Subject: Compulsory Maths
Triangles are threesided closed figures which have three straight sides joined at three vertices and have three angles enclosed within the figure at the vertices. There are several types of triangles based on the lengths of its sides and the angles they contain. What many do not know is that a triangle is a threesided polygon.
The basic concepts of geometry which have been dealt with previous classes and necessary to study in this class are listed below. A definite image which has no length, breadth and thickness is called a point. It is denoted by symbol dot (.) and capital letters A, B, C etc.
Straight line: The shortest locus which passes through two fixed points is called the straight line. It can be extended on both sides.
Line segment: A certain portion of a straight line is called the line segment. It has a fixed measurement.
Curved line: The line joining two fixed points without fixed direction is called the curved line.
Parallel lines: Any two or more than two straight lines which intersect each other after extending or the length of the perpendicular distance between them is always equal are parallel lines.
Angle: When any two straight lines or line segments meet at a point, they form a corner which is called an angle.
1. Adjacent angles
A pair of angles having same vertex and a common side is called adjacent angles. If the exterior sides of both angles lie on a straight line then their sum is two right angles.
2 . Vertically opposite angles
When two straight line segments intersect at a point then a pair of angles formed in opposite each other is called vertically opposite angles.
1. Equilateral triangles
A triangle whose all sides are equal is known as an equilateral triangle. Each angle of an equilateral triangle are 60^{o}. So, an equilateral triangle is also called as equiangular i.e all there internal angles are also congruent to each other and are each 60^{o}. They are regular polygons.
2. Isosceles triangle
A triangle whose (at least) two sides are equal is called an isosceles triangle. Base angles of an isoceles triangle are equal. An isoceles triangle, therefore, has both two equal sides and two equal angles. In \(\triangle\)ABC, ∠B = ∠C ∴ \(\triangle\)ABC is an isosceles triangle.
3. Scalene triangle
A triangle whose all three sides are unequal is called a scalene triangle. In \(\triangle\)XYZ, none of the sides are equal. ∴ \(\triangle\)XYZ is a scalene triangle.
1. Acuteangled triangle: A triangle whose all angles are acute angles (less than 90^{o}) is called an acuteangled triangle. In the given figure, all angles are less than 90^{o} so it is an acuteangled triangle.
2. Obtuseangled triangle: A triangle whose one angle is obtuse (greater than 90^{o}) is called an obtuseangled triangle. In the given figure, ∠Y is 120^{o }(greater than 90^{o}) ∴ \(\triangle\)XYZ is an obtuseangled triangle.
3. Rightangled triangle: A triangle whose one angle is right angle (90^{o}) is called rightangled triangle. In \(\triangle\)ABC, ∠B = 90^{o}, so \(\triangle\)ABC, is a rightangled triangle.
Properties of triangles
Triangles are threesided closed figures which have three straight sides joined at three vertices and have three angles enclosed within the figure at the vertices. There are several types of triangles based on the lengths of its sides and the angles they contain. What many do not know is that a triangle is a threesided polygon.
Axoims
An axiom is a selfevident truth which is wellestablished, that accepted without controversy or question. Some of the axioms are presented in the following table with their conditions:
Axoims  Conditons 
Equality of addition 
When the equal quantities are added to both sides of equal quantities, the sum is also equal. e.g if a=b then a+c=b+c 
Equality of subtraction 
When the equal quantites are subtracted from both sides of equal quantities, the difference is also equal. e.g. if a=b then ac=bc 
Equality of multiplication 
When the equal quantites are multiplied by the same quantity, the product is also equal. e.g. if a=b then a*c=b*c 
Equality of division 
When the equal quantites are divided by the same quantity, the quotient is also equal. e.g. if a=b then a/c = b/c 
Equality axiom  When two separate quantites are equal to a quantity the they is also equal to each other, If a=c and b=c then a=c 
Whole part of axiom 
A whole quantity is always equal to the sum of its all parts and the whole quantity is always greater than each of its parts. e.g. AD=AB+BC+CD and AD>AB or AD>BC or AD>CD 
Substitution axiom 
A quantity can be replaced by another equal quantity. This doesn't alter the final result. If, a=b then ax+c can be expressed as bx+c. 
