SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, LM=MN, ∠BNM=2*x-5°, the center of circle B is B, the diameter of ⊙B is NL. Find the value of x.
问题：如图所示，LM=MN，∠BNM=2*x-5°，B是⊙B的圆心，NL是⊙B的直径。求x的值。
Construction (CDL):
Shape(BML,LM)
Shape(BLN,NB,BL)
Shape(ML,LB,BN,NM)
Shape(BNM,MN)
Collinear(LBN)
Cocircular(B,LNM)
Conditions (CDL):
Equal(LengthOfLine(LM),LengthOfLine(MN))
Equal(MeasureOfAngle(BNM),2*x-5)
IsCentreOfCircle(B,B)
IsDiameterOfCircle(NL,B)
Goal: Value(x)
Answer: 25
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,MLN)
isosceles_triangle_property_angle_equal(1,MLN)
diameter_of_circle_property_right_angle(1,NML,B)
triangle_property_angle_sum(1,MLN)
Problem: As shown in the diagram, ∠WXH=130°, ∠YZI=20°, Mul(LengthOfLine(IJ)=LengthOfLine(YJ)). Find the measure of ∠HIJ.
问题：如图所示，∠WXH=130°，∠YZI=20°，Mul(LengthOfLine(IJ)=LengthOfLine(YJ))。求∠HIJ的大小。
Construction (CDL):
Shape(JH,HI,IJ)
Shape(HX,XY,YI,IH)
Shape(IY,YZ,ZI)
Shape(WX,XH)
Collinear(WXYZ)
Collinear(JHX)
Collinear(JIY)
Collinear(HIZ)
Conditions (CDL):
Equal(MeasureOfAngle(WXH),130)
Equal(MeasureOfAngle(YZI),20)
Equal(Mul(LengthOfLine(IJ),LengthOfLine(YJ)),Mul(LengthOfLine(HJ),LengthOfLine(XJ)))
Goal: Value(MeasureOfAngle(HIJ))
Answer: 50
Theorem sequence:
adjacent_complementary_angle(1,WXH,HXY)
mirror_similar_triangle_judgment_sas(1,JHI,JXY)
mirror_similar_triangle_property_angle_equal(1,IJH,XYJ)
Problem: As shown in the diagram, ∠BVC=52°, ∠VCB=6*x+14*y°, ∠ZXY=15*x-8*y°, ∠ZXY=∠BVC, CB⊥VB, XY⊥ZY. Find the value of y.
问题：如图所示，∠BVC=52°，∠VCB=6*x+14*y°，∠ZXY=15*x-8*y°，∠ZXY=∠BVC，CB垂直于VB，XY⊥ZY。求y的值。
Construction (CDL):
Shape(XY,YZ,ZX)
Shape(VC,CB,BV)
Conditions (CDL):
Equal(MeasureOfAngle(BVC),52)
Equal(MeasureOfAngle(VCB),6*x+14*y)
Equal(MeasureOfAngle(ZXY),15*x-8*y)
Equal(MeasureOfAngle(ZXY),MeasureOfAngle(BVC))
PerpendicularBetweenLine(CB,VB)
PerpendicularBetweenLine(XY,ZY)
Goal: Value(y)
Answer: 1
Theorem sequence:
triangle_property_angle_sum(1,VCB)
Problem: As shown in the diagram, AB=16, AC=5, BC=x, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=16，AC=5，BC=x，BC垂直于AC。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: sqrt(231)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, the measure of ⌒BRS is 103, the center of ⊙B is B, TR is the tangent to circle B, the tangent to circle B is TS. Find the measure of ∠RTS.
问题：如图所示，弧BRS的角度为103，B是⊙B的圆心，圆O的切线为TR，⊙O的切线为TS。求∠RTS的大小。
Construction (CDL):
Shape(BS,BSR,RB)
Shape(BR,BRS,SB)
Shape(BRS,RT,TS)
Cocircular(B,RS)
Conditions (CDL):
Equal(MeasureOfArc(BRS),103)
IsCentreOfCircle(B,B)
IsTangentOfCircle(TR,B)
IsTangentOfCircle(TS,B)
Goal: Value(MeasureOfAngle(RTS))
Answer: 77
Theorem sequence:
tangent_of_circle_property_perpendicular(1,TR,B,B)
tangent_of_circle_property_perpendicular(2,TS,B,B)
arc_property_center_angle(1,BRS,B)
quadrilateral_property_angle_sum(1,BRTS)
Problem: As shown in the diagram, ∠ABC=76°, ∠CAB=Mul(MeasureOfAngle(ABC), 1/2)°. Find the measure of ∠BCA.
