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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. Acute-angled triangle: A triangle whose all angles are acute angles (less than 90^{o}) is called an acute-angled triangle. In the given figure, all angles are less than 90^{o} so it is an acute-angled triangle.
2. Obtuse-angled triangle: A triangle whose one angle is obtuse (greater than 90^{o}) is called an obtuse-angled triangle. In the given figure, ∠Y is 120^{o }(greater than 90^{o}) ∴ \(\triangle\)XYZ is an obtuse-angled triangle.
3. Right-angled triangle: A triangle whose one angle is right angle (90^{o}) is called right-angled triangle. In \(\triangle\)ABC, ∠B = 90^{o}, so \(\triangle\)ABC, is a right-angled triangle.
Properties of triangles
Triangles are three-sided 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 three-sided polygon.
Axoims
An axiom is a self-evident truth which is well-established, 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 a-c=b-c |
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 co-interior angles is 180^{o}: If two parallel lines are intersected by a transversal line, then the sum of the co-interior angles so formed is 180^{o}.
∴ ∠AEF + ∠EFX = 180^{o}
∠BEF + ∠EFY = 180^{o}
d) The sum of co-exterior angles is 180^{o}: If two parallel lines are intersected by a transversl line, then the sum of the co-exterior 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 whole-part 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 non-adjacent angles inside the triangle.
Step 2: Measure the size of exterior angle and other two non-adjacent 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 right-angled 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 mid-point 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 mid-point 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 right-angled 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 right-angled 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 right-angled 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:
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}
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 }
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}
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}
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}
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 non-adjacent 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}
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
Here , ODC = OAB [alternate angles and AB||CD]
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.
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.
Here , EAB = AEC [in aec , exterior angle = sum of two non-adjacent interior angle]
or , x = 37
Again , DCA = AEC + CAE [ In? AEC , exterior angle = sum of two non-adjacent 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
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) |
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 |
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 |
Given :D is the mid point of a line segment AB , EF is the line passing through the mid-point 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. |
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. |
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)) |
On the basis of sides, how many triangles are classified into?
If a^{0}, 60^{0 }and 40^{0} are the angles of a triangle, find the value of a^{0}.
The angles of a triangle are in the ratio 4:5:6. Find each angle in degrees.
If 90^{0}, 45^{0} and x^{0} are the angles of a triangle, find the value of x^{0}.
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puskar
how to find tsa of triangular prism
Mar 25, 2017
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ninirpa bk
Ask any queries on this note.edd
Mar 05, 2017
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Anup
In the diagram AB=AC , BX=CY and KX parallel AY then prove that :Triangle XKP congruent triangle YCP
Mar 04, 2017
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Funsuk Wangdu
What is triangle?
Jan 31, 2017
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