SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠ADC=23*x+12°, ∠CBA=21*x-8°, the center of ⊙G is G. Find the measure of ∠ADC.
问题：如图所示，∠ADC=23*x+12°，∠CBA=21*x-8°，⊙G的圆心为G。求∠ADC的大小。
Construction (CDL):
Shape(AB,GBA)
Shape(DA,GAD)
Shape(CD,GDC)
Shape(BC,GCB)
Shape(BA,AD,DC,CB)
Cocircular(G,ADCB)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),23*x+12)
Equal(MeasureOfAngle(CBA),21*x-8)
IsCentreOfCircle(G,G)
Goal: Value(MeasureOfAngle(ADC))
Answer: 104
Theorem sequence:
arc_property_circumference_angle_external(1,GAC,B)
arc_property_circumference_angle_internal(1,GAC,D)
Problem: As shown in the diagram, CF=4*x+1, DF=x+13, quadrilateral ADCB is a rectangle. Find the value of x.
问题：如图所示，CF=4*x+1，DF=x+13，ADCB是矩形。求x的值。
Construction (CDL):
Shape(BA,AF,FB)
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(BF,FC,CB)
Collinear(AFC)
Collinear(DFB)
Conditions (CDL):
Equal(LengthOfLine(CF),4*x+1)
Equal(LengthOfLine(DF),x+13)
Rectangle(ADCB)
Goal: Value(x)
Answer: 4
Theorem sequence:
line_addition(1,AF,FC)
line_addition(1,DF,FB)
rectangle_property_diagonal_equal(1,ADCB)
parallelogram_property_diagonal_bisection(1,ADCB,F)
parallelogram_property_diagonal_bisection(1,DCBA,F)
Problem: As shown in the diagram, ∠ABC=2*x°, ∠BCA=4*x°, ∠CAB=3*x°. Find the value of x.
问题：如图所示，∠ABC=2*x°，∠BCA=4*x°，∠CAB=3*x°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),2*x)
Equal(MeasureOfAngle(BCA),4*x)
Equal(MeasureOfAngle(CAB),3*x)
Goal: Value(x)
Answer: 20
Theorem sequence:
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, BC=4*x+8, DC=9*x-7, EC is the angle bisector of ∠BED, CD⊥ND, FB is perpendicular to CB. Find the length of line BC.
问题：如图所示，BC=4*x+8，DC=9*x-7，EC平分∠BED，CD⊥ND，FB垂直于CB。求直线BC的长度。
Construction (CDL):
Shape(ED,DC,CE)
Shape(EC,CB,BE)
Shape(BC,CA,AF,FB)
Shape(CD,DN,NA,AC)
Collinear(EDN)
Collinear(ECA)
Collinear(EBF)
Conditions (CDL):
Equal(LengthOfLine(BC),4*x+8)
Equal(LengthOfLine(DC),9*x-7)
IsBisectorOfAngle(EC,BED)
PerpendicularBetweenLine(CD,ND)
PerpendicularBetweenLine(FB,CB)
Goal: Value(LengthOfLine(BC))
Answer: 20
Theorem sequence:
adjacent_complementary_angle(1,EDC,CDN)
adjacent_complementary_angle(1,FBC,CBE)
bisector_of_angle_property_distance_equal(1,EC,BED)
Problem: As shown in the diagram, AB=3, AC=1, BC=x, ∠CAB=120°. Find the value of x.
问题：如图所示，AB=3，AC=1，BC=x，∠CAB=120°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),1)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(CAB),120)
Goal: Value(x)
Answer: sqrt(13)
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, AB=x, AC=9*sqrt(2), BC=9*sqrt(2), AC⊥BC. Find the value of x.
问题：如图所示，AB=x，AC=9*sqrt(2)，BC=9*sqrt(2)，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),9*sqrt(2))
Equal(LengthOfLine(BC),9*sqrt(2))
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 18
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, AC=32, BC=x, ∠BCA=53°, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AC=32，BC=x，∠BCA=53°，CA垂直于BA。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AC),32)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(BCA),53)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 32/sin(37*pi/180)
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,CAB)
Problem: As shown in the diagram, ∠CAD=2*x°, ∠DAE=3*x-20°, ∠EAB=x-4°. Find the measure of ∠EAB.
