2. A Triangle is formed by three lines segments obtained by
joining three pairs of points taken from a set of three non
collinear points in the plane.
In fig 1.1 , three non collinear points A, B, C have been joined
and the figure ABC, enclosed by three line segments ,
AB,BC, and CA is called a triangle.
The symbol ▲ (delta) is used to denote a triangle. The three
given points are called the vertices of a triangle. The three
line segments are called the sides of a triangle. The angle
made by the line segment at the vertices are called the
angles of a triangle.
If the sides of a ▲ABC are extended in order, then the angle
between the extended and the adjoining side is called the
exterior angles of the triangle.
In the fig. 1.2
1 ,2 and 3 are the exterior angles while 4 ,5,and 6 are the
interior angles of the triangles.
1 4
5
2
6 3
A
B
B C
A
CB
Fig. 1.1
Fig. 1.2
3. • ON THE BASIS OF SIDES
1. SCALENE TRIANGLE – If all the sides of triangle are unequal
then it is a scalene tirangle.
2. ISOSCELES TRIANGLE –If any two sides of a triangle are
equal then it is a isosceles triangle.
3. EQUILATERAL TRIANGLE- If all the sides of a triangle are
equal then it is an equilateral triangle.
• ON THE BASIS OF ANGLES
1. ACUTE ANGLED TRIANGLE –If all the three angles of a
triangle are less than 900
then it is an acute anglesd triangle.
2. RIGHT ANGLED TRIANGLE- If one angle of a triangle is equal
to 900
then it is a right angled triangle.
3. OBTUSE ANGLED TRIANGLE- If one angle of a triangle is
greater then 900
then it is a obtuse angled triangle.
4. Two triangles are congruent if three sides
and three angles of one triangle are
equal to the corresponding sides and
angles of other triangle.
The congruence of two triangles follows
immediately from the congruence of
three lines segments and three angles.
5. 1. SIDE – ANGLE – SIDE RULE (SAS RULE)
Two triangles are congruent if any two sides and the includes angle of
one triangle is equal to the two sides and the included angle of other
triangle.
EXAMPLE :- (in fig 1.3)
GIVEN: AB=DE, BC=EF ,
B= E
SOLUTION: IF AB=DE, BC=EF , B= E then by SAS Rule
▲ABS = ▲DEF
4 cm4 cm
600
600
A
B C
D
FE
Fig. 1.3
6. 2. ANGLE – SIDE – ANGLE RULE (ASA RULE )
Two triangles are congruent if any two angles and the included side
of one triangle is equal to the two angles and the included side of the
other triangle.
EXAMPLE : (in fig. 1.4)
GIVEN: ABC= DEF,
ACB= DFE,
BC = EF
TO PROVE : ▲ABC = ▲DEF
ABC = DEF, (GIVEN)
ACB = DFE, (GIVEN)
BS = EF (GIVEN)
▲ABC = ▲DEF (BY ASA RULE)
A
B C
D
E F
Fig. 1.4
7. 3. ANGLE – ANGLE – SIDE RULE (AAS RULE)
Two triangles are congruent if two angles and a side
of one triangle is equal to the two angles and one a side of the
other.
EXAMPLE: (in fig. 1.5)
GIVEN: IN ▲ ABC & ▲DEF
B = E
A= D
BC = EF
TO PROVE :▲ABC = ▲DEF
B = E
A = D
BC = EF
D
E F
A
B C
Fig. 1.5
8. 4. SIDE – SIDE – SIDE RULE (SSS RULE)
Two triangles are congruent if all the three
sides of one triangle are equal to the three sides of other
triangle.
Example: (in fig. 1.6)
Given: IN ▲ ABC & ▲DEF
AB = DE , BC = EF , AC = DF
TO PROVE : ▲ABC = ▲DEF
AB = DE (GIVEN )
BC = EF (GIVEN )
AC = DF (GIVEN )
▲ABC = ▲DEF (BY SSS RULE)
D
E F
A
B C
Fig. 1.6
9. 5. RIGHT – HYPOTENUSE – SIDE RULE (RHS RULE )
Two triangles are congruent if the hypotenuse and the
side of one triangle are equal to the hypotenuse and the side of other
triangle.
EXAMPLE : (in fig 1.7)
GIVEN: IN ▲ ABC & ▲DEF
B = E = 900
, AC = DF , AB = DE
TO PROVE : ▲ABC = ▲DEF
B = E = 900 (GIVEN)
AC = DF (GIVEN)
AB = DE (GIVEN)
▲ABC = ▲DEF (BY RHS RULE)
D
E F
A
B C
900
900
Fig. 1.7
10. 1. The angles opposite to equal sides are always equal.
