Detailed Chapter Notes - Triangles, Class 9 Mathematics Class 9 Notes | EduRev

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Class 9 : Detailed Chapter Notes - Triangles, Class 9 Mathematics Class 9 Notes | EduRev

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INTRODUCTION
You have learnt about triangles and some of their properties in your previous classes. In this chapter, we will study about the congruency of triangles and some more properties. We may had some idea about these properties in lower classes but here, we will study these properties in greater details.

CONGRUENCE OF TRIANGLES
Congruent figures : Two geometrical figures, having exactly the same shape and size are known as congruent fig. For congruence, we use the symbol NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Thus, two line segments are congruent, if they have the same length. Two angles are congruent, if they have the same measure.

Congruent Triangles : Two triangles are congruent, if and only if one of them can be made to superpose on 
the other, so as to cover it exactly.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Thus, congruent triangles are exactly identical, i.e., their corresponding three sides and the three angles are equal.
If ΔABC is congruent to ΔDEF, we write ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF. This happens when AB = DE, BC = EF, AC = DF and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D,NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9F. In this case, we say that the sides corresponding to AB, BC and AC are DE, EF and DF respectively. And, the angles corresponding to NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C are NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9F respectively. Thus, the corresponding parts of two congruent triangles are equal. We show it by the abbreviation C.P.C.T., which means corresponding parts of congruent triangles.

Congruence Relation in the Set of All Triangles : From the definition of congruence of two triangles, we obtain the following results :
(i) Every triangle is congruent to itself i.e. ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔABC.
(ii) If ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF, then ΔDEF NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ΔABC.
(iii) if ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF, and ΔDEF NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔPQR, then ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔPQR.

CRITERIA FOR CONGRUENCE OF TRIANGLES
In earlier classes, we have learnt some criteria for congruence of triangles. Here, in this class we will learn the truth of these either experimentally or by deductive proof. In the previous section, we have studied that two triangles are congruent if and only if there exists a correspondence between their vertices such that the corresponding sides and the corresponding angles of two triangles are equal i.e.
six equalities hold good, three of the corresponding sides and three of the corresponding angles. In this section, we shall prove that if three properly chosen conditions out of the six conditions are satisfied, then the other three are automatically satisfied. Let us now discuss those three conditions which ensure the congruence of two angles.

I . SIDE-ANGLES-SIDE (SAS) CONGRUENCE CRITERION
AXIOM : Two triangles are congruent if two sides and included angle of one triangle are equal to the sides and the included angle of the other triangle. In the given figure, in ΔABC and ΔDEF, we have :
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
AB = DE, AC = DF and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF [By SAS-criteria]

REMARK : SAS Congruence rule holds but ASS or SSA rule does not hold.

II. ANGLE-SIDE-ANGLE (ASA) CONGRUENCE CRITERION
THEOREM-1 : Two triangles are congruent, if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.
Given : Two triangles ABC and DEF

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
such that NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9F and BC = EF.
To prove : ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ΔDEF.
Proof : To prove ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF, we need to consider three possible situations :
Case 1. Let AB = DE.

STATEMENT REASON

1.

 

 

 

2.

In ΔABC and ΔDEF

(i) AB = DE

(ii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E

(iii)  BC = EF Given

ΔABCNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ΔDEF 

 

Supposed

Given

Given

By SAS criteria

Case 2. Let AB > DE.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Construction : Take a point H on AB such that HB = DE.

STATEMENT REASON

1.

 

 

 

2.

3.

4.

5.

In ΔHBC and ΔDEF

(i) HB = DE

(ii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E

(iii) BC = EF 

 ΔHBC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF

 ΔHCB NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDFE

 ΔACB NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDFE

 ΔHCB NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔACB

 

By Construction

Given

Given

By SAS criteria

C.P.C.T.

Given

From (3) and (4)

It can be possible only if H coincides with A. In other words, AB = DE.

Case 3. Let AB < DE.
Construction : Take a point M on DE such that ME = AB.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
By repeating the same arguments as in Case 2, we can prove that AB = DE, and so ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF.
Hence, proved.

