95 Holt McDougal Geometry

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1 1. It is given that KN is the perpendicular bisector of J and N is the perpendicular bisector of K. B the Perpendicular Bisector Theorem, JK = K and K =. Thus JK = b the Trans. Prop. of =. B the definition of segs., JK. B the Seg. Add. Post., JR + R = J and.t + TK = K. B the definition of the perpendicular bisector, R is the midpoint of J and T is the midpoint of K. Thus JR R and T TK. B the definition of cong segs., JR = R and T = TK. B Subst., JR + JR = J and T + T = K. It is given that JR T. So JR = T b definition of segs. B Subst., JR + JR = K. B the Trans. Prop. of =, J = K, so J K b the definition of segs. B the Reflex. Prop. of, J J. Therefore JK J b SSS, and JK J b CPCTC. 5- bisectors of triangles check it out! 1a. G = J = 1.5 b. GK = KH = 1.6 c. Z is circumcenter of GHJ, B the Circumcenter Theorem, Z is equidistant from the vertices of GHJ. JZ = GZ = = G x = H x (, -.5) Step 1 Graph. Step Find equations for two perpendicular bisectors. Since two sides of lie along the axes, use the graph to find the perpendicular bisectors of these two sides. the perpendicular bisector of G is = -.5, and the perpendicular bisector of H is x =. Step Find the intersection of the two equations. The lines = -.5 and x = intersect at (, -.5), the circumcenter of GH. a. X is the incenter of PQR. B the Incenter Theorem, X is euqidistant from the sides of PQR. The distance from X to PR is 19., so the distance from X to PQ is also 19.. b. m PRQ = m PRX m PRQ = (1 ) = m RQP + m PRQ + m QPR = m RQP = 10 m RQP = 10 m PQX = 1_ m RQP m PQX = 1_ (10 ) = 5. B the Incenter Theorem, the incenter of a is equidistant from the sides of. Draw formed b the streets and draw the bisectors to find the incenter, point. The cit should place the monument at point. think and discuss 1. Possible answer:. Q; P. Possible answer: the incenter is alwas inside, so Q cannot be the incenter. Therefore P must be the incenter, and Q must be the circumcenter.. Definition Distance ocation (Inside, utside, or n) exercises guided Practice Circumcenter the bisectors Equidistant from the vertices of Can be inside, outside, or on Incenter the bisectors Equidistant from the sides of Inside 1. The do not intersect at a single point.. circumscribed about. N is the circumcenter of PQR. B the Circumcenter Theorem, N is equidistant from vertices of PQR. NR = NP = 5.6. RV = PV = TR = QT = N is the circumcenter of PQR. B the Circumcenter Theorem, N is equidistant from vertices of PQR. QN = NP = Holt cdougal Geometr