Example: (in fig 1.8)
Given: ▲ABC is an isosceles triangle in which AB = AC
TO PROVE: B = C
CONSTRUCTION : Draw AD bisector of BAC which meets BC
at D
PROOF: IN ▲ABC & ▲ACD
AB = AD (GIVEN)
BAD = CAD (GIVEN)
AD = AD (COMMON)
▲ABD = ▲ ACD (BY SAS RULE)
B = C (BY CPCT)
A
B D C
Fig. 1.8
11. 2. The sides opposite to equal angles of a triangle are always
equal.
Example : (in fig. 1.9)
Given : ▲ ABC is an isosceles triangle in which B = C
TO PROVE: AB = AC
CONSTRUCTION : Draw AD the bisector of BAC which meets
BC at D
Proof : IN ▲ ABD & ▲ ACD
B = C (GIVEN)
AD = AD (GIVEN)
BAD = CAD (GIVEN)
▲ ABD = ▲ ACD (BY ASA RULE)
AB = AC (BY CPCT)
A
B D C
Fig. 1.9
12. When two quantities are unequal then on comparing
these quantities we obtain a relation between their
measures called “ inequality “ relation.
13. Theorem 1 . If two sides of a triangle are unequal the larger side
has the greater angle opposite to it. Example: (in fig. 2.1)
Given : IN ▲ABC , AB>AC
TO PROVE : C = B
Draw a line segment CD from vertex such that AC = AD
Proof : IN ▲ACD , AC = AD
ACD = ADC --- (1)
But ADC is an exterior angle of ▲BDC
ADC > B --- (2)
From (1) &(2)
ACD > B --- (3)
ACB > ACD ---4
From (3) & (4)
ACB > ACD > B , ACB > B ,
C > B
A
B
D
C
Fig. 2.1
14. THEOREM 2. In a triangle the greater angle has a large side opposite to it
Example: (in fig. 2.2)
Given: IN ▲ ABC B > C
TO PROVE : AC > AB
PROOF : We have the three possibility for sides AB and AC of ▲ABC
(i) AC = AB
If AC = AB then opposite angles of the equal sides are equal than
B = C
AC ≠ AB
(ii) If AC < AB
We know that larger side has greater angles opposite to it.
AC < AB , C > B
AC is not greater then AB
(iii) If AC > AB
We have left only this possibility AC > AB
A
CB
Fig. 2.2
15. THEOREM 3. The sum of any two angles is greater than its third side
Example (in fig. 2.3) TO PROVE : AB + BC > AC
BC + AC > AB
AC + AB > BC
CONSTRUCTION: Produce BA to D such that AD + AC .
Proof: AD = AC (GIVEN)
ACD = ADC (Angles opposite to equal sides are equal )
ACD = ADC --- (1)
BCD > ACD ----(2)
From (1) & (2) BCD > ADC = BDC
BD > AC (Greater angles have larger opposite sides )
BA + AD > BC ( BD = BA + AD)
BA + AC > BC (By construction)
AB + BC > AC
BC + AC >AB
A
CB
D
Fig. 2.3
16. THEOREM 4. Of all the line segments that can be drawn to a given line from an
external point , the perpendicular line segment is the shortest.
Example: (in fig 2.4)
Given : A line AB and an external point. Join CD and draw CE AB
TO PROVE CE < CD
PROOF : IN ▲CED, CED = 900
THEN CDE < CED
CD < CE ( Greater angles have larger side opposite to them. )
B
A
C
ED
Fig. 2.4
17. 1. If the altitude from one vertex of a triangle bisects the opposite side, then the
triangle is isosceles triangle.
Example : (in fig.2.5)
Given : A ▲ABC such that the altitude AD from A on the opposite side BC
bisects BC i. e. BD = DC
To prove : AB = AC
SOLUTION : IN ▲ ADB & ▲ADC
BD = DC
ADB = ADC = 900
AD = AD (COMMON )
▲ADB = ▲ ADC (BY SAS RULE )
AB = AC (BY CPCT)
A
CDB
Fig. 2.5
18. THEOREM 2. In a isosceles triangle altitude from the vertex bisects the base .
EXAMPLE: (in fig. 2.6)
GIVEN: An isosceles triangle
AB = AC
To prove : D bisects BC i.e. BD = DC
Proof:
IN ▲ ADB & ▲ADC
ADB = ADC
AD = AD
B = C ( AB = AC ; B = C)
▲ADB = ▲ ADC
BD = DC (BY CPCT)
A
CDB
Fig. 2.6
19. THEOREM 3. If the bisector of the vertical angle of a triangle bisects the base
of the triangle, then the triangle is isosceles. EXAMPLE: (in fig. 2.7)
GIVEN: A ▲ABC in which AD bisects A meeting BC in D such that BD = DC,
AD = DE
To prove : ▲ABC is isosceles triangle .
