CIRCLES & COORDINATE GEOMETRY
Week 1 Session 1
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CENTRAL ANGLES,
ARCS AND CHORDS
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Objectives
At the end of the session, you can…
find the measure of an arc
find the measure of a central angle
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My Bag of Ideas 2
A central angle of a circle is an angle whose vertex is the center of the circle.
An arc is a part of a circle. It is measured in degrees.
A minor arc is an arc which is smaller than a semicircle. Its measure and is less than
180o .
A major arc is an arc which is bigger than a semicircle. Its measure is greater than 180°.
A semicircle is an arc that is half of the circle. It measures 180o and its endpoints form a
diameter.
Congruent circles are circles with the same radius.
Congruent arcs are arcs with the same measure.
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Illustration: Refer to circle 𝑂 below. Identify the Central Angles, Minor Arcs, Major Arcs,
and Semicircles.
Central Angles: ∠𝐴𝑂𝐵, ∠𝐶𝑂𝐷, ∠𝐴𝑂𝐷, and ∠𝐵𝑂𝐶
Minor Arcs: 𝐴𝐵 , 𝐵𝐶 , 𝐶𝐷 , and 𝐴𝐷 .
Major Arcs: 𝐴𝐵𝐷 , 𝐴𝐶𝐵 , 𝐶𝐴𝐵 , and 𝐶𝐴𝐷 .
Semicircles: 𝐴𝐵𝐶 and 𝐵𝐶𝐷 .
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Illustration: Refer to circle 𝑂 below. Identify the figure.
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The figure above shows the
central angle 𝐴𝑂𝐵. In symbols,
∠𝐴𝑂𝐵.
The figure above shows the
minor arc 𝐴𝐵. In symbols, 𝐴𝐵 . It
can also be named as 𝐵𝐴 . It is
smaller than a semicircle.
Illustration: Refer to circle 𝑂 below. Identify the figure.
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The figure above shows the major
arc 𝐵𝐴𝐶 . In symbols, 𝐵𝐴𝐶 . It can
also be named as 𝐵𝐷𝐶 , 𝐶𝐷𝐵 , or
𝐶𝐴𝐵 . It is greater than a semicircle.
The figure above shows the
semicircle 𝐴𝐷𝐶 . In symbols,
𝐴𝐷𝐶 . It can also be named as
𝐶𝐷𝐴 . It is half of the circle.
REMEMBER:
Arcs are named after its endpoints with curving line above
them (e.g 𝑋𝑌).
We only use two letters in naming a minor arc where the
order of endpoints does not matter (e.g 𝐴𝐵 or 𝐵𝐴).
To name a major arc and a semicircle, we use three letters
by naming the two endpoints, as well as an intermediate
point that the arc passes through (e.g 𝐴𝐷𝐵 or 𝐵𝐷𝐴).
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Practice Now 2
Use the given figure to identify the different the arcs as minor arc, major arc or semicircle.
Mark the appropriate box for each.
Minor Arc
1. 𝑀𝑁
2. 𝑀𝑆𝑄
3. 𝑃𝑀
✓
✓
4. 𝑃𝑆𝑅
5. 𝑀𝑅𝑄
6. 𝑃𝑄
7. 𝑁𝑃𝑅
8. 𝑀𝑄𝑁
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9. 𝑁𝑅𝑄
10. 𝑄𝑃𝑆
Major Arc
Semicircle
✓
✓
✓
✓
✓
✓
✓
✓
Postulate: The Arc Addition Postulate
The measure of the arc formed by two adjacent arcs is
the sum of the measures of the two arcs.
In the figure at the right,
𝑚 𝐴𝐶 = 𝑚 𝐴𝐵 + 𝑚 𝐵𝐶
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Example: Find the measure of each arc in the figure at the right.
Given: 𝑚 𝐴𝐵 = 52 and 𝑚 𝐵𝐶 = 148.
a. Find 𝑚 𝐴𝐵𝐶.
b. Find 𝑚 𝐴𝐶.
Solution:
a. Based on the figure, observe that 𝐴𝐵𝐶 is
composed of 𝐴𝐵 and 𝐵𝐶 . Applying the Arc
Addition Postulate gives
𝑚 𝐴𝐵𝐶 = 𝑚 𝐴𝐵 + 𝑚 𝐵𝐶
𝑚 𝐴𝐵𝐶 = 52 + 148
𝑚 𝐴𝐵𝐶 = 200.
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Therefore 𝑨𝑩𝑪 measures 𝟐𝟎𝟎𝐨 .
Example: Find the measure of each arc in the figure at the right.
Given: 𝑚 𝐴𝐵 = 52 and 𝑚 𝐵𝐶 = 148.
a. Find 𝑚 𝐴𝐵𝐶.
b. Find 𝑚 𝐴𝐶.
Solution:
b. Notice that ⊙ 𝑂 is composed of 𝐴𝐵𝐶 and
𝐴𝐶 . Note that the degree measure of a circle is
360. Applying the Arc Addition Postulate gives
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𝑚 𝐴𝐵𝐶 + 𝑚 𝐴𝐶 = 360.
Based on the previous item, 𝑚 𝐴𝐵𝐶 = 200,
substitute it in the equation.
200 + 𝑚 𝐴𝐶 = 360.
𝑚 𝐴𝐶 = 360 − 200
Therefore,
𝑚 𝐴𝐶 = 160.
𝑨𝑪 measures 𝟏𝟔𝟎𝐨 .
Postulate: The Central AngleIntercepted Arc Postulate
The measure of a central angle of a circle is equal to the
measure of its intercepted arc.
In the figure at the right,
𝑚∠𝐴𝑂𝐵 = 𝑚 𝐴𝐵
𝑟o
𝑟o
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Example: In ⊙ 𝑂 at the right, 𝐴𝐷 is a diameter. Find the measure of each arc.
a. Find 𝑚 𝐴𝐵.
. 𝐴𝐸𝐷 .
c. Find 𝑚
b. Find 𝑚 𝐴𝐶 .
d. Find 𝑚 𝐶𝐷 .
e. Find 𝑚 𝐶𝐷𝐴 .
Solution:
a. 𝐴𝐵 is intercepted by the central angle ∠𝐴𝑂𝐵. Since
the measure of the central angle is equal to the
measure of its intercepted arc, thus
𝑚∠𝐴𝑂𝐵 = 𝑚 𝐴𝐵.
Since 𝑚∠𝐴𝑂𝐵 = 55, 𝑚 𝐴𝐵 = 55.
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Therefore, 𝑨𝑩 measures 𝟓𝟓𝐨 .
