SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠LWX=53°, GN is parallel to IK, ML is parallel to HE. Find the measure of ∠KZH.
问题：如图所示，∠LWX=53°，GN平行于IK，ML平行于HE。求∠KZH的大小。
Construction (CDL):
Shape(GW,WL)
Shape(LW,WX)
Shape(XW,WY)
Shape(YW,WG)
Shape(WX,XE)
Shape(EX,XN)
Shape(NX,XZ)
Shape(ZX,XW)
Shape(IY,YW)
Shape(WY,YZ)
Shape(ZY,YM)
Shape(MY,YI)
Shape(YZ,ZX)
Shape(XZ,ZK)
Shape(KZ,ZH)
Shape(HZ,ZY)
Collinear(GWXN)
Collinear(IYZK)
Collinear(MYWL)
Collinear(HZXE)
Conditions (CDL):
Equal(MeasureOfAngle(LWX),53)
ParallelBetweenLine(GN,IK)
ParallelBetweenLine(ML,HE)
Goal: Value(MeasureOfAngle(KZH))
Answer: 127
Theorem sequence:
parallel_property_collinear_extend(3,GN,IK,W)
parallel_property_collinear_extend(3,KI,NW,Y)
parallel_property_collinear_extend(3,ML,HE,Y)
parallel_property_collinear_extend(3,EH,LY,Z)
parallel_property_corresponding_angle(1,WN,YK,L)
parallel_property_ipsilateral_internal_angle(1,YL,ZE)
vertical_angle(1,YZX,KZH)
Problem: As shown in the diagram, AE=3*x+7, BC=5*x-9, the measure of arc DAE is 142, the measure of ⌒DBC is 142. Find the value of x.
问题：如图所示，AE=3*x+7，BC=5*x-9，弧DAE的角度为142，⌒DBC的角度为142。求x的值。
Construction (CDL):
Shape(EA,DAE)
Shape(CB,DBC)
Shape(AE,DEB,BC,DCA)
Cocircular(D,AEBC)
Conditions (CDL):
Equal(LengthOfLine(AE),3*x+7)
Equal(LengthOfLine(BC),5*x-9)
Equal(MeasureOfArc(DAE),142)
Equal(MeasureOfArc(DBC),142)
Goal: Value(x)
Answer: 8
Theorem sequence:
congruent_arc_judgment_measure_equal(1,DAE,DBC)
congruent_arc_property_chord_equal(1,DAE,DBC)
Problem: As shown in the diagram, the area of the sector KJL is 94, LK=14, the measure of ⌒KJL is x, K is the center of circle K. Find the value of x.
问题：如图所示，扇形KJL的面积为94，LK=14，弧KJL的角度为x，圆K的圆心为K。求x的值。
Construction (CDL):
Shape(LK,KJ,KJL)
Cocircular(K,JL)
Conditions (CDL):
Equal(AreaOfSector(KJL),94)
Equal(LengthOfLine(LK),14)
Equal(MeasureOfArc(KJL),x)
IsCentreOfCircle(K,K)
Goal: Value(x)
Answer: 8460/(49*pi)
Theorem sequence:
radius_of_circle_property_length_equal(1,KL,K)
sector_area_formula(1,KJL)
Problem: As shown in the diagram, BD=12, E is the center of ⊙E, DB is the diameter of circle E. Find the radius of circle E.
问题：如图所示，BD=12，圆E的圆心为E，DB是⊙E的直径。求⊙E的半径。
Construction (CDL):
Shape(AB,EBA)
Shape(AE,EB,BA)
Shape(DE,EA,EAD)
Shape(CE,ED,EDC)
Shape(BE,EC,ECB)
Collinear(AEC)
Collinear(DEB)
Cocircular(E,ADCB)
Conditions (CDL):
Equal(LengthOfLine(BD),12)
IsCentreOfCircle(E,E)
IsDiameterOfCircle(DB,E)
Goal: Value(RadiusOfCircle(E))
Answer: 6
Theorem sequence:
diameter_of_circle_property_length_equal(1,DB,E)
circle_property_length_of_radius_and_diameter(1,E)
Problem: As shown in the diagram, ∠BVC=52°, ∠BVC=∠ZXY, ∠VCB=6*x+14*y°, ∠ZXY=15*x-8*y°, CB is perpendicular to VB, XY is perpendicular to ZY. Find the value of x.