Proof: In ▲ ADB & ▲ EDC
BD = DC
AD = DE
ADB = EDC
▲ADB = ▲EDC
AB = EC
BAD = CED (BY CPCT)
BAD = CAD (GIVEN)
CAD = CED
AC = EC (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL)
AC = AB , HENCE ▲ABC IS AN ISOSCELES TRIANGLE.
E
D CB
A
Fig. 2.7
20. QUES. IN FIG. AB = AC, D is the point in interior of ▲ABC such that
DBC = DCB. Prove that AD bisects BAC of ▲ ABC
SOLUTION: IN ▲BDC,
DBC = DCB (GIVEN)
DC = DB (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL.)
IN ▲ ABD & ▲ ACD
AB = AC (GIVEN)
BD = CD (GIVEN)
AD = AD (COMMON)
▲ABD = ▲ACD (BY SSS RULE)
BAD = CAD (BY CPCT)
Hence, AD is the bisector of ▲BAC
A
D
B C
21. QUES. In fig. PA AB , QB AB & PA = QB.If PQ intersects AB
at O, show that o is the mid point of AB as well as that of PQ
SOLUTION: In ▲OAP & ▲OBQ
PA = QB (GIVEN)
PAO = OBQ = 900
& AOP = BOQ
▲ OAP = ▲OBQ (BY AAS RULE)
OA = OB ( BY CPCT)
OP = OQ ( BY CPCT)
O
B
A
P
Q
22. QUES. In fig. PS = PR, TPS = QPR.Prove that PT = PQ
SOLUTION: IN ▲PRS
PS = PR
PRS = PSR (ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL)
1800
- PRS = 1800
- PSR
PRQ = PST
Thus , in ▲ PST & ▲PRQ
TPS = QTR
PS = PR
PST = PRQ
▲PRT = ▲PSQ ( BY ASA RULE)
PT = PQ (BY CPCT)
T
S R
Q
P
23. QUES. In fig. BM and DN are both perpendicular to the line segment AC and
BM = DN. Prove that AC bisects BD.
SOLUTION IN ▲BMR and ▲DNR
BMR = DNR (GIVEN)
BRM = DRN (VERTICALLY OPPOSITE ANGLES)
BM = DN (GIVEN)
▲ BMR = ▲DNR (BY AAS RULE)
BR = DR
R is the mid point of BD
Hence AC bisects BD
B A
C
D
M
N
R
24. Deduction 1: All angles at the circumference on the same arc are equal
in measure.
To prove: ∠bac = ∠bdc
Proof: ∠3 = 2 ∠1 Angle at the centre is twice the angle on
the circumference (both on the arc bc)
∠3 = 2 ∠2 Angle at the centre is twice the angle
on the circumference (both on arc bc)
∴ 2 ∠1 = 2 ∠2
∴ ∠1 = ∠2
i.e. ∠bac = Q.E.D.
a
b
c
d
3
1 2
.o
25. QUES. In fig . It is given that BC = CE and 1 = 2. Prove that
▲ GCB = ▲ DCE
SOLUTION:
IN ▲ GCB & ▲ DCE
1 + GBC = 1800
(LINEAR PAIR )
2 + DEC = 1800
( LINEAR PAIR)
1 + GBC = 2 + DEC -- (1)
GBC = DEC (From (1) )
GBC = DEC (GIVEN)
BC = CE (GIVEN)
GCB = DCE (VERTICALLY OPPOSITE ANGLES )
▲GCB = ▲ DCE (BY ASA RULE)
*----- *---- *------ * -----* ----- * ----* ---- *
A
B
C
F
G
E
D
1 2
26. a
b
c
Ld
∟
∟.
Theorem: A line through the centre of a circle perpendicular to a chord bisects
the chord.
Given
:
Circle, centre c, a line L containing c, chord [ab],
such that L ⊥ab and L∩ ab = d.
To
prove:
ad = bd
Constructio
n:
Label right angles 1 and 2.