100o
55o
Example: In ⊙ 𝑂 at the right, 𝐴𝐷 is a diameter. Find the measure of each arc.
a. Find 𝑚 𝐴𝐵.
. 𝐴𝐸𝐷 .
c. Find 𝑚
b. Find 𝑚 𝐴𝐶 .
d. Find 𝑚 𝐶𝐷 .
e. Find 𝑚 𝐶𝐷𝐴 .
Solution:
b. 𝐴𝐶 is intercepted by the central angle ∠𝐴𝑂𝐶. So,
𝑚∠𝐴𝑂𝐶 = 𝑚 𝐴𝐶. Solving for 𝑚∠𝐴𝑂𝐶,
100o
𝑚∠𝐴𝑂𝐶 = 𝑚∠𝐴𝑂𝐵 + 𝑚∠𝐵𝑂𝐶.
By substitution,
𝑚∠𝐴𝑂𝐶 = 55 + 100
𝑚∠𝐴𝑂𝐶 = 155.
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Since 𝑚∠𝐴𝑂𝐶 = 𝑚 𝐴𝐶 , it follows that 𝑚 𝐴𝐶 = 155.
Hence, 𝑨𝑪 measures 𝟏𝟓𝟓𝐨 .
55o
Example: In ⊙ 𝑂 at the right, 𝐴𝐷 is a diameter. Find the measure of each arc.
a. Find 𝑚 𝐴𝐵.
. 𝐴𝐸𝐷 .
c. Find 𝑚
b. Find 𝑚 𝐴𝐶 .
d. Find 𝑚 𝐶𝐷 .
e. Find 𝑚 𝐶𝐷𝐴 .
Solution:
c. 𝐴𝐸𝐷 is a semicircle. Note that the degree measure of a
circle is 360. Since a semicircle is half of a circle,
𝑚 𝐴𝐸𝐷 = 180.
100o
55o
d. 𝐴𝐵𝐷, a semicircle, is composed of 𝐴𝐵 , 𝐵𝐶, and 𝐶𝐷 .
Applying the Arc Addition Postulate,
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𝑚 𝐴𝐵 + 𝑚 𝐵𝐶 + 𝑚 𝐶𝐷 = 𝑚 𝐴𝐵𝐷
55 + 100 + 𝑚 𝐶𝐷 = 180
𝑚 𝐶𝐷 = 180 − 155
Therefore,
𝑚 𝐶𝐷 = 25
𝑪𝑫 measures 𝟐𝟓𝒐 .
Example: In ⊙ 𝑂 at the right, 𝐴𝐷 is a diameter. Find the measure of each arc.
a. Find 𝑚 𝐴𝐵.
. 𝐴𝐸𝐷 .
c. Find 𝑚
b. Find 𝑚 𝐴𝐶 .
d. Find 𝑚 𝐶𝐷 .
e. Find 𝑚 𝐶𝐷𝐴 .
Solution:
e. 𝑚 𝐶𝐷𝐴 is a major arc composed of 𝐶𝐷 and semicircle
𝐷𝐸𝐴. Applying the Arc Addition Postulate,
𝑚 𝐶𝐷 + 𝑚 𝐷𝐸𝐴 = 𝑚 𝐶𝐷𝐴
25 + 180 = 𝑚 𝐶𝐷𝐴
205 = 𝑚 𝐶𝐷𝐴
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Thus, 𝑪𝑫𝑨 is 𝟐𝟎𝟓𝐨 .
100o
55o
REMEMBER:
The measure of a circle is 360°.
The measure of a semicircle is 180°.
The measure of a central angle is equal to the measure
of its intercepted arc.
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Practice Now 3
In ⊙ 𝑂 at the right, 𝑚∠𝑄𝑂𝑅 = 40. Find the measure of the following:
1. 𝑚 𝑅𝑄 =
2.
𝑚 𝑀𝑆 =
3.
𝑚 𝑀𝑅𝑄 =
4.
𝑚∠𝑀𝑂𝑆 =
5.
𝑚∠𝑀𝑂𝑅 =
𝟒𝟎
𝟒𝟎
𝟏𝟖𝟎
𝟒𝟎
𝟏𝟒𝟎
6.
7.
8.
9.
𝑚 𝑀𝑆𝑄 =
𝑚 𝑀𝑅 =
𝑚 𝑆𝑄 =
𝑚 𝑅𝑀𝑆 =
10. 𝑚∠𝑆𝑂𝑄 =
𝟏𝟖𝟎
𝟏𝟒𝟎
𝟏𝟒𝟎
𝟏𝟖𝟎
𝟏𝟒𝟎
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INSCRIBED ANGLES AND
INTERCEPTED ARCS
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Objectives
At the end of the session, you can…
identify an inscribed angle and its intercepted arc
find the measure of an inscribed angle and intercepted arc
using the theorems on inscribed angles and intercepted arcs
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My Bag of Ideas 3
An inscribed angle in a circle is an angle whose vertex is on the circle and whose sides
contain chords of the circle.
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inscribed angle
not an inscribed angle
not an inscribed angle
Theorem 9
In a circle, an inscribed angle is half the measure of its
intercepted arc.
1
𝑚∠𝐴 = 𝑚 𝐵𝐶
2
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Example: Using the figure at the right. Complete the table by identifying the corresponding intercepted
arc of each inscribed angle.
Inscribed Angle
Intercepted Arc
1.
∠𝑊𝑇𝑉
𝑾𝑽
2.
∠𝑈𝑊𝑉
3.
∠𝑊𝑈𝑉
4.
∠𝑇𝑊𝑈
5.
∠𝑇𝑈𝑉
6.
∠𝑇𝑈𝑊
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Example: Using the figure at the right. Complete the table by identifying the corresponding intercepted
arc of each inscribed angle.
Inscribed Angle
Intercepted Arc
1.
∠𝑊𝑇𝑉
𝑾𝑽
2.
∠𝑈𝑊𝑉
𝑼𝑽
3.
∠𝑊𝑈𝑉
4.
∠𝑇𝑊𝑈
5.
∠𝑇𝑈𝑉
6.
∠𝑇𝑈𝑊
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Example: Using the figure at the right. Complete the table by identifying the corresponding intercepted
arc of each inscribed angle.
Inscribed Angle
Intercepted Arc
1.
∠𝑊𝑇𝑉
𝑾𝑽
2.
∠𝑈𝑊𝑉
𝑼𝑽
3.
∠𝑊𝑈𝑉
𝑾𝑽
4.
∠𝑇𝑊𝑈
5.