2 7.. = 6 1 K (, 6) x x = Step 1 Graph. Step Find equations for the two perpendicular bisectors. Since the two sides of lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of K is = 6, and the perpendicular bisector of is x =. Step Find the intersection of the two equations. The lines = 6 and x = intersect at (, 6), the circumcenter of K. x = -.5 A (-.5, -5) = -5 B x Step 1 Graph. Step Find equations for the two perpendicular bisectors. Since the two sides of lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of A is x = -.5, and the perpendicular bisector of is = -5. Step Find the intersection of the two equations. The lines x = -.5 and = -5 intersect at (-.5, -5), the circumcenter of AB. 9. F is the incenter of CDE. B the Incenter Theorem, F is equaldistant from the sides of CDE. The distance from F to DE is.1, so the distance from F to CD is also m DCE = m FCD m DCE = (17 ) = m DCE + m CDE + m CED = m CED = 10 m CED = 9 m FED = 1_ m CED m FED = 1_ (9 ) = The largest possible 〇 in the int. of is its inscribed 〇, and the center of the inscribed 〇 is the incenter. Draw and its bisectors. Center the 〇 at E, the point of the bisectors. 5 Q P 50 0 practice and problem solving 1. CF = FA = YC = YB = 6.9 N 1. DB = AD = AY = YB = Step 1 Write equations of the perpendicular bisectors of and N. The perpendicular bisector of is x = -.5; the perpendicular bisector of N is = 7. Step Find the circumcenter of. The circumcenter is at the intersection of the perpendicular bisectors, (-.5, 7). 17. Step 1 Write equations of the perpendicular bisectors of V and W. The perpendicular bisector of V is = 9.5; the perpendicular bisector of W is x = Step Find the circumcenter of. The circumcenter is at the intersection of the perpendicular bisectors, (-1.5, 9.5). 1. J is the incenter of RST. B the Incenter Theorem, J is equaldistant from the sides of RST. The distance from J to ST is.7, so the distance from J to RS is also m TSR = m JSR m TSR = (1 ) = m TSR + m SRT + m RTS = m TSR = 10 m TSR = 110 m RTJ = 1_ m TSR m RTJ = 1_ (110 ) = B the Circumcenter Theorem, the circumcenter of is equidistant from the vertices. Draw formed b the cities, and draw the perpendicular bisectors of the sides. The main office should be located at, the circumcenter. 1. Possible answer: if J is a rt., then m J + m J = 90 because the acute of a rt. are comp. Since is the incenter of JK, J and are bisectors of JK. So b the def. of bisector, m KJ = m J and m KJ = m J. B subst., m KJ + m KJ = (m J + m J) = (90 ) = 10. But b Sum Theorem, m K = 10 - (m KJ + m KJ) = = 0. This would mean that JK is not a. Therefore J cannot be a rt.. P S H. The angle bisector; m BAE = m EAC. The perpendicular bisector; AD = BD, AD DG and BD DG. The angle bisector; m ABG = m GBC 5. The angle bisector; since AE and BG are bisectors, R is the incenter. CR intersects the incenter, so it is an the bisector. 6. neither 7. neither. never 96 Holt cdougal Geometr

3 9. sometimes 1. never. sometimes 0. sometimes. The slope of A is ; the midpoint of A is (, ). The perpendicular bisector of A is - = - 1_ ( x - ). The perpendicular bisector of B is x =. At the intersection, x = and - = - 1_ ( - ) = -1, so =. The circumcenter is at (, ).. The perpendicular bisector of Y is = 6. The slope of Z is = 1; the midpoint of Z is (, ). The perpendicular bisector of Z is - = -(x - ). At the intersection, = 6 and 6 - = = -x +, so x = 0. The circumcenter is at (0, 6). 5a. Bisector Theorem b. the bisector of B c. PX = PZ 6. Statements Reasons 1. QS bisects PQR, PQ RQ. 1. Given. PQS RQS. Def. of bisector. QS QS. Reflex. Prop. of. PQS RQS. SAS 5. PSQ RSQ 5. CPCTC 6. PSQ and RSQ are supp. 6. in. Pair Thm. 7. PSQ and RSQ are rt.. 7. supp. rt.. PSQ = RSQ = 90. Def. of rt. 9. QS È PR 9. Def. of 10. PS RS 10. CPCTC 11. S is midpoint of PR. 1. QS È is the perpendicular bisector of PR. 11. Def. of midpoint 1. Def. of the perpendicular bisector 7a. The new store is at the circumcenter of ABC. The perpendicular bisector of AB is x =. The slope of AC is _ ; the midpoint of AC is (, _ ). The perpendicular bisector of AC is - _ = - _ ( x - ). At the intersection, x = and - = -, so = 9 _ - 16 = - 7_ 6 6. = - _ ( - ) The new store is located at (, - 7_ 6 ). b. outside, since > 0 for all int. poins. of, but - 7_ 6 < 0 c. distance from each store = distance from store C = - ( - 7_ 6 ) = 1_. mi 6. Possible answers: Similarities: Both are circles. Both intersect the triangle in exactl points. Differences: The inscribed circle is smaller than the circumscribed circle. Except for the points of intersection, the inscribed circle lies inside the triangle, while the circumscribed circle lies outside. The center of the inscribed circle alwas lies inside the triangle, while the center of the circumscribed circle ma be inside, outside, or on the triangle. The center of the inscribed circle is the point of the angle bisectors, while the center of the circumscribed circle is the point of concurrenc of the perpendicular bisectors. 9a. Check students constructions. b. Check students constructions. test prep 0. B; PX = PY b the Incenter Theorem. 1. F; m = 1, + = x - 5, or = x KN = N 5z - = z + 11 z = 15 z =.75 N = N = = 1.75 challenge and extend a. Possible answer: Given: is the midpoint of QR. Prove: P = Q = R Proof: The coordinates of are + a ( 0 _, _ b + 0 ) = (a, b). B the Distance Formula, P = ÇÇÇÇÇÇÇÇ (a - 0 ) + (b - 0 ) = ÇÇÇÇ a + b, Q = ÇÇÇÇÇÇÇÇ (a - 0 ) + (b - b ) = ÇÇÇÇÇ a + (-b ) = ÇÇÇÇ a + b, and R = ÇÇÇÇÇÇÇÇ (a - a ) + (b - 0 ) = ÇÇÇÇÇ (-a ) + b = ÇÇÇÇ a + b. Therefore, P = Q = R. b. Possible answer: The circumcenter of a rt. is the midpoint of the hp. 97 Holt cdougal Geometr