Proof
: ∠1 = ∠2 =
90°
Given
ca =
cb
Both radii
cd = cd commo
nR H S
Corresponding sides
Consider cda and
cdb:
cda cdb∴
ad =
bd
∴
1
2
Q.E.D.
27. b
a
c
d
e f
2
1
2
3
1 3
Theorem: If two triangles are equiangular, the lengths of the corresponding
sides are in proportion.
Given
:
Two triangles with equal
angles.
To
prove: |df|
|ac|
=
|de|
|ab|
|ef|
|bc|
=
Constructio
n:
On ab mark off ax equal in length to
de. On ac mark off ay equal to df and
label the angles 4 and 5.
Proof: ∠1 = ∠4
[xy] is parallel to [bc]
|ay|
|ac|
=
|ax|
|ab|
As xy is parallel to
bc.
|df|
|ac|
=
|de|
|ab|
Similarly
|ef|
|bc|
=
x y4 5
Q.E.D.
28. Theorem: In a right-angled triangle, the square of the length of the
side opposite to the right angle is equal to the sum of the squares of
the other two sides.
Q.E.D.
c
b
a
c
b
a
b
a c
b
a
c
1
2
3
45
To prove that angle 1 is 90º
Proof:
∠3+ ∠4+ ∠5 = 180º ……Angles in a triangle
But ∠5 = 90º => ∠3+ ∠4 = 90º
=> ∠3+ ∠2 = 90º ……Since ∠2 = ∠4
Now ∠1+ ∠2+ ∠3 = 180º ……Straight line
=> ∠1 = 180º - ( ∠3+ ∠2 )
=> ∠1 = 180º - ( 90º ) ……Since ∠3+ ∠2 already
proved to be 90º
=> ∠1 = 90º
29. The HypotenuseThe Hypotenuse
The hypotenuse of aThe hypotenuse of a
right triangle is theright triangle is the
triangle's longest side,triangle's longest side,
i.e., the side oppositei.e., the side opposite
the right angle.the right angle.
30. To day we are TeachingTo day we are Teaching
abouTabouT
31. Pythagorean Theorem
Pythagoras (~580-500 B.C.)
He was a Greek philosopher responsible for important
developments in mathematics, astronomy and the theory
of music.
32. 1. cut a triangle with
base 4 cm and
height 3 cm
0 1 2 3 4 5
4 cm
012345
3cm2. measure the length
of the hypotenuse
0
1
2
3
4
5
Now take out a square paper and a
ruler.
5 cm
33. Consider a square PQRS with sides a + b
a
a
a
a
b
b
b
b
c
c
c
c
Now, the square is cut into
- 4 congruent right-angled triangles
and
- 1 smaller square with
sides c
Proof of Pythagoras’ Theorem
P Q
R S
34. a +
b
a + b
A B
CD
Area of square
ABCD
= (a + b) 2
b
b
a b
b
a
a
a
c
c
c
c
P Q
RS
Area of square
PQRS
= 4 + c 2
ab
2
a 2
+ 2ab +
b 2
= 2ab +
c 2
a
2
+ b
2
= c
2
35. Theorem states that:
"The area of the square built upon the hypotenuse of a
right triangle is equal to the sum of the areas of the
squares upon the remaining sides."
The Pythagorean Theorem asserts that for a right triangle, the
square of the hypotenuse is equal to the sum of the squares of the
other two sides: a2
+ b2
= c2
The figure above at the right is a visual display of the theorem's
conclusion. The figure at the left contains a proof of the theorem,
because the area of the big, outer, green square is equal to the
sum of the areas of the four red triangles and the little, inner white
square:
c2
= 4(ab/2) + (a - b)2
= 2ab + (a2
- 2ab + b2
) = a2
+ b2
36. Animated Proof of the Pythagorean Theorem
Below is an animated proof of the Pythagorean Theorem. Starting with a right triangle and squares
on each side, the middle size square is cut into congruent quadrilaterals (the cuts through the
center and parallel to the sides of the biggest square). Then the quadrilaterals are hinged and
rotated and shifted to the big square. Finally the smallest square is translated to cover the
remaining middle part of the biggest square. A perfect fit! Thus the sum of the squares on the
smaller two sides equals the square on the biggest side.
Afterward, the small square is translated back and the four
quadrilaterals are directly translated back to their original position.
The process is repeated forever.
37. Pythagorean Theorem
Over 2,500 years ago, a Greek mathematician named Pythagoras developed a proof that
the relationship between the hypotenuse and the legs is true for all right triangles.