问题：如图所示，∠BVC=52°，∠BVC=∠ZXY，∠VCB=6*x+14*y°，∠ZXY=15*x-8*y°，CB垂直于VB，XY垂直于ZY。求x的值。
Construction (CDL):
Shape(XY,YZ,ZX)
Shape(BV,VC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(BVC),52)
Equal(MeasureOfAngle(BVC),MeasureOfAngle(ZXY))
Equal(MeasureOfAngle(VCB),6*x+14*y)
Equal(MeasureOfAngle(ZXY),15*x-8*y)
PerpendicularBetweenLine(CB,VB)
PerpendicularBetweenLine(XY,ZY)
Goal: Value(x)
Answer: 4
Theorem sequence:
triangle_property_angle_sum(1,BVC)
Problem: As shown in the diagram, SU=20, YW=20, the measure of arc RYX is 45, the center of circle R is R, RV⊥SV, YZ⊥RZ. Find the length of line UV.
问题：如图所示，SU=20，YW=20，⌒RYX的角度为45，R是⊙R的圆心，RV垂直于SV，YZ⊥RZ。求直线UV的长度。
Construction (CDL):
Shape(SV,VT,RTS)
Shape(TV,VU,SUT)
Shape(YZ,ZR,RV,VS,RSY)
Shape(UV,VR,RZ,ZW,RWU)
Shape(XZ,ZY,RYX)
Shape(WZ,ZX,RXW)
Collinear(YZW)
Collinear(RZX)
Collinear(RVT)
Collinear(SVU)
Cocircular(R,SYXWUT)
Conditions (CDL):
Equal(LengthOfLine(SU),20)
Equal(LengthOfLine(YW),20)
Equal(MeasureOfArc(RYX),45)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(RV,SV)
PerpendicularBetweenLine(YZ,RZ)
Goal: Value(LengthOfLine(UV))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,UVR,RVS)
circle_property_chord_perpendicular_bisect_chord(1,R,RV,US)
line_addition(1,SV,VU)
Problem: As shown in the diagram, DI=7/2, IF=3, NG=11/2, the center of circle G is G, DI is perpendicular to FI, EN is perpendicular to GN, GD is perpendicular to ID, JE⊥NE, JG is perpendicular to FG. Find the sum of the area of the sector GFJ, the area of DIFG and the area of quadrilateral NGJE.
问题：如图所示，DI=7/2，IF=3，NG=11/2，圆G的圆心为G，DI⊥FI，EN垂直于GN，GD垂直于ID，JE⊥NE，JG⊥FG。求扇形GFJ的面积、四边形DIFG的面积与四边形NGJE的面积之和。
Construction (CDL):
Shape(DI,IF,FG,GD)
Shape(NG,GJ,JE,EN)
Shape(JG,GF,GFJ)
Collinear(DGJ)
Collinear(FGN)
Cocircular(G,FJ)
Conditions (CDL):
Equal(LengthOfLine(DI),7/2)
Equal(LengthOfLine(IF),3)
Equal(LengthOfLine(NG),11/2)
IsCentreOfCircle(G,G)
PerpendicularBetweenLine(DI,FI)
PerpendicularBetweenLine(EN,GN)
PerpendicularBetweenLine(GD,ID)
PerpendicularBetweenLine(JE,NE)
PerpendicularBetweenLine(JG,FG)
Goal: Value(Add(AreaOfSector(GFJ),AreaOfQuadrilateral(DIFG),AreaOfQuadrilateral(NGJE)))
Answer: 49*pi/16+119/4
Theorem sequence:
adjacent_complementary_angle(1,JGF,FGD)
adjacent_complementary_angle(1,NGJ,JGF)
perpendicular_judgment_angle(1,FG,DG)
perpendicular_judgment_angle(1,NG,JG)
parallel_judgment_per_per(1,DG,IF)
parallel_judgment_per_per(1,GF,DI)
parallel_judgment_per_per(1,EJ,NG)
parallel_judgment_per_per(1,NE,GJ)
parallelogram_judgment_parallel_and_parallel(1,IFGD)
parallelogram_judgment_parallel_and_parallel(1,NGJE)
parallelogram_property_opposite_line_equal(1,DIFG)
parallelogram_property_opposite_line_equal(1,GJEN)
parallelogram_area_formula_sine(1,DIFG)
parallelogram_area_formula_sine(1,ENGJ)
radius_of_circle_property_length_equal(1,GF,G)
radius_of_circle_property_length_equal(1,GJ,G)
arc_property_center_angle(1,GFJ,G)
sector_area_formula(1,GFJ)
Problem: As shown in the diagram, AB=y, AC=x, BC=12, ∠CBA=45°, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=y，AC=x，BC=12，∠CBA=45°，BA⊥CA。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),12)
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 6*sqrt(2)
Theorem sequence:
sine_theorem(1,CBA)
Problem: As shown in the diagram, ∠MQP=6*x°, ∠QPN=7*x+11°, quadrilateral MQPN is a ▱. Find the value of x.