Postulates
A statement which is taken to be true without proof is called a postulate. Postulates are the basic structure from which lemmas and theorems are derived. Some of the postulates that we need in our geometry are listed below.
1. There is only one straight line that you can draw between any two points.
2. An infinite number of straight lines can be drawn through a point.
3. A line contains exactly at least two points.
4. Through any three noncollinear points, there exist exactly one plane.
5. Only one line can be the bisector of a given angle.
6. A straight line can be produced on either side infinitely.
7. Through one point, there is only line parallel to the first line.
8. The length of perpendicular means the distance between a point and a line.
Let's draw at least two parallel lines AB and XY. The transversal line CD intersects AB at E and XY at F. We can establish the following relations among the angles formed:
a) Alternate angles are equal: If two parallel lines are intersected by a transversal line, then the alternate angles so formed are equal. ∴ ∠AEF = ∠EFY and ∠BEF = ∠EFX
b) Corresponding angles are equal: If two parallel lines are intersected by a transversal line, then the corresponding angles so formed are equal.
∴ ∠AED = ∠EFX, ∠AEF = ∠XFC
∠DEB = ∠EFY, ∠BEF = ∠YFC
c) The sum of cointerior angles is 180^{o}: If two parallel lines are intersected by a transversal line, then the sum of the cointerior angles so formed is 180^{o}.
∴ ∠AEF + ∠EFX = 180^{o}
∠BEF + ∠EFY = 180^{o}
d) The sum of coexterior angles is 180^{o}: If two parallel lines are intersected by a transversl line, then the sum of the coexterior angles so formed is 180^{o}.
∴ ∠AED + ∠XEC = 180^{o }and ∠DEB + ∠YEC = 180^{o}
Experimental Verification:
Step1: Let's draw three triangles of different orientations sizes. And name them as \(\triangle\)ABC each. (Draw the triangles in such a way that you can measure each angle by using a protractor.)
Step 2: Measure each angle of these three triangles with the help of protractor and fill up the following table.
Fig.  ∠A  ∠B  ∠C  Results 
(i)  ∠A +∠B +∠C =  
(ii)  ∠A +∠B +∠C =  
(iii)  ∠A +∠B +∠C = 
Conclusion :
Theoretical proof:
Given: ∠ABC, ∠BCA and ∠BAC are three angles of triangle ABC.
To proof: ∠ABC + ∠BCA + ∠BAC = 180^{o}
Construction: Let's draw a straight line parallel to BC through A.
Proof:
S.N.  Statements  Reasons 
1  ∠XAB = ∠ABC  XY \(\parallel\) BC and being alternate angles. 
2  ∠YAC = ∠ACB  XY \(\parallel\) BC and being alternate angles. 
3  ∠XAB + ∠BAC + ∠YAC = ∠XAY  According to wholepart axiom. 
4  ∠ABC + ∠BAC + ∠ACB = ∠XAY  From statement (1), (2) and (3). 
5  ∠XAY = 180^{0}  A straight angle constitutes 180^{o }(two right angles) 
6  ∴ ∠ABC + ∠BAC + ∠ACB = 180^{o }or 2 right angles.  Substituting the value of ∠XAY into a statement (4). 
Proved
Experimental verification:
Step 1: Let's draw three triangles of different size in different orientations. Produce side BC to D in each figure. In each figure, ∠ACD is an exterior angle and ∠A and ∠B are two nonadjacent angles inside the triangle.
Step 2: Measure the size of exterior angle and other two nonadjacent angles in each figure and complete the table.