In any right triangle, the square of the length of the hypotenuse is equal to the sum
of the squares of the lengths of the legs."
This relationship can be stated as:
and is known as the
Pythagorean Theorem
a, b are legs.
c is the hypotenuse (across
from the right angle).
There are certain sets of numbers that have a very special property. Not only do
these numbers satisfy the Pythagorean Theorem, but any multiples of these numbers
also satisfy the Pythagorean Theorem
For example: the numbers 3, 4, and 5 satisfy the Pythagorean Theorem. If you
multiply all three numbers by 2 (6, 8, and 10), these new numbers ALSO satisfy the
Pythagorean theorem.
38. If we think about a right triangle we know of course that one of the angles is a right angle. We also know that the
other two angles are acute angles (why?). In fact we know that the other two angles are complementary angles.
Therefore there is a relationship between the sizes of the angles that the two acute angles have measures that
add up to ninety degrees.
What about sides? Is there a relationship between the sides of a right triangle? We know from previous lessons
that if we have the lengths of just two of the sides we can construct the triangle so it is enough to know the
lengths of two sides to determine the length of the third side. We shall now try to figure out the relationship. We
shall, to make it easy to communicate assume that the length of the hypotenuse is c units and that the two legs
are of length a and b units.So according to the Pythagorean Theorem, the area of square A, plus the
area of square B should equal the area of square C.
The special sets of numbers that possess this property are called
Pythagorean Triples.
The most common Pythagorean Triples are:
3, 4, 5
5, 12, 13
8, 15, 17
The Pythagorean Theorem
The Pythagorean Theorem is one of Euclidean Geometry's most beautiful theorems. It is simple, yet obscure,
and is used continuously in mathematics and physics. In short, it is really cool.
This first method is one of the ways the Pythagoreans would have proved the theorem. Unfortunately, it
lacks glamour. In the following picture let ABC be a right triangle and BD be a segment drawn perpendicular
to AC.
Since the triangles are similar, the sides must be of proportional lengths.
•AB/AD=AC/AB, or AB x AB = AD x AC
•BC/CD=AC/BC, or BC x BC= AC x CD
•Then, adding the two together, BC^2 + AB^2 = (AD + DC) x AC= AC^2
•
39.
40. Pythagorean Theorem in text bookPythagorean Theorem in text book
of 10of 10thth
classclass
Given:- ABC is a right angle TriangleGiven:- ABC is a right angle Triangle
..
angle B =90angle B =9000
R.T.P:- ACR.T.P:- AC22
= AB= AB22
+BC+BC22
Construction:- To draw BDConstruction:- To draw BD ⊥⊥ AC .AC .
A
B C
D
Proof:- In Triangles ADB and ABC
AngleA=Angle A (common)
Angle ADB=Angle ABC (each 900
)
ADB ~ ABC ( A.A corollary )
So that AD/AB=AB/AC (In similar triangles corresponding
sides are proportional )
AB2
= AD X AC _________(1)
Similarly BC2
= DCXAC _________(2)
Adding (1)&(2) we get
AB2
+BC2
= AD X AC + DCXAC
= AC (AD +DC)
= AC . AC
=AC2
There fore
AB2
+BC2
=AC2
42. Example 1. Find the length of AC.
Hypotenus
e
AC2
= 122
+ 162
(Pythagoras’ Theorem)
AC2
= 144 + 256
AC2
= 400
AC = 20
A
CB
16
12
Solution :
43. Example 2. Find the length of diagonal d .
10
24
d
Solution:
d2
= 102
+ 242
(Pythagoras’ Theorem)
d = +
=
=
10 24
26
2 2
676
44. 16km
12km
1.A car travels 16 km from east to west. Then it turns
left and travels a further 12 km. Find the displacement
between the starting point and the destination point of
the car.
N
?
Application of Pythagoras’ Theorem
45. 16 km
12 km
A
B
C
Solution :
In the figure,
AB = 16
BC = 12
AC2
= AB2
+ BC2
(Pythagoras’ Theorem)
AC2
= 162
+ 122
AC2
= 400
AC = 20
The displacement between the starting point and the
destination point of the car is 20 km
46. 2. The height of a tree is 5 m. The distance between
the top of it and the tip of its shadow is 13 m.
Solution:
132
= 52
+ L2
(Pythagoras’ Theorem)
L2
= 132
- 52
L2
= 144
L = 12
Find the length of the shadow L.
5 m
13 m
L