III. ANGLE-ANGLE-SIDE (AAS) CONGRUENCE CRITERION
THEOREM-2 : Two triangles are congruent, if any two pairs of angles and one pair of corresponding sides are equal.
Given : Two Δs ABC and DEF such that NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E, BC = EF.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
To prove : ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF. B C E F
Proof :

STATEMENT REASON

1.

2.

3.

 

 

 

4.

 

 

 

5.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D

 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9180° – NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 180° – NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9F

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C =NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9F

 

In ΔABC and ΔDEF,
(i) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E
(ii) BC = EF Given
(iii)NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9F

ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF

Given

Given

Adding (1) and (2)

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 180°                         NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 180° – NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C
Similarly, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9D + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9E = 180° – NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9F

Given

Given

From (3) 

By ASA criteria

Hence, proved.

IV. SIDE-SIDE-SIDE (SSS) CONGRUENCE CRITERION

Two triangles are congruent, if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
In the given figure, in ΔABC and ΔDEF, we have :
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
AB = DE, BC = EF and AC = DF.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ΔDEF [By SSS-criteria]

V. RIGHT ANGLE-HYPOTENUSE-SIDE (RHS) CONGRUENCE CRITERION
Two right angles are congruent, if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle.
In the given figure, ΔABC and ΔDEF are right-angled triangles in which Hyp. AC = Hyp. DF and BC = EF.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔDEF [By RHS-criteria]

Ex.1 In fig. OA = OB and OD = OC. Show that [NCERT]
(i) Δ AOD NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 Δ BOC and (ii) AD || BC. 
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Sol. Given : In fig. OA = OB and OD = OC.
To prove : (i)ΔAODNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ΔBOC and (ii) AD || BC.
Proof :

STATEMENT REASON

1.

 

 

 

 

2.

3.

4.

In ΔAOD and ΔBOC

(i) OA = OB 

(ii) OD = OC

(iii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AODNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BOC

 ΔAOD NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔBOC

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OBC

 AD || BC.

 

Given

Given

Vertically opposite angles

 

By SAS criteria

C.P.C.T.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OAD andNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OBC form a pair of
alternate angles and are equal.

Hence, proved.

Ex.2 In ΔABC, AB = AC. If P is a point on AB and Q is a point on AC such that AP = AQ.
Prove that
(i) ΔAPC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔAQB (ii) ΔBPC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCQB.
Sol. Given : In ΔABC, AB = AC. P is a point on AB and Q is a point on AC such that AP = AQ.
To prove : (i) ΔAPC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔAQB (ii) ΔBPC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCQB.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Proof :

STATEMENT REASON

1.

 

 

 

 

2.

3.

4.

 

 

5.

In ΔAPC and ΔAQB

(i) AP = AQ 
(ii) AC = AB 
(iii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CAP = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAQ

ΔAPC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔAQB

 

AB = AC and AP = AQ
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9(AB – AP) = (AC – AQ)
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BP = CQ

In ΔBPC and ΔCQB
(i) BP = CQ
(ii) PC = QB
(iii) BC = CB

 ΔBPC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCQB

 

Given

Given

Common

 

By SAS criteria

Given

 

From (3)

C.P.C.T. APC and AQB

Common

By SSS criteria

Hence, proved.

Ex.3 In figure, AC = AE, AB = AD andNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EAC, prove that BC = DE. [NCERT]
Sol. Given : In figure, AC = AE, AB = AD andNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EAC.
To prove : BC = DE.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Construction : Joint DE.

STATEMENT REASON

1.

 

 

 

 

2.

3.

In ΔABC and ΔADE

(i) AB = AD Given
(ii) AC = AE Given
(iii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EAC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAD + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DAC = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DAC + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EAC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAC = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DAE

ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔADE

BC = DE

 

Given

Given

Given

Adding NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DAC to both sides

By SAS criteria

C.P.C.T.

Hence, proved.

Ex.4 In figure, diagonal AC of a quadrilateral ABCD bisects the angles A and C. Prove that AB = AD and CB = CD.
Sol. Given : In figure, diagonal AC of a quadrilateral ABCD
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
bisects the angles A and C.