问题：如图所示，∠MQP=6*x°，∠QPN=7*x+11°，四边形MQPN是平行四边形。求x的值。
Construction (CDL):
Shape(MQ,QP,PN,NM)
Conditions (CDL):
Equal(MeasureOfAngle(MQP),6*x)
Equal(MeasureOfAngle(QPN),7*x+11)
Parallelogram(MQPN)
Goal: Value(x)
Answer: 13
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,QM,PN)
Problem: As shown in the diagram, SU=x, TU=15*sqrt(3), ∠SUT=63°, UT is perpendicular to ST. Find the value of x.
问题：如图所示，SU=x，TU=15*sqrt(3)，∠SUT=63°，UT⊥ST。求x的值。
Construction (CDL):
Shape(UT,TS,SU)
Conditions (CDL):
Equal(LengthOfLine(SU),x)
Equal(LengthOfLine(TU),15*sqrt(3))
Equal(MeasureOfAngle(SUT),63)
PerpendicularBetweenLine(UT,ST)
Goal: Value(x)
Answer: 60*sqrt(6)/(-sqrt(5)+1+sqrt(2)*sqrt(sqrt(5)+5))
Theorem sequence:
triangle_property_angle_sum(1,UTS)
sine_theorem(1,UTS)
Problem: As shown in the diagram, AB=AD, CB=4, DC is perpendicular to AC. Find the length of line CD.
问题：如图所示，AB=AD，CB=4，DC⊥AC。求直线CD的长度。
Construction (CDL):
Shape(AD,DC,CA)
Shape(CB,BA,AC)
Collinear(BCD)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AD))
Equal(LengthOfLine(CB),4)
PerpendicularBetweenLine(DC,AC)
Goal: Value(LengthOfLine(CD))
Answer: 4
Theorem sequence:
perpendicular_bisector_judgment_distance_equal(1,AC,DB)
Problem: As shown in the diagram, AB=18, CA=x, CB=8, ∠CBA=140°. Find the value of x.
问题：如图所示，AB=18，CA=x，CB=8，∠CBA=140°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(CA),x)
Equal(LengthOfLine(CB),8)
Equal(MeasureOfAngle(CBA),140)
Goal: Value(x)
Answer: 2*sqrt(72*cos(2*pi/9)+97)
Theorem sequence:
cosine_theorem(1,BAC)
Problem: As shown in the diagram, AB=15, BC=x, ∠BAC=45°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=15，BC=x，∠BAC=45°，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(BAC),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 15*sqrt(2)/2
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, CA=1, CB=x+3, DC=6, EC=11*x-2, AB is parallel to ED. Find the value of x.
问题：如图所示，CA=1，CB=x+3，DC=6，EC=11*x-2，AB∥ED。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Shape(DC,CE,ED)
Collinear(BCE)
Collinear(ACD)
Conditions (CDL):
Equal(LengthOfLine(CA),1)
Equal(LengthOfLine(CB),x+3)
Equal(LengthOfLine(DC),6)
Equal(LengthOfLine(EC),11*x-2)
ParallelBetweenLine(AB,ED)
Goal: Value(x)
Answer: 4
Theorem sequence:
parallel_property_alternate_interior_angle(1,AB,ED)
parallel_property_alternate_interior_angle(2,AB,ED)
similar_triangle_judgment_aa(1,CED,CBA)
similar_triangle_property_line_ratio(1,EDC,BAC)
similar_triangle_property_line_ratio(1,DCE,ACB)
Problem: As shown in the diagram, ∠ADE=47°, ∠BCA=55°, ∠DEA=57°. Find the measure of ∠ABC.
问题：如图所示，∠ADE=47°，∠BCA=55°，∠DEA=57°。求∠ABC的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(CA,AB,BC)
Collinear(EAC)
Collinear(DAB)
Conditions (CDL):
Equal(MeasureOfAngle(ADE),47)
Equal(MeasureOfAngle(BCA),55)
Equal(MeasureOfAngle(DEA),57)
Goal: Value(MeasureOfAngle(ABC))
Answer: 49
Theorem sequence:
triangle_property_angle_sum(1,ADE)
vertical_angle(1,EAD,CAB)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, BD=56, ∠BDC=42°, ADCB is a rectangle. Find the length of line AB.
