Answer Key

8.4 Proportionality Theorems Exercise 1 Parallel Triangle proportionality theorem Exercise 2 RC/BR=AC/AB Exercise 3 BC/AB=CD/DE (Triangle proportionality theorem) 3/AB=4/12 (substitute) 3/AB=1/3 3(3)=AB 9=AB The length of AB is 9 units. Exercise 4 AB/BC=AE/ED (Triangle proportionality theorem) AB/18=14/12 (substitute) AB/18=7/6 AB(6)=7(18) AB=21 The length of AB is 21 units. Exercise 5 LK/KJ=8/5 LM/MN=12/7.5=8/5 Because 8/5=12/7.5, KM is parallel to JN by the converse of the triangle proportionality theorem. Exercise 6 LK/KJ=25/22.5=10/9 LM/MN=20/18=10/9 Because 25/22.5=20/18, KM is parallel to JN by the converse of the triangle proportionality theorem. Exercise 7 LK/KJ=18/10=9/5 LM/MN=24/15=8/5 Because 18/10 is not equal to 24/15, KM is not parallel to JN by the contrapositive of the triangle proportionality theorem. Exercise 8 LK/KJ=16/35 LM/MN=15/34 Because 16/35 is not equal to 15/34, KM is not parallel to JN by the contrapositive of the triangle proportionality theorem. Exercise 9 1) Draw a line segment AB of length 3 in. Choose any point C not on AB. Draw AC. 2) Place the point of a compass at A and make an arc of any radius intersecting AC.

Label the point of intersection D. Using the same compass setting, make four more arcs on AC. Label the point of intersection E, F, G and H. 3) Join H and B. Copy ∠ AHB and construct congruent angles at D, E, F and G with sides that intersect AB at J, K, L and M. Sides DJ, EK, FL GM are all parallel and they divide AB equally. So, AJ = JK = KL = LM = MB. Point J divide AB in the ratio 1 to 4. The image of the construction, Exercise 10 1) Draw a line segment AB of length 2 in. Choose any point C not on AB. Draw AC. 2) Place the point of a compass at A and make an arc of any radius intersecting AC. Label the point of intersection D. Using the same compass setting, make four more arcs on AC. Label the point of intersection E, F, G and H. 3) Join H and B. Copy ∠ AHB and construct congruent angles at D, E, F and G with sides that intersect AB at J, K, L and M. Sides DJ, EK, FL GM are all parallel and they divide AB equally. So, AJ = JK = KL = LM = MB. Point K divide AB in the ratio 2 to 3. The image of the construction, Exercise 11 1) Draw a line segment AB of length 12 cm. Choose any point C not on AB. Draw AC. 2) Place the point of a compass at A and make an arc of any radius intersecting AC.

Label the point of intersection D. Using the same compass setting, make three more arcs on AC. Label the point of intersection E, F, and G. 3) Join G and B. Copy ∠ AGB and construct congruent angles at D, E, and F with sides that intersect AB at J, K, and L. Sides DJ, EK, and FL are all parallel and they divide AB equally. So, AJ = JK = KL = LB. Point J divide AB in the ratio 1 to 3. The image of the construction, Exercise 12 1) Draw a line segment AB of length 9 cm. Choose any point C not on AB. Draw AC. 2) Place the point of a compass at A and make an arc of any radius intersecting AC. Label the point of intersection D. Using the same compass setting, make six more arcs on AC. Label the point of intersection E, F, G, H, I and J. 3) Join J and B. Copy ∠ AJB and construct congruent angles at D, E, F, G, H, and I with sides that intersect AB at K, L, M, N, O and P. Sides DK, EL, FM, GN, HO and IP are all parallel and they divide AB equally. So, AK = KL = LM = MN = NO = OP = PB. Point L divide AB in the ratio 2 to 5. The image of the construction, Exercise 13 BD/BF=CE/CG Exercise 14 CG/EG=BF/DF

Exercise 15 EG/CE=DF/BD Exercise 16 BF/BD=CG/CE Exercise 17 UW/WY=VX/XZ (Three parallel lines theorem) 8/20=VX/15 2/5=VX/15 (2/5)15=VX 6=VX Exercise 18 NS/SU=PR/RT (Three parallel lines theorem) 10/SU=8/12 10/SU=2/3 10(3)=SU(2) 30/2=SU 15=SU Exercise 19 4/6=8/y (Triangle angle bisector theorem) 4(y)=8(6) 4y=48 y=12 Exercise 20 z/1.5=3/4.5 (Triangle angle bisector theorem) z(4.5)=3(1.5) z=1 Exercise 21 11/(29-11)=16.5/ p (Triangle angle bisector theorem) 11/18=16.5/p 11p=16.5(18) p=27 Exercise 22 q/(16-q)=36/28 (Triangle angle bisector theorem) q/(16-q)=9/7 7q=9(16-q) 7q=144-9q 16q=144 q=9 Exercise 23 AB/BC=AD/DC 10/16=x/14 5/8=x/14 5(14)=x(8)

