powerpoint presentation

3. In the figure, ∠𝐴𝐵𝐶 is inscribed in the
arc 𝐴𝐵𝐶. So, ∠𝐴𝐵𝐶 is called inscribed angle.
Note These:
• The vertex of an inscribed angle is on the
circle. (vertex is pt. B)
• Its sides contain chords of the circle.
• An inscribed angle intercepts an arc.
• The intercepted arc lies in the interior of an
inscribed angle.
• The end points of the intercepted arc lie on
the angle.
∠𝐴𝐵𝐶 = ½ 𝐴𝐶
The Inscribed angle theorem 1:
• If an angle is inscribed in a circle,
then its measure is half the
measure of the intercepted arc.

4. Given 𝐶𝐴 = 82°. Find ∠CBA.
Since an inscribed angle is one-half of its
intercepted arc, thus,
∠𝐶𝐵𝐴 =
1
2
𝐶𝐴
=
1
2
(82°)
∠𝑪𝑩𝑨 = 41°

5. Corollary 1: Any angle inscribed in a semicircle is a right angle.
By the Inscribed Angle Theorem, the inscribed angle is equal to
one-half of the measure of a half-circle or semicircle.
∠ABC = 90°
𝑚∠ABC =
1
2
𝐴𝐶

6. Example 1. Given ∠CBA = 35°, find the following:
1. m𝐴𝐶 = 2 . (m∠𝐶𝐵𝐴) = 2(35°) = 70
2. m𝐶𝐵 = 180 - m𝐴𝐶 = 180 – 70 = 110°
3. m𝐴𝐵 = 180° (since it is a semicircle)
4. m∠𝐶𝐴𝐵 or w hat is x? = 180 – (∠𝐶 + ∠𝐵)
= 180 – (90° + 35°)
= 180 - 125° = 55°
5. m∠𝐴𝐶𝐵 = 90° (an angle inscribed in a semicircle or half circle

7. Example 2. Given ∠𝑃𝐵𝐴 = 72°, with diameter AB. Find the measurement
of the following:
1. m𝐴𝑃 = (72° . 2) = 144°
2. m𝑃𝐵 = (180° - 144°) = 36°
3. m𝐴𝐵 = 180° (semicircle)
4. m∠𝐴𝑃𝐵 = 90° (angle inscribed a semicircle
5. m∠𝑃𝐴𝐵 or what is a° = 180 - (∠𝑃 + ∠𝐵)
= 180 – (90° + 72°)
= 180 - 162° = 18°

8. Activity 1. Determine the following:
1. m𝑃𝑄 =
2. m𝑄𝑆𝑅 =
3. m∠𝑃𝑄𝑅 =
4. m∠𝑃𝑅𝑄 =

9. Corollary 2: Every two angles inscribed in the same arc are congruent.

10. Example 1.
Given ∠𝐴𝐵𝐷 or ∠𝐵 ≅ ∠𝐴𝐶𝐷 or ∠𝐶 = 49°.
m𝐴𝐷 = 98°

11. Example 2.
x = 52°
Example 3.
∠𝐴𝐸𝐵 =
1
2
(60°) = 30°
∠𝐴𝐷𝐵 =
1
2
(60°) = 30°
∠𝐴𝐶𝐵 =
1
2
(60°) = 30°

12. Activity 2. Give what is being ask.
1. m𝐴𝐶 =
2. m∠𝐴𝐷𝐶 =
3. m∠𝐴𝑂𝐶 =