9.4. Inscribed Angles
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1 9.4. Inscribed Angles Inscribed Angles Learning Objectives Find the measure of inscribed angles and the arcs they intercept. Review Queue We are going to use #14 from the homework in the previous section. 1. What is the measure of each angle in the triangle? How do you know? 2. What do you know about the three arcs? 3. What is the measure of each arc? 4. What is the relationship between the angles in the triangles and the measure of each arc? Know What? Your family went to Washington DC over the summer and saw the White House. The closest you can get to the White House are the walking trails on the far right. You got as close as you could (on the trail) to the fence to take a picture (you were not allowed to walk on the grass). Where else could you have taken your picture from to get the same frame of the White House? Where do you think the best place to stand would be? Your line of sight in the camera is marked in the picture as the grey lines. The white dotted arcs do not actually exist, but were added to help with this problem. 518
2 Chapter 9. Circles Inscribed Angles We have discussed central angles so far in this chapter. We will now introduce another type of angle, the inscribed angle. Inscribed Angle: An angle with its vertex is the circle and its sides contain chords. Intercepted Arc: The arc that is on the interior of the inscribed angle and whose endpoints are on the angle. The vertex of an inscribed angle can be anywhere on the circle as long as its sides intersect the circle to form an intercepted arc. Now, we will investigation the relationship between the inscribed angle, the central angle and the arc they intercept. Investigation 9-4: Measuring an Inscribed Angle Tools Needed: pencil, paper, compass, ruler, protractor 1. Draw three circles with three different inscribed angles. For A, make one side of the inscribed angle a diameter, for B, make B inside the angle and for C make C outside the angle. Try to make all the angles different sizes. 2. Using your ruler, draw in the corresponding central angle for each angle and label each set of endpoints. 519
3 9.4. Inscribed Angles 3. Using your protractor measure the six angles and determine if there is a relationship between the central angle, the inscribed angle, and the intercepted arc. m LAM = m NBP = m QCR = m LM = m NP = m QR = m LKM = m NOP = m QSR = Inscribed Angle Theorem: The measure of an inscribed angle is half the measure of its intercepted arc. In the picture, m ADC = 1 2 mâc. If we had drawn in the central angle ABC, we could also say that m ADC = 1 2 m ABC because the measure of the central angle is equal to the measure of the intercepted arc. To prove the Inscribed Angle Theorem, you would need to split it up into three cases, like the three different angles drawn from Investigation 9-4. We will touch on the algebraic proofs in the review exercises. Example 1: Find m DC and m ADB. Solution: From the Inscribed Angle Theorem, m DC = 2 45 = 90. m ADB = = 38. Example 2: Find m ADB and m ACB. 520
4 Chapter 9. Circles Solution: The intercepted arc for both angles is ÂB. Therefore, m ADB = m ACB = = 62 This example leads us to our next theorem. Theorem 9-8: Inscribed angles that intercept the same arc are congruent. To prove Theorem 9-8, you would use the similar triangles that are formed by the chords. Example 3: Find m DAB in C. Solution: Because C is the center, DB is a diameter. Therefore, DAB inscribes semicircle, or 180. m DAB = = 90. Theorem 9-9: An angle that intercepts a semicircle is a right angle. In Theorem 9-9 we could also say that the angle is inscribed in a semicircle. Anytime a right angle is inscribed in a circle, the endpoints of the angle are the endpoints of a diameter. Therefore, the converse of Theorem 9-9 is also true. When the three vertices of a triangle are on the circle, like in Example 3, we say that the triangle is inscribed in the circle. We can also say that the circle is circumscribed around (or about) the triangle. Any polygon can be inscribed in a circle. Example 4: Find m PMN, m PN, m MNP, m LNP, and m LN. Solution: m PMN = m PLN = 68 by Theorem 9-8. m PN = 2 68 = 136 from the Inscribed Angle Theorem. m MNP = 90 by Theorem 9-9. m LNP = = 46 from the Inscribed Angle Theorem. To find m LN, we need to find m LPN. LPN is the third angle in LPN, so m LPN = 180. m LPN = 66, which means that m LN = 2 66 = 132. Inscribed Quadrilaterals The last theorem for this section involves inscribing a quadrilateral in a circle. 521