4 . et C be the circumcenter of. Given: AC = cm; so b the properties of , BC = 1_ AC. So AB = AC + BC = _ AC 5- = _ () = cm. medians and altitudes of Triangles check it out! 1a. KZ + ZW = KW _ KW + ZW = KW ZW = 1_ KW 7 = 1_ KW 1 = KW b. Z = _ X = _ (.1) = 5.. ; ; possible answer: the x-coordinate of the centroid is the average of the x-coordinates of the vertices of, and the -coordinate of the centroid is the average of the -coordinates of the vertices of.. Possible answer: An equation of altitude to JK is = - 1_ x +. It is true that = - 1_ ( -) +, so (-, ) is a solution of this equation. Therefore this altitude passes through the orthocenter. think and discuss 1. Possible answer: is isosc. B A D. Possible answer: is a rt.. K J C. The ratio of the length of the longer segment to the length of the shorter segment is : 1.. Centroid rthocenter Definition ocation (Inside, utside, or n) the medians Inside the altitudes Can be inside, outside, or on 5. RW = _ RY 10 = _ RY _ (10) = RY RY = WY = 1_ RW = 1_ (10) = Understand the Problem Answer will be the coordinates of the centroid of. Important information is the location of vertices, A(0, ), B(7, ), and C(5. 0). ake a Plan The centroid of is the point of intersection of the three medians. So write the equations for two medians and find their point of intersection. Solve et be the midpoint of AB and N be the midpoint of BC. = ( _ 0 + 7, _ + ) = (.5, ) N = ( _ 5 + 7, _ + 0 ) = (6, ) AN is horizontal. Its equation is =. Slope of C = _ - 0 = -. Its equation is.5-5 = -(x - 5). At the centroid, = = -(x - 5), so x = 5 + (-1) =. The coordinates of the centroid are D(, ). ook Back et be the midpoint of AC. Equation for B is - = _ ( x - 7), which intersects = at (, ) x = K (, -1) + = 1_ (x- ) x 10 Step 1 Graph. Step Find an equation of the line containing the altitude from to K. Since K is horizontal, the altitude is vertical, so the equation is x =. Step Find an equation of the line containing the altitude from K to. Slope of = _ = -. Equation is + = 1_ ( x - ). Step Solve the sstem to find the coordinates of the orthocenter. x = and + = 1_ ( - ) = 1, so = -1. The coordinates of the orthocenter are (, -1). exercises guided Practice 1. centroid. altitude. VW = _ VX. WX = 1_ = _ VX (0) = 16 = 1_ (0) = 6 9 Holt cdougal Geometr

