Transcript Theorems 0011 0010 1010 1101 0001 0100 1011 Leaving Certificate - Ordinary level Developed by Pádraic Kavanagh.

Theorems
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0011 0010 1010 1101 0001 0100 1011
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Leaving Certificate - Ordinary level
Developed by Pádraic Kavanagh
Click on the theorem you what to revise!
Theorem 1
Theorem 6
Theorem 2
Theorem 7
Theorem 3
Theorem 8
0011 0010 1010 1101 0001 0100 1011
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Theorem 9
Theorem 5
Theorem 10
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Theorem 4
Theorem 1:
The sum of the degree measures of the angles of a triangle is
1800
Given:
Triangle
To Prove:
1 + 2 + 3 = 1800
Construction: Draw line through 3 parallel to the base
4 3 5
Proof:

1
2
3 + 4 + 5 = 1800
Straight line
1 = 4 and 2 = 5
Alternate angles
3 + 1 + 2 = 1800
1 + 2 + 3 = 1800
Q.E.D.
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Theorem 2:
The opposite sides of a parallelogram have equal lengths.
Given:
Parallelogram abcd
To Prove:
|ab| = |cd| and |ad| = |bc|
Construction: Diagonal |ac|
Proof:
b
a
3
Quit
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c
2
d
Alternate angles
|ac| = |ac|
Given
2 = 3
Alternate angles
ASA D abc = D acd

1
1 = 4
|ab| = |cd| and |ad| = |bc|
Q.E.D.
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Theorem 3:
If three parallel lines make intercepts of equal length on a
transversal, then they will also make equal length on any other
transversal.
d
b1
2
c 3
4
Given:
Diagram as shown with |ab| = |bc|
To Prove:
|db| = |be|
Construction:
Another transversal through b.
Proof:
1 = 2
Verticially opposite
|ab| = |bc|
Given
3 = 4
Alternate angles
a
ASA D dab = D bec
e

|db| = |be|
Q.E.D.
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Theorem 4:
A line which is parallel to one side-line of a triangle, and cuts a
second side, will cut the third side in the same proportion as the
second.
Given:
Diagram as shown with line |xy| parallel
to base.
|ax |
To Prove:
|xb|
a
b
y
c
m equal parts n equal parts
x
=
|ay|
|yc|
Construction:
Draw in division lines
Proof:
|ax | is divided in m equal parts

|ay| is also
|xb| is divided in n equal parts

|yc| is also
|ax |
|xb|
=
Theorem 2
|ay|
|yc|
Q.E.D.
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Theorem 5:
If the three angles of one triangle have degree-measures equal,
respectively, to the degree-measures of the angles of a second
triangle, then the lengths of the corresponding sides of the two
triangles are proportional.
Given:
Two Triangles with equal angles
To Prove:
|ab|
=
|de|
x
4
b
Quit
|df |
=
|bc|
|ef |
a
d
Construction: Map D def onto D axy, draw in [xy]
2
2
Proof:
5
y e
1
D def mapped onto D axy
1 = 4
3

f

1
|ac|
3
[xy] is parallel to [bc]
|ab|
|ax|
|ab|
c
|de|
=
=
Q.E.D.
|ac|
Theorem 3
|ay|
|ac|
|df |
Similarly =
|bc|
|ef |
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Theorem 6:
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 lengths of the other two sides. (Pythagoras)
b
a
a
c
c
c
b
a
Quit
Given:
Triangle abc
To Prove:
a2 + b2 = c2
Construction: Three right angled triangles as shown
b
Proof:
Area of large sq. = area of small sq. + 4(area D)
(a + b)2 = c2 + 4(½ab)
a2 + 2ab +b2 = c2 + 2ab
c
b
a
a2 + b2 = c2
Q.E.D.
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Theorem 7:
If the square of the length of one side of a triangle is equal to the
sum of the squares of the lengths of the other two sides, then the
triangle has a right-angle and this is opposite the longest side.
(Converse of Pythagoras’ Theorem)
Given:
Triangle abc with |ac|2 = |ab|2 + |bc|2
Construction: Triangle def with |de| = |ab|
|ef | = |bc|
&
d
a
b
1
c
e
2
2 = 900
To Prove:
To prove 1 = 900
Proof:
|ac|2 = |ab|2 + |bc|2
given
|ac|2 = |de|2 + |ef|2
from construction
|ac|2 = |df|2
|df|2 = |de|2 + |ef|2
f

|ac| = |df|
SSS  D abc = D def
1 = 900
Q.E.D.
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Theorem 8:
The products of the lengths of the sides of a triangle by
the corresponding altitudes are equal.
Given:
Triangle abc
To Prove:
|bc||ad| = |ac||be|
Construction: Altitudes [be] and [ad]
Compare D bce to D acd
Proof:
2
a
b
5
2
b
3
d
4
2 = 4

3 = 5
d
4
1
c
Similar triangles
c
|bc|
|ac|

Quit
c
1 = 1
e
1
5
1
3
a
e
=
|be|
|ad|
|bc||ad| = |ac||be|
Q.E.D.
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Theorem 9:
If lengths of two sides of a triangle are unequal, then the
degree-measures of the angles opposite to them are unequal,
with the greater angle opposite to the longer side.
Given:
Triangle abc
To Prove:
abc > acb
Construction: Draw [bd] such that |ab| = |ad|
1 = 2
Proof:
a
2
b
1
Quit
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
1 + 3 > 2
But
2 > 4

1 + 3 > 4

abc > acb
As |ab|= |ad|
and the greater angle is
d
opposite the longer side.
4
c
Q.E.D
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Theorem 10:
The sum of the lengths of any two sides of a triangle is greater
than that of the third side.
Given:
Triangle abc
To Prove:
|ba| + |ac| > |bc|
Construction: Draw D acd such that |ad| = |ac|
In D acd |ad| = |ac|
Proof:
1 = 2
d
2
a
b

1 + 3 > 2

|bd| > |bc|
But
|bd| = |ba| + |ad|
1
3 c
|bd| = |ba| + |ac|

as |ad| = |bc|
|ba| + |ac| > |bc|
Q.E.D.
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