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42510011 0010 1010 1101 0001 0100 1011
TheoremsLeaving Certificate - Ordinary level
Developed by Pádraic Kavanagh
42510011 0010 1010 1101 0001 0100 1011
Click on the theorem you what to revise!
Theorem 1
Theorem 7
Theorem 6
Theorem 3
Theorem 4
Theorem 5
Theorem 8
Theorem 10
Theorem 9
Theorem 2
Quit
Proof: 3 + 4 + 5 = 1800Straight line
1 = 4 and 2 = 5Alternate angles
3 + 1 + 2 = 1800
1 + 2 + 3 = 1800
Q.E.D.
4 5
Given: Triangle
1 2
3Construction: Draw line through 3 parallel to the base
Theorem 1: The sum of the degree measures of the angles of a triangle is 1800
To Prove: 1 + 2 + 3 = 1800
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2
31
4
Given: Parallelogram abcd
cb
a d
Construction: Diagonal |ac|
Proof: 1 = 4 Alternate angles
|ac| = |ac| Given
2 = 3 Alternate angles
ASA abc = acd
|ab| = |cd| and |ad| = |bc|
Q.E.D.
Theorem 2: The opposite sides of a parallelogram have equal lengths.
To Prove: |ab| = |cd| and |ad| = |bc|
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Given: Diagram as shown with |ab| = |bc|
Proof: 1 = 2 Verticially
opposite
|ab| = |bc| Given
3 = 4 Alternate angles
ASA dab = bec
|db| = |be|
Q.E.D.
a
b
c
d
e
1
2
3
4
Construction: Another transversal through b.
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.
To Prove: |db| = |be|
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Given: Diagram as shown with line |xy| parallel to base.
x y
a
b c
Proof: |ax | is divided in m equal parts
|ay| is also
|xb| is divided in n equal parts
|yc| is also Theorem 2
m e
qu
al
parts
n e
qual
parts
|yc|
|ay|=
|xb|
|ax |
Construction: Draw in division lines
Q.E.D.
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.
To Prove:|yc|
|ay|=
|xb|
|ax |
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Given: Two Triangles with equal angles
Proof: def mapped onto axy
1 = 4
[xy] is parallel to [bc]
Construction: Map def onto axy, draw in [xy]
|df |
|ac|=
|de|
|ab|Similarly
|ef |
|bc|=
Q.E.D.
a
cb
d
fe1
2
3
1 3
2
x y4 5
|ay|
|ac|=
|ax|
|ab| Theorem 3
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.
To Prove:|df |
|ac|=
|de|
|ab|
|ef |
|bc|=
Quit Menu
Given: Triangle abc
Proof: Area of large sq. = area of small sq. + 4(area )
(a + b)2 = c2 + 4(½ab)
a2 + 2ab +b2 = c2 + 2ab
a2 + b2 = c2
Q.E.D.
a
b
c
a
bc
a
b
c
a
b c
Construction: Three right angled triangles as shown
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)
To Prove: a2 + b2 = c2
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Given: Triangle abc with |ac|2 = |ab|2 + |bc|2
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
|ac| = |df|
SSS abc = def
1 = 900
Q.E.D.
e
d
f
a
b c1 2
Construction:Triangle def with|de| = |ab|
|ef | = |bc|
& 2 = 900
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)
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Given: Triangle abc
Construction: Altitudes [be] and [ad]
Proof: Compare bce to acd
1 = 1
2 = 4
3 = 5
Similar triangles
|ad|
|be|=
|ac|
|bc|
|bc||ad| = |ac||be|
Q.E.D.
a
b c4
5
13
2
d
e
Theorem 8: The products of the lengths of the sides of a triangle by the corresponding altitudes are equal.
4
5
1cd
a
13
2
b c
e
To Prove: |bc||ad| = |ac||be|
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Given: Triangle abc
Construction:Draw [bd] such that |ab| = |ad|
a
b c
d
31
2
4
Proof: 1 = 2 As |ab|= |ad|
1 + 3 > 2
But 2 > 4
1 + 3 > 4
abc > acb
and the greater angle is
opposite the longer side.
Q.E.D
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.
To Prove: abc > acb
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Given: Triangle abc
Construction:Draw acd such that |ad| = |ac|
Proof: In acd |ad| = |ac|
1 = 2
1 + 3 > 2
|bd| > |bc|
But |bd| = |ba|+ |ad|
|bd| = |ba| + |ac| as |ad| = |
bc|
|ba| + |ac| > |bc|
Q.E.D.
Theorem 10: The sum of the lengths of any two sides of a triangle is greater than that of the third side.
b
a
c
To Prove: |ba| + |ac| > |bc|
3
d
1
2
Quit Menu
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