问题：如图所示，∠ABC=76°，∠CAB=Mul(MeasureOfAngle(ABC), 1/2)°。求∠BCA的大小。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),76)
Equal(MeasureOfAngle(CAB),Mul(MeasureOfAngle(ABC), 1/2))
Goal: Value(MeasureOfAngle(BCA))
Answer: 66
Theorem sequence:
triangle_property_angle_sum(1,CAB)
Problem: As shown in the diagram, FJ=FH, GF=GH, ∠HFJ=34°. Find the measure of ∠FJH.
问题：如图所示，FJ=FH，GF=GH，∠HFJ=34°。求∠FJH的大小。
Construction (CDL):
Shape(FJ,JH,HF)
Shape(FH,HG,GF)
Conditions (CDL):
Equal(LengthOfLine(FJ),LengthOfLine(FH))
Equal(LengthOfLine(GF),LengthOfLine(GH))
Equal(MeasureOfAngle(HFJ),34)
Goal: Value(MeasureOfAngle(FJH))
Answer: 73
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,FJH)
isosceles_triangle_property_angle_equal(1,FJH)
triangle_property_angle_sum(1,FJH)
Problem: As shown in the diagram, FN=1/4*x+6, IP=PD, NA=2*x-29, PD=16-5*y, PI=12-3*y, IF is parallel to PN, PN is parallel to DA. Find the value of x.
问题：如图所示，FN=1/4*x+6，IP=PD，NA=2*x-29，PD=16-5*y，PI=12-3*y，IF平行于PN，PN平行于DA。求x的值。
Construction (CDL):
Shape(IP,PN,NF,FI)
Shape(PD,DA,AN,NP)
Collinear(IPD)
Collinear(FNA)
Conditions (CDL):
Equal(LengthOfLine(FN),1/4*x+6)
Equal(LengthOfLine(IP),LengthOfLine(PD))
Equal(LengthOfLine(NA),2*x-29)
Equal(LengthOfLine(PD),16-5*y)
Equal(LengthOfLine(PI),12-3*y)
ParallelBetweenLine(IF,PN)
ParallelBetweenLine(PN,DA)
Goal: Value(x)
Answer: 20
Theorem sequence:
parallel_judgment_par_par(1,IF,PN,DA)
trapezoid_judgment_parallel(1,IDAF)
midsegment_of_quadrilateral_judgment_parallel(1,PN,IDAF)
Problem: As shown in the diagram, TV=TU, ∠VTU=74°. Find the measure of ∠TUV.
问题：如图所示，TV=TU，∠VTU=74°。求∠TUV的大小。
Construction (CDL):
Shape(TU,UV,VT)
Conditions (CDL):
Equal(LengthOfLine(TV),LengthOfLine(TU))
Equal(MeasureOfAngle(VTU),74)
Goal: Value(MeasureOfAngle(TUV))
Answer: 53
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,TUV)
isosceles_triangle_property_angle_equal(1,TUV)
triangle_property_angle_sum(1,TUV)
Problem: As shown in the diagram, AB=22, AC=14, BD=14, CD=22, the center of circle E is E, the center of ⊙O is O, AC is the diameter of circle O, BD is the diameter of circle E, OC is perpendicular to DC. Find the area of quadrilateral ACDB minus the sum of the areas of sectors OCA and EBD.
问题：如图所示，AB=22，AC=14，BD=14，CD=22，⊙E的圆心为E，圆O的圆心为O，AC是⊙O的直径，圆E的直径为BD，OC垂直于DC。求四边形ACDB的面积减去扇形OCA和EBD的面积和。
Construction (CDL):
Shape(AO,OC,OCA)
Shape(OCA,CD,EBD,BA)
Shape(EBD,DE,EB)
Collinear(AOC)
Collinear(BED)
Cocircular(O,AC)
Cocircular(E,BD)
Conditions (CDL):
Equal(LengthOfLine(AB),22)
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(BD),14)
Equal(LengthOfLine(CD),22)
IsCentreOfCircle(E,E)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AC,O)
IsDiameterOfCircle(BD,E)
PerpendicularBetweenLine(OC,DC)
Goal: Value(Sub(AreaOfQuadrilateral(ACDB),Add(AreaOfSector(OCA),AreaOfSector(EBD))))
Answer: 308-49*pi
Theorem sequence:
parallelogram_judgment_equal_and_equal(1,ACDB)
parallelogram_area_formula_sine(1,ACDB)
diameter_of_circle_property_length_equal(1,AC,O)
diameter_of_circle_property_length_equal(1,BD,E)
circle_property_length_of_radius_and_diameter(1,O)
circle_property_length_of_radius_and_diameter(1,E)
flat_angle(1,AOC)
flat_angle(1,DEB)
arc_property_center_angle(1,OCA,O)
arc_property_center_angle(1,EBD,E)
sector_area_formula(1,OCA)
sector_area_formula(1,EBD)
Problem: As shown in the diagram, ∠BDE=109°, ∠CBE=24°, ∠EAC=95°, ∠ECB=33°. Find the measure of ∠EBD.