问题：如图所示，∠CAD=2*x°，∠DAE=3*x-20°，∠EAB=x-4°。求∠EAB的大小。
Construction (CDL):
Shape(CA,AD)
Shape(DA,AE)
Shape(EA,AB)
Shape(BA,AF)
Shape(FA,AC)
Collinear(CAB)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),2*x)
Equal(MeasureOfAngle(DAE),3*x-20)
Equal(MeasureOfAngle(EAB),x-4)
Goal: Value(MeasureOfAngle(EAB))
Answer: 30
Theorem sequence:
angle_addition(1,CDA,DAE)
adjacent_complementary_angle(1,CAE,EAB)
Problem: As shown in the diagram, DG=FG, FG=FH, ∠FDG=28°, △FGH is an isosceles △, triangleGFD is an isosceles triangle. Find the measure of ∠FGH.
问题：如图所示，DG=FG，FG=FH，∠FDG=28°，∠FGH和∠GHF是等腰△FGH的底角，∠GFD和∠FDG是等腰三角形GFD的底角。求∠FGH的大小。
Construction (CDL):
Shape(FD,DG,GF)
Shape(FG,GH,HF)
Collinear(DGH)
Conditions (CDL):
Equal(LengthOfLine(DG),LengthOfLine(FG))
Equal(LengthOfLine(FG),LengthOfLine(FH))
Equal(MeasureOfAngle(FDG),28)
IsoscelesTriangle(FGH)
IsoscelesTriangle(GFD)
Goal: Value(MeasureOfAngle(FGH))
Answer: 56
Theorem sequence:
isosceles_triangle_property_angle_equal(1,GFD)
triangle_property_angle_sum(1,FDG)
adjacent_complementary_angle(1,DGF,FGH)
Problem: As shown in the diagram, BE=BC, ED=7, BD is perpendicular to CD. Find the length of line EC.
问题：如图所示，BE=BC，ED=7，BD⊥CD。求直线EC的长度。
Construction (CDL):
Shape(EF,FD,DE)
Shape(FC,CD,DF)
Shape(ED,DB,BE)
Shape(DC,CB,BD)
Collinear(EDC)
Collinear(FDB)
Conditions (CDL):
Equal(LengthOfLine(BE),LengthOfLine(BC))
Equal(LengthOfLine(ED),7)
PerpendicularBetweenLine(BD,CD)
Goal: Value(LengthOfLine(EC))
Answer: 14
Theorem sequence:
line_addition(1,ED,DC)
adjacent_complementary_angle(1,EDB,BDC)
isosceles_triangle_judgment_line_equal(1,BEC)
altitude_of_triangle_judgment(1,BD,BEC)
isosceles_triangle_property_line_coincidence(1,BEC,D)
Problem: As shown in the diagram, AB=12, ∠CAB=x°, Sin(x)=1/2, BC⊥AC. Find the area of △ABC.
问题：如图所示，AB=12，∠CAB=x°，Sin(x)=1/2，BC垂直于AC。求三角形ABC的面积。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(MeasureOfAngle(CAB),x)
Equal(Sin(x),1/2)
PerpendicularBetweenLine(BC,AC)
Goal: Value(AreaOfTriangle(ABC))
Answer: 18*sqrt(3)
Theorem sequence:
sine_theorem(1,BCA)
triangle_property_angle_sum(1,ABC)
triangle_area_formula_sine(1,BCA)
Problem: As shown in the diagram, LM=2*y+6, LP=5*x-3, MP=4, ∠LMP=60°, ∠MPL=60°. Find the value of y.
问题：如图所示，LM=2*y+6，LP=5*x-3，MP=4，∠LMP=60°，∠MPL=60°。求y的值。
Construction (CDL):
Shape(MP,PL,LM)
Conditions (CDL):
Equal(LengthOfLine(LM),2*y+6)
Equal(LengthOfLine(LP),5*x-3)
Equal(LengthOfLine(MP),4)
Equal(MeasureOfAngle(LMP),60)
Equal(MeasureOfAngle(MPL),60)
Goal: Value(y)
Answer: -1
Theorem sequence:
triangle_property_angle_sum(1,MPL)
isosceles_triangle_judgment_angle_equal(1,MPL)
Problem: As shown in the diagram, ∠RTS=2*x°, ∠SRT=x°, B is the center of ⊙B, the diameter of ⊙B is RT. Find the measure of ∠RTS.
问题：如图所示，∠RTS=2*x°，∠SRT=x°，B是⊙B的圆心，圆B的直径为RT。求∠RTS的大小。
Construction (CDL):
Shape(RS,BSR)
Shape(SR,RB,BT,TS)
Shape(ST,BTS)
Shape(TB,BR,BRT)
Collinear(RBT)
Cocircular(B,SRT)
Conditions (CDL):
Equal(MeasureOfAngle(RTS),2*x)
Equal(MeasureOfAngle(SRT),x)
IsCentreOfCircle(B,B)
IsDiameterOfCircle(RT,B)
Goal: Value(MeasureOfAngle(RTS))
Answer: 60
Theorem sequence:
diameter_of_circle_property_right_angle(1,TSR,B)
triangle_property_angle_sum(1,SRT)
Problem: As shown in the diagram, ∠TSR=42°, S is the center of ⊙S. Find the measure of arc SET.
