Download Theorem 1: The sum of the degree measures of the angles of a

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.
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
4
c
2
∠1 = ∠4
Alternate angles
|ac| = |ac|
Given
∠2 = ∠3
Alternate angles
ASA ⇒ ∆ abc = ∆ acd
a
3
⇒
1
d
|ab| = |cd| and |ad| = |bc|
Q.E.D.
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 ⇒ ∆ dab = ∆ bec
e
⇒
|db| = |be|
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.
Given:
Diagram as shown with line |xy| parallel
to base.
To Prove:
|xb|
a
b
y
c
m equal parts n equal parts
x
|ax |
=
|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|
=
Q.E.D.
|ay|
|yc|
Theorem 2
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
|df |
=
|bc|
|ef |
a
d
Construction: Map ∆ def onto ∆ axy, draw in [xy]
2
2
Proof:
∆ def mapped onto ∆ axy
∠1 = ∠4
5
y e
1
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 |
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
b
c
a
Given:
Triangle abc
To Prove:
a 2 + b2 = c 2
Construction: Three right angled triangles as shown
b
Proof:
Area of large sq. = area of small sq. + 4(area ∆)
(a + b)2 = c2 + 4(½ab)
a2 + 2ab +b2 = c2 + 2ab
c
b
a
a 2 + b2 = c 2
Q.E.D.
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)
Triangle abc with |ac|2 = |ab|2 + |bc|2
Given:
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 ⇒ ∆ abc = ∆ def
∠1 = 900
Q.E.D.
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 ∆ bce to ∆ acd
Proof:
2
a
b
5
2
b
3
4
d
c
∠1 = ∠1
e
∠2 = ∠4
⇒
1
5
1
3
a
e
∠3 = ∠5
Similar triangles
c
|bc|
|ac|
⇒
=
|be|
|ad|
|bc||ad| = |ac||be|
Q.E.D.
d
4
1
c
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
3
⇒
∠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
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 ∆ acd such that |ad| = |ac|
In ∆ 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|
⇒
|ba| + |ac| > |bc|
Q.E.D.
as |ad| = |bc|