问题：如图所示，∠BDE=109°，∠CBE=24°，∠EAC=95°，∠ECB=33°。求∠EBD的大小。
Construction (CDL):
Shape(AC,CE,EA)
Shape(DE,EB,BD)
Shape(EC,CB,BE)
Collinear(AEB)
Collinear(CED)
Conditions (CDL):
Equal(MeasureOfAngle(BDE),109)
Equal(MeasureOfAngle(CBE),24)
Equal(MeasureOfAngle(EAC),95)
Equal(MeasureOfAngle(ECB),33)
Goal: Value(MeasureOfAngle(EBD))
Answer: 14
Theorem sequence:
triangle_property_angle_sum(1,ECB)
flat_angle(1,DEC)
angle_addition(1,DEB,BEC)
triangle_property_angle_sum(1,EBD)
Problem: As shown in the diagram, AM=MP, AP=sqrt(13), PD=3*sqrt(13), PN=ND, ∠MAB=∠NDC, the perimeter of triangle BPA is 12, CP⊥NP. Find the perimeter of △CPD.
问题：如图所示，AM=MP，AP=sqrt(13)，PD=3*sqrt(13)，PN=ND，∠MAB=∠NDC，△BPA的周长为12，CP垂直于NP。求三角形CPD的周长。
Construction (CDL):
Shape(AB,BM,MA)
Shape(MB,BP,PM)
Shape(CP,PN,NC)
Shape(CN,ND,DC)
Collinear(AMPND)
Collinear(CPB)
Conditions (CDL):
Equal(LengthOfLine(AM),LengthOfLine(MP))
Equal(LengthOfLine(AP),sqrt(13))
Equal(LengthOfLine(PD),3*sqrt(13))
Equal(LengthOfLine(PN),LengthOfLine(ND))
Equal(MeasureOfAngle(MAB),MeasureOfAngle(NDC))
Equal(PerimeterOfTriangle(BPA),12)
PerpendicularBetweenLine(CP,NP)
Goal: Value(PerimeterOfTriangle(CPD))
Answer: 36
Theorem sequence:
vertical_angle(1,BPM,CPN)
similar_triangle_judgment_aa(1,BPA,CPD)
similar_triangle_property_line_ratio(1,BPA,CPD)
similar_triangle_property_perimeter_ratio(1,PAB,PDC)
Problem: As shown in the diagram, AD=15, PF=6, the centroid of △ACE is P. Find the length of line AP.
问题：如图所示，AD=15，PF=6，P是△ACE的重心。求直线AP的长度。
Construction (CDL):
Shape(AB,BP,PA)
Shape(BC,CP,PB)
Shape(PC,CD,DP)
Shape(PD,DE,EP)
Shape(PE,EF,FP)
Shape(PF,FA,AP)
Collinear(ABC)
Collinear(CDE)
Collinear(EFA)
Collinear(BPE)
Collinear(CPF)
Collinear(APD)
Conditions (CDL):
Equal(LengthOfLine(AD),15)
Equal(LengthOfLine(PF),6)
IsCentroidOfTriangle(P,ACE)
Goal: Value(LengthOfLine(AP))
Answer: 10
Theorem sequence:
centroid_of_triangle_property_line_ratio(1,P,ACE,D)
line_addition(1,AP,PD)
Problem: As shown in the diagram, ∠PSR=x+10°, ∠QPS=x°, ∠RQP=2*x-16°, ∠SRQ=2*x°. Find the measure of ∠PSR.
问题：如图所示，∠PSR=x+10°，∠QPS=x°，∠RQP=2*x-16°，∠SRQ=2*x°。求∠PSR的大小。
Construction (CDL):
Shape(QP,PS,SR,RQ)
Conditions (CDL):
Equal(MeasureOfAngle(PSR),x+10)
Equal(MeasureOfAngle(QPS),x)
Equal(MeasureOfAngle(RQP),2*x-16)
Equal(MeasureOfAngle(SRQ),2*x)
Goal: Value(MeasureOfAngle(PSR))
Answer: 71
Theorem sequence:
quadrilateral_property_angle_sum(1,QPSR)
Problem: As shown in the diagram, KL=10, ∠LKJ=85°, K is the center of circle K. Find the area of the sector KLJ.