问题：如图所示，BD=56，∠BDC=42°，四边形ADCB是矩形。求直线AB的长度。
Construction (CDL):
Shape(AD,DB,BA)
Shape(DC,CB,BD)
Conditions (CDL):
Equal(LengthOfLine(BD),56)
Equal(MeasureOfAngle(BDC),42)
Rectangle(ADCB)
Goal: Value(LengthOfLine(AB))
Answer: -7*sqrt(3)+7*sqrt(2*sqrt(5)+10)+7*sqrt(15)
Theorem sequence:
triangle_property_angle_sum(1,DCB)
sine_theorem(1,DCB)
parallelogram_property_opposite_line_equal(1,DCBA)
Problem: As shown in the diagram, PQ=6-x, QR=6+x, SQ=3+x, TQ=3, TP is parallel to RS. Find the length of line QS.
问题：如图所示，PQ=6-x，QR=6+x，SQ=3+x，TQ=3，TP∥RS。求直线QS的长度。
Construction (CDL):
Shape(PT,TQ,QP)
Shape(QR,RS,SQ)
Collinear(TQS)
Collinear(PQR)
Conditions (CDL):
Equal(LengthOfLine(PQ),6-x)
Equal(LengthOfLine(QR),6+x)
Equal(LengthOfLine(SQ),3+x)
Equal(LengthOfLine(TQ),3)
ParallelBetweenLine(TP,RS)
Goal: Value(LengthOfLine(QS))
Answer: 3
Theorem sequence:
parallel_property_alternate_interior_angle(1,TP,RS)
parallel_property_alternate_interior_angle(2,TP,RS)
similar_triangle_judgment_aa(1,QPT,QRS)
similar_triangle_property_line_ratio(1,TQP,SQR)
similar_triangle_property_line_ratio(1,PTQ,RSQ)
Problem: As shown in the diagram, AC=AD, AE=20, AF=8, BE=15, CA is perpendicular to FA, quadrilateral BEAC is a rectangle. Find the sum of the area of quadrilateral BEAC, the area of △CAF and the area of triangle FAD.
问题：如图所示，AC=AD，AE=20，AF=8，BE=15，CA⊥FA，四边形BEAC是长方形。求四边形BEAC的面积、三角形CAF的面积与△FAD的面积之和。
Construction (CDL):
Shape(CB,BE,EA,AC)
Shape(FC,CA,AF)
Shape(AD,DF,FA)
Collinear(EAF)
Collinear(CAD)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(AD))
Equal(LengthOfLine(AE),20)
Equal(LengthOfLine(AF),8)
Equal(LengthOfLine(BE),15)
PerpendicularBetweenLine(CA,FA)
Rectangle(BEAC)
Goal: Value(Add(AreaOfQuadrilateral(BEAC),AreaOfTriangle(CAF),AreaOfTriangle(FAD)))
Answer: 420
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BEAC)
parallelogram_area_formula_sine(1,BEAC)
adjacent_complementary_angle(1,CAF,FAD)
triangle_area_formula_sine(1,AFC)
triangle_area_formula_sine(1,ADF)
Problem: As shown in the diagram, AC=20, CF=12, DB=2*x+1, FB=2*x-1, ∠ACB=22°, ∠CBD=22°, CF⊥AF. Find the length of line CB.
问题：如图所示，AC=20，CF=12，DB=2*x+1，FB=2*x-1，∠ACB=22°，∠CBD=22°，CF⊥AF。求直线CB的长度。
Construction (CDL):
Shape(AC,CF,FA)
Shape(DF,FB,BD)
Collinear(ADF)
Collinear(BFC)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(CF),12)
Equal(LengthOfLine(DB),2*x+1)
Equal(LengthOfLine(FB),2*x-1)
Equal(MeasureOfAngle(ACB),22)
Equal(MeasureOfAngle(CBD),22)
PerpendicularBetweenLine(CF,AF)
Goal: Value(LengthOfLine(CB))
Answer: 15
Theorem sequence:
adjacent_complementary_angle(1,CFA,AFB)
mirror_similar_triangle_judgment_aa(1,ACF,DFB)
mirror_similar_triangle_property_line_ratio(1,ACF,DFB)
mirror_similar_triangle_property_line_ratio(1,FAC,FBD)
line_addition(1,BF,FC)
Problem: As shown in the diagram, BC=x, GJ=6, HJ=9, VB=8, ∠BAC=∠CVB, ∠HGJ=∠CVB, ∠HGJ=∠HIJ, BC is perpendicular to VC, JH⊥GH. Find the value of x.