Fig.  Exterior angle ∠ACD  ∠A  ∠B  Results 
(i)  
(ii)  
(iii) 
Conclusion:
Theoretical proof:
Given: ABC is a triangle whose side BC is extended upto D such that ∠ACD ia an external angle.
To prove: ∠ACD = ∠ABC + ∠BAC
Proof:
S.N.  Statement  Reasons 
1  ∠ACD + ∠BCA = 180^{o}  The sum of adjacent angles in a straight line is 180^{o}. 
2  ∠ABC + ∠BCA + ∠BAC = 180^{o}  The sum of angles in a triangle is 180^{o}. 
3  ∠ACD + ∠BCA = ∠ABC + ∠BCA + ∠BAC  From statements (1) and (2). 
4  ∠ACD = ∠ABC + ∠BAC  Cancelling ∠BCA on both sides of the statement (3). 
Proved
Alternative method:
To prove: ∠ACD = ∠ABC + ∠BAC
Construction: Let us draw CE parallel to BA.
Proof:
S.N.  Statements  Results 
1  ∠ACD = ∠ACE + ∠DCE  Whole part axiom. 
2  ∠DCE = ∠ABC  CE \(\parallel\) BA being corresponding angles. 
3  ∠ACE = ∠BAC  CE \(\parallel\) BA being alternate angles. 
4  ∠ACD = ∠BAC + ∠ABC  From statements (1), (2) and (3). 
Proved
Isosceles triangle is that triangle whose two sides are equal. Triangle is symmetric being two sides equal hence it possesses different properties. For example, the base angles of an isosceles triangle are equal. This kind of properties is proved as theoretical proof here which duly needs the conditions of congruency of triangles. Before this, we discuss the different conditions of congruency of triangles in a brief.
There are 3 sides and 3 angles in a triangle. Two triangles are congruent when 3 parts (out of 6 parts) of one triangle are equal to 3 corresponding parts of another triangle
as per the following conditions. We take these conditions as axioms.
The two triangles are said to be congruent when three sides of a triangle are respectively equal to three corresponding sides of another triangle under S.S.S. axiom.
In \(\triangle\)ABC and \(\triangle\)MNO ,
a) AB = MN (S)
b) BC = NO (S)
c) AC = MO (S)
∴ \(\triangle\)ABC ≅ \(\triangle\)MNO [S.S.S. axiom]
Corresponding parts of congruent triangles are also equal. i.e. ∠A =∠M, ∠B = ∠N and ∠C =∠O.
The two triangles are said to be congruent when two sides and an angle made by them of a triangle are respectively equal to the corresponding sides and an angle of another triangle, under S.A.S. axiom.
In \(\triangle\)ABC and \(\triangle\)PQR,
a) AB = PQ (S)
b) ∠B = ∠Q (A)
c) BC = QR (S)
∴ \(\triangle\)ABC = \(\triangle\)PQR [S.A.S. axiom]
Now, ∠C = ∠R and ∠A = ∠P [corresponding angles of congruent triangles]
AC = PR [Corresponding sides of congruent triangles]
The two of are said to be congruent when two angles and their adjacent side of one triangle are respectively equal to the corresponding angles and side of another triangle, under A.S.A. axiom.
In \(\triangle\)ABC and \(\triangle\)DEF,
a) ∠B = ∠E (A)
b) BC = EF (S)
c) ∠C = ∠F (A)
∴ \(\triangle\)ABC ≅ \(\triangle\)DEF [A.S.A. axiom]
Now, AC = DF and AB = DE
[Correspondong sides of congruent triangles]
∠A =∠D [Corresponding angles of congruent triangles]
The two rightangled triangles are said to be congruent when the hypotenuse and one of the remaining sides of both triangles are respectively equal, under R.H.S. axiom.
