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 . SIDEANGLESSIDE (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 SAScriteria]
REMARK : SAS Congruence rule holds but ASS or SSA rule does not hold.
II. ANGLESIDEANGLE (ASA) CONGRUENCE CRITERION
THEOREM1 : 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
such that B = E, C = F and BC = EF.
To prove : ΔABC ΔDEF.
Proof : To prove ΔABC 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) B = E (iii) BC = EF Given ΔABCΔDEF 
Supposed Given Given By SAS criteria 
Case 2. Let AB > DE.
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) B = E (iii) BC = EF ΔHBC ΔDEF ΔHCB ΔDFE ΔACB ΔDFE ΔHCB 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.
By repeating the same arguments as in Case 2, we can prove that AB = DE, and so ΔABC ΔDEF.
Hence, proved.
III. ANGLEANGLESIDE (AAS) CONGRUENCE CRITERION
THEOREM2 : 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 A = D, B = E, BC = EF.
To prove : ΔABC ΔDEF. B C E F
Proof :
STATEMENT  REASON  
1. 2. 3.
4.
5.  A = D B = E A B = D = E C =F
In ΔABC and ΔDEF, ΔABC ΔDEF  Given Given Adding (1) and (2) A + B + C = 180° A + B = 180° – C Given Given From (3) By ASA criteria 
Hence, proved.
IV. SIDESIDESIDE (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 SSScriteria]
V. RIGHT ANGLEHYPOTENUSESIDE (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 rightangled triangles in which Hyp. AC = Hyp. DF and BC = EF.
ΔABC ΔDEF [By RHScriteria]
Ex.1 In fig. OA = OB and OD = OC. Show that [NCERT]
(i) Δ AOD Δ BOC and (ii) AD  BC.
Sol. Given : In fig. OA = OB and OD = OC.
To prove : (i)ΔAODΔBOC and (ii) AD  BC.
Proof :
STATEMENT  REASON  
1.
2. 3. 4.  In ΔAOD and ΔBOC (i) OA = OB (ii) OD = OC (iii) AOD BOC ΔAOD ΔBOC OAD = OBC AD  BC. 
Given Given Vertically opposite angles
By SAS criteria C.P.C.T. OAD andOBC form a pair of 
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 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.
Proof :
STATEMENT  REASON  
1.
2. 3. 4.
5.  In ΔAPC and ΔAQB (i) AP = AQ ΔAPC ΔAQB
AB = AC and AP = AQ In ΔBPC and ΔCQB ΔBPC 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 andBAD = EAC, prove that BC = DE. [NCERT]
Sol. Given : In figure, AC = AE, AB = AD andBAD = EAC.
To prove : BC = DE.
Construction : Joint DE.
STATEMENT  REASON  
1.
2. 3.  In ΔABC and ΔADE (i) AB = AD Given ΔABC ΔADE BC = DE 
Given Given Given Adding DAC 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
bisects the angles A and C.
To prove : AB = AD and CB = CD
STATEMENT  REASON  
1.
2. 3.  In ΔABC and ΔADC (i) BAC = DAC or 1 = 2 (ii) ACB = ACD or 3 = 4 (iii) AC = AC ΔABC 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 midpoint. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB. Show that (i) ΔDAP ΔEBP (ii) AD = BE. [NCERT]
Sol. Given : AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that BAD = ABE andEPA = DPB. E D
To prove : (i) ΔDAP ΔEBP
(ii) AD = BE
Proof :
STATEMENT  REASON  
1.
2. 3.  In ΔDAP and ΔEBP, (i) AP = BP (ii) DAP =EBP (iii) EPA = DPB ΔDAP ΔEBP AD = BE 
P is the mid point of AB Given Given Adding 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.
THEOREM3 : 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 :
STATEMENT  REASON  
1.
2. 3.  1. In ΔBAD and ΔCAD (i) AB = AC (ii) BAD = CAD (iii) AD = AD ΔBAD ΔCAD B = C 
Given Construction Common. By ASA criteria C.P.C.T. 
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 :
STATEMENT  REASON  
1. 2. 3. 4. 5.  AB = AC = BC B = C A = C A = B = C A + B +C = 180°  ΔABC is an equilateral triangle. AB = AC AB = BC From (2) and (3) Angle sum property.

Hence, proved.
THEOREM4 : 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 :
STATEMENT  REASON  
1.
2. 3.  InΔABD and ΔACD (i) ABD = ACD (ii) ADB = ADC (iii) AD = AD ΔABD 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
.
Ex.17 ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that BCD is a right angle. [NCERT]
Sol. Given : ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB
To prove : BCD = 90°
Proof :
STATEMENT  REASON  
1.
2. 3.
.4.
5.
 AB = AC ABC = ACB AB = AD AC = AD CDA =ACD ABC+CDB=ACB+ACD In ΔBCD  Given s opposite to equal sides of a Δ are equal Given From (1) & (2) s opposite to equal sides of a Δ are equal
Adding (1) & (3)
Angle sum property
Using (4)

Hence, proved.
INEQUALITIES IN A TRIANGLE
THEOREM5 : 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 : ABC > ACB.
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 ABD = BDA BDA > DCB ABD > DCB ABC > ABD ABC > DCB ABC > ACB  By construction s opposite to equal sides of a Δ are equal (Ext. of ΔBCD) > (Each of its int. Opp. s) Using (2) ABD is a part of ABC. Using (5) DCB = ACB. 
THEOREM6 (Converse of Theorem5) : If two angles of a triangle are unequal, then the greater
angle has greater side opposite to it.
Given : A ΔABC in which ABC > ACB.
To prove : AC > AB.
Proof :
STATEMENT  REASON  
 We may have three possibilities only : CaseI. ABC = ACB CaseII. AC < AB. 
AC = AB.
Greater side has greater angle opp. to it. 
Hence, proved.
THEOREM7 : The sum of any two sides of a triangle is greater than its third side.
Given : A ΔABC.
To prove : (i) AB + AC > BC
(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 BCD > ACD BCD > ADC Similarly, AB + BC > AC  By construction s 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 B = 40° and ACD = 100°.
To prove : (i) AB > AC
(ii) AB > BC D
(iii) BC > AC.
Proof :
STATEMENT  REASON  
1.
2.
3.
4. 5.
 A + B = 100°  Ext. = sum of int. opt. s Linear pair of angles. C = 80° and B = 40° Greater angle has greater side opp. to it. C = 80° and A = 60° Greater angle has greater side opp. to it A = 60° and B = 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 P, B Qand C R, then symbolically, it is expressed as Δ ABC 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. 