问题：如图所示，BC=x，GJ=6，HJ=9，VB=8，∠BAC=∠CVB，∠HGJ=∠CVB，∠HGJ=∠HIJ，BC垂直于VC，JH⊥GH。求x的值。
Construction (CDL):
Shape(BC,CV,VB)
Shape(AC,CB,BA)
Shape(HG,GJ,JH)
Shape(HJ,JI,IH)
Collinear(ACV)
Collinear(GHI)
Conditions (CDL):
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(GJ),6)
Equal(LengthOfLine(HJ),9)
Equal(LengthOfLine(VB),8)
Equal(MeasureOfAngle(BAC),MeasureOfAngle(CVB))
Equal(MeasureOfAngle(HGJ),MeasureOfAngle(CVB))
Equal(MeasureOfAngle(HGJ),MeasureOfAngle(HIJ))
PerpendicularBetweenLine(BC,VC)
PerpendicularBetweenLine(JH,GH)
Goal: Value(x)
Answer: 12
Theorem sequence:
similar_triangle_judgment_aa(1,BCV,JHG)
similar_triangle_property_line_ratio(1,CVB,HGJ)
similar_triangle_property_line_ratio(1,VBC,GJH)
Problem: As shown in the diagram, QR=12, UT=22, S is the midpoint of segment RT, V bisects segment QU, QUTR is a trapezoid. Find the length of line VS.
问题：如图所示，QR=12，UT=22，S是线段RT的中点，V平分线段QU，QU和TR是梯形QUTR的腰。求直线VS的长度。
Construction (CDL):
Shape(QV,VS,SR,RQ)
Shape(VU,UT,TS,SV)
Collinear(QVU)
Collinear(RST)
Conditions (CDL):
Equal(LengthOfLine(QR),12)
Equal(LengthOfLine(UT),22)
IsMidpointOfLine(S,RT)
IsMidpointOfLine(V,QU)
Trapezoid(QUTR)
Goal: Value(LengthOfLine(VS))
Answer: 17
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,VS,QUTR)
midsegment_of_quadrilateral_property_length(1,VS,QUTR)
Problem: As shown in the diagram, ∠AFE=x-4°, ∠BAF=x+2°, ∠CBA=x-8°, ∠DCB=x+7°, ∠EDC=x-3°, ∠FED=x+6°. Find the measure of ∠FED.
问题：如图所示，∠AFE=x-4°，∠BAF=x+2°，∠CBA=x-8°，∠DCB=x+7°，∠EDC=x-3°，∠FED=x+6°。求∠FED的大小。
Construction (CDL):
Shape(AF,FE,ED,DA)
Shape(BA,AD,DC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(AFE),x-4)
Equal(MeasureOfAngle(BAF),x+2)
Equal(MeasureOfAngle(CBA),x-8)
Equal(MeasureOfAngle(DCB),x+7)
Equal(MeasureOfAngle(EDC),x-3)
Equal(MeasureOfAngle(FED),x+6)
Goal: Value(MeasureOfAngle(FED))
Answer: 126
Theorem sequence:
angle_addition(1,BAD,DAF)
angle_addition(1,EDA,ADC)
quadrilateral_property_angle_sum(1,AFED)
quadrilateral_property_angle_sum(1,BADC)
Problem: As shown in the diagram, KJ=24, LJ=17, JL⊥KL. Find the measure of ∠KJL.
问题：如图所示，KJ=24，LJ=17，JL垂直于KL。求∠KJL的大小。
Construction (CDL):
Shape(JL,LK,KJ)
Conditions (CDL):
Equal(LengthOfLine(KJ),24)
Equal(LengthOfLine(LJ),17)
PerpendicularBetweenLine(JL,KL)
Goal: Value(MeasureOfAngle(KJL))
Answer: 180*acos(17/24)/pi
Theorem sequence:
cosine_theorem(1,JLK)
cosine_theorem(1,LKJ)
Problem: As shown in the diagram, PS=4, ∠RPS=130°, ∠TPQ=112°, P is the center of circle P. Find the length of arc PRQ.