70=8x 8.75=x Exercise 24 AD is not a bisector of ∠ BAC, we can't apply the triangle angle bisector theorem. Exercise 25 If a line divides two sides of a triangle proportionally, then it is parallel to the third side. So for PQ||RS, RT/QR=ST/PS 7/(3x+5)=5/(2x+4) 7(2x+4)=5(3x+5) 14x+28=15x+25 3=x Exercise 26 If a line divides two sides of a triangle proportionally, then it is parallel to the third side. So for PQ||RS, RT/PR=ST/QS (2x-2)/12=(3x-1)/21 21(2x-2)=12(3x-1) 42x-42=36x-12 6x=30 x=5 Exercise 27 Since QS||TU, ∠ Q~= ∠ RTU (corresponding angles) ∠ S~= ∠ RUT (corresponding angles) So, by angle-angle similarity theorem, triangle RQS~=triangle RTU. Then, the ratios of the corresponding side lengths are congruent. RQ/RT=RS/RU (RT+TQ)/RT=(RU+US)/RU (segment addition postulate) 1+(TQ/RT)=1+(US/RU) TQ/RT=US/RU (subtract 1 by both sides) RT/TQ=RU/US (take reciprocal in both sides) Exercise 28 ZY/YW=ZX/XV (given) YW/ZY=XV/ZX (take reciprocal on both sides) 1+ (YW/ZY)=1+(XV/ZX) (ZY/ZY)+(YW/ZY)=(ZX/ZX)+(XV/ZX) (ZY+YW)/ZY=(ZX+XV)/ZX ZW/ZY=ZV/ZX Also, ∠ Z~= ∠ Z (transitive property of equality) Then by the SAS similarity theorem, triangle XYZ~=triangle ZWV So, the corresponding angles of two triangles are congruent. i.e. ∠ W~= ∠ ZYX

But ∠ W and ∠ ZYX are corresponding angles. Therefore, YX||WV Exercise 29 a. Let lake frontage of lot A is x. By the three parallel lines theorem, 48/(55+61)=x/(174-x) 48/116=x/(174-x) 12/29=x/(174-x) 12(174-x)=29x 2088-12x=29x 2088=41x 50.9268=x 50.9 yd=x Let lake frontage of lot B is y. By the three parallel lines theorem, 50.93/48=y/55 (50.93/48) 55=y 58.3572=y 58.4 yd=y Let lake frontage of lot C is z. z=174 - (50.9+58.4)= 64.7 yd b. Lot C has the most lake frontage, therefore lot C should be listed for the highest price. c. Least expensive lot si lot A, because A has the least lake frontage. 50.9 yd ---------> $250 000 1 yd ---------------> $250 000/ 50.9 58.36 yd ---------->($250 000/50.9) 58.4=$286 836.9 (lot B) 64.7 yd ------------>($250 000/50.9) 64.7=$317 779.9 (lot C) Exercise 30 By angle-angle similarity theorem, 2/(1.5)=(2+5)/x 2/1.5=7/x 2x=10.5 x=5.25 2/1.5=10/y 2y=15 y=7.5 Exercise 31 Sides DJ, EK, FL and GB are all parallel. Also, AD, DE, EF and FG are all equal. Consider the AEK triangle, Since DJ||EK, AD/DE=AJ/JK

1=AJ/JK (since AD/DE=1) JK=AJ Similarly, we can show that AJ=JK=KL=LB. Exercise 32 Let the intersection point of the auxiliary line and line K2 be G. Consider triangle ACD, Since K1||K2, by triangle proportionality theorem, BA/CB=AG/GD CB/BA=GD/AG (take reciprocal on both sides)----------(1) Now consider triangle ADF, Since K2||K3, by triangle proportionality theorem, DE/EF=DG/GA ------------(2) The right and side of both equations are equal, so CB/BA=DE/EF Exercise 33 Let the intersection point of the bisector of ∠ M and side LN be P. By triangle angle bisector theorem, LP/PN=ML/MN Since the bisector of ∠ M bisects LN, LP=LN. Then LN/LP=1. 1=ML/MN MN=ML Hence triangle LMN is an isosceles triangle. Exercise 34 P1-player 1 P2-player 2 Q-quarterback R-receiver By triangle angle bisector theorem, P1R/RP2=QP1/QP2 Since QP1<QP2, P1R/RP2<1 P1R<RP2 Since both defensive players move at the same speed and P1R<RP2, player 1 reaches the receiver first. Exercise 35 Consider triangle AYZ, Since XW||AZ, by triangle proportionality theorem, YW/WZ=XY/XA ---------(1) ∠ YXW~= ∠ XAZ (corresponding angles in XW||AZ)--------(2) ∠ WXZ~= ∠ XZA (alternative angles in XW||AZ)--------(3)