To prove : AB = AD and CB = CD
 

STATEMENT REASON

1.

 

 

 

2.

3.

In ΔABC and ΔADC

(i) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAC = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DAC or NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 91 = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

(ii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD or NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 93 = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

(iii) AC = AC

ΔABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔADC

AB  AD .

BC = CD 

 

Given

Given

Common

ASA Axiom

C.P.C.T.

C.P.C.T

Hence, proved.

Ex.5 AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABE and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EPA = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DPB. Show that (i) ΔDAP NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔEBP (ii) AD = BE. [NCERT]
Sol. Given : AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABE andNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EPA = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DPB. E D
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
To prove : (i) ΔDAP NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔEBP
(ii) AD = BE
Proof :

STATEMENT REASON

1.

 

 

 

 

 

2.

3.

In ΔDAP and ΔEBP,

(i) AP = BP

(ii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DAP =NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EBP

(iii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EPA = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DPB
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EPA + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EPD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9EPD + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DPB
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9APD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BPE

ΔDAP NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔEBP

AD = BE

 

P is the mid point of AB

Given

Given

Adding NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9​EPD to both sides.

 

By ASA criteria

C.P.C.T.

Hence, proved.

 

PROPERTIES OF AN ISOSCELES TRIANGLE In this section, we will learn some properties related to a triangle whose two sides are
equal. We know that a triangle whose two sides are equal is called an isosceles triangle. Here, we will apply SAS congruence criteria and ASA (or AAS) congruence criteria to study some properties of an isosceles triangle.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
THEOREM-3 : Angles opposite to equal sides of an isosceles triangle are equal.
Given : ΔABC is an isosceles triangle and AB = AC.
To prove : NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C.
Construction : Draw AD the bisector of NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A. AD meets BC at D.
Proof :

STATEMENT REASON

1.

 

 

 

2.

3.

1. In ΔBAD and ΔCAD

(i) AB = AC

(ii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BAD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CAD

(iii) AD = AD

ΔBAD NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔCAD

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

 

Given

Construction

Common.

By ASA criteria

C.P.C.T.

Hence, proved.

COROLLARY : Each angle of an equilateral triangle is of 60°.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Given : ΔABC is an equilateral triangle.
To prove : NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 60°.
Proof :

STATEMENT REASON

1.

2.

3.

4.

5.

AB = AC = BC

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B +NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 180°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A=NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B=NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C= 13 × 180°=60°

ΔABC is an equilateral triangle.

AB = AC

AB = BC

From (2) and (3)

Angle sum property.

 

Hence, proved.

THEOREM-4 : The sides opposite to equal angles of a triangle are equal.
Given : ΔABC in which NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
To prove : AB = AC.
Construction : Draw AD NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 BC. AD meets BC in D.
Proof :

STATEMENT REASON

1.

 

 

 

2.

3.

InΔABD and ΔACD

(i) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD 

(ii) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ADB = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ADC

(iii) AD = AD 

ΔABD NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔACD 

 AB = AC 

 

Given

Each = 90°

Common.

By ASA criteria

C.P.C.T.

Hence, proved.

Ex.16 In the adjoining fig, find the value of x. B
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9.

Ex.17 ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD is a right angle. [NCERT]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Sol. Given : ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB
To prove : NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD = 90°
Proof :

STATEMENT REASON

1.

 

2.

3.

 

 

.4.

 

5.

 

 

 

 

 

AB = AC

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB

AB = AD

AC = AD

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CDA =NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD
or NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CDB = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC+NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CDB=NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB+NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CDB = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD

In ΔBCD
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DBC + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CDB = 180°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CDB = 180°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD = 180°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 92NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD = 180°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD = 90°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD is a right angle.

Given

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9s opposite to equal sides of a Δ are equal

Given

From (1) & (2)

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9s opposite to equal sides of a Δ are equal

 

Adding (1) & (3)

 

Angle sum property

 

Using (4)

 

Hence, proved.

