Chapter 7 - Triangles, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 PDF Download

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 .
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. 

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  ΔDEF. This happens when AB = DE, BC = EF, AC =
DF and A = D, B = E, C = F.
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 A, B and C are D, E and F 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  ΔABC.
(ii) If ΔABC  ΔDEF, then ΔDEF  ΔABC.
(iii) if ΔABC ΔDEF, and ΔDEF  ΔPQR, then ΔABC  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 :

AB = DE, AC = DF and A = D.
 ΔABC  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.

Proof : To prove ΔABC  ΔDEF, we need to consider three possible situations :

Case 1. Let AB = DE

Case 2. Let AB > DE.

Construction : take a point H on AB such that HB = DE

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.

By repeating the same arguments as in Case 2, we can
prove that AB = DE, and so ΔABC  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.

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 :

AB = DE, BC = EF and AC = DF.
 ΔABC   ΔDEF [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.

Ex.1 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 ΔAQB (ii) ΔBPC  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 ΔAQB

(ii) ΔBPC ΔCQB.

Ex.2 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 bisects the angles A and C.

To prove : AB = AD and CB = CD

Ex. 3 AB is a line-segment. AX and BY are two equal line-segments drawn on opposite sides of line AB such that AX || BY. If AB and XY intersect each other at P.
 Prove that : 

(i) ΔAPX  ΔBPY 

(ii) AB and XY bisect each other at P.

 

Sol. Given : AB is a line-segment. AX and BY are two equal line-segments drawn on opposite
sides of line AB such that AX || BY.
To prove :

(i) ΔAPX  ΔBPY

(ii) AB and XY bisect each other at P.

Ex.4 In fig., AD is a median and BL, CM are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BL = CM.
 Sol. Given : In fig., AD is a median and BL, CM are perpendiculars drawn from B and C respectively on AD and AD produced. D

Ex.5 In fig. BCD = ADC and ACB = BDA.

Prove that AD = BC and A = B.
 Sol. Given : In fig. BCD = ADC and ACB = BDA
To prove : (i) AD = BC (ii) A = B

Hence, proved.

Ex.6 In the fig. ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that :
 (i) AC bisects each of the angles A and C.
 (ii) BE = ED.
 (iii) ABC = ADC. Can we say that AE = EC? 

Sol. Given : In the fig. ABCD is a quadrilateral in which
AB = AD and BC = DC.
 

To prove : (i) AC bisects each of the angles A and C.
(ii) BE = ED.
(iii) ABC = ADC. Can we say that AE = EC?

No, AE  EC. For AE to be equal to EC, it is necessary that ΔAED  ΔCED for which 2 must be equal to 4.
Hence, proved.

Ex. 7 ΔABC and ΔPBC are two isosceles triangles on the same base BC and vertices A and P are on the same side of BC.
 A and P are joined, show that :
 (i) ΔABP  ΔACP and 

(ii) AP bisects A of ΔABC.
 

Sol. Given : ΔABC and ΔPBC are two isosceles triangles on the same base BC and vertices A and P are on
the same side of BC.
To prove : (i) ΔABP ΔACP and (ii) AP bisects A of ΔABC.

Ex.8 AD, BE and CF, the altitudes of ΔABC are equal. Prove that ΔABC is an equilateral triangle.

Sol. Given : AD, BE and CF are the altitudes of ABC and are equal. AP = BE = CF
To prove : ΔABC is an equilateral triangle

Ex. 9 In a ΔABC, the internal bisectors of B and C meet at O. Prove that OA is the internal bisector of A.

Sol. Given : In a ΔABC, the internal bisectors of B and C meet at O.
To prove : OA bisects A.
Construction : Draw OD  BC, OE  CA and OF  AB.
Proof :

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.

THEOREM-3 : Angles opposite to equal sides of an isosceles triangle are equal.
Given : ΔABC is an isosceles triangle and AB = AC.
To prove :B = C.
Construction : Draw AD the bisector of A. AD meets BC at D.
Proof :

Hence, proved.

COROLLARY : Each angle of an equilateral triangle is of 60°.

Given : ΔABC is an equilateral triangle.
To prove : A = B = C = 60°.
Proof :

THEOREM-4 : The sides opposite to equal angles of a triangle are equal.
 Given : ΔABC in which B = C.

To prove : AB = AC.
Construction : Draw AD  BC. AD meets BC in D.
Proof :

Hence, proved.

Ex.10 In the adjoining fig, find the value of x. 

 

Sol. We have,
CAD + ADC + DCA = 180° [Angle sum property]

But CD = CA ⇒ CAD = ADC            [s opposite to equal sides of a Δ are equal]

Now, ADC = ABD + DAB [Ext.  of a Δ = sum of int. opp. s]