Inscribed Angle Theorem

Properties of Central and Inscribed Angles

3rd Quarter Lesson - Mathematics

Inscribed Angle Theorem

• The measure of an inscribed angle is always half the measure of its intercepted arc or the central angle.

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Semicircle Theorem

• If an inscribed angle intercepts a semicircle, then it is a right angle.

Example 1:

• In the given figure, m∠VUW = 94°. Find m∠VXW.

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

• Notice that ∠VXW is an inscribed angle. Thus, by virtue of the Inscribed Angle Theorem, it must be half the measure of its intercepted arc, VŴ , which is the same in measure as the central angle ∠VUW.

• Thus, m∠VUW = 94° ÷ 2 = 47°.

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Example 2:

• In the given figure, suppose m∠KML = (3x − 4)° and m∠KNL = (x + 12)°. What is m∠KNL?

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 1: Set up an equation relating the two angles.

• Since ∠KML is a central angle, and ∠KNL is an inscribed angle that intercepts the same arc, KL̂ , we have the following by virtue of the Inscribed Angle Theorem:

3x − 4

---------- = x + 12

2​

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 2: Solve for x.

3x − 4

---------- = x + 12

2

3x − 4

2 ( ---------- ) = (x + 12)(2)

2

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 2: Solve for x.

​

3x − 4 = 2x + 24

3x − 2x = 24 + 4

x = 28

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 3: Determine m∠KNL.

• Substituting x = 28 into the expression for m∠KNL, we have m∠KNL = (x + 12)° ​

  = (28 + 12)°

  = 40°

• Therefore, m∠KNL = 40°.

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Example 3:

• In the given figure, CE is a diameter. If EF = x − 3, CF = x + 4, and CE = x + 5, how long is CE?

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 1: Determine m∠CFE.

• Since CE is a diameter, CÊ is a semicircle. This arc is also the arc intercepted by inscribed angle ∠CFE. Thus, according to the Semicircle Theorem, m∠CFE is a right angle, that is, m∠CFE = 90°.

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 2: Set up an equation relating CF,EF, and CE.

• Since ∠CFE is a right angle, ∆CFE is a right triangle with CE as its hypotenuse. Thus, the sides of ∆CFE satisfy the Pythagorean Theorem.

  (CF)^2 + (EF)^2 = (CE)^2

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 3: Solve for x.

(CF)^2 + (EF)^2 = (CE)^2

(x + 4)^2 + (x − 3)^6 = (x + 5)^2

(x^2 + 8x + 16) + (x^2 − 6x + 9) = x^2 + 10x + 25

2x^2 + 2x + 25 = x^2 + 10x + 25 x^2 − 8x = 0

x(x − 8) = 0

x = 8

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 3: Solve for x.

• Note that we reject x = 0 because it will make EF = x − 3 negative and dimensions of a triangle must not be negative.​

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles

Solution:

Step 4: Determine the length of CE.

• Substituting x into the expression for CE, we have

  CE = x + 5

  = 8 + 5

  = 13

• Therefore, CE is 13 units long.

3rd Quarter Lesson - Mathematics

Properties of Central and Inscribed Angles