∠𝑇𝑈𝑉
6.
∠𝑇𝑈𝑊
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Example: Using the figure at the right. Complete the table by identifying the corresponding intercepted
arc of each inscribed angle.
Inscribed Angle
Intercepted Arc
1.
∠𝑊𝑇𝑉
𝑾𝑽
2.
∠𝑈𝑊𝑉
𝑼𝑽
3.
∠𝑊𝑈𝑉
𝑾𝑽
4.
∠𝑇𝑊𝑈
𝑻𝑼
5.
∠𝑇𝑈𝑉
6.
∠𝑇𝑈𝑊
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Example: Using the figure at the right. Complete the table by identifying the corresponding intercepted
arc of each inscribed angle.
Inscribed Angle
Intercepted Arc
1.
∠𝑊𝑇𝑉
𝑾𝑽
2.
∠𝑈𝑊𝑉
𝑼𝑽
3.
∠𝑊𝑈𝑉
𝑾𝑽
4.
∠𝑇𝑊𝑈
𝑻𝑼
5.
∠𝑇𝑈𝑉
𝑻𝑾𝑽
6.
∠𝑇𝑈𝑊
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Example: Using the figure at the right. Complete the table by identifying the corresponding intercepted
arc of each inscribed angle.
Inscribed Angle
Intercepted Arc
1.
∠𝑊𝑇𝑉
𝑾𝑽
2.
∠𝑈𝑊𝑉
𝑼𝑽
3.
∠𝑊𝑈𝑉
𝑾𝑽
4.
∠𝑇𝑊𝑈
𝑻𝑼
5.
∠𝑇𝑈𝑉
𝑻𝑾𝑽
6.
∠𝑇𝑈𝑊
𝑻𝑾
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REMEMBER:
A minor arc is named using its endpoints where the
order does not matter.
Semicircles and major arcs are named using 3 letters; its
endpoints and a point in between. The order of the
endpoints does not matter.
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Example: The circle below has center O and ∠𝐵𝑂𝐴 = 40o . Find the measures of the following:
1. ∠𝐴𝐷𝐵
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Solution:
Looking at the figure, see that ∠𝐴𝐷𝐵 is an inscribed angle with
intercept arc 𝐴𝐵. Knowing that the measure of an inscribed angle
is half of its intercepted arc, first find the measure of 𝐴𝐵 . Now,
since 𝐴𝐵 is also the intercepted arc of the central angle ∠𝐵𝑂𝐴, it
follows that 𝑚 𝐴𝐵 = 𝑚∠𝐵𝑂𝐴. Since it is given that 𝑚∠𝐵𝑂𝐴 = 40,
then 𝑚 𝐴𝐵 = 40. Thus,
1
𝑚∠𝐴𝐷𝐵 = 𝑚 𝐴𝐵
2
1
𝑚∠𝐴𝐷𝐵 = 40
2
𝑚∠𝐴𝐷𝐵 = 20.
Therefore, the measure of ∠𝑨𝑫𝑩 is 𝟐𝟎𝐨 .
40o
Example: The circle below has center O and ∠𝐵𝑂𝐴 = 40o . Find the measures of the following:
2. 𝐴𝐵
Solution:
In the figure, 𝐴𝐵 is the intercepted arc of the central ∠𝐵𝑂𝐴. Then it
follows that 𝑚∠𝐵𝑂𝐴 = 𝑚 𝐴𝐵 . Since 𝑚∠𝐵𝑂𝐴 = 40 then 𝐴𝐵
measures 40o .
3. 𝑚∠𝐴𝐶𝐵
Solution:
The figure shows that ∠𝐴𝐶𝐵 is an inscribed angle with intercepted
1
arc 𝐴𝐵. Hence, 𝑚∠𝐴𝐶𝐵 = 2 𝐴𝐵 . From item number 2, we see that
𝑚 𝐴𝐵 = 40. Then it follows that 𝑚∠𝐴𝐶𝐵 = 20.
40o
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REMEMBER:
The measure of a central angle is equal to its
intercepted arc.
The measure of an inscribed angle is one-half the
measure of its intercepted arc.
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Theorem 10
An angle inscribed in a semicircle is a right angle.
● In ⊙O, ∠𝐶 is inscribed in semicircle 𝐴𝑃𝐵, then ∠𝐴𝐶𝐵
is a right angle.
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Theorem 11
Two or more angles inscribed in the same arc are
congruent.
● In ⊙O, both ∠𝐵 and ∠𝐶 are inscribed in 𝐴𝐷, then
∠𝐵 ≅ ∠𝐶.
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Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
b.
𝑚∠𝐵𝐶𝐴 =
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
c.
𝑚∠𝐵𝑂𝐴 =
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
i.
𝑚 𝐷𝐸𝐵 =
e.
𝑚∠𝐷𝐸𝐴 =
j.
𝑚∠𝐵𝑂𝐷 =
Solution:
1
a. 𝐴𝐵 is the intercepted arc of the inscribed angle ∠𝐵𝐸𝐴. Since 𝑚∠𝐵𝐸𝐴 = 2 𝐴𝐵 and
𝑚∠𝐵𝐸𝐴 = 40, then 𝑚 𝐴𝐵 = 80.
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Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
b.
𝑚∠𝐵𝐶𝐴 =
c.
𝑚∠𝐵𝑂𝐴 =
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
i.
𝑚 𝐷𝐸𝐵
𝐷𝐸𝐴 =
𝑚
e.
𝑚∠𝐷𝐸𝐴 =
j.
𝑚∠𝐵𝑂𝐷 =
Solution:
1
b. ∠𝐵𝐶𝐴 is an inscribed angle with intercepted arc 𝐴𝐵 . Hence, 𝑚∠𝐵𝐶𝐴 = 2 𝐴𝐵 . From
letter a above, 𝑚 𝐴𝐵 = 40. Then it follows that 𝑚∠𝐵𝐶𝐴 = 40.
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Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
b.
𝑚∠𝐵𝐶𝐴 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
c.
𝑚∠𝐵𝑂𝐴 =
80
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
i.
𝑚 𝐷𝐸𝐵 =
e.
𝑚∠𝐷𝐸𝐴 =
j.
𝑚∠𝐵𝑂𝐷 =
Solution:
c. ∠𝐵𝑂𝐴 is a central angle with intercepted arc 𝐴𝐵 . Then 𝑚∠𝐵𝑂𝐴 = 𝑚 𝐴𝐵. From
letter a above, 𝑚 𝐴𝐵 = 80. Therefore, 𝑚∠𝐵𝑂𝐴 = 80.
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Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
b.
