangle PQR$, $PQ = PR$. $S$ is a point on $QR$ such that $PS \perp QR$. Prove that $QS = SR$.

Answer: In $\triangle PQS$ and $\triangle PRS$: $PQ = PR$ (given), $PS = PS$ (common), $\angle PSQ = \angle PSR = 90^\circ$. By RHS: $\triangle PQS \cong \triangle PRS$. By CPCT: $QS = SR$.

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Q4. $ABCD$ is a quadrilateral with $AB = CD$ and $BC = DA$. Show that $\angle A = \angle C$.

Answer: Draw diagonal $BD$. In $\triangle ABD$ and $\triangle CDB$: $AB = CD$, $DA = BC$, $BD = DB$ (common). By SSS: $\triangle ABD \cong \triangle CDB$. By CPCT: $\angle A = \angle C$.

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Q5. Arrange the sides of $\triangle ABC$ in ascending order if $\angle A = 72^\circ$, $\angle B = 53^\circ$, $\angle C = 55^\circ$.

Answer: Since $\angle B < \angle C < \angle A$, the sides opposite them are in ascending order: $AC < AB < BC$.

• Five valid congruence criteria: SAS, ASA, AAS, SSS, RHS. AAA and SSA are NOT valid.
  - Always state the congruence rule and use CPCT to extract results.
  - In an isosceles triangle, base angles are equal (and conversely, equal angles imply equal opposite sides).
  - The triangle inequality theorem ($a + b > c$) must hold for all three pairs of sides.
  - The side opposite the larger angle is always the longer side, and vice versa.
  - The hypotenuse is always the longest side in a right-angled triangle.
  - Proof-writing requires a clear, structured format: list corresponding parts, state the rule, then derive the conclusion.
  - Most problems in this chapter involve identifying a common side or a pair of vertically opposite angles to set up the congruence.

Practice triangle problems on SparkEd for step-by-step feedback!