问题：如图所示，PS=4，∠RPS=130°，∠TPQ=112°，P是圆P的圆心。求⌒PRQ的长度。
Construction (CDL):
Shape(QP,PR,PRQ)
Shape(TP,PQ,PQT)
Shape(SP,PT,PTS)
Shape(RP,PS,PSR)
Collinear(RPT)
Cocircular(P,RQTS)
Conditions (CDL):
Equal(LengthOfLine(PS),4)
Equal(MeasureOfAngle(RPS),130)
Equal(MeasureOfAngle(TPQ),112)
IsCentreOfCircle(P,P)
Goal: Value(LengthOfArc(PRQ))
Answer: 68*pi/45
Theorem sequence:
adjacent_complementary_angle(1,TPQ,QPR)
arc_property_center_angle(1,PRQ,P)
radius_of_circle_property_length_equal(1,PS,P)
arc_length_formula(1,PRQ)
Problem: As shown in the diagram, AB=26, AD=12, DG=9/2, EF=8, GF=14, ∠AGF=108°, quadrilateral ADCB is similar to quadrilateral AGFE. Find The ratio of similar quadrilaterals ADCB and AGFE.
问题：如图所示，AB=26，AD=12，DG=9/2，EF=8，GF=14，∠AGF=108°，四边形ADCB与四边形AGFE相似。求四边形ADCB和四边形AGFE的相似比。
Construction (CDL):
Shape(AG,GF,FE,EA)
Shape(GD,DC,CB,BE,EF,FG)
Collinear(AGD)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DG),9/2)
Equal(LengthOfLine(EF),8)
Equal(LengthOfLine(GF),14)
Equal(MeasureOfAngle(AGF),108)
SimilarBetweenQuadrilateral(ADCB,AGFE)
Goal: Value(RatioOfSimilarQuadrilateral(ADCB,AGFE))
Answer: 8/5
Theorem sequence:
line_addition(1,AG,GD)
similar_quadrilateral_property_line_ratio(1,ADCB,AGFE)
Problem: As shown in the diagram, AB=x, BC=15, ∠BAC=25°, CB⊥AB. Find the value of x.
问题：如图所示，AB=x，BC=15，∠BAC=25°，CB垂直于AB。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),15)
Equal(MeasureOfAngle(BAC),25)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 15/tan(5*pi/36)
Theorem sequence:
sine_theorem(1,BAC)
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, WT=3, WX is perpendicular to YX, XY⊥ZY, YZ⊥WZ, ZW⊥XW, quadrilateral WXYZ is a square. Find the measure of ∠WTZ.
问题：如图所示，WT=3，WX垂直于YX，XY垂直于ZY，YZ⊥WZ，ZW⊥XW，四边形WXYZ是正方形。求∠WTZ的大小。
Construction (CDL):
Shape(WX,XT,TW)
Shape(XY,YT,TX)
Shape(TY,YZ,ZT)
Shape(WT,TZ,ZW)
Collinear(XTZ)
Collinear(YTW)
Conditions (CDL):
Equal(LengthOfLine(WT),3)
PerpendicularBetweenLine(WX,YX)
PerpendicularBetweenLine(XY,ZY)
PerpendicularBetweenLine(YZ,WZ)
PerpendicularBetweenLine(ZW,XW)
Square(WXYZ)
Goal: Value(MeasureOfAngle(WTZ))
Answer: 90
Theorem sequence:
line_addition(1,XT,TZ)
line_addition(1,YT,TW)
isosceles_triangle_judgment_line_equal(1,WXZ)
isosceles_triangle_property_angle_equal(1,WXZ)
triangle_property_angle_sum(1,WXZ)
parallelogram_property_diagonal_bisection(1,XYZW,T)
parallelogram_property_diagonal_bisection(1,YZWX,T)
rectangle_property_diagonal_equal(1,XYZW)
isosceles_triangle_judgment_line_equal(1,TZW)
isosceles_triangle_property_angle_equal(1,TZW)
triangle_property_angle_sum(1,TZW)
Problem: As shown in the diagram, ∠ANE=30°, ∠CEB=110°, ∠ECA=∠AFN, ∠ENJ=130°. Find the measure of ∠NAF.
问题：如图所示，∠ANE=30°，∠CEB=110°，∠ECA=∠AFN，∠ENJ=130°。求∠NAF的大小。
Construction (CDL):
Shape(FA,AL)
Shape(LA,AG)
Shape(GA,AC)
Shape(CA,AE,EC)
Shape(AF,FN,NA)
Shape(EA,AN,NE)
Shape(CE,EB)
Shape(BE,EN)
Shape(EN,NJ)
Shape(JN,NM)
Shape(MN,NF)
Collinear(CAF)
Collinear(CEN)
Collinear(GANM)
Collinear(FNJ)
Collinear(LAEB)
Conditions (CDL):
Equal(MeasureOfAngle(ANE),30)
Equal(MeasureOfAngle(CEB),110)
Equal(MeasureOfAngle(ECA),MeasureOfAngle(AFN))
Equal(MeasureOfAngle(ENJ),130)
Goal: Value(MeasureOfAngle(NAF))
Answer: 95
Theorem sequence:
angle_addition(1,FNA,ANE)
adjacent_complementary_angle(1,FNE,ENJ)
triangle_property_angle_sum(1,FNC)
triangle_property_angle_sum(1,FNA)
Problem: As shown in the diagram, ∠AGC=40°, ∠DGF=53°, CB is perpendicular to GB, FG⊥CG, GF is perpendicular to DF. Find the measure of ∠GAF.