In right angled \(\triangle\)ABC and \(\triangle\)MNO,
a)∠B = ∠N (R)
b) AC = MO (H)
c) BC = NO (S)
∴ \(\triangle\)ABC ≅ \(\triangle\)MNO [ R.H.S. axiom]
Now, ∠C = ∠O and ∠A = ∠M [Corresponding angles of congruent triangles]
AB = MN [Corresponding sides of congruent triangles]
The two triangles are said to be congruent when two angles and a side of one triangle are respectively equal to the corresponding angles and side of another triangle, under S.A.A. axiom. This axiom can be verified by using A.S.A. axiom.
Here,
a) ∠A = ∠D [Given]
b) ∠B = ∠E [Given]
c) ∠A + ∠B + ∠C = ∠D + ∠E + ∠F [Sum of the angles of any triangle is 180^{o}]
d) ∠C = ∠F [From (a), (b) and (c)]
Now, in \(\triangle\)ABC and \(\triangle\)DEF,
e) ∠B = ∠F (A) [Given]
f) BC = EF (S) [Given]
g) ∠C = ∠F (A) [From (d)]
h) \(\triangle\)ABC ≅ \(\triangle\)DEF [A.S.A. axiom]
Now, AB = DE and AC = DF [Corresponding sides of congruent triangles]
Step 1: Draw three isosceles triangle ABC with different shapes and sizes in different orientation where AB = AC.
Step 2: Measure the angles opposite to the equal sides of each triangle and tabulate them in the table.figureConclusion:
Theoretical proof:
Given: \(\triangle\)ABC is an isosceles triangle where AB = AC.To prove: ∠ABC = ∠ACB
Construction: Let us draw AD⊥BC, from the vertex A.
Proof:
S.N.  Statement  Reasons 
1.  In \(\triangle\)ABD and \(\triangle\)ACD  
i)  ∠ADB = ∠ADC (R)  Both angles are right angle. 
ii)  AB = AC (H)  Given 
iii)  AD = AD (S)  Common side 
2.  ∴ \(\triangle\)ABD ≅ \(\triangle\)ACD  By RHS axiom. 
3. 
∠ABD = ∠ACD i.e. ∠ABC = ∠ACB 
Corresponding angles of congruent triangles are equal. 
Proved
Experimental verification:
Step 1: Draw three line segments BC of different lengths in different positions.
Step2: Equal sizes of angles at both points B and C on each line segment is drawn. Mark the points as A where the arms of these angles meet. Now three \(\triangle\)ABC are formed.
Step 3: Measure the length of each side opposite to the equal angles in each triangle (i.e. AB and AC) and tabulate them in the table.figure
Conclusion:
Theoretical proof:
Given: In \(\triangle\)ABC, base angles are equal i.e. ∠B =∠C.
To prove: AB = AC, \(\triangle\)ABC is an isosceles triangle.
Construction: From the vertex A, let us draw AD⊥BC.
Proof:
S.N.  Statement  Reasons 
1.  In \(\triangle\)ABD and \(\triangle\)ACD  
i)  AD = AD (S)  Common side 
ii)  ∠ADB = ∠ADC (A)  By construction, AD⊥BC, both angles are equal. 
iii)  ∠ABD = ∠ACD (A)  Given 
2.  ∴ \(\triangle\)ABD ≅ \(\triangle\)ACD  By S.A.A. axiom. 
3. 
AB = AC i.e. \(\triangle\)ABC is an isosceles triangle. 
Corresponding sides of the congruent triangles are equal. 
Proved
Experimental verification:
Step 1: Draw three isosceles triangle ABC with different positions and sizes such that AB = AC in each triangle.
Step 2: Draw the bisector of the vertex angle ∠A in each triangle. The bisector meets BC at D.
Step 3: Measure the lengths of BD and DC and the angles ADB and ADC, then tabulate them in the table.Figure  BD  DC  Result  ∠ADB  ∠ADC  Result 
i)  
ii)  
iii) 
Conclusion:
Theoretical proof:
Given: \(\triangle\)ABC is an isosceles triangle where AB = AC. AD is the bisector of∠BAC.
To prove: AD⊥BC and BD = DC.
