Chapter 5- Introduction to Deductive Method

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Chapter 5
-

Introduction to Deductive Method


5.1 Intuitive Method and Deductive Method

Intuitive method

-

the way of arriving at conclusions through
observations

and
experiments
.

Example


Ancient people observed that the sun rose in the east everyday.



Conclusion: ‘the sun rises in the east’

This method
-

not reliable


-

the cases we observed are limited

may be wrong

-

Request students to give another example.

To disprove a conclusions drawn by intuitive method

counter example is needed.

Example

Fact: Sparrow is a bird and it can fly.



Parrot is a bird and it can fly.


Conclusion: All birds can fly.


Group discussion: Try to find an counter example to disprove conclusion


Another method:
Deductive method




Example







Deductive me
thod is better than intuitive method.


5.2 The origin of deductive geometry

Three elements of deductive geometry:

I.

Definitions

-

give everyone a common understanding of geometrical objects.

Example
1. A
point

only shows a location & has no size.


2. A line h
as length but no width.


3. The ends of a
line segment

are points.


4. A surface only has length and width.


II.

Axioms

(Request students to present the 5 Axioms to the whole class.)

-

are some generally accepted and correct geometrical facts.
It is the common k
nowledge
of people that can be understood without proofs.

-

are the foundations of deductive proofs of geometry.

5 axioms chosed by Euclid (a Greek mathematician who put geometrical theorems with
proofs in a logical order.):

wrong

Given conditions

Logical reasoning

Conclusion

Given conditions

Logical reasoning

a)

TYT students are good students.

b)

_______ is TYT students.

_______ is a good student.

Conclusion

a)

Al
l the positive numbers are
greater than 0.

b)

5 is a positive number.

5 is greater than 0.

The study of
geometry in a
deductive way

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Axiom 1: Only one st. line can
be drawn between two points

Axiom 2: A st. line can be extended towards two ends infinitely.

Axiom 3: A circle can be fromed by taking any point as the center and any length as the radius.

Axiom 4: All right angles are equal.

Axiom 5: When two st. line ar
e cut by a transversal on the same plane, if the sum of 2 interior
angles on the same side is smaller than 2 right angles (i.e.
), then the 2 lines must
meet when they are produced on that side.


III.

Theorems

Euclid used the deductive met
hod to deduce
more than 460 theorems about geometry.

Ask students to prepare 5.3 in advance.


5.3 Theorems related to straight lines
(Presented by students)


Theorem 1:

If the sum of two adjacent angles
and
is
, the AOB is a straight line.

i.e. If


then AOB is a straight line

[adj.
supp.]






Note: “
and
are supplementary” means that

Also: “
and
are complementary” means that


Example

1

In the figure,
, prove that AOB is a straight line.











AOB is a straight line [adj.
supp.]

O

B

A

b

a

O

B

A

D

C

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Theorem 2:

The sum of all angles at a point is

i.e.
at a pt.]










Example

2


Find x in the figure.









at a pt.)



Theorem 3:

If two straight lines intersect, the vertically opposite angles are equal.

i.e. If AOB and COD are two st. lines,


then a = b, and c = d (vert. opp.
)




Example
In the
figure, AOB, COD and EOF are st. lines, prove that


[vert. opp.


[vert. opp.
]


[vert. opp.


at a pt.)



Classwork: (Ex5A,no.5 and 8)

Solution: No.5








No.8


(vert. opp.
)









A

B

D

C

a

b

c

d


x

A

C

D

B

c

a

d

b

E

D

B

F

C

A

O

z

y

x

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5.4 Theorems related to triangles

A.

Theorems for congruent triangles

1.

SAS

If AB = PQ, BC = QR and
,

Then
[S.A.S.]









2.

S.S.S

If AB = PQ, AC = PR and BC = QR,

Then
(S.S.S)







3.

A.S.A

If
and BC = QR

Then
(A.S.A)



4.

R.H.S

If AB = PQ, AC =
PR and

Then
(R.H.S)






* Note:


When 2 triangles are congruent, then

a)

the corresponding angles are equal. [corr.
]

b)

the corresponding sides are equal. [corr. sides,
]

Should be an included angle.

A

C

B

P

R

Q

A

C

B

P

R

Q

A

C

B

P

R

Q

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Example
Refer to the above No.3,
,


If
,


Then
[corr.
]

* Distribute worksheet 5.1 and ask students to complete it within 10 minutes.

Answers f
or worksheet 5.1


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Example
In the figure, AX = AC and AB = AY

a)

Prove that

b)

If
and
, find

Solution:

a)

Consider
and

A
B = AY (given)

(common angle)

AC = AX (given)

(S.A.S)

b)

sum of
)











Or common

C

Y

X

B

A

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B.

Theorems for isosceles triangles

Basic knowledge:

When two sides of a triangle are equal, the triangle is called an
isosceles triangle
.



Theorem of isosceles triangles:

Theorem 4:

The angles at the base (or base angl
es) of an isosceles triangle are equal.

i.e. In


if AB = AC


then


Theorem 5:

If two angles of a triangles are equal, then their opposite sides are equal.

i.e. In


if


then AB = AC (sides opp. equal
)



Example

In the figure, ABD is a straight line.

and AB = BC. Find a, b and c.


Solution:


on st. line)






AB =BC(given)






i.e.




c = a (proved)



A

B

C

A

C

B

C

A

B

D


b

a

c

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Example
In the figure, BDEC is a straight line. BD = EC and
. Show that

is an isosceles triangle.




(sides opp. equal
)


on st. line)


on st. line)


But




BD = EC (given)






is an isosceles triangle.


Classwork

(textbook P.187,Ex5B No.6)

Solution:

a)

Consider
and

CB = BC (common side)

AC = AB (given)

and AY = AX (given)






=
BX














b)

(proved in (a))




(sides opp. equal
)


is an isosceles triangle.


Appendix E5

Construction

I.

Construction of angle bisector of an angle


Step 1

Take B as the center.

Choose an appropriate radius,
and draw an arc to intersect
BA and BC at P and Q
re
spectively.


A

B

C

D

E

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Step 2

Take P and Q as centers, select
a radius longer than
PQ and
draw two arcs such that these
two arcs intersect at D.


Step 3

Join BD. BD is the angle
bisector of



II.

Construction of perpend
icular bisector of a line segment


Step 1

Take A as the center. Choose a
radius longer than
and
draw an arc above and below
the line segment AB.


Step 2

Take B as the center. Use the
same radius and draw another
two ones to intersec
t the arcs at
C and D.


Step 3

Join CD. CD intersects AB at
M, M is the mid
-
point of AB.
CD is the perpendicular
bisector of AB


III.

Construction of a line passing through a given point on a line segment and perpendicular to
that line segment


Step 1

Take
P as the center. Choose
an appropriate radius and draw
an arc to intersect AB at H and
K.


Step 2

Take H and K as centers. Use a
radius longer than
and
draw two arcs such that they
meet at Q.


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Step 3

Join PQ. Then
.



IV.

Construction of a line passing through a point lying outside a line segment and
perpendicular to that line segment


Step 1

Take P as the center. Choose
an appropriate radius and draw
an arc to intersect AB at H and
K.


Step 2

Take H and K as

centers.
Choose a radius longer than
and draw two arcs such
that the two arcs meet at Q.


Step 3

Join PQ. The line PQ
intersects AB at M, and