𝑚∠𝐵𝐶𝐴 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
c.
𝑚∠𝐵𝑂𝐴 =
80
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
180
i.
𝑚 𝐷𝐸𝐵 =
e.
𝑚∠𝐷𝐸𝐴 =
j.
𝑚∠𝐵𝑂𝐷 =
Solution:
d. 𝐴𝐵𝐷 is a semicircle. Therefore, 𝑚 𝐴𝐵𝐷 = 180.
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Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
b.
𝑚∠𝐵𝐶𝐴 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
c.
𝑚∠𝐵𝑂𝐴 =
80
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
180
i.
𝑚 𝐷𝐸𝐵 =
e.
𝑚∠𝐷𝐸𝐴 =
90
j.
𝑚∠𝐵𝑂𝐷 =
Solution:
e. ∠𝐷𝐸𝐴 is inscribed in the semicircle 𝐴𝐵𝐷. From Theorem 10, ∠𝐷𝐸𝐴 is a right
angle. Hence, 𝑚∠𝐷𝐸𝐴 = 90.
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Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
b.
𝑚∠𝐵𝐶𝐴 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
c.
𝑚∠𝐵𝑂𝐴 =
80
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
180
i.
𝑚 𝐷𝐸𝐵 =
e.
𝑚∠𝐷𝐸𝐴 =
90
j.
𝑚∠𝐵𝑂𝐷 =
50
Solution:
f. 𝑚∠𝐷𝐸𝐵 + 𝑚∠𝐵𝐸𝐴 = 𝑚∠𝐷𝐸𝐴. From letter e above, it follows that
𝑚∠𝐷𝐸𝐵 + 𝑚∠𝐵𝐸𝐴 = 90. Since 𝑚∠𝐵𝐸𝐴 = 40, then
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𝑚∠𝐷𝐸𝐵 + 40 = 90
Thus 𝑚∠𝐷𝐸𝐵 = 50.
Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
b.
𝑚∠𝐵𝐶𝐴 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
c.
𝑚∠𝐵𝑂𝐴 =
80
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
180
i.
𝑚 𝐷𝐸𝐵 =
e.
𝑚∠𝐷𝐸𝐴 =
90
j.
𝑚∠𝐵𝑂𝐷 =
50
100
Solution:
1
g. 𝐵𝐷 is the intercepted arc of the inscribed angle ∠𝐷𝐸𝐵. Since 𝑚∠𝐷𝐸𝐵 = 2 𝐵𝐷 and
𝑚∠𝐷𝐸𝐵 = 50. Then 𝑚 𝐵𝐷 = 100.
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Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
50
b.
𝑚∠𝐵𝐶𝐴 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
100
c.
𝑚∠𝐵𝑂𝐴 =
40
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
180
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
180
i.
𝑚 𝐷𝐸𝐵 =
e.
𝑚∠𝐷𝐸𝐴 =
90
j.
𝑚∠𝐵𝑂𝐷 =
Solution:
h. 𝐷𝐸𝐴 is a semicircle. Hence, 𝑚 𝐷𝐸𝐴 = 180.
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Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
50
b.
𝑚∠𝐵𝐶𝐴 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
100
c.
𝑚∠𝐵𝑂𝐴 =
40
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
180
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
180
i.
𝑚 𝐷𝐸𝐵 =
260
e.
𝑚∠𝐷𝐸𝐴 =
90
j.
𝑚∠𝐵𝑂𝐷 =
Solution:
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i. 𝐷𝐸𝐵 is a major arc that is composed of the semicircle 𝐷𝐸𝐴 and 𝐴𝐵.
That is, 𝑚 𝐷𝐸𝐴 + 𝑚 𝐴𝐵 = 𝑚 𝐷𝐸𝐵
From letters a and h above, 𝑚 𝐷𝐸𝐵 = 180 + 80. Therefore, 𝑚 𝐷𝐸𝐵 = 260.
Example 1: In ⊙O below, 𝑚∠𝐴𝐸𝐵 = 40 and 𝐴𝐷 is a diameter. Find the following measures.
a. 𝑚
𝑚 𝐴𝐵
𝐴𝐵 =
=
80
f.
𝑚∠𝐷𝐸𝐵 =
50
b.
𝑚∠𝐵𝐶𝐴 =
40
g.
𝑚 𝐵𝐷
𝐵𝐷 =
=
𝑚
100
c.
𝑚∠𝐵𝑂𝐴 =
40
h.
𝑚 𝐷𝐸𝐴
𝐷𝐸𝐴 =
=
𝑚
180
d.
𝑚 𝐴𝐵𝐷
𝐴𝐵𝐷 =
=
𝑚
180
i.
𝑚 𝐷𝐸𝐵 =
260
e.
𝑚∠𝐷𝐸𝐴 =
90
j.
𝑚∠𝐵𝑂𝐷 =
100
Solution:
j. ∠𝐵𝑂𝐷 is a central angle with intercepted arc 𝐵𝐷 . Hence, 𝑚∠𝐵𝑂𝐷 = 𝑚 𝐵𝐷. From
letter g above, 𝑚 𝐵𝐷 = 100, then it follows that 𝑚∠𝐵𝑂𝐷 = 100.
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Example 2: Given circle 𝐼 with diameter 𝑂𝐹, 𝑚∠𝐺𝑂𝐹 = 7𝑥 + 10, 𝑚∠𝑂𝐹𝐺 = 8𝑥 + 20,
𝑚∠𝐷𝑂𝐹 = 3𝑦 − 12, 𝑚∠𝐿𝐹𝑂 = 2𝑦 − 3 and 𝑂𝐿 ≅ 𝐷𝐹. Find the following:
1. 𝑚∠𝐺𝑂𝐹
Solution:
Because ∠𝑂𝐺𝐹 is inscribed in a semicircle, by the semicircle
theorem, ∠𝑂𝐺𝐹 is a right angle, that is 𝑚∠𝑂𝐺𝐹 = 90. Hence,
∆𝑂𝐺𝐹 is a right triangle. Note that the sum of the angles in a
triangle is 180. Thus,
𝑚∠𝐺𝑂𝐹 + 𝑚∠𝑂𝐹𝐺 + 𝑚∠𝑂𝐺𝐹 = 180
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Since 𝑚∠𝑂𝐺𝐹 = 90,
𝑚∠𝐺𝑂𝐹 + 𝑚∠𝑂𝐹𝐺 + 90 = 180
𝑚∠𝐺𝑂𝐹 + 𝑚∠𝑂𝐹𝐺 = 90
By substitution,
7𝑥 + 10 + 8𝑥 + 20 = 90
15𝑥 + 30 = 90
15𝑥 = 60
𝑥=4
Therefore,
𝑚∠𝐺𝑂𝐹 = 7𝑥 + 10
𝑚∠𝐺𝑂𝐹 = 7 4 + 10
𝑚∠𝐺𝑂𝐹 = 38
Thus, ∠𝑮𝑶𝑭 measures 𝟑𝟖𝐨 .