Proof:
S.N.  Statement  Reasons 
1.  In \(\triangle\)ADB and \(\triangle\)ACD  
i)  AB = AC (S)  Given 
ii)  ∠BAD = ∠CAD (A)  Given 
iii)  AD = AD (S)  Common side 
2.  ∴ \(\triangle\)ABD ≅ \(\triangle\)ACD  S.A.S. axiom 
3.  ∠ADB =∠ADC  Corresponding angles of congruent triangles are equal. 
4.  AD⊥BC  Equal adjacent angles in a linear pair mean the line is perpendicular. 
5.  BD = DC  Corresponding sides of congruent triangles are equal. 
6.  AD is the perpendicular bisector of BC.  From statement 4 and 5. 
Proved
Experimental verification:
Step 1: Draw three isosceles triangles ABC with different positions and size in a different position such that AB = AC in each triangle.Step 2: Mark the midpoint of BC at D. Join A and D in each triangle.
Step 3: Measure ∠ADB, ∠ADC, ∠BAD and ∠CAD and tabulate them below:
Figure  ∠ADB  ∠ADC  Result  ∠BAD  ∠DAC  Result 
i)  
ii)  
iii) 
Conclusion:
Theoretical proof:
Given: \(\triangle\)ABC is an isosceles triangle where AB = AC.
AD joins the vertex A and midpoint D of the base BC.
To prove: AD⊥BC and ∠BAD = ∠CAD.
S.N.  Statements  Reasons 
1.  In \(\triangle\)ADB and \(\triangle\)ACD  
i)  AB = AC (S)  Given 
ii)  AD = AD (S)  Common side 
iii)  BD = DC (S)  Given 
2.  ∴ \(\triangle\)ABD ≅ \(\triangle\)ACD  By S.S.S. axiom 
3.  ∠ADB = ∠ADC  Corresponding angles of congruent triangles are equal. 
4.  AD⊥ BC  Each angle of adjacent angles in a linear pair are 90^{0} both. 
5.  ∠BAD = ∠CAD  Corresponding angles of congruent triangles are equal. 
Proved
Experimental verification:
Step 1: Three triangles ABC with different positions and sizes in different orientation are drawn.
Step 2: Measure all three sides of each triangle and fill up the table.
Figure  AB  BC  CA  AB + BC  BC + CA  AB + CA  Result 
i)  
ii)  
iii) 
Conclusion:The sum of two sides of a triangle is greater than the third side.
Experimental verification:
Step 1: Three triangles ABC with different positions and size in different orientations are drawn in such a way that BC is the longest and CA is the shortest side in each triangle.
Step 2: The angle opposite to the greater side BC (i.e.∠B) are measured and tabulate in the table:
Figure  Angle opposite to BC:∠A  Angle opposite to CA:∠B  Result 
i)  
ii)  
iii) 
Conclusion: In any triangle, the angle opposite to the longer side is greater than the angle opposite to the shorter side.
Experimental verification:
Step 1: Three triangle ABC with different positions and sizes in different orientations are drawn in such a way that∠A is the greatest and∠B is the smallest angles in each triangle.
Step 2: Measure the side BC opposite to the greatest single angle A and the side CA opposite to the smallest angle B and fill up the table below:Figure  Angle opposite to ∠A (BC)  Angle opposite to ∠B (CA)  Result 
i)  
ii)  
iii) 
Conclusion: In any triangle, the side opposite to the greater angle is longer than the side opposite to the smaller angle.
Experimental verification:
Step 1: Three straight line segments XY of different length in different orientations are drawn.A point P is taken outside of each line segment. Three line segments PA, PB, PC and a perpendicular line PM are drawn from P to XY.
Step 2: Measure the lengths of each line segment PA, PB, PC and PM. Then tabulate them below:
Figure  PA  PB  PC  PM 
i  
ii  
iii 
Conclusion: Among all straight line segments down to a given line form a point outside it, the perpendicular is the shortest line.
