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. In ΔBAD and ΔCAD 

2. ΔBAD ≌ ΔCAD  By SAS criteria 
3. ∠B = ∠C  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. AB = AC = BC  ΔABC is an equilateral triangle. 
2. ∠B = ∠C  AB = AC 
3. ∠A = ∠C  AB = BC 
4. ∠A = ∠B = ∠C  From (2) and (3) 
5. ∠A + ∠B + ∠C = 180°  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. In ΔABD and ΔACD  Given 
2. ΔABD ≌ ΔACD  By AAS criteria 
3. AB = A C  C.P.C.T. 
Ex.16 In the adjoining fig, find the value of x.
Sol. We have, ∠CAD + ∠ADC + ∠DCA = 180° [Angle sum property] ⇒ ∠CAD + ∠ADC + 64° = 180°
⇒ ∠CAD + ∠ADC = (180° – 64°) = 116°
But CD = CA ⇒ ∠CAD = ∠ADC [∠s opposite to equal sides of a D are equal]
So, ∠CAD = ∠ADC
Now, ∠AD C = ∠AB D + ∠DA B [Ext. ∠ of a D = sum of int. opp. ∠s]
But, AD = BD ⇒ ∠ABD = ∠DAB.
So, ∠ADC = 2 ∠DAB ⇒ ∠DAB = 1/2 ∠AD C
Hence, x = 29.
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.
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. AB = A C ⇒ ∠ABC = ∠ACB  Given ∠s opposite to equal sides of a D are equal 
2. A B = AD  Given 
3. AC = A D  From (1) & (2) ∠s opposite to equal sides of a D are equal 
4. ∠A B C + ∠C D B = ∠A C B + ∠A C D  Adding (1) & (3) 
5. In ΔBCD ∠BCD + ∠DBC + ∠CDB = 180°  Angle sum property Using (4) 
Hence, proved.
Ex.18 In a right angled triangle, one acute angle is double the other.
Prove that the hypotenuse is double the smallest side.
Sol. Given : ΔABC in which ∠B = 90° and ∠ACB = 2 ∠CAB.
To prove : AC = 2BC.
Construction : Produce CB to D such that BD = BC. Join AD.
Proof :
STATEMENT  REASON 
1. In ΔABD and ΔABC  By construction. Each = 90° Common 
2. ΔABD ≌ ΔA B C  By SAS criteria 
3. AD = AC, ∠DAB = ∠CAB = x°  C.P.C.T 
4. A D = D C  Sides opposite to equal ∠s are equal. 
5. A C = D C  AD = AC, from (3) 
6. AC = 2 BC  DC = BD + BC = 2BC, as BD = BC. 
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 : ΔABC in which AC > AB.
To prove : ∠ABC > ∠ACB.
Construction : Mark a point D on AC such that AD = AB. Join BD.
Proof : ST
STATEMENT  REASON 
1. A B = A D  By construction 
2. ∠A B D = ∠B D A  ∠s opposite to equal sides of a D are equal 
3. ∠B D A > ∠D C B  (Ext. ∠ of ΔBCD) > (Each of its int. Opp. ∠s) 
4. ∠A B D > ∠D C B  Using (2) 
5. ∠AB C > ∠AB D  ∠ABD is a part of ∠ABC. 
6. ∠AB C > ∠DC B  Using (5) 
7. ∠ABC > ∠AC B  ∠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 : ΔABC in which ∠ABC > ∠ACB.
To prove : AC > AB.
Proof :
STATEMENT  REASON 
We may have three possibilities only : (i) AC = AB  AC = AB. 
⇒ ∠ABC = ∠ACB. This is contrary to what is given.∴ AC = AB is not true.  ∠s opposite to equal sides of a D are equal 
CaseII. AC < AB. ⇒ AB > AC This is contrary to what is given. ∴ AC < AB is not true.  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 : 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. A D = A C  By construction 
2. ∠B C D > ∠A C D  ∠s opposite to equal sides of a D are equal 
3. ∠B C D > ∠A D C  Using (1) & (2) Greater angle has greater side opp. to it. BAD is a straight line, BD = BA + AD. 
4. Similarly, AB + BC > AC and BC + AC > AB.  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 (iii) BC > AC.
Proof :
STATEMENT  REASON 
1. ∠A + ∠B = 100°  Ext. ∠ = sum of int. opt. ∠s 
2. ∠C + 100° = 180°  ∠C = 80° and ∠B = 40° 
3. ∠C > ∠B  ∠C = 80° and ∠A = 60° 
4. ∠C > ∠A  ∠A = 60° and ∠B = 40° 
5. ∠A > ∠B  Greater angle has greater side opp. to it. 
Hence, proved.
Ex.20 Show that of all the line segments that can be drawn to a given straight line from a given point outside it, the perpendicular is the shortest.
Sol. Given : A straight line AB and a point P outside it; PM ⊥ AB and N is any other point on AB.
To prove : PM < PN.
Proof :
STATEMENT  REASON 
1. ∠PMN = 90°  P M ⊥ A B 
2. ∠PNM < 90°  In a right D, one angle measures 90° and each one of the remaining two is acute. 
3. ∠P N M < ∠P M N  From (1) and (2) 
4. P M < P N  Side opp. the smaller angle is smaller. 
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 ⇔ Q and 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. 
1 videos228 docs21 tests

1. What is an isosceles triangle? 
2. What are the properties of an isosceles triangle? 
3. Can an isosceles triangle have all angles equal? 
4. How can we find the measure of the angles in an isosceles triangle? 
5. Are all equilateral triangles isosceles? 
1 videos228 docs21 tests


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