5 9.4. Inscribed Angles Inscribed Polygon: A polygon where every vertex is on a circle. Note, that not every quadrilateral or polygon can be inscribed in a circle. Inscribed quadrilaterals are also called cyclic quadrilaterals. For these types of quadrilaterals, they must have one special property. We will investigate it here. Investigation 9-5: Inscribing Quadrilaterals Tools Needed: pencil, paper, compass, ruler, colored pencils, scissors 1. Draw a circle. Mark the center point A. 2. Place four points on the circle. Connect them to form a quadrilateral. Color the 4 angles of the quadrilateral 4 different colors. 3. Cut out the quadrilateral. Then cut the quadrilateral into two triangles, by cutting on a diagonal. 4. Line up B and D so that they are adjacent angles. What do you notice? What does this show? This investigation shows that the opposite angles in an inscribed quadrilateral are supplementary. By cutting the quadrilateral in half, through the diagonal, we were able to show that the other two angles (that we did not cut through) formed a linear pair when matched up. 522
6 Chapter 9. Circles Theorem 9-10: A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. Example 5: Find the value of the missing variables. a) b) Solution: a) x + 80 = 180 by Theorem x = 100. y + 71 = 180 by Theorem y = 109. b) It is easiest to figure out z first. It is supplementary with 93, so z = 87. Second, we can find x. x is an inscribed angle that intercepts the arc = 164. Therefore, by the Inscribed Angle Theorem, x = 82. y is supplementary with x, so y = 98. Example 6: Algebra Connection Find x and y in the picture below. Solution: The opposite angles are supplementary. Set up an equation for x and y. (7x + 1) = 180 (4y + 14) + (7y + 1) = 180 7x = y + 15 = 180 7x = 84 11y = 165 x = 12 y = 15 Example 7: Find x and y in the picture below. 523
7 9.4. Inscribed Angles Solution: To find x, use ACE. m ACE = 14 because it is half of m BE by the Inscribed Angle Theorem x = 180 x = 134 To find y, we will use EFD. m m FED = 180 x FED = = 46 m BDE = m ACE = 14 because they intercept the same arc, Theorem 9-8. Let s solve for y in EFD, using the Triangle Sum Theorem y = 180 y = 120 Know What? Revisited You can take the picture from anywhere on the semicircular walking path. The best place to take the picture is subjective, but most would think the pale green frame, straight-on, would be the best view. Review Questions Quadrilateral ABCD is inscribed in E. Find: 524
8 Chapter 9. Circles 1. m DBC 2. m BC 3. mâb 4. m ACD 5. m ADC 6. m ACB Quadrilateral ABCD is inscribed in E. Find: 7. m A 8. m B 9. m C 10. m D Find the value of x and/or y in A
9 9.4. Inscribed Angles
10 Chapter 9. Circles 19. Algebra Connection Solve for the variables Use the diagram below to find the measures of the indicated angles and arcs in problems
11 9.4. Inscribed Angles m EBO 25. m EOB 26. m BC 27. m ABO 28. m A 29. m EDC 30. Fill in the blanks of one proof of the Inscribed Angle Theorem. Given: Inscribed ABC and diameter BDProve: m ABC = 1 2 mâc TABLE 9.2: Statement Reason 1. Inscribed ABC and diameter BD m ABE = x and m CBE = y 2. x + y = m ABC 3. All radii are congruent 4. Definition of an isosceles triangle 5. m EAB = x and m ECB = y 6. m AED = 2x and m CED = 2y 7. mâd = 2x and m DC = 2y 8. Arc Addition Postulate 9. mâc = 2x + 2y 10. Distributive PoE 11. mâc = 2m ABC 12. m ABC = 1 2 mâc Use the diagram below to write a proof of Theorem 9-8.
12 Chapter 9. Circles 32. Suppose that AB is a diameter of a circle centered at O, and C is any other point on the circle. Draw the line through O that is parallel to AC, and let D be the point where it meets BC. Prove that D is the midpoint of BC. Review Queue Answers 1. 60, it is an equilateral triangle. 2. They are congruent because the chords are congruent = The arcs are double each angle. 529
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