Example 2: Given circle 𝐼 with diameter 𝑂𝐹, 𝑚∠𝐺𝑂𝐹 = 7𝑥 + 10, 𝑚∠𝑂𝐹𝐺 = 8𝑥 + 20,
𝑚∠𝐷𝑂𝐹 = 3𝑦 − 12, 𝑚∠𝐿𝐹𝑂 = 2𝑦 − 3 and 𝑂𝐿 ≅ 𝐷𝐹. Find the following:
2. 𝑚∠𝐷𝑂𝐹
Solution:
Based on the given, 𝑂𝐿 ≅ 𝐷𝐹. Since 𝑂𝐿 and 𝐷𝐹 are congruent,
from Theorem 11, the angles that inscribe these arcs are also
congruent. That is,
∠𝐷𝑂𝐹 ≅ ∠𝐿𝐹𝑂
or
By substitution,
3𝑦 − 12 = 2𝑦 − 3
𝑦 = 9.
𝑚∠𝐷𝑂𝐹 = 𝑚∠𝐿𝐹𝑂
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Substituting the value of 𝑦,
𝑚∠𝐷𝑂𝐹 = 3𝑦 − 12
𝑚∠𝐷𝑂𝐹 = 3(9) − 12
𝑚∠𝐷𝑂𝐹 = 15.
Thus, ∠𝑫𝑶𝑭 measures 𝟏𝟓𝐨 .
Example 2: Given circle 𝐼 with diameter 𝑂𝐹, 𝑚∠𝐺𝑂𝐹 = 7𝑥 + 10, 𝑚∠𝑂𝐹𝐺 = 8𝑥 + 20,
𝑚∠𝐷𝑂𝐹 = 3𝑦 − 12, 𝑚∠𝐿𝐹𝑂 = 2𝑦 − 3 and 𝑂𝐿 ≅ 𝐷𝐹. Find the following:
3. 𝑚∠𝐿𝐹𝑂
Solution:
From item number 2 above, since 𝑚∠𝐷𝑂𝐹 = 𝑚∠𝐿𝐹𝑂, then it
follows that ∠𝐿𝐹𝑂 also measures 15o .
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Theorem 12
Opposite angles of an inscribed quadrilateral are
supplementary.
If ABCD is an inscribed quadrilateral, then:
a. ∠𝐴 and ∠𝐶 are supplementary angles. That is,
𝑚∠𝐴 + 𝑚∠𝐶 = 180.
b. ∠𝐵 and ∠𝐷 are supplementary angles. That is,
𝑚∠𝐵 + 𝑚∠𝐷 = 180
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Example: In ⊙O below, ∠𝐷𝐶𝐴 = 25𝑜 and ∠𝐴𝐷𝐶 = 105𝑜 . Find the measures of the following:
1. 𝑚∠𝐷𝐵𝐴
Solution:
Based on ⊙O, ∠𝐷𝐵𝐴 and ∠𝐷𝐶𝐴 both intercept 𝐴𝐷. From
Theorem 11, it follows that ∠𝐷𝐵𝐴 ≅ ∠𝐷𝐶𝐴.
Since 𝑚∠𝐷𝐶𝐴 = 25o , then ∠𝐷𝐵𝐴 also measures 25o
Thus, ∠𝑫𝑩𝑨 measures 𝟐𝟓𝐨 .
2. 𝑚∠𝐷𝐶𝐵
Solution:
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Since ∠𝐷𝐶𝐵 is inscribed in a semicircle 𝐵𝐶𝐷, from
Theorem 10, ∠𝐷𝐶𝐵 is a right angle. Hence, 𝑚∠𝐷𝐶𝐵 = 90.
Thus, ∠𝑫𝑪𝑩 measures 𝟗𝟎𝐨 .
Example: In ⊙O below, ∠𝐷𝐶𝐴 = 25𝑜 and ∠𝐴𝐷𝐶 = 105𝑜 . Find the measures of the following:
3. 𝑚∠𝐴𝐵𝐶
Solution:
The figure shows quadrilateral 𝐴𝐵𝐶𝐷 inscribed in a circle. From
Theorem 12, ∠𝐴𝐵𝐶 and ∠𝐴𝐷𝐶 are supplementary angles. That is,
𝑚∠𝐴𝐵𝐶 + 𝑚∠𝐴𝐷𝐶 = 180.
𝑚∠𝐴𝐵𝐶 + 105 = 180
𝑚∠𝐴𝐵𝐶 = 75.
Thus, ∠𝑨𝑩𝑪 measures 𝟕𝟓𝐨 .
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Practice Now 4
A. Refer to the illustration at the right to complete the table by identifying the corresponding
inscribed angles or intercepted arcs.
Inscribed Angle
Intercepted Arc
1.
∠𝑀𝑍𝑈
𝑴𝑼
2.
∠𝑴𝒁𝑼
𝑀𝑆
3.
∠𝑨𝑼𝒁
𝐴𝑍
4.
∠𝑆𝑍𝑈
𝑺𝑼
5.
∠𝑈𝐴𝑆
𝑺𝑼
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Practice Now 4
B. In circle , 𝐴𝑅 ≅ 𝑅𝑂 ≅ 𝑂𝑆 ≅ 𝐴𝑆, 𝑚∠𝐴𝑀𝑅 = 3𝑥 + 20, and 𝑚∠𝑂𝑀𝑅 = 𝑥 + 30.
Find the following measures.
1. 𝑥
5
6.
𝑚 𝐴𝑀
=
110
2. 𝑚∠𝐴𝑀𝑅 =
35
7.
𝑚 𝑂𝑀
𝑂𝑀
𝑚
=
110
3. 𝑚∠𝑅𝐴𝑀 =
90
8.
𝑚∠𝑂𝑅𝑀 =
55
=
70
9.
𝑚∠𝐴𝑅𝑂 =
110
5. 𝑚∠𝑅𝑂𝑀 =
90
10. 𝑚∠𝐴𝑀𝑂 =
4. 𝑚 𝐴𝑅
=
70
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ANGLES FORMED BY
SECANTS AND TANGENTS
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Objectives
At the end of the session, you can…
find the measure of an angle formed by secants and tangents
by applying the different theorems.