Step 1: Draw three rightangled triangles of different dimensions.
Step 2: Fill up the given table.
Figure  AB  BC  CA  AB^{2}  BC^{2}  CA^{2}  AB^{2} + BC^{2}  Results 
i)  
ii)  
iii) 
Conclusion:The square of hypotenuse in rightangled triangle is equal to the sum of squares of it's base and perpendicular.
Experiment no.5 is known as Pythagoras Theorem. Around 2500 years ago, Pythagoras was a Greek Mathematician. He invented a fact on the rightangled triangle which is named as Pythagoras Theorem. This theorem is used in all other fields of mathematics not only used in geometry.
The angle is mainly divided into the following types:
Find the unknown angles of the traingles given below :
(i).
Here , y = 56^{o} + 4^{o}
\(\therefore\) y = 105^{o }[exterior angle in a traiangle is equal to sum of two no  adjacent interior angles]
Again , x + 56^{o}+49^{o} = 180^{o }[sum of three interior angle of a triangle is 180^{o}]
\(\therefore\) x = 180^{o}  56^{o} = 49^{o} = 180^{o}  105^{o} = 75^{o }
Hence , x = 75^{o}and y = 105^{o}
Find the unknown angles of the traingles:
Here , a + 75^{o}+ 60^{o}= 180^{o }[sum of three angles of traingles]
or , a + 135^{o} = 180^{o}
or , a = 180^{o}  135 = 45^{o}
Here , b + 60^{o} = 180 [straight angle]
\(\therefore\) b = 180^{o}  60^{o} = 120
Hence , a = 45^{o} and b = 120^{o }
Find the unknown angles of the triangles :
Here , a + 51^{o} + 60^{o} = 180^{o} [ sum of three angle of a triangle is 180^{o}]
or , a = 180^{o}  51^{o}  60^{o} = 180^{0}  111^{o} = 69^{o }
Again , b = 60^{o} + 51^{o} = 111^{o} [being exterior angle and sum of opposite interior angles of a triangle]
Hence , a = 69^{o} and b = 111^{o}
Find the unkown angles of triangles:
Here , AOB = DOC = 75^{o}[∴AOB and DOC are vertically opposite angles]
Again , p+ 68^{o}+75^{o =}180^{o}
or , p + 143^{o} = 180^{o}
or , p = 180^{o}  143^{o} = 37^{o}Ans.
Now , in ABO
or , q+ 75^{o} + P = 180^{o}
or , q + 75^{o} + 37^{o} = 108^{o}
\(\therefore\) q = 180^{o}  112^{o} = 68^{o}
Find the unknown angles of the triangle:
Here , x = 61^{o} + 82^{o} = 143^{o }[being exterior angle and sum of two opposite interior angle]
Again , y + 82^{o} = 180^{o}\(\therefore\) y = 180^{o}  82^{o} = 98^{o}
Hence , x = 143^{o}and y = 98^{o}
Find the unknown angles of triangle
InΔ BCE
p = 85^{o} + 30^{o} = 180^{o} [sum of three interior angles of a triangle]
or , p + 115^{o} = 180^{o}
or , p = 180^{o}  115^{o} = 65^{o}
Here , ABE = 85^{o} + 30^{o} = 115^{o} [In , BCe exterior angle = sum of two nonadjacent interior angle]
Now , inΔABO ,
q + ABO + 46^{o} = 180 [The sum of three interior angle of a triangle]
or , q + 115^{o} + 46^{o} = 180^{o}[ABO = ABE = 115]
or , q + 161^{o} = 180^{o}
or , q = 180^{o}  161^{0} = 19^{o}
Hence , p = 65^{o} and q = 19^{o}
Find the unknown angle of triangle :
Here , AED = ACB [corresponding angles DE  BC]
or , x = 35
Now , in ΔABC
y+80+x= 180 [sum of all interior angle of a triangle]
or , y + 35 + 80 = 180
or , y = 180  80  35 = 180  115 = 65
Hence , x = 35 and y = 65
Find the unknown angles of the triangles :
Here , ODC = OAB [alternate angles and ABCD]
or , OCD = 38
Now , in DCO
x + 81+ODC = 180 ]sum of all three angles of a triangle]
or , x + 81 + 38 = 180
or , x + 119 = 180
or , x = 180  119 = 61 Ans.