问题：如图所示，∠AGC=40°，∠DGF=53°，CB垂直于GB，FG⊥CG，GF⊥DF。求∠GAF的大小。
Construction (CDL):
Shape(FD,DG,GF)
Shape(AF,FG,GB,BA)
Shape(BG,GC,CB)
Shape(AB,BC,CA)
Collinear(DFA)
Collinear(GBA)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),40)
Equal(MeasureOfAngle(DGF),53)
PerpendicularBetweenLine(CB,GB)
PerpendicularBetweenLine(FG,CG)
PerpendicularBetweenLine(GF,DF)
Goal: Value(MeasureOfAngle(GAF))
Answer: 40
Theorem sequence:
angle_addition(1,FGA,AGC)
adjacent_complementary_angle(1,AFG,GFD)
triangle_property_angle_sum(1,FGA)
Problem: As shown in the diagram, ∠GJC=50°, ∠IAH=x°, ∠JLA=72°, IO∥PL, PL∥CN. Find the value of x.
问题：如图所示，∠GJC=50°，∠IAH=x°，∠JLA=72°，IO∥PL，PL∥CN。求x的值。
Construction (CDL):
Shape(IA,AH)
Shape(HA,AO)
Shape(OA,AL)
Shape(LA,AI)
Shape(PL,LA)
Shape(JL,LP)
Shape(CJ,JL)
Shape(LJ,JN)
Shape(NJ,JG)
Shape(GJ,JC)
Collinear(HAL)
Collinear(LJG)
Collinear(IAO)
Collinear(CJN)
Conditions (CDL):
Equal(MeasureOfAngle(GJC),50)
Equal(MeasureOfAngle(IAH),x)
Equal(MeasureOfAngle(JLA),72)
ParallelBetweenLine(IO,PL)
ParallelBetweenLine(PL,CN)
Goal: Value(x)
Answer: 22
Theorem sequence:
angle_addition(1,JLP,PLA)
parallel_property_collinear_extend(3,NC,LP,J)
parallel_property_corresponding_angle(1,JC,LP,G)
parallel_property_collinear_extend(3,IO,PL,A)
parallel_property_corresponding_angle(2,LP,AI,H)
Problem: As shown in the diagram, AB=8, XY=20, ∠XYB=70°, the midsegment of quadrilateral AXYB is EF, AX and YB are the non-parallel sides of isosceles trapezoid AXYB. Find the length of line EF.
问题：如图所示，AB=8，XY=20，∠XYB=70°，四边形AXYB的中位线为EF，AX和YB是等腰梯形AXYB的腰。求直线EF的长度。
Construction (CDL):
Shape(BA,AE,EF,FB)
Shape(EX,XY,YF,FE)
Collinear(AEX)
Collinear(BFY)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(XY),20)
Equal(MeasureOfAngle(XYB),70)
IsMidsegmentOfQuadrilateral(EF,AXYB)
IsoscelesTrapezoid(AXYB)
Goal: Value(LengthOfLine(EF))
Answer: 14
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,EF,AXYB)
Problem: As shown in the diagram, AB=y, AC=10*sqrt(3), BC=x, ∠CAB=30°, BC⊥AC. Find the value of y.
问题：如图所示，AB=y，AC=10*sqrt(3)，BC=x，∠CAB=30°，BC⊥AC。求y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),10*sqrt(3))
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(CAB),30)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 20
Theorem sequence:
triangle_property_angle_sum(1,BCA)
sine_theorem(1,ABC)
Problem: As shown in the diagram, SM=3*x+5, SP=6*x-7, ∠MJS=∠SJP, SM⊥JM, SP is perpendicular to TP. Find the length of line SP.