问题：如图所示，KL=10，∠LKJ=85°，K是⊙K的圆心。求扇形KLJ的面积。
Construction (CDL):
Shape(KJL,LK,KJ)
Shape(KLJ,JK,KL)
Cocircular(K,JL)
Conditions (CDL):
Equal(LengthOfLine(KL),10)
Equal(MeasureOfAngle(LKJ),85)
IsCentreOfCircle(K,K)
Goal: Value(AreaOfSector(KLJ))
Answer: 1375*pi/18
Theorem sequence:
radius_of_circle_property_length_equal(1,KL,K)
arc_property_center_angle(1,KLJ,K)
round_angle(1,LKJ,JKL)
sector_area_formula(1,KLJ)
Problem: As shown in the diagram, XW=ZW, XY=ZY, ∠XWZ=70°, ∠ZYX=56°. Find the measure of ∠YXW.
问题：如图所示，XW=ZW，XY=ZY，∠XWZ=70°，∠ZYX=56°。求∠YXW的大小。
Construction (CDL):
Shape(XW,WZ,ZY,YX)
Conditions (CDL):
Equal(LengthOfLine(XW),LengthOfLine(ZW))
Equal(LengthOfLine(XY),LengthOfLine(ZY))
Equal(MeasureOfAngle(XWZ),70)
Equal(MeasureOfAngle(ZYX),56)
Goal: Value(MeasureOfAngle(YXW))
Answer: 117
Theorem sequence:
kite_judgment_equal_and_equal(1,WZYX)
kite_property_opposite_angle_equal(1,WZYX)
quadrilateral_property_angle_sum(1,WZYX)
Problem: As shown in the diagram, ∠HPM=4*y°, ∠MPR=68°, ∠PRC=x°, ∠SCR=5*z+2°, CM∥RP, CR is parallel to MP. Find the value of y.
问题：如图所示，∠HPM=4*y°，∠MPR=68°，∠PRC=x°，∠SCR=5*z+2°，CM∥RP，CR∥MP。求y的值。
Construction (CDL):
Shape(CM,MP,PR,RC)
Collinear(NCRI)
Collinear(DMPL)
Collinear(SCME)
Collinear(GRPH)
Shape(HP,PM)
Conditions (CDL):
Equal(MeasureOfAngle(HPM),4*y)
Equal(MeasureOfAngle(MPR),68)
Equal(MeasureOfAngle(PRC),x)
Equal(MeasureOfAngle(SCR),5*z+2)
ParallelBetweenLine(CM,RP)
ParallelBetweenLine(CR,MP)
Goal: Value(y)
Answer: 28
Theorem sequence:
flat_angle(1,HPR)
angle_addition(1,HPM,MPR)
Problem: As shown in the diagram, CN=8, N is the center of circle N. Find the length of line DN.
问题：如图所示，CN=8，圆N的圆心为N。求直线DN的长度。
Construction (CDL):
Shape(NDC,CN,ND)
Shape(NCF,FN,NC)
Shape(NF,FE,EN)
Shape(NED,DN,NE)
Shape(NFE,EN,NF)
Cocircular(N,DCFE)
Conditions (CDL):
Equal(LengthOfLine(CN),8)
IsCentreOfCircle(N,N)
Goal: Value(LengthOfLine(DN))
Answer: 8
Theorem sequence:
radius_of_circle_property_length_equal(1,ND,N)
radius_of_circle_property_length_equal(1,NC,N)
Problem: As shown in the diagram, AB=20, AC=10, BC=x, ∠ABC=y°, BC is perpendicular to AC. Find the value of y.
问题：如图所示，AB=20，AC=10，BC=x，∠ABC=y°，BC⊥AC。求y的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ABC),y)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 30
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, CB=40, DB=38, SD=28, quadrilateral ACBS is a parallelogram, BD is perpendicular to SD. Find the area of ACBS.
问题：如图所示，CB=40，DB=38，SD=28，四边形ACBS是平行四边形，BD垂直于SD。求四边形ACBS的面积。
Construction (CDL):
Shape(AC,CB,BS,SA)
Shape(BD,DS,SB)
Collinear(ASD)
Conditions (CDL):
Equal(LengthOfLine(CB),40)
Equal(LengthOfLine(DB),38)
Equal(LengthOfLine(SD),28)
Parallelogram(ACBS)
PerpendicularBetweenLine(BD,SD)
Goal: Value(AreaOfQuadrilateral(ACBS))
Answer: 1520
Theorem sequence:
altitude_of_quadrilateral_judgment_left_vertex(3,BD,BSAC)
parallelogram_property_opposite_line_equal(1,CBSA)
parallelogram_area_formula_common(1,BSAC)
Problem: As shown in the diagram, AB=CB, DA=5*x-6, DC=3*x+4, BD⊥AD. Find the length of line AC.
