CBSE Class 9 Maths Notes Chapter 7 Triangles

Congruent geometric figures share the same shape and size. To verify if two plane figures are congruent, one can superimpose them and compare their alignment. For instance, if the shaded portion of one figure aligns perfectly with the unshaded portion of another, they are considered congruent. When comparing two line segments or angles, if they have the same length or measure respectively, they are congruent. This is denoted as AB ≅ CD for line segments and ∠AOB ≅ ∠POQ for angles, where AOB and POQ are angle names. For example, in Figure line segments AB and CD are congruent because they have the same length. In Figure  angles a and b, as well as angles m and n, are congruent because they share the same measure.Angles m and n are considered congruent as they are vertically opposite angles.

Congruent Circles

Two circles are said to be congruent if they have the same radius.

In the above figure, both the circles have the same radius

Congruent Triangles

The definition of congruent triangles states that two triangles are congruent if and only if all their sides and angles match. This means that if all the sides and angles of one triangle are equal to the corresponding sides and angles of another triangle, then the triangles are congruent.

Sufficient Condition for Triangle Congruence

In the activity comparing triangles ΔABC and ΔDEF, we observed that when two sides and the included angle of one triangle match those of another, the triangles are congruent. This principle is known as the Side-Angle-Side (SAS) congruency condition. It simplifies the process of determining triangle congruence by requiring only three conditions to be met instead of all six.

Theorem 9

"Two triangles are congruent if any two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle".

In the two given triangles, Δ ABC ΔABC and Δ DEF ΔDEF , AB = DE AB = DE

AC = DF AC = DF and

∠ BAC = ∠ EDF ∠BAC = ∠EDF .

To prove: Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF

Proof:

If we rotate and drag Δ ABC ΔABC on Δ DEF ΔDEF , such the vertices B falls on the vertex of the other triangle E and place BC along EF, we will find that, since AB = DE AB = DE , C falls on F.

Also, ∠ B = ∠ E ∠B = ∠E .

AB falls on DE, A will coincide with the vertex D and C with F.

So, AC coincides with DF.

∴ Δ ABC ∴ΔABC coincides with Δ DEF ΔDEF .

∴ Δ ABC ≅ Δ DEF

Application of SAS Congruency

Theorem 10

"Angles opposite to equal sides are equal".

In Δ ABC ΔABC , AB = AC AB = AC

To prove:

∠ ABC = ∠ ACB ∠ABC = ∠ACB

Construction: Draw the angle bisector of A, AD.

Proof:

Comparing both the triangles,

Given that AB = AC AB = AC

AD is the common side.

∠ BAD = ∠ DAC ∠BAD = ∠DAC (as AD is the angle bisector)

∴ Δ BAD = Δ DAC ∴ΔBAD = ΔDAC (by SAS congruency rule)

∴ ∠ ABD = ∠ ACD ∴∠ABD = ∠ACD (by CPCT)

∴ ∠ B = ∠ C ∴∠B = ∠C

Hence, proved.

ASA Congruence Condition

The Angle-Side-Angle (ASA) congruence condition states that two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle. This condition simplifies the process of determining triangle congruence by requiring the equality of two angles and the side between them, thereby ensuring that the triangles have the same shape and size.

Theorem 11

Given:

In Δ ABC ΔABC and Δ DEF ΔDEF ,

∠ B = ∠ E ∠B = ∠E and ∠ C = ∠ F ∠C = ∠F

BC = EF BC = EF .

To prove:

Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF

Proof:

There are three possibilities

Case I:

Case II:

Case III:

Case I: In addition to data, if

then

Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF (by SAS congruence postulate)

Case II: If

and let K is any point on DE such that

.

Join KF.

Now compare triangles ABC and KCF.

(given)

∠ B = ∠ E ∠B = ∠E (given)

Let

(SAS criterion)

Hence,

But,

(given)

Hence, K coincides with D.

Therefore, AB must be equal to DE.

Case III: If

, then a similar argument applies.

AB must be equal to DE.

Hence the only possibility is that AB must be equal to DE and from SAS congruence condition

Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF

Hence the theorem is proved.

Theorem 12

"In a triangle the sides opposite to equal angles are equal".

This theorem can also be stated as

"The sides opposite to equal angles of a triangle are equal".