5 9. The midpoint of AB is (1, 0); slope of AB = _ -10 = -1, so the slope of the 10 perpendicular bisector is 1; the equation of the perpendicular bisector is = x The midpoint of XY is (, 6); slope of XY = _ =, so the slope of the perpendicular bisector is -0.5; the equation of the perpendicular bisector is - 6 = -0.5(x - ). 11. No; to appl the Converse of the Angle Bisector Theorem, ou need to know that AP AB and CP CB. 1. Yes; since AP AB, CP CB, and AP CP, P is on the bisector of ABC b the Converse of the Angle Bisector Theorem. 5- Bisectors of Triangles 1. GY = HY =. 1. GP = JP = GJ = GX 16. PH = JP = 6 = (.) = distance from A to UV = distance from A to UW = 1 1. m WVU + m VUW + m UWV = 10 m WVA + (0)+ 66 = 10 m WVA = 7 m WVA = is vertical, so the equation of the horizontal perpendicular bisector is = ; N is horizontal, so the equation of the vertical perpendicular bisector is x =. The circumcenter is at (, ). 0. R is vertical, so the equation of the horizontal perpendicular bisector is = -.5; 5- S is horizontal, so the equation of the vertical perpendicular bisector is x = -6. The circumcenter is at (-6, -.5). edians and Altitudes of Triangles 1. DZ = _ DB = _. EZ = ZC 11.6 = ZC ZC = 5. (.6) = 16.. DB = ZB.6 = ZB ZB =.. EC = ZC = (5.) = JK is vertical, so the equation of the altitude from is = 0; K is horizontal, so the equation of the altitude from J is x = -6. The orthocenter is at (-6, 0). 6. AB is horizontal, so the equation of the altitude from C is x = 1; AC is vertical, so the equation of the altitude from B is =. The orthocenter is at (1, ). 7. RT is horizontal, so the equation of the altitude from S is x = 7; RS has slope 5_ = 1, so the equation of the altitude 5 from T is - = -(x - ). At the orthocenter, x = 7 and - = -(7 - ) = 1 =, so the orthocenter is at (7, ).. XY is horizontal, so the equation of the altitude from Z is x = ; XZ has slope 6_ = -1, so the equation of the altitude -6 from Y is - = x - 5 or = x -. At the orthocenter, x = and = x - = 0, so the orthocenter is at (, 0). 9. G = ( 1_ ( ), 1_ ( + + 0) ) = (, ) The Triangle idsegment Theorem 5-0. BC = 1_ XY = 1_ (70.) = 5.1. XC = 1_ XZ = AB =.. m BAX = 10 - m ABC = 10 - = 1 1. XZ = AB = (.) = 6.. m BCZ = m ABC = 5. m YXZ = m BCZ = 6. V = (-1, -1); W = (6, 1); slope of VW = _ 7 ; slope of GJ = _ 1 = _ ; since the slopes are the 7 same, VW ǁ GJ. VW = ÇÇÇÇ + 7 = 5 ÇÇ ; GJ = ÇÇÇÇ + 1 = 5 ÇÇ, so VW = 1_ GJ. 5-5 indirect Proof and Inequalities in ne Triangle 7. A is the smallest, so BC is the shortest side; C is the largest, so AB is the longest side; From shortest to longest, the order is BC, AC, AB.. GH is the shortest side, so F is the smallest ; FH is the longest side, so G is the largest ; From smallest to largest, the order is F, H, G. 9. x +.5 > > x x > 9 1 > x Range of the values: > 9 cm and < 1 cm > 1. Yes; possible answer: the sum of each pair of lengths is greater than the third length. 11 Holt cdougal Geometr

) = (3.5, 3) 5-3. check it out!

) = (3.5, 3) 5-3. check it out! 44. Let be the irumenter of the. Given: = m; so by the properties of -6-9,. So = + = _ 5- = _ () = 4 m. medians and altitudes of Triangles hek it out! 1a. KZ + ZW = KW _ KW + ZW = KW ZW KW 7 KW 1 = KW