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Exam 7: Locus and Concurrence

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-Use the drawing provided to explain the following theorem. The three angle bisectors of the angles of a triangle are concurrent. Given: , , and are the angle bisectors of the angles of Prove: , , and are concurrent at point Z -Use the drawing provided to explain the following theorem. "The three angle bisectors of the angles of a triangle are concurrent." Given: -Use the drawing provided to explain the following theorem. The three angle bisectors of the angles of a triangle are concurrent. Given: , , and are the angle bisectors of the angles of Prove: , , and are concurrent at point Z , -Use the drawing provided to explain the following theorem. The three angle bisectors of the angles of a triangle are concurrent. Given: , , and are the angle bisectors of the angles of Prove: , , and are concurrent at point Z , and -Use the drawing provided to explain the following theorem. The three angle bisectors of the angles of a triangle are concurrent. Given: , , and are the angle bisectors of the angles of Prove: , , and are concurrent at point Z are the angle bisectors of the angles of -Use the drawing provided to explain the following theorem. The three angle bisectors of the angles of a triangle are concurrent. Given: , , and are the angle bisectors of the angles of Prove: , , and are concurrent at point Z Prove: -Use the drawing provided to explain the following theorem. The three angle bisectors of the angles of a triangle are concurrent. Given: , , and are the angle bisectors of the angles of Prove: , , and are concurrent at point Z , -Use the drawing provided to explain the following theorem. The three angle bisectors of the angles of a triangle are concurrent. Given: , , and are the angle bisectors of the angles of Prove: , , and are concurrent at point Z , and -Use the drawing provided to explain the following theorem. The three angle bisectors of the angles of a triangle are concurrent. Given: , , and are the angle bisectors of the angles of Prove: , , and are concurrent at point Z are concurrent at point Z

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Let point Z name the point of intersection of the angle bisectors Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. and Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. . Because Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. bisects Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. , every point on Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. (including Z) is equidistant from the sides of Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. .
With Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. and Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. , it follows that Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. . With respect to Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. , also note
that Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. . Because Z lies on angle bisector Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. , we also have Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. . By the Transitive Property of Congruence, Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. . With point Z being equidistant from the sides of Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. , Z must also lie on angle bisector Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. . Thus, Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. , Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. , and Let point Z name the point of intersection of the angle bisectors and . Because bisects , every point on (including Z) is equidistant from the sides of . With and , it follows that . With respect to , also note that . Because Z lies on angle bisector , we also have . By the Transitive Property of Congruence, . With point Z being equidistant from the sides of , Z must also lie on angle bisector . Thus, , , and are concurrent at point Z. are concurrent at point Z.

-Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. -Supply missing reasons for the following proof. Given: -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. bisects -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. ; -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. and -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. Prove: -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. S1. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. bisects -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. ; R1. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. and -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. S2. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. R2. S3. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. and -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. are rt. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. R3. S4. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. R4. S5. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. R5. S6. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. R6. S7. -Supply missing reasons for the following proof. Given: bisects ; and Prove: S1. bisects ; R1. and S2. R2. S3. and are rt. R3. S4. R4. S5. R5. S6. R6. S7. R7. R7.

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R1. Given
R2. Definition of angle-bisector
R3. Perpendicular lines form right angles.
R4. All right angles are congruent.
R5. Identity
R6. AAS
R7. CPCTC

-Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment -Supply missing statements and missing reasons for the following proof. Given: Point X not on -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment so that -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment Prove: X lies on the perpendicular-bisector of -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment S1. R1. S2. -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment R2. S3. With M the midpoint of -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment , R3. draw -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment S4. Then -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment R4. S5. Also, -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment R5. S6. R6. SSS S7. -Supply missing statements and missing reasons for the following proof. Given: Point X not on so that Prove: X lies on the perpendicular-bisector of S1. R1. S2. R2. S3. With M the midpoint of , R3. draw S4. Then R4. S5. Also, R5. S6. R6. SSS S7. R7. S8. R8. Definition of perpendicular-bisector of a line segment R7. S8. R8. Definition of perpendicular-bisector of a line segment

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S1. Point X not on S1. Point X not on so that R1. Given R2. Definition of congruent line segments R3. Through 2 points, there is exactly one line. R4. Definition of midpoint of line segment R5. Identity S6. R7. CPCTC S8. X lies on the perpendicular-bisector of so that S1. Point X not on so that R1. Given R2. Definition of congruent line segments R3. Through 2 points, there is exactly one line. R4. Definition of midpoint of line segment R5. Identity S6. R7. CPCTC S8. X lies on the perpendicular-bisector of R1. Given
R2. Definition of congruent line segments
R3. Through 2 points, there is exactly one line.
R4. Definition of midpoint of line segment
R5. Identity
S6. S1. Point X not on so that R1. Given R2. Definition of congruent line segments R3. Through 2 points, there is exactly one line. R4. Definition of midpoint of line segment R5. Identity S6. R7. CPCTC S8. X lies on the perpendicular-bisector of R7. CPCTC
S8. X lies on the perpendicular-bisector of S1. Point X not on so that R1. Given R2. Definition of congruent line segments R3. Through 2 points, there is exactly one line. R4. Definition of midpoint of line segment R5. Identity S6. R7. CPCTC S8. X lies on the perpendicular-bisector of

-Use the drawing provided to explain the following. Given: , , and are the medians of Prove: -Use the drawing provided to explain the following. Given: -Use the drawing provided to explain the following. Given: , , and are the medians of Prove: , -Use the drawing provided to explain the following. Given: , , and are the medians of Prove: , and -Use the drawing provided to explain the following. Given: , , and are the medians of Prove: are the medians of -Use the drawing provided to explain the following. Given: , , and are the medians of Prove: Prove: -Use the drawing provided to explain the following. Given: , , and are the medians of Prove:

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