Given:

In

To prove:

AB ¯¯¯¯¯¯¯ = AC ¯¯¯¯¯¯¯ AB¯ = AC¯

Construction:

Draw AD ¯¯¯¯¯¯¯¯ ⊥ BC ¯¯¯¯¯¯¯ AD¯ ⊥ BC¯

Proof:

Construct two right angle triangles, ADB and ADC, right angled at D.

Here,

∠ ADB = ∠ ADC = 90 ∘ ∠ADB = ∠ADC = 90∘ (from the construction)

AD ¯¯¯¯¯¯¯¯ AD¯ is common for both the triangles.

∴ Δ ADB ≅ Δ ADC ∴ΔADB ≅ ΔADC (by ASA postulate)

AB ¯¯¯¯¯¯¯ = AC ¯¯¯¯¯¯¯ AB¯ = AC¯ (corresponding sides)

AAS Congruence Condition

The Angle-Angle-Side (AAS) congruence condition states that two triangles are congruent if two angles and a non-included side of one triangle are equal to the corresponding two angles and the side of the other triangle. This condition provides another method for determining triangle congruence by requiring the equality of two angles and a side not included between them, ensuring that the triangles have the same shape and size.

Given:

In triangles ABC and DEF,

(non-included sides)

To prove:

Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF

Proof:

(given)

(given)

Now, adding both,

…(1)

Since,

, considering (1) we can say that,

…(2)

Now in triangle ABC and DEF,

(given)

(proved in (2))

BC = EF BC = EF (given)

Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF (by SAS congruency)

SSS Congruence Condition

"Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle".

In triangles ABC and DEF,

Note:

Let BC and EF be the longest sides of triangles ABC and DEF respectively.

To prove:

Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF

Construction: If BC is the longest side, draw EG such that EG = AB EG = AB and ∠ GEF = ∠ ABC ∠GEF = ∠ABC .

Join GF and DG.

Proof:

In triangles ABC and GEF,

(by construction)

(given)

∠ ABC = ∠ GEF ∠ABC = ∠GEF (by construction)

∴ Δ ABC = Δ GEF ∴ΔABC = ΔGEF (SAS congruence condition)

∠ BAC = ∠ EGF ∠BAC = ∠EGF (by CPCT)

and AC = GF AC = GF (by construction)

But AB = DE AB = DE (given)

∴ DE = GE ∴DE = GE

Similarly, DF = GF DF = GF

In Δ EDG ΔEDG ,

DE = GE DE = GE (Proved)

∴ ∠ 1 = ∠ 2 ∴∠1=∠2 …(1) (angles opposite equal sides)

In Δ EGF ΔEGF ,

DF = GF DF = GF (Proved)

∴ ∠ 3 = ∠ 4 ∴∠3=∠4 …(2) (angles opposite equal sides)

By adding (1) and (2), we get,

∴ ∠ 1 + ∠ 3 = ∠ 4 + ∠ 2 ∴∠1+∠3=∠4+∠2

∠ EDF = ∠ EGF ∠EDF = ∠EGF but we have proved that ∠ BAC = ∠ EGF ∠BAC = ∠EGF

Therefore, ∠ EDF = ∠ BAC ∠EDF = ∠BAC

Now in triangles ABC and DEF,

(given)

(given)

∠ EDF = ∠ BAC ∠EDF = ∠BAC (proved)

Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF (by SAS congruency)

Theorem of RHS (Right Angle Hypotenuse Side) Congruence

If the hypotenuse and one side of one triangle are equal to the hypotenuse and corresponding side of the other triangle, the two triangles are congruent.

ABC and DEF are two right-angled triangles such that

1. ∠ B = ∠ E = 90 ∘ ∠B = ∠E = 90∘

2. Hypotenuse AC ¯¯¯¯¯¯¯¯¯ = AC ¯ = Hypotenuse DF ¯¯¯¯¯¯¯ DF¯ and

3. Side BC ¯¯¯¯¯¯¯¯ = BC ¯= Side EF ¯¯¯¯¯¯¯ EF¯

To prove:

Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF

Construction

Produce DE to M so that

. Join MF.

