Download Proportions: A ratio is the quotient of two

Proportions:
2
is a ratio of 2 and 3.
3
An equality of two ratios is a proportion. For example,
3 15
=
is a proportion.
7 45
If two sets of numbers (none of which is 0) are in a proportion, then we say that
the numbers are proportional.
a
c
Example, if = , then a and b are proportional to c and d.
b
d
a
c
Remember from algebra that, if = , then ad = bc.
b
d
The following proportions are equivalent:
c
a
=
b
d
a
b
=
c
d
d
b
=
a
c
c
d
=
a
b
Theorem: Three or more parallel lines divide trasversals into proportion.
A ratio is the quotient of two numbers. For example,
l
a
c
m
b
d
n
In the picture above, lines l, m, and n are parallel to each other. This means
that they divide the two transversals into proportion. In other words,
a
c
=
b
d
a
b
=
c
d
b
d
=
a
c
c
d
=
a
b
Theorem: In any triangle, a line that is paralle to one of the sides divides the
other two sides in a proportion.
E
a
c
F
G
b
d
A
B
In picture above, F G is parallel to AB, so it divides the other two sides of 4AEB,
EA and EB into proportions. In particular,
a
c
=
b
d
b
d
=
a
c
a
b
=
c
d
c
d
=
a
b
Theorem: The bisector of an angle in a triangle divides the opposite side into
segments that are proportional to the adjacent side.
C
a
b
A
c
D
d
B
In 4ACB above, CD is the bisector of ∠ACB. We have:
a b
=
d c
d c
=
a b
d a
=
c
b
c
b
=
d a
Similar Triangles:
Two triangles are said to be similar to each other if they have the same shape
(not necessarily the same size). If two triangles are similar, then all three of
their angles are congruent to each other, and their corresponding sides are in
proportion. This means that the ratio of their corresponding sides are equal
to each other. We use the ∼ symbol to mean similar to. So we write 4ABC ∼
4DEF if triangle ABC is similar triangle DEF .
In the picture below, 4ABC ∼ 4A0 B 0 C 0 . This means that ∠A ∼
=
= ∠A0 , ∠B ∼
0
0
∼
∠B , and ∠C = ∠C . Further more, their corresponding sides are in proportion,
so
AB
A0 B 0
= 0 0
AC
AC
AB
A0 B 0
= 0 0
BC
BC
BC
B 0C 0
= 0 0
AC
AC
C0
C
A
B
A0
B0
Theorem: If 4ABC ∼ 4DEF , and 4DEF ∼ GHI, then 4ABC ∼ GHI. In
other words, similarity is transitive.
AA (Angle-Angle) Similarity Theorem:
If two angles of a triangle is congruent to two angles of another triangle, then the
two triangles are similar.
C0
C
A
B
A0
B0
In picture above, ∠A ∼
= ∠A0 , ∠B ∼
= ∠B 0 , therefore, 4ABC ∼ 4A0 B 0 C 0
SAS (Side Angle Side) Similarity Theorem: If an angle is congruent to an
angle of another triangle the the sides that make up the angles are in proportion,
then the two triangles are similar.
C0
C
b0
a0
a
b
A
A0
B
B0
In picture above, if ∠C ∼
= ∠C 0 , and AC, BC are in proportion with A0 C 0 , B 0 C 0 ,
0
a a
meaning that = 0 , then 4ABC ∼ 4A0 B 0 C 0
b
b
SSS (Side-Side-Side) Similarity: If all three sides of a triangle are in proportion with three sides of another triangle, then the two triangles are similar.
C0
C
b0
b
A
c
a0
a
A0
B
c
0
B0
In picture above, if the three sides of 4ABC is proportional to the three sides
of 4A0 B 0 C 0 , meaning that:
a
b
c
=
=
,
a0
b0
c0
then 4ABC ∼ 4A0 B 0 C 0
Theorem: A line parallel to one of the sides of a triangle divides the triangle
into a smaller triangle similar to the original one.
E
F
A
G
B
In picture above, if F G||AB, then 4F EG ∼ 4AEB
Theorem: The altitude of a right triangle (through the right angle) divides the
triangle into two similar right triangles that are also similar to the original right
triangle.
C
A
D
B
In picture above, 4ABC is a right triangle with ∠ACB being the right angle.
CD is the altitude throught C perpendicular to AB. We will prove that
4ACB ∼ 4ADC ∼ 4CDB
Proof:
Statements
1.
CD is altitude to 4ACB
Reasons
1.
given
2.
3.
4.
Def. of Altitude
Def. of ⊥ lines
Angle Addition Postulate
5.
6.
7.
8.
9.
Def. of Comp. Angles
Sum of Interior Angles of 4
Def. of Comp. Angles
∠’s Comp. to same ∠ are ∼
=
Sum of Interior Angles of 4
2.
3.
4.