问题：如图所示，SM=3*x+5，SP=6*x-7，∠MJS=∠SJP，SM垂直于JM，SP垂直于TP。求直线SP的长度。
Construction (CDL):
Shape(MJ,JS,SM)
Shape(SJ,JP,PS)
Collinear(JMC)
Collinear(JSB)
Collinear(JPT)
Conditions (CDL):
Equal(LengthOfLine(SM),3*x+5)
Equal(LengthOfLine(SP),6*x-7)
Equal(MeasureOfAngle(MJS),MeasureOfAngle(SJP))
PerpendicularBetweenLine(SM,JM)
PerpendicularBetweenLine(SP,TP)
Goal: Value(LengthOfLine(SP))
Answer: 17
Theorem sequence:
adjacent_complementary_angle(1,JPS,SPT)
triangle_property_angle_sum(1,MJS)
triangle_property_angle_sum(1,SJP)
mirror_congruent_triangle_judgment_aas(1,JSM,JPS)
mirror_congruent_triangle_property_line_equal(1,JSM,JPS)
Problem: As shown in the diagram, AB=x, BC=39/2, ∠ABC=45°, ∠CAB=45°, BC⊥AC. Find the value of x.
问题：如图所示，AB=x，BC=39/2，∠ABC=45°，∠CAB=45°，BC⊥AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),39/2)
Equal(MeasureOfAngle(ABC),45)
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 39*sqrt(2)/2
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, KJ=5, KL=12, LJ=13, LK is perpendicular to JK. Find the value of sin(JL).
问题：如图所示，KJ=5，KL=12，LJ=13，LK⊥JK。求sin(JL)的值。
Construction (CDL):
Shape(LK,KJ,JL)
Conditions (CDL):
Equal(LengthOfLine(KJ),5)
Equal(LengthOfLine(KL),12)
Equal(LengthOfLine(LJ),13)
PerpendicularBetweenLine(LK,JK)
Goal: Value(Sin(MeasureOfAngle(JLK)))
Answer: 5/13
Theorem sequence:
sine_theorem(1,JLK)
Problem: As shown in the diagram, BA=6, DA=9, FC=7, ∠BAF=32°, ∠CBF=40°, ∠FAD=20°, quadrilateral ADCB is a ▱. Find the length of line AF.
问题：如图所示，BA=6，DA=9，FC=7，∠BAF=32°，∠CBF=40°，∠FAD=20°，ADCB是▱。求直线AF的长度。
Construction (CDL):
Shape(AD,DF,FA)
Shape(AF,FB,BA)
Shape(FD,DC,CF)
Shape(FC,CB,BF)
Collinear(DFB)
Collinear(AFC)
Conditions (CDL):
Equal(LengthOfLine(BA),6)
Equal(LengthOfLine(DA),9)
Equal(LengthOfLine(FC),7)
Equal(MeasureOfAngle(BAF),32)
Equal(MeasureOfAngle(CBF),40)
Equal(MeasureOfAngle(FAD),20)
Parallelogram(ADCB)
Goal: Value(LengthOfLine(AF))
Answer: 7
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ADCB,F)
Problem: As shown in the diagram, ∠XYZ=128°, XW and XY are the legs of the isosceles △ XWY, YZ and YX are the legs of the isosceles triangle YZX. Find the measure of ∠YZX.
问题：如图所示，∠XYZ=128°，∠XWY和∠WYX是等腰三角形XWY的底角，YZ和YX是等腰△YZX的腰。求∠YZX的大小。
Construction (CDL):
Shape(XW,WY,YX)
Shape(XY,YZ,ZX)
Collinear(WYZ)
Conditions (CDL):
Equal(MeasureOfAngle(XYZ),128)
IsoscelesTriangle(XWY)
IsoscelesTriangle(YZX)
Goal: Value(MeasureOfAngle(YZX))
Answer: 26
Theorem sequence:
isosceles_triangle_property_angle_equal(1,YZX)
triangle_property_angle_sum(1,YZX)
Problem: As shown in the diagram, ∠ADB=4*y+x°, ∠BCA=2*x+2*y°, ∠CAD=x+y°, quadrilateral CADB is a ▱. Find the value of y.
问题：如图所示，∠ADB=4*y+x°，∠BCA=2*x+2*y°，∠CAD=x+y°，AC和DB是▱CADB的一组对边。求y的值。
Construction (CDL):
Shape(BC,CA,AD,DB)
Conditions (CDL):
Equal(MeasureOfAngle(ADB),4*y+x)
Equal(MeasureOfAngle(BCA),2*x+2*y)
Equal(MeasureOfAngle(CAD),x+y)
Parallelogram(CADB)
Goal: Value(y)
Answer: 20
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,CADB)
parallel_property_ipsilateral_internal_angle(1,AC,DB)