Proof:

In triangles ABC and MEF,

(construction)

(given) ∠ ABC = ∠ MEF = 90 ∘ ∠ABC = ∠MEF = 90∘

Therefore, Δ ABC ≅ Δ MEF ΔABC ≅ ΔMEF (SAS congruency)

Hence, ∠ A = ∠ M ∠A = ∠M …(1) (by CPCT)

AC ¯¯¯¯¯¯¯ = MF ¯¯¯¯¯¯¯¯ AC¯=MF¯ …(2) (by CPCT)

Also, AC = DF AC = DF … (3) (given)

From (1) and (3), DF = MF DF = MF

Therefore, ∠ D = ∠ M ∠D = ∠M ...(4) (Angles opposite to equal sides of

)

From (2) and (4),

∠ A = ∠ D ∠A = ∠D …(5)

Now compare triangles ABC and DEF,

∠ A = ∠ D ∠A = ∠D (from 5)

∠ B = ∠ E = 90 ∘ ∠B = ∠E = 90∘ (given)

∴ ∠ C= ∠ F ∴∠C=∠F …(6)

Compare triangles ABC and DEF,

BC ¯¯¯¯¯¯¯ = EF ¯¯¯¯¯¯¯ BC¯=EF¯ (given)

AC ¯¯¯¯¯¯¯ = DF ¯¯¯¯¯¯¯ AC¯=DF¯ (given)

∠ C= ∠ F ∠C=∠F (from 6)

Therefore, Δ ABC ≅ Δ DEF ΔABC ≅ ΔDEF (by SAS congruency)

Inequality of Angles

Here we can see that, both these angles are not equal, as 135 ∘ < 160 ∘ 135∘<160∘

Inequality in a Triangle

Construct a triangle ABC as shown in the figure.

Observe that in triangle ABC, AC ¯¯¯¯¯¯¯ AC¯ is the smallest side (2 cm)

B is the angle opposite to AC ¯¯¯¯¯¯¯ AC¯ and ∠ B = 20 ∘ ∠B = 20∘

BC ¯¯¯¯¯¯¯ BC¯ is the greatest side (6 cm)

A is the angle opposite to BC ¯¯¯¯¯¯¯ BC¯ and ∠ A = 100 ∘ ∠A = 100∘

From the measurements made above of side and angle opposite to it, we can write the relation in the form of a statement.

"If two sides of a triangle are unequal then the longer side has the greater angle opposite to it”.

Theorem on Inequalities

Theorem 1:

If two sides of a triangle are unequal, the longer side has the greater angle opposite to it.

Read the statement and draw a triangle as per data.

Draw

, such that

.

Data:

To Prove:

∠ ABC = ∠ ACB ∠ABC = ∠ACB

Construction:

Take a point D on AC ¯¯¯¯¯¯¯ AC¯ such that AB ¯¯¯¯¯¯¯ = AD ¯¯¯¯¯¯¯¯ AB¯ = AD¯ . Join B to D.

Proof:

In Δ ABC ΔABC , AB ¯¯¯¯¯¯¯ = AD ¯¯¯¯¯¯¯¯ AB¯ = AD¯ (by construction)

∴ ∠ ABD = ∠ ADB ∴∠ABD = ∠ADB …(1)

but ∠ ADB ∠ADB is the exterior angle with reference to Δ DBC ΔDBC .

∠ ADB ∠ DCB ∠ADB ∠DCB … (2)

From relation (1) and (2) we can write ∠ ABD ∠ DCB ∠ABD ∠DCB

But ∠ ABD ∠ABD is a part of ∠ ACB ∠ACB .

∴ ∠ ACB ∠ DCB ∴∠ACB ∠DCB or ∴ ∠ ABC ∠ ACB ∴∠ABC ∠ACB

Hence, proved.

Angle Side Relation

Theorem 2:

In a triangle, if two angles are unequal, the side opposite to greater angle is longer than the side opposite to the smaller angle.

Theorem 3

In a triangle, the greater angle has the longer side opposite to it.

In Δ ABC ΔABC , ∠ ABC ∠ ACB ∠ABC ∠ACB

To prove:

Proof:

In

, AB and AC are two line segments. So the following are the three possibilities of which

exactly one must be true.

1. either

   AB = AC AB = AC

   , then ∠ B = ∠ C ∠B = ∠C which is contrary to the hypothesis.

2. AB  AC AB  AC

   , then ∠ B ∠ C ∠B ∠C which is contrary to the hypothesis.