问题：如图所示，AB=CB，DA=5*x-6，DC=3*x+4，BD垂直于AD。求直线AC的长度。
Construction (CDL):
Shape(AB,BD,DA)
Shape(BC,CD,DB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(CB))
Equal(LengthOfLine(DA),5*x-6)
Equal(LengthOfLine(DC),3*x+4)
PerpendicularBetweenLine(BD,AD)
Goal: Value(LengthOfLine(AC))
Answer: 38
Theorem sequence:
adjacent_complementary_angle(1,CDB,BDA)
perpendicular_bisector_judgment_distance_equal(1,BD,CA)
line_addition(1,AD,DC)
Problem: As shown in the diagram, ∠ACB=29°, ∠CBA=x°, ∠DAB=4*x°. Find the value of x.
问题：如图所示，∠ACB=29°，∠CBA=x°，∠DAB=4*x°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Shape(DA,AB)
Collinear(DAC)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),29)
Equal(MeasureOfAngle(CBA),x)
Equal(MeasureOfAngle(DAB),4*x)
Goal: Value(x)
Answer: 29/3
Theorem sequence:
triangle_property_angle_sum(1,ACB)
adjacent_complementary_angle(1,DAB,BAC)
Problem: As shown in the diagram, NZ=9, XM=4, XN=6, NM∥ZY. Find the length of line XY.
问题：如图所示，NZ=9，XM=4，XN=6，NM平行于ZY。求直线XY的长度。
Construction (CDL):
Shape(MX,XN,NM)
Shape(MN,NZ,ZY,YM)
Collinear(XNZ)
Collinear(XMY)
Conditions (CDL):
Equal(LengthOfLine(NZ),9)
Equal(LengthOfLine(XM),4)
Equal(LengthOfLine(XN),6)
ParallelBetweenLine(NM,ZY)
Goal: Value(LengthOfLine(XY))
Answer: 10
Theorem sequence:
parallel_property_corresponding_angle(1,NM,ZY,X)
similar_triangle_judgment_aa(1,MXN,YXZ)
line_addition(1,XN,NZ)
similar_triangle_property_line_ratio(1,MXN,YXZ)
similar_triangle_property_line_ratio(1,NMX,ZYX)
Problem: As shown in the diagram, the measure of ⌒AHG is 78, GB is perpendicular to HB, JH is perpendicular to GH. Find the measure of ∠HGB.
问题：如图所示，⌒AHG的角度为78，GB垂直于HB，JH垂直于GH。求∠HGB的大小。
Construction (CDL):
Shape(AGF,FB,BG)
Shape(AGJ,JB,BF)
Shape(GB,BH,HG)
Shape(BA,AH,HB)
Shape(HA,AJ,JH)
Shape(AHG,GH)
Shape(AJH,HJ)
Collinear(FBH)
Collinear(GBAJ)
Cocircular(A,GFJH)
Conditions (CDL):
Equal(MeasureOfArc(AHG),78)
PerpendicularBetweenLine(GB,HB)
PerpendicularBetweenLine(JH,GH)
Goal: Value(MeasureOfAngle(HGB))
Answer: 51
Theorem sequence:
arc_property_circumference_angle_external(1,AHG,J)
triangle_property_angle_sum(1,GJH)
Problem: As shown in the diagram, AD=12, BD=4, AD⊥CD, DE⊥CE, EC⊥AC. Find the length of line DE.
问题：如图所示，AD=12，BD=4，AD垂直于CD，DE⊥CE，EC⊥AC。求直线DE的长度。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DE,EC)
Shape(ED,DB,BE)
Collinear(ADB)
Collinear(CEB)
Conditions (CDL):
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(BD),4)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(DE,CE)
PerpendicularBetweenLine(EC,AC)
Goal: Value(LengthOfLine(DE))
Answer: 2*sqrt(3)
Theorem sequence:
line_addition(1,AD,DB)
adjacent_complementary_angle(1,BED,DEC)
mirror_similar_triangle_judgment_aa(1,CAD,BCA)
mirror_similar_triangle_property_line_ratio(1,CAD,BCA)
mirror_similar_triangle_property_line_ratio(1,DCA,CAB)
similar_triangle_judgment_aa(1,DBE,ABC)
similar_triangle_property_line_ratio(1,BED,BCA)
similar_triangle_property_line_ratio(1,EDB,CAB)
Problem: As shown in the diagram, WZ=4, XW=4, ∠WYX=23°, WZ is perpendicular to YZ, YX is perpendicular to WX. Find the length of line XY.