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My Bag of Ideas 4
A secant to a circle is a line which intersects the circle in exactly two points.
It is a line that contains a chord.
A secant segment is a segment which intersects a circle in two points.
A tangent to a circle is a line which intersects the circle at exactly one point.
A point of tangency is the intersection point of the circle and the line
tangent to the circle.
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A tangent segment is a segment which is part of a tangent line and one of
its endpoints is the point of tangency.
Illustration:
In ⊙𝑃,
𝐴𝐵 is a secant to ⊙𝑃.
𝐷𝐵 is a secant segment of ⊙𝑃.
𝐹𝐺 is a tangent to ⊙𝑃.
𝐻𝐵 is a tangent segment to ⊙𝑃.
Points E and B are points of tangency.
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Note: The symbol
↔ above the letters denote that it is a line while
the symbol − above the letters denote that it is a segment.
Illustration:
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Figure 1
Figure
1
shows
secant
𝐴𝐷
intersecting circle 𝑃 at points 𝐵 and
𝐶. Notice that 𝐵𝐶 is a chord of circle
𝑃. 𝐴𝐷 is a secant segment of circle 𝑃.
Figure 2
Figure 2 shows tangent 𝐶𝐷
intersecting circle 𝑃 at point 𝐶. 𝐶𝐷 is
a tangent segment of circle 𝑃.
Theorem 13
The measure of an angle formed by two
secants that intersect outside of a circle is half
the difference of its intercepted arcs.
In ⊙O, with secants 𝐴𝐵 and 𝐴𝐶,
1
𝑚∠𝐴 = (𝑚 𝐵𝐶 − 𝑚 𝐷𝐸)
2
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Example
Given: 𝐴𝐶 and 𝐴𝐸 are secants, 𝑚 𝐵𝐷 = 60𝑜 and
𝑚 𝐶𝐸 = 160o . Find 𝑚∠𝐴.
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Solution:
From Theorem 13, the angle formed by two
secants intersecting outside the circle measures
half the difference of the intercepted arcs. In the
figure, ∠𝐴 is the angle formed by secants 𝐴𝐶 and
𝐴𝐸. The corresponding arcs are 𝐶𝐸 (larger arc)
and 𝐵𝐷 smaller arc. Thus,
1
𝑚∠𝐴 = (𝑚 𝐶𝐸 − 𝑚 𝐵𝐷)
2
Example
Given: 𝐴𝐶 and 𝐴𝐸 are secants, 𝑚 𝐵𝐷 = 60𝑜 and
𝑚 𝐶𝐸 = 160o . Find 𝑚∠𝐴.
Solution:
By substitution,
1
𝑚∠𝐴 = 160 − 60
2
1
𝑚∠𝐴 = 100
2
𝑚∠𝐴 = 50.
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Therefore, ∠𝑨 measures 𝟓𝟎𝐨 .
Example
Given: 𝐴𝑅 and 𝐴𝑀 are secants, 𝑚∠𝐴 = 36, and
𝑚 𝑅𝑀 = 100o . Find 𝑚 𝐸𝐿.
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Solution:
From Theorem 13, the angle formed by two
secants intersecting outside the circle measures
half the difference of the intercepted arcs. In the
figure, ∠𝐴 is the angle formed by secants 𝐴𝑅 and
𝐴𝑀. The intercepted arcs are 𝑅𝑀 (larger arc) and
𝐸𝐿 smaller arc. Thus,
1
𝑚∠𝐴 = (𝑚 𝑅𝑀 − 𝑚 𝐸𝐿)
2
Example
Given: 𝐴𝑅 and 𝐴𝑀 are secants, 𝑚∠𝐴 = 36, and
𝑚 𝑅𝑀 = 100o . Find 𝑚 𝐸𝐿.
Solution:
By substitution,
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1
36 = (100 − 𝑚 𝐸𝐿)
2
72 = 100 − 𝑚 𝐸𝐿
𝑚 𝐸𝐿 = 100 − 72
𝑚 𝐸𝐿 = 28
Therefore, 𝑬𝑳 measures 𝟐𝟖𝐨 .
Multiplying both
sides of the
equation by 2.
Theorem 14
The measure of an angle formed by a tangent
and a secant outside the circle, is half the
difference of the measures of the two arcs
intercepted by this angle.
With secant 𝐴𝐵 and tangent 𝐴𝐷,
1
𝑚∠𝐴 = 𝑚 𝐵𝐷 − m 𝐶𝐷 .
2
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Example
Given: 𝑅𝐶 is a secant segment and 𝑅𝐿 is a tangent segment,
𝑚 𝐴𝐿 = 50o , 𝑚∠𝑅 = 𝑥 + 3
o
and 𝑚 𝐶𝐿 = (4𝑥 + 8)o . Find 𝑚 𝐴𝐶.
Solution:
The figure at the right shows ∠𝑅 formed by secant 𝑅𝐶 and tangent 𝑅𝐿
outside the circle where 𝐶𝐿 (larger arc) and 𝐴𝐿 (smaller arc) are the
intercepted arcs.
From Theorem 14, the angle formed by a secant and a tangent line outside the circle measures
half the difference of the intercepted arcs. Thus,
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𝑚∠𝑅 =
1
(𝑚 𝐶𝐿 − 𝑚 𝐴𝐿)
2
Example
Given: 𝑅𝐶 is a secant segment and 𝑅𝐿 is a tangent segment,
𝑚 𝐴𝐿 = 50o , 𝑚∠𝑅 = 𝑥 + 3
o
and 𝑚 𝐶𝐿 = (4𝑥 + 8)o . Find 𝑚 𝐴𝐶.
Solution:
By substitution,
1
𝑥 + 3 = [ 4𝑥 + 8 − 50 ]
2
2(𝑥 + 3) = (4𝑥 + 8 − 50)
Solving for 𝑥,
2𝑥 + 6 = 4𝑥 − 42
6 + 42 = 4𝑥 − 2𝑥
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48 = 2𝑥
24 = 𝑥
Multiplying both
sides of the
equation by 2.
Example
Given: 𝑅𝐶 and 𝑅𝐿 are secants, 𝑚 𝐴𝐿 = 50o ,
𝑚∠𝑅 = 𝑥 + 3 o and 𝑚 𝐶𝐿 = (4𝑥 + 8)o . Find 𝑚 𝐴𝐶.