3. AB  AC AB  AC

   , this is the only condition we are left with, so

   AB  AC AB  AC

   must be true.

Hence, proved.

Theorem 4

Prove that in any triangle the sum of the lengths of any two sides of a triangle is greater than the length of its third side. Draw Δ ABC Δ ABC .

Data:

ABC is a triangle.

To prove:

AB ¯¯¯¯¯¯¯ + AC ¯¯¯¯¯¯¯ > BC ¯¯¯¯¯¯¯ AB¯+AC¯>BC¯

AB ¯¯¯¯¯¯¯ + BC ¯¯¯¯¯¯¯ > AC ¯¯¯¯¯¯¯ AB¯+BC¯>AC¯

AC ¯¯¯¯¯¯¯ + BC ¯¯¯¯¯¯¯ > BA ¯¯¯¯¯¯¯ AC¯+BC¯>BA¯

Construction:

Produce BC ¯¯¯¯¯¯¯ BC¯ to D such that

. Join A to D.

Proof:

(by construction)

∠ 1 = ∠ 2 ∠1=∠2 …(1)

From the figure,

∠ BAD = ∠ 2+ ∠ 3 ∠BAD = ∠2+∠3 …(2)

∠ BAD ∠ 2 ∠BAD ∠2

∴ ∠ BAD ∠ 1 ∴∠BAD ∠1 (proved from 1)

AB ¯¯¯¯¯¯¯ AB¯ is opposite to ∠ 1 ∠1 and BD ¯¯¯¯¯¯¯ BD¯ is opposite to ∠ BAD ∠BAD .

In Δ BAD ΔBAD , ∠ 1 ∠ 2+ ∠ 3 ∠1 ∠2+∠3

(side opposite to greater angle is greater)

From figure

Since, CD = AC CD = AC , hence, sum of two sides of a triangle is greater than the third side.

Similarly, AB ¯¯¯¯¯¯¯ + BC ¯¯¯¯¯¯¯ > AC ¯¯¯¯¯¯¯ AB¯+BC¯>AC¯ and AC ¯¯¯¯¯¯¯ + BC ¯¯¯¯¯¯¯ > BA ¯¯¯¯¯¯¯ AC¯+BC¯>BA¯ .

Hence, proved.

Theorem 5

Of all the line segments that can be drawn to a given line, from a point not lying on it, the perpendicular line segment is the shortest.

l 𝑙 is a line and P is a point not lying on it.

. N is any point on l other than M.

To prove:

Proof:

In Δ PMN Δ PMN , ∠ M ∠M is the right angle.

∴ ∠ N ∴∠N is an acute angle, from angle sum property.

∵ ∠ M ∠ N ∵∠M ∠N

PN  PM PN  PM (side opposite to greater angle)

∴ PMPN ∴PMPN .

Benefits of CBSE Class 9 Maths Notes Chapter 7 Triangles

• Concept Clarity : These notes provide clear explanations of various concepts related to triangles, making it easier for students to understand the fundamentals.
• Structured Learning : The notes follow a structured format, covering each topic comprehensively, ensuring that students don't miss out on any essential concepts.
• Exam Preparation : The notes include important formulas, theorems, and properties related to triangles, which are crucial for exam preparation. Students can use these notes for quick revision before exams.
• Visual Aid: Diagrams and illustrations included in the notes help students visualize geometric concepts and understand them better.

CBSE Maths Notes For Class 9

Chapter 1 – Number System

Chapter 2 – Polynomials

Chapter 3 – Coordinate Geometry

Chapter 4 – Linear Equations in Two Variables

Chapter 5 – Introduction to Euclid’s Geometry

Chapter 6 – Lines and Angles

Chapter 7 – Triangles

Chapter 8 – Quadrilaterals

Chapter 9 - Areas of Parallelograms and Triangles

CBSE Class 9 Maths Notes Chapter 10 - Circles Notes

CBSE Class 9 Maths Notes Chapter 11 - Constructions Notes

CBSE Class 9 Maths Notes Chapter 12 - Herons Formula Notes

CBSE Class 9 Maths Notes Chapter 13 - Surface Areas and Volumes Notes

CBSE Class 9 Maths Notes Chapter 14 - Statistics Notes

CBSE Class 9 Maths Notes Chapter 15 - Probability Notes