问题：如图所示，WZ=4，XW=4，∠WYX=23°，WZ垂直于YZ，YX⊥WX。求直线XY的长度。
Construction (CDL):
Shape(XW,WY,YX)
Shape(WZ,ZY,YW)
Conditions (CDL):
Equal(LengthOfLine(WZ),4)
Equal(LengthOfLine(XW),4)
Equal(MeasureOfAngle(WYX),23)
PerpendicularBetweenLine(WZ,YZ)
PerpendicularBetweenLine(YX,WX)
Goal: Value(LengthOfLine(XY))
Answer: 4/tan(23*pi/180)
Theorem sequence:
triangle_property_angle_sum(1,XWY)
sine_theorem(1,XWY)
Problem: As shown in the diagram, ∠BEC=57°, DF is parallel to HE, FE∥DH. Find the measure of ∠IFA.
问题：如图所示，∠BEC=57°，DF∥HE，FE平行于DH。求∠IFA的大小。
Construction (CDL):
Shape(FD,DH,HE,EF)
Collinear(AFDL)
Collinear(BEHK)
Collinear(IFEC)
Collinear(GDHJ)
Shape(BE,EC)
Shape(IF,FA)
Conditions (CDL):
Equal(MeasureOfAngle(BEC),57)
ParallelBetweenLine(DF,HE)
ParallelBetweenLine(FE,DH)
Goal: Value(MeasureOfAngle(IFA))
Answer: 123
Theorem sequence:
vertical_angle(1,BEC,HEI)
vertical_angle(1,IFA,EFD)
parallel_property_ipsilateral_internal_angle(1,EH,FD)
Problem: As shown in the diagram, ∠NAP=120°, ∠PAQ=100°, the center of circle A is A. Find the measure of ∠QPN.
问题：如图所示，∠NAP=120°，∠PAQ=100°，A是⊙A的圆心。求∠QPN的大小。
Construction (CDL):
Shape(APN,NP)
Shape(AP,PN,NA)
Shape(PA,AQ,QP)
Shape(AQP,PQ)
Shape(ANQ,QA,AN)
Cocircular(A,PNQ)
Conditions (CDL):
Equal(MeasureOfAngle(NAP),120)
Equal(MeasureOfAngle(PAQ),100)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfAngle(QPN))
Answer: 70
Theorem sequence:
radius_of_circle_property_length_equal(1,AN,A)
radius_of_circle_property_length_equal(1,AP,A)
radius_of_circle_property_length_equal(1,AQ,A)
isosceles_triangle_judgment_line_equal(1,APN)
isosceles_triangle_judgment_line_equal(1,AQP)
isosceles_triangle_property_angle_equal(1,APN)
isosceles_triangle_property_angle_equal(1,AQP)
triangle_property_angle_sum(1,APN)
triangle_property_angle_sum(1,AQP)
angle_addition(1,QPA,APN)
Problem: As shown in the diagram, ∠UTV=47°, RU⊥SU, TV is perpendicular to UV, VS is perpendicular to RS. Find the measure of ∠USR.
问题：如图所示，∠UTV=47°，RU垂直于SU，TV⊥UV，VS垂直于RS。求∠USR的大小。
Construction (CDL):
Shape(RU,US,SR)
Shape(SU,UV,VS)
Shape(VU,UT,TV)
Collinear(RUT)
Collinear(SVT)
Conditions (CDL):
Equal(MeasureOfAngle(UTV),47)
PerpendicularBetweenLine(RU,SU)
PerpendicularBetweenLine(TV,UV)
PerpendicularBetweenLine(VS,RS)
Goal: Value(MeasureOfAngle(USR))
Answer: 47
Theorem sequence:
triangle_property_angle_sum(1,RUS)
triangle_property_angle_sum(1,TSR)
Problem: As shown in the diagram, FB=9, FC=6, FD=6, FE=x. Find the value of x.
问题：如图所示，FB=9，FC=6，FD=6，FE=x。求x的值。
Construction (CDL):
Shape(AEC,CF,FE)
Shape(ADE,EF,FD)
Shape(ACB,BF,FC)
Shape(ABD,DF,FB)
Collinear(CFD)
Collinear(EFAB)
Cocircular(A,ECBD)
Conditions (CDL):
Equal(LengthOfLine(FB),9)
Equal(LengthOfLine(FC),6)
Equal(LengthOfLine(FD),6)
Equal(LengthOfLine(FE),x)
Goal: Value(x)
Answer: 4
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,EFB,CFD,A)
Problem: As shown in the diagram, KR=RJ, KS=SL, LT=TJ, PT=2. Find the length of line KP.