Solution:
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Note that 𝐴𝐶 , 𝐴𝐿 and 𝐶𝐿 make up the circle and that a circle
measures 360o . By substitution,
𝑚 𝐴𝐶 + 𝑚 𝐴𝐿 + 𝑚 𝐶𝐿 = 360
𝑚 𝐴𝐶 + 50 + 4𝑥 + 8 = 360
𝑚 𝐴𝐶 + 50 + 4 24 + 8 = 360
𝑚 𝐴𝐶 + 50 + 96 + 8 = 360
𝑚 𝐴𝐶 + 154 = 360
𝑚 𝐴𝐶 = 360 − 154
𝑚 𝐴𝐶 = 206
Therefore, 𝑨𝑪 measures 𝟐𝟎𝟔𝐨 .
Theorem 15
The measure of an angle formed by a tangent
and a secant through the point of tangency is
half the measure of the intercepted arc.
In ⊙O, with 𝐵𝐶 as tangent and intersecting
secant 𝐴𝐵 at point B,
1
𝑚∠𝐴𝐵𝐶 = 𝑚 𝐴𝐵 .
2
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Example
Find 𝑚∠𝐵𝐴𝐶 if 𝐵𝐷𝐴 = 270o .
Solution:
The figure shows ∠𝐵𝐴𝐶 formed by the secant 𝐴𝐵 and tangent
line 𝐴𝐶 intersecting at the point of tangency 𝐴 where the
intercepted arc is the major arc 𝐵𝐷𝐴.
Applying the Theorem 15,
1
𝑚∠𝐵𝐴𝐶 = 𝑚 𝐵𝐷𝐴
2
Substituting 𝑚 𝐵𝐷𝐴 = 270,
1
𝑚∠𝐵𝐴𝐶 = 270
2
𝑚∠𝐵𝐴𝐶 = 135
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Therefore, ∠𝑩𝑨𝑪 measures 𝟏𝟑𝟓𝐨 .
Theorem 16
The measure of an angle formed by two
intersecting tangents is half of the difference of
the measures of the two intercepted arcs.
With tangents 𝐵𝑃and 𝐴𝑃 intersecting at 𝑃,
1
𝑚∠𝑃 = 𝑚 𝐴𝑈𝐵 − 𝑚 𝐴𝐵 .
2
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Example
Refer to the circle on the right.
1. If 𝑀𝑃𝑁 = 254o and 𝑀𝑁 = 106o , find ∠𝑀𝐿𝑁.
2. If 𝑀𝑃𝑁 = 270o , find 𝑚∠𝑀𝐿𝑁.
3. If ∠𝑀𝐿𝑁 = 50o , find 𝑀𝑃𝑁 and 𝑀𝑁.
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Example
Refer to the circle on the right.
1.
If 𝑀𝑃𝑁 = 254o and 𝑀𝑁 = 106o , find ∠𝑀𝐿𝑁.
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Solution:
The figure shows tangent lines 𝑀𝐿 and 𝑁𝐿 intersecting at point 𝐿.
Note that the intercepted arcs are the major arc 𝑀𝑃𝑁 and the
minor arc 𝑀𝑁. Theorem 16 states that the measure of the angle formed by two tangents is half
the difference of the intercepted arcs. Hence,
1
𝑚∠𝑀𝐿𝑁 = 𝑚 𝑀𝑃𝑁 − 𝑚 𝑀𝑁
2
1
𝑚∠𝑀𝐿𝑁 = 254 − 106
2
𝑚∠𝑀𝐿𝑁 = 74
Therefore, ∠𝑴𝑳𝑵 measures 𝟕𝟒𝐨 .
Example
Refer to the circle on the right.
2.
If 𝑀𝑃𝑁 = 270o , find 𝑚∠𝑀𝐿𝑁.
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Solution:
Before applying Theorem 16, find first the measure of
the minor arc 𝑀𝑁. Remember that the measure of a
circle is 360o and note that the circle above is
composed of the major arc 𝑀𝑃𝑁 and the minor arc
𝑀𝑁 . That is,
𝑚 𝑀𝑃𝑁 + 𝑚 𝑀𝑁 = 360
270 + 𝑚 𝑀𝑁 = 360
𝑚 𝑀𝑁 = 90
Now, applying Theorem 15,
1
𝑚∠𝑀𝐿𝑁 = 𝑚 𝑀𝑃𝑁 − 𝑚 𝑀𝑁
2
1
𝑚∠𝑀𝐿𝑁 = 270 − 90
2
𝑚∠𝑀𝐿𝑁 = 90
Therefore, ∠𝑴𝑳𝑵 measures 𝟗𝟎𝐨 .
Example
Refer to the circle on the right.
3.
If ∠𝑀𝐿𝑁 = 50o , find 𝑀𝑃𝑁 and 𝑀𝑁.
Solution:
Applying Theorem 16,
1
𝑚∠𝑀𝐿𝑁 = 𝑚 𝑀𝑃𝑁 − 𝑚 𝑀𝑁
2
By substitution,
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1
50 = 𝑚 𝑀𝑃𝑁 − 𝑚 𝑀𝑁
2
100 = 𝑚 𝑀𝑃𝑁 − 𝑚 𝑀𝑁
Again, remember that 𝑀𝑁
and 𝑀𝑃𝑁
together make up the entire circle, and that
a circle measures 360o . In symbols,
𝑚 𝑀𝑁 + 𝑚 𝑀𝑃𝑁 = 360
Example
Refer to the circle on the right.
3.
If ∠𝑀𝐿𝑁 = 50o , find 𝑀𝑃𝑁 and 𝑀𝑁.
Solution:
If we let 𝑚 𝑀𝑁 = 𝑥, then it follows that 𝑚 𝑀𝑃𝑁 = 360 − 𝑥 .
Substituting this in the previous equation,
100 = 𝑚 𝑀𝑃𝑁 − 𝑚 𝑀𝑁
100 = 360 − 𝑥 − 𝑥
Solving for 𝑚 𝑀𝑃𝑁 and 𝑀𝑁 using this value of 𝑥,
it gives 𝑚 𝑀𝑁 = 130 and 𝑚 𝑀𝑃𝑁 = 230.
100 = 360 − 2𝑥
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−260 = −2𝑥
𝑥 = 130
Therefore, 𝑴𝑷𝑵 and 𝑴𝑵 measure 𝟐𝟑𝟎𝐨 and
𝟏𝟑𝟎𝐨 respectively.
Theorem 17
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The measure of an angle formed by two secants intersecting
in the interior of the circle is equal to one-half the sum of the
measures of the intercepted arcs.