AB ⊥ CD
∠CDB and ∠CDA are right ∠’s
5.
6.
7.
8.
9.
∠ACD complementary to ∠BCD
m∠CAD + m∠ACD = 90◦
∠ACD complementary to ∠CAD
∠CAD ∼
= ∠CAD
m∠CBD + m∠BCD = 90◦
10.
11.
12.
13.
∠CBD complementary to ∠BCD 10. Def. of Comp. Angles
∠CBD ∼
11. ∠’s Comp. to same ∠ are ∼
= ∠ACD
=
4ADC ∼ 4CDB
12. AA ∼
4ACB ∼ 4ADC
13. AA ∼
m∠ACD + m∠BCD = 90◦
The previous theorem about the altitude of a right triangle dividing the right
triangle into three similar triangles allows us to prove one of the most important
theorem in geometry (and mathematics in general):
Pythagorean Theorem:
In any right triangle, the sum of the square of the length of the two legs is equal
to the square of the length of the hypotenuse.
C
b
a
h
d
A
e
B
D
c
D
D
h
d
A
b
e
h
C
C
a
To prove the pythagorean theorem, note from the previous theorem that
4ACB ∼ 4ADC ∼ 4CDB.
Therefore, their corresponding sides are in proportion. In particular,
b d
a
e
= ;
=
c
b
c
a
The first equation gives us:
b2
b d
= ⇒d=
c
b
c
The second equation gives us:
a
e
a2
= ⇒e=
c
a
c
Notice that e + d = c, therefore, we have:
a2 b2
a2 + b2
e+d=
+ =c⇒
= c ⇒ a2 + b 2 = c 2
c
c
c
B
Example:
5
7
2
h
√
c
6
√
√
In first triangle, 52 + c2 = 72 ⇒ 25 + c2 = 49 ⇒ c2 = 24 ⇒ c = 24 = 2 6
√
√ 2
6 + (2)2 = h2 ⇒ h2 = 6 + 4 = 10 ⇒ h = 10
In second triangle,
Special Triangles:
A right triangle where one of the acute angle is 30◦ and the other acute angle is
60◦ is called a 30-60-90 triangle.
Theorem: In a 30-60-90 triangle, if the side opposite √
the 30◦ angle has length
of a, then the side opposite the 60◦ angle has length of 3a, and the hypotenuse
has length 2a.
B
60◦
2a
A
30◦
√
a
C
3a
In 4ABC, BC is the side opposite the 30◦ angle ∠A, if length of √
BC = a, then
◦
the side opposite the 60 angle, side AC, will have a length of 3a, and the
hypotenuse has length 2a.
We can prove the 30◦ − 60◦ − 90◦ triangle by drawing the altitude CD to the
equilateral triangle ABC as above. Let b be the length of each of its side. Notice
that the altitude CD divides the eqilateral triangle into two congruent triangles
(why?). Each of the triangle is a 30 − 60 − 90 triangle and the length of AD is
half of b. Using Pythagorean theorem, we see that
!2
b2
b2
2
2
2
2
+h =b ⇒ +h =b ⇒h =b − ⇒
4
4
√
4b2 b2
3b2
3b
b√
h2 =
− ⇒ h2 =
⇒h=
=
3
4
4
4
2
2
b
2
2
2
C
30◦
b
60◦
A
b
h
D
b/2
b/2
B
b
A 45-45-90 triangle is an isoceles right triangle. In an isoceles right triangle,
each acute angle is the same, therefore, each acute angle is 45◦ .
Theorem: In a 45◦ − 45◦ − 90◦ triangle, if the length of one of the leg is a,
√then
the length of the other side is also a, and the length of the hypotenuse is 2a
B
45◦
√
A
2a
a
45◦
a
C
The 45◦ − 45◦ − 90◦ triangle can be proved just using the Pythagorean theorem.
In 4ABC, if one of the leg has length a, the the Isoceles triangle theorem tells us
the other leg must also has length a since the base angles are congruent. Using
the Pythagorean theorem, the length of the hypotenuse, h, is:
√
2
2
2
2
h = a + a = 2a ⇒ h = 2a
If two chords intersect inside a circle, the product of the lengths of the segments
of one chord is equal to the product of the lengths of the segments of the other.
B
C
A
E
D
In the circle above, AC, BD are chords intersect at E, the theorem tells us that:
AE · EC = BE · ED
If two secant lines intersect outside a circle, the product of the length of one of
the secants with the length outside the circle is equal to the product of the length
of the other secant with the length outside the circle.
D
C
A
E
B
In circle above, secants AC, BC intersect the circle at points D and E, respectively, and we have:
CA · CD = CB · CE
If a tangent and a secant to a circle intersect outside the circle, the square of the
length of the tangent is equal to the product of the length of the secant times
the length of the secant outside the circle.
B
C
D
A
In circle above, tangent BC intersects AC outside the circle at C, we have:
CB · CB = CA · CD