问题：如图所示，KR=RJ，KS=SL，LT=TJ，PT=2。求直线KP的长度。
Construction (CDL):
Shape(KR,RP,PK)
Shape(PR,RJ,JP)
Shape(PJ,JT,TP)
Shape(PT,TL,LP)
Shape(PL,LS,SP)
Shape(PS,SK,KP)
Collinear(KRJ)
Collinear(JTL)
Collinear(LSK)
Collinear(KPT)
Collinear(RPL)
Collinear(JPS)
Conditions (CDL):
Equal(LengthOfLine(KR),LengthOfLine(RJ))
Equal(LengthOfLine(KS),LengthOfLine(SL))
Equal(LengthOfLine(LT),LengthOfLine(TJ))
Equal(LengthOfLine(PT),2)
Goal: Value(LengthOfLine(KP))
Answer: 4
Theorem sequence:
median_of_triangle_judgment(1,LR,LKJ)
median_of_triangle_judgment(1,KT,KJL)
centroid_of_triangle_judgment_intersection(1,P,JLK,T,R)
centroid_of_triangle_property_line_ratio(1,P,KJL,T)
Problem: As shown in the diagram, TR=ST, ∠STR=50°. Find the measure of ∠TRS.
问题：如图所示，TR=ST，∠STR=50°。求∠TRS的大小。
Construction (CDL):
Shape(RS,ST,TR)
Conditions (CDL):
Equal(LengthOfLine(TR),LengthOfLine(ST))
Equal(MeasureOfAngle(STR),50)
Goal: Value(MeasureOfAngle(TRS))
Answer: 65
Theorem sequence:
triangle_property_angle_sum(1,RST)
isosceles_triangle_judgment_line_equal(1,TRS)
isosceles_triangle_property_angle_equal(1,TRS)
Problem: As shown in the diagram, AC=9*x-1, AF=2*x+7, quadrilateral ADCB is a rectangle. Find the length of line AF.
问题：如图所示，AC=9*x-1，AF=2*x+7，四边形ADCB是矩形。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(AFC)
Collinear(DFB)
Conditions (CDL):
Equal(LengthOfLine(AC),9*x-1)
Equal(LengthOfLine(AF),2*x+7)
Rectangle(ADCB)
Goal: Value(LengthOfLine(AF))
Answer: 13
Theorem sequence:
line_addition(1,AF,FC)
parallelogram_property_diagonal_bisection(1,ADCB,F)
Problem: As shown in the diagram, LR=10, PM=Mul(LengthOfLine(KP),2), PR is parallel to KL, KN is perpendicular to MN, RM⊥PM. Find the length of line RM.
问题：如图所示，LR=10，PM=Mul(LengthOfLine(KP),2)，PR∥KL，KN垂直于MN，RM垂直于PM。求直线RM的长度。
Construction (CDL):
Shape(LR,RQ,QN,NL)
Shape(RM,MQ,QR)
Shape(QM,MP,PQ)
Shape(NQ,QP,PK,KN)
Collinear(LRM)
Collinear(MPK)
Collinear(LNK)
Collinear(RQP)
Collinear(NQM)
Conditions (CDL):
Equal(LengthOfLine(LR),10)
Equal(LengthOfLine(PM),Mul(LengthOfLine(KP),2))
ParallelBetweenLine(PR,KL)
PerpendicularBetweenLine(KN,MN)
PerpendicularBetweenLine(RM,PM)
Goal: Value(LengthOfLine(RM))
Answer: 20
Theorem sequence:
parallel_property_corresponding_angle(1,PR,KL,M)
similar_triangle_judgment_aa(1,RMP,LMK)
line_addition(1,MP,PK)
line_addition(1,MR,RL)
similar_triangle_property_line_ratio(1,RMP,LMK)
similar_triangle_property_line_ratio(1,PRM,KLM)
Problem: As shown in the diagram, LR=3, RW=6, TR=8, TS is parallel to LW. Find the length of line WS.
问题：如图所示，LR=3，RW=6，TR=8，TS平行于LW。求直线WS的长度。
Construction (CDL):
Shape(RW,WL,LR)
Shape(LW,WS,ST,TL)
Collinear(RLT)
Collinear(RWS)
Conditions (CDL):
Equal(LengthOfLine(LR),3)
Equal(LengthOfLine(RW),6)
Equal(LengthOfLine(TR),8)
ParallelBetweenLine(TS,LW)
Goal: Value(LengthOfLine(WS))
Answer: 10
Theorem sequence:
parallel_property_corresponding_angle(1,WL,ST,R)
similar_triangle_judgment_aa(1,LRW,TRS)
line_addition(1,RW,WS)
similar_triangle_property_line_ratio(1,LRW,TRS)
similar_triangle_property_line_ratio(1,WLR,STR)
Problem: As shown in the diagram, AC=AY, AD=x+2, CB=5/3*x+11, DB=3*y-9, DB=YD, YD=2*y+6. Find the value of x.