Find the angles of triangle:
In DOC ,
70 + DOC = DCE [ being exterior angle and sum of two opposite interior angle]
or , DOC = 150  70 = 80
Now , AOB = DOC = 80 [Vertically oppostie angle]
Again , in ABO ,
x + 63 + AOB = 180[sum of three angles of a triangle]
or , x + 63 + 80 = 180
x + 143 = 180
or , x = 180  143 = 37 Ans.
Find the uknown angles of triangle :
Here , EAB = AEC [in aec , exterior angle = sum of two nonadjacent interior angle]
or , x = 37
Again , DCA = AEC + CAE [ In? AEC , exterior angle = sum of two nonadjacent interior angle]
or , 63 = 37 + y
or , y = 63  37 = 26.
Here , ACB = 90 [from given]
Now , In?ACB
CAB+ACB+CBA = 180 [sum of three interior angle of triangle]
or , x + y + 90 + z = 180
or , 37 + 26 + 90 + z = 180,
or , 153 + z = 180
or , z = 180  153 = 27
Hence , x = 37 , y = 26 and z = 27
In the given figure ,
<PAB =<QAC , AP = AQ and PB = QC
Prove that:ΔABC is an isoceles triangle.
Given : AP = AQ and PB = QC
To prove ABC is an isoceles triangle
i.e. AB = AC
Statements  Reasons 
In APB and AQC 1. PB = QC 2 . APB = AQC 3. AP = AQ 4. APB≅AQC 5. AB = AC 6. ABC is an isoceles traingle.  1. Given 2. Base anglen of an isoceles angle APQ 3.Given 4. S.A.S Axiom 5. Corresponding sides of congruent triangles are equal. 6. From statement (5) 
ΔABC is an isoceles triangle , BO and CO are bisectors of ABC and ACB respectively. Prove that BOC is an isoceles triangle.
Given: In isoceles triangle ABC , BO and CO bisect ABC and ACB respectively.
To prove: BOC is an isoceles triangle i.e. BO = CO
Proof :
Statements  Reasons 
1. ABC = ACB 2. OBC = \(\frac{1}{2}\) ABC and OCB = \(\frac{1}{2}\)ACB 3. OBC = OCB 4. OB = OC \(\therefore\) BOC is an isoceles triangle.  1. Base angle of an isoceles triangle. 2. OB and OC bisect ABC and ACB respectively. 3. From (1) and (2) , i.e. half of the equal angles 4. From statement (3) , i.e. base angles are equal 
In the following figures ,
PS⊥QR ,
QPS = 45
and QPR = 90
Prove that PQR is an isoceles triangle.
Given :
In PQR ,PS⊥QR , QPS = 45 and QPR = 90
To prove: PQR is an isoceles triangle.
Statements  Reasons 
1. PQS + QPS = 90 or , PQS + 45 = 90 or , PQS = 90  45 = 45 2. QPR = 90 3. QRP +PQR =90 or , QRP + 45 =90 QRP = 90  45 = 45 4. PQ = PR 5. PQR is an isoceles triangle.  1. In right angled triangle PSQ , sum of acute angle is one right angle.

In the given figure find the value of angle a , b , c , d , e and f.
Statements  Reasons 
1. a = 60  Each angles of a equilateral triangle is 60. 
2. b + a + 53 = 180 or , b + 60 + 52 = 180 or , b = 180  60  53 = 67  2. Sum of the all angles made at a point E of st.line AD and lying on one side or AD. 