∠ YXW~= ∠ WXZ (given)--------(4) By (2), (3) and (4), ∠ XAZ~= ∠ XZA Then AX=XZ (since ∠ XAZ~= ∠ XZA) by (1), YW/WZ=XY/XZ (since AX=XZ) Exercise 36 If a ray that throughs an angle and the opposite side to the angle divides the opposite side into segments whose lengths are proportional to the lengths of the other two sides, then it bisects the angle. Exercise 37 The Triangle Mid-segment theorem is the special case of the Triangle Proportionality theorem, when the segment parallel to one side of the triangle connects the midpoints of the other two sides. Exercise 38 No! the friend is wrong. By the triangle proportionality theorem, CD/DA=CE/EB CD/0.6=CE/0.9 CD/CE=6/9 CD/CE=2/3 Let CD=2x then CE=3x. AC=0.6+2x and BC=0.9+3x Let the speed of the person who leaves point B is v. Since two people meet at point C at the same time, AC/3=BC/v (0.6+2x)/3=(0.9+3x)/v 2(0.3+x)/3=3(0.3+x)/v 2/3=3/v ( divide both sides by (0.3+x)) 2v=9 v=4.5 miles per hour Exercise 39 1) Using a ruler, measure the lengths of all the given three line segments. This gives us the values of r, s and t. 2) Using the equation 𝑟𝑟𝑠𝑠 = 𝑡𝑡 𝑥𝑥 , find the value of x. 3) Using a ruler, construct a line segment of length x. Exercise 40 Consider triangle ACM, By the triangle proportionality theorem, Since YP||CM, AY/YC=AP/PM-------(1) Consider triangles APN and MPC,

∠ APN~= ∠ CPM (vertically opposite angles) ∠ NAP~= ∠ PMC (alternative angles in AN||CM) Then AA similarity theorem, triangle APN~triangle MPC. So, AP/PM=PN/PC=AN/CM---------(2) By (1) and (2), AY/YC=AP/PM=PN/PC=AN/CM------> AY/YC=AN/CM-------(3) Consider triangles CXM and BXP, ∠ CXM~= ∠ BXP (vertically opposite angles) ∠ CMX~= ∠ BPX (alternative angles in PB||CM) Then AA similarity theorem, triangle CXM~triangle BXP. So, CM/BP=MX/XP=CX/XB-------> CX/XB=CM/BP----------(4) Consider triangles AZN and PZB, ∠ AZN~= ∠ BZP (vertically opposite angles) ∠ ANZ~= ∠ ZPB (alternative angles in AN||CM) Then AA similarity theorem, triangle AZN~triangle PZB. So, AN/PB=AZ/ZB=NZ/ZP--------> BZ/ZA=PB/AN----------(5) By (3). (4). (5), (AY/YC)(CX/XB)(BZ/ZA)=(AN/CM)(CM/BP)(PB/AN) (AY/YC)(CX/XB)(BZ/ZA)=1 Exercise 41 a and b Exercise 42 c Exercise 43 𝑥𝑥 2 = 121 𝑥𝑥 2 = (+ − 11) 2 𝑥𝑥 = +11 𝑜𝑜𝑜𝑜 − 11 Exercise 44 𝑥𝑥 2 + 16 = 25 𝑥𝑥 2 = 25 − 16 𝑥𝑥 2 = 9 𝑥𝑥 2 = (+ − 3) 2 𝑥𝑥 = +3 𝑜𝑜𝑜𝑜 − 3 Exercise 45 36 + 𝑥𝑥 2 = 85 𝑥𝑥 2 = 85 − 36 𝑥𝑥 2 = 49 𝑥𝑥 2 = (+ − 7) 2 𝑥𝑥 = +7 𝑜𝑜𝑜𝑜 − 7

Chapter 8.4 Proportionality Theorems

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