问题：如图所示，AC=AY，AD=x+2，CB=5/3*x+11，DB=3*y-9，DB=YD，YD=2*y+6。求x的值。
Construction (CDL):
Shape(CA,AD,DB,BC)
Shape(AY,YD,DA)
Collinear(CAY)
Collinear(BDY)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(AY))
Equal(LengthOfLine(AD),x+2)
Equal(LengthOfLine(CB),5/3*x+11)
Equal(LengthOfLine(DB),3*y-9)
Equal(LengthOfLine(DB),LengthOfLine(YD))
Equal(LengthOfLine(YD),2*y+6)
Goal: Value(x)
Answer: 21
Theorem sequence:
line_addition(1,CA,AY)
line_addition(1,BD,DY)
similar_triangle_judgment_sas(1,YDA,YBC)
similar_triangle_property_line_ratio(1,YDA,YBC)
similar_triangle_property_line_ratio(1,AYD,CYB)
Problem: As shown in the diagram, AB=2*x+1, BC=3*x-7, the tangent to ⊙O is BA, the tangent to ⊙O is BC. Find the value of x.
问题：如图所示，AB=2*x+1，BC=3*x-7，BA是圆O的切线，BC是圆O的切线。求x的值。
Construction (CDL):
Shape(OCA,CB,BA)
Shape(OCA,OAC)
Cocircular(O,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x+1)
Equal(LengthOfLine(BC),3*x-7)
IsTangentOfCircle(BA,O)
IsTangentOfCircle(BC,O)
Goal: Value(x)
Answer: 8
Theorem sequence:
tangent_of_circle_property_length_equal(1,BA,BC,O)
Problem: As shown in the diagram, ∠AVS=x°, ∠QVA=167°, ∠SVT=77°, ∠TVU=x°, ∠UVQ=26°. Find the value of x.
问题：如图所示，∠AVS=x°，∠QVA=167°，∠SVT=77°，∠TVU=x°，∠UVQ=26°。求x的值。
Construction (CDL):
Shape(VQU,UV,VQ)
Shape(VUT,TV,VU)
Shape(VTS,SV,VT)
Shape(VSA,AV,VS)
Shape(VAQ,QV,VA)
Cocircular(V,QUTSA)
Conditions (CDL):
Equal(MeasureOfAngle(AVS),x)
Equal(MeasureOfAngle(QVA),167)
Equal(MeasureOfAngle(SVT),77)
Equal(MeasureOfAngle(TVU),x)
Equal(MeasureOfAngle(UVQ),26)
Goal: Value(x)
Answer: 45
Theorem sequence:
angle_addition(1,UVQ,QVA)
round_angle(1,UVA,AVU)
angle_addition(1,SVT,TVU)
angle_addition(1,AVS,SVU)
Problem: As shown in the diagram, BC=4, BF=x, ∠ABD=30°, ∠CBE=30°, ∠DBF=30°, ∠EBA=30°, BA is perpendicular to EA, BD is perpendicular to AD, BE is perpendicular to CE, BF⊥DF. Find the value of x.
问题：如图所示，BC=4，BF=x，∠ABD=30°，∠CBE=30°，∠DBF=30°，∠EBA=30°，BA⊥EA，BD⊥AD，BE⊥CE，BF⊥DF。求x的值。
Construction (CDL):
Shape(CB,BE,EC)
Shape(EB,BA,AE)
Shape(AB,BD,DA)
Shape(DB,BF,FD)
Conditions (CDL):
Equal(LengthOfLine(BC),4)
Equal(LengthOfLine(BF),x)
Equal(MeasureOfAngle(ABD),30)
Equal(MeasureOfAngle(CBE),30)
Equal(MeasureOfAngle(DBF),30)
Equal(MeasureOfAngle(EBA),30)
PerpendicularBetweenLine(BA,EA)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(BE,CE)
PerpendicularBetweenLine(BF,DF)
Goal: Value(x)
Answer: 9/4
Theorem sequence:
triangle_property_angle_sum(1,BEC)
sine_theorem(1,BEC)
triangle_property_angle_sum(1,BAE)
sine_theorem(1,BAE)
triangle_property_angle_sum(1,BDA)
sine_theorem(1,BDA)
triangle_property_angle_sum(1,BFD)
sine_theorem(1,BFD)
Problem: As shown in the diagram, AB=x, AC=6, BC=15, BC⊥AC. Find the value of x.
问题：如图所示，AB=x，AC=6，BC=15，BC垂直于AC。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(BC),15)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 3*sqrt(29)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)