If the secant 𝐴𝐵, intersects the secant 𝐶𝐷 at P, then
1
𝑚∠𝐷𝑃𝐵 = (𝑚 𝐵𝐷 + 𝑚 𝐴𝐶)
2
Similarly,
1
𝑚∠𝐶𝑃𝐴 = 𝑚 𝐵𝐷 + 𝑚 𝐴𝐶
2
1
𝑚∠𝐵𝑃𝐶 = 𝑚 𝐵𝐶 + 𝑚 𝐴𝐷
2
1
𝑚∠𝐴𝑃𝐷 = (𝑚 𝐴𝐷 + 𝑚 𝐵𝐶
2
Example
Refer to the circle on the right.
1. If 𝑚 𝐵𝑁 = 60 and 𝑚 𝐴𝑅 = 70, find 𝑚∠𝐵𝐷𝑁 and 𝑚∠𝑅𝐷𝐴.
2. If 𝑚 𝑁𝐵𝐴 = 260 and 𝑚 𝐵𝑅 = 80, find 𝑚∠𝐴𝐷𝑁 and 𝑚∠𝐵𝐷𝑁.
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Example
Refer to the circle on the right.
1.
If 𝑚 𝐵𝑁 = 60 and 𝑚 𝐴𝑅 = 70, find 𝑚∠𝐵𝐷𝑁 and 𝑚∠𝑅𝐷𝐴.
Solution:
The intersection of the two secants is the point 𝐷 inside the circle and the
intercepted arcs of ∠𝐵𝐷𝑁 are 𝐵𝑁 and 𝐴𝑅. By Theorem 17, the measure
of the angle formed by two intersecting lines inside the circle is half the
sum of the corresponding intercepted arcs.
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Hence, by substitution,
1
𝑚∠𝐵𝐷𝑁 = 𝑚 𝐵𝑁 + 𝑚 𝐴𝑅
2
1
𝑚∠𝐵𝐷𝑁 = 60 + 70
2
1
𝑚∠𝐵𝐷𝑁 = 130
2
𝑚∠𝐵𝐷𝑁 = 65
Therefore, ∠𝑩𝑫𝑵 is 𝟔𝟓𝐨 .
Note that ∠𝑅𝐷𝐴 and ∠𝐵𝐷𝑁 are vertical angles.
Remember that vertical angles are congruent.
Hence, ∠𝑹𝑫𝑨 also measures 𝟔𝟓𝐎 .
Example
Refer to the circle on the right.
2.
If 𝑚 𝑁𝐵𝐴 = 260 and 𝑚 𝐵𝑅 = 80, find 𝑚∠𝐴𝐷𝑁 and 𝑚∠𝐵𝐷𝑁.
Solution:
Secants 𝐵𝐴 and 𝑁𝑅 form angle ∠𝐴𝐷𝑁 and intercept the arcs 𝐵𝑅 and
𝐴𝑁. From Theorem 17,
1
𝑚∠𝐴𝐷𝑁 = 𝑚 𝐵𝑅 + 𝑚 𝐴𝑁 .
2
First, find the 𝑚 𝐴𝑁. Based on the figure, 𝐴𝑁 and 𝑁𝐵𝐴 makeup ⊙D.
In symbols,
Solving for 𝐴𝑁,
𝑚 𝐴𝑁 + 260 = 360
𝑚 𝐴𝑁 = 360 − 260
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𝑚 𝐴𝑁 = 100
𝑚 𝐴𝑁 + 𝑚 𝑁𝐵𝐴 = 360
Example
Refer to the circle on the right.
2.
If 𝑚 𝑁𝐵𝐴 = 260 and 𝑚 𝐵𝑅 = 80, find 𝑚∠𝐴𝐷𝑁 and 𝑚∠𝐵𝐷𝑁.
Solution:
By substitution,
1
𝐵𝑅 + 𝑚 𝐴𝑁
2
1
𝑚∠𝐴𝐷𝑁 = 80 + 100
2
1
𝑚∠𝐴𝐷𝑁 = (180)
2
𝑚∠𝐴𝐷𝑁 = 90
𝑚∠𝐴𝐷𝑁 =
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Therefore, ∠𝑨𝑫𝑵 is 𝟗𝟎𝐨 .
In the figure, note that ∠𝐵𝐷𝑁 and ∠𝐴𝐷𝑁 form a linear pair. Remember
that linear pairs are supplementary. Hence, by substitution,
𝑚∠𝐵𝐷𝑁 + 𝑚∠𝐴𝐷𝑁 = 180
𝑚∠𝐵𝐷𝑁 + 90 = 180
𝑚∠𝐵𝐷𝑁 = 90
Therefore, ∠𝑩𝑫𝑵 measures 𝟗𝟎𝐨 .
Vertical Angles and Linear Pairs
Vertical Angles are opposite angles formed by intersecting lines.
∠𝐵𝑃𝑁 and ∠𝑅𝑃𝐷 are vertical angles.
∠𝐷𝑃𝑁 and ∠𝐵𝑃𝑅 are also vertical angles
Linear Pairs are adjacent angles formed by
intersecting lines.
∠𝑁𝑃𝐵 and ∠𝐵𝑃𝑅 are linear pairs
∠𝐵𝑃𝑁 and ∠𝑁𝑃𝐷 are linear pairs
∠𝑁𝑃𝐷 and ∠𝐷𝑃𝑅 are linear pairs
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∠𝐷𝑃𝑅 and ∠𝑅𝑃𝐵 are linear pairs
Practice Now 5
In circle 𝑀 below, 𝐴𝑆 and 𝐿𝑆 are tangent lines at points 𝐴 and 𝐿 respectively. Given
that 𝑚 𝑈𝐿 = 110o , 𝑚 𝐿𝐴 = 150o , and 𝑚 𝐶𝐴 = 70o . Find the following measures.
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𝟑𝟎°
1. 𝑚 𝑈𝐶 = _______
𝟗𝟎°
6. 𝑚∠𝐿𝑁𝐴 = _______
𝟕𝟓°
2. 𝑚∠𝐿𝐶𝐴 = _______
𝟏𝟓°
7. 𝑚∠𝐶𝐴𝑈 = _______
𝟕𝟓°
3. 𝑚∠𝐿𝑈𝐴 = _______
𝟏𝟑𝟎°
8. 𝑚∠𝑈𝐴𝑆 = _______
𝟔𝟎°
4. 𝑚∠𝐵 = _______
𝟑𝟎°
9. 𝑚∠𝐴𝑆𝐿 = _______
𝟗𝟎°
5. 𝑚∠𝑈𝑁𝐿 = _______
𝟏𝟏𝟎°
10. 𝑚∠𝐶𝐿𝑆 = _______