3 . EBA = b = 67 or , B = 76  3. Being base angles of an isoceles triangle EAB. 
4. In EAB , c + b + B = 180 or , c + 67 + 67 = 180  4. Sum of interior angles of a triangle. 
5. C = DCE = 53  5. Base angle of all of a isoceles triangle i.e DE = DC 
6. d + 53 = 180  6. The sum of all three interior angles of a triagle is 180 
In the given figure fin the value of a , b , c , d , e and f.
Statements  Reasons 
1. BEH + 140 = 180 BEH = 180  140 = 140  1.Sum of linear pairs. 
2. f = BEH = 40  2. Alternate angles and ACGI 
3. d = BHE = 56  3. ALternate angles an ACEI 
4. f + e + d = 180  4. From the reason 1 
5. 40 + e + 56 = 180 e = 180  96 = 84 e = 84 , f = 40 and d = 56  5. From the statement 2 and 3 
If theb perpendiculars drawn from any two vertices of a triangle to heir oppsite sides are equal then the traingle is isoceles.,
Given : In ABC , CD and BE are perpendicular drawn from the point B an C on Ab and Ac respectively as well as BE = CD
To proveABC is an isoceles triangle , i.e , AB= AC.
Statements  Reasons 
1. BE = CD  1. Given 
2. AEB = ADC  2. Both are right angles 
3. EAB = DAc  3. Common angles 
4. AEB≅ ADC  4. By S. A. S facts 
5. Ab = AC  5. Correspondind sides of congruent triangles AEB an ADC 
The extremities of a line segment are equidistant from any line passing through the mid point of the given line segment.
Given :D is the mid point of a line segment AB , EF is the line passing through the midpoint D. AX and BY are perpendiculars drawn from A and B on EF respectively i.e. AX ⊥ EF and BY ⊥ EF.
Statements  Reason 
1. In ΔAXD and ΔBYD (i). AD = BD (ii). ADX = BDY (iii). AXD = BYD  1 (i) being D as mid point (ii). vertically opposite angle (iii). Both are right angles 
2. ΔAXD≅ΔBYD  2. BY S.A.A fact 
3. AX = BY  3. Corresponding sides of congruent triangle. 
If the perpendicular draw from the midpoint of one side of a triangle to other two sides are equal , then the triangle is isoceles.
Given , InΔABC , D is midpoint of BC. From the midpoint D , DF⊥AB and DE⊥AC are drawn where Df = DE.
To prove , Δ ABC is an isoceles triangle.
or ,AB = AC
Statements  Reasons 
DFB = DEC  both are right angle 
BD = DC  From given , being D is mid point. 
BF = DE  Given 
ΔBFD≅CED  By RHS axiom 
DBF = DCE  Corresponding angles of congruent triangles 
AB = AC  Fom statement (5) , base angles of ΔABC are equal. 
In the adjoining figure , how can one find thw width from A to B without crossing the river ?
GIven : AB⊥AD , AO = OD and CD⊥AD
To prove: AB = CD
Statements  Reasons 
1. InΔAOB and ΔOCD (i) OAB = OCD (ii) AO = OD (iii) BOA = COD  1. (i). Both are right angle (ii) Given (iii) Vertivally Opposite angles are equal. 
2. ΔABO≅ΔOCD  2. A.S.A fact 
3. AB = CD \(\therefore\) Breadth of river = CD  3. Corresponding sides of congruent. 
In the given figure PQRS , SP = RQ and RP = SQ.
Prive that RT = ST.
Given ,
SP = RQ , RP = SQ
To prove , RT = ST
Statements  Reasons 
1. In ΔQRS and ΔPSR (i) SR = SR (ii)RQ = SP (iii) SQ = RP  (i) Common side (ii) From given (iii) From given 
2. ΔQRS≅ ΔPSR  By S.S.S fact 
3. QSR = PSR  3. corresponding angles of congruent triangles. 
4. RT = ST  4. Being TSR = TRS (From(3)) 
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