INEQUALITIES IN A TRIANGLE

THEOREM-5 : If two sides of a triangle are unequal, then the greater side has greater angle opposite to it.
Given : A ΔABC in which AC > AB.
To prove : NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Construction : Mark a point D on AC such that AD = AB. Join BD. D
Proof :

STATEMENT REASON

1.

2.

3.

4.

5.

6.

7.

AB = AD

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BDA

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BDA > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCB

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABD > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCB

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABD

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCB

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB

By construction

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9s opposite to equal sides of a Δ are equal

(Ext. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 of ΔBCD) > (Each of its int. Opp. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9s)

Using (2)

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABD is a part of NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC.

Using (5)

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9DCB = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB.

 

THEOREM-6 (Converse of Theorem-5) : If two angles of a triangle are unequal, then the greater
angle has greater side opposite to it.

Given : A ΔABC in which NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB.
To prove : AC > AB.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Proof :

STATEMENT REASON

 

We may have three possibilities only :
(i) AC = AB
(ii) AC < AB
(iii) AC > AB
Out of these exactly one must be true.

Case-I.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB
This is contrary to what is given.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AC = AB is not true.

Case-II. AC < AB.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB > AC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACB > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC
This is contrary to what is given.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AC < AB is not true.
Thus, we are left with the only possibility
AC > AB, which must be true.

 

 

AC = AB.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9s opposite to equal sides of a Δ are equal

 

 

 

Greater side has greater angle opp. to it.

Hence, proved.


THEOREM-7 : The sum of any two sides of a triangle is greater than its third side.
Given : A ΔABC.
To prove : (i) AB + AC > BC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
(ii) AB + BC > AC
(iii) BC + AC > AB.
Construction : Produce BA to D such that AD = AC. Join CD.
Proof :

STATEMENT REASON

1.

 

2.

3.

 

 

4.

 

AD = AC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ADC

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BCD > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ADC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BD > BC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BA + AD > BC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BA + AC > BC
or AB + AC > BC.

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

By construction

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9s opposite to equal sides of a Δ are equa

Using (1) & (2)

Greater angle has greater side opp. to it.

BAD is a straight line, BD = BA + AD.

AD = AC, by construction

 

REMARK :(i) The largest side of a triangle has the greatest angle opposite to it and converse is also true.
(ii) The smallest side of a triangle has the smallest angle opposite to it and converse in also true.
Ex.19 In fig, show that :
(i) AB > AC (ii) AB > BC and (iii) BC > AC.
Sol. Given : A ΔABC in which NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 40° and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ACD = 100°.
To prove : (i) AB > AC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
(ii) AB > BC D
(iii) BC > AC.
Proof :

STATEMENT REASON

1.

 

2.

 

3.

 

4.

5.

 

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 100°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A + 40° = 100°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = 60°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C + 100° = 180°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 80°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AB > AC
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AB > BC .
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A > NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9BC > AC

Ext. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 = sum of int. opt. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9s

Linear pair of angles.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 80° and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 40°

Greater angle has greater side opp. to it.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9C = 80° and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = 60°

Greater angle has greater side opp. to it

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9A = 60° and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9B = 40°

Greater angle has greater side opp. to it.

 

THINGS TO REMEMBER

1. Two figures are congruent, if they are of the same shape and of the same size.

2. Two circles of the same radii are congruent.

3. Two squares of the same sides are congruent.

4. If two triangles ABC and PQR are congruent under the correspondence A NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 P, B NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 Qand C NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 R, then symbolically, it is expressed as Δ ABC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 Δ PQR.

5. If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, then the two triangles are congruent (SAS congruence rule).

6. If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent (ASA congruence rule).

7. If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent (AAS Congruence rule).

8. Angle opposite to equal sides of a triangle are equal.

9. Sides opposite to equal angles of a triangle are equal.

10. Each angle of an equilateral triangle is of 60°.

11. If three sides of one triangle are equal to three sides of the other triangle, then the two triangles are congruent (SSS congruence rule).

12. If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent (RHS congruence rule).

13. In a triangle, angle opposite to the longer side is larger (greater).

14. In a triangle, side opposite to the larger (greater) angle is longer.

15. Sum of any two sides of a triangle is greater than the third side.

 

 

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