问题：如图所示，∠TSR=42°，S是圆S的圆心。求⌒SET的角度。
Construction (CDL):
Shape(SR,SRT,TS)
Shape(ST,STK,KS)
Shape(SK,SKE,ES)
Shape(SE,SER,RS)
Collinear(TSE)
Collinear(RSK)
Cocircular(S,RTKE)
Conditions (CDL):
Equal(MeasureOfAngle(TSR),42)
IsCentreOfCircle(S,S)
Goal: Value(MeasureOfArc(SET))
Answer: 180
Theorem sequence:
flat_angle(1,TSE)
arc_property_center_angle(1,SET,S)
Problem: As shown in the diagram, ∠VYZ=45°, the measure of ⌒AVZ is 10*x, A is the center of circle A, the tangent to circle A is YV. Find the value of x.
问题：如图所示，∠VYZ=45°，⌒AVZ的角度为10*x，A是圆A的圆心，圆O的切线为YV。求x的值。
Construction (CDL):
Shape(WV,VY,YW)
Shape(AW,AWV,VA)
Collinear(WAV)
Collinear(WZY)
Cocircular(A,ZWV)
Conditions (CDL):
Equal(MeasureOfAngle(VYZ),45)
Equal(MeasureOfArc(AVZ),10*x)
IsCentreOfCircle(A,A)
IsTangentOfCircle(YV,A)
Goal: Value(x)
Answer: 9
Theorem sequence:
flat_angle(1,VAW)
arc_property_center_angle(1,AWV,A)
circle_property_circular_power_tangent_and_segment_angle(1,YV,YZW,A)
Problem: As shown in the diagram, AB=13, AC=12, CB=15. Find the measure of ∠CBA.
问题：如图所示，AB=13，AC=12，CB=15。求∠CBA的大小。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(CB),15)
Goal: Value(MeasureOfAngle(CBA))
Answer: 180*acos(25/39)/pi
Theorem sequence:
cosine_theorem(1,BAC)
Problem: As shown in the diagram, AC=12, AS=x, BC=9, SB=7, SR=6, AC is perpendicular to BC, RS is perpendicular to AS. Find the length of line AB.
问题：如图所示，AC=12，AS=x，BC=9，SB=7，SR=6，AC垂直于BC，RS⊥AS。求直线AB的长度。
Construction (CDL):
Shape(SA,AR,RS)
Shape(RC,CB,BS,SR)
Collinear(ASB)
Collinear(ARC)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(AS),x)
Equal(LengthOfLine(BC),9)
Equal(LengthOfLine(SB),7)
Equal(LengthOfLine(SR),6)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(RS,AS)
Goal: Value(LengthOfLine(AB))
Answer: 15
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,RSA,BAC)
mirror_similar_triangle_property_line_ratio(1,ARS,ACB)
mirror_similar_triangle_property_line_ratio(1,RSA,BAC)
line_addition(1,AS,SB)
Problem: As shown in the diagram, AB=z, AC=y, AN=x, NB=12, NC=8, BA⊥NA, NC⊥AC. Find the value of y.
问题：如图所示，AB=z，AC=y，AN=x，NB=12，NC=8，BA⊥NA，NC⊥AC。求y的值。
Construction (CDL):
Shape(BA,AC,CB)
Shape(AN,NC,CA)
Collinear(BCN)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AN),x)
Equal(LengthOfLine(NB),12)
Equal(LengthOfLine(NC),8)
PerpendicularBetweenLine(BA,NA)
PerpendicularBetweenLine(NC,AC)
Goal: Value(y)
Answer: 4*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,NCA)
mirror_similar_triangle_judgment_aa(1,ANC,BAN)
mirror_similar_triangle_property_line_ratio(1,CAN,ANB)
mirror_similar_triangle_property_line_ratio(1,ANC,BAN)
right_triangle_property_pythagorean(1,NCA)
Problem: As shown in the diagram, LM=16, MS=SN, PN=4*x, QS=7, RS=7, the radius of circle S is equal to the length of line MS, the radius of ⊙S is equal to the length of line SN, S is the center of circle S, MQ is perpendicular to SQ, PR⊥SR. Find the value of x.
问题：如图所示，LM=16，MS=SN，PN=4*x，QS=7，RS=7，圆S的半径与直线MS的长度相等，圆S的半径与直线SN的长度相等，圆S的圆心为S，MQ⊥SQ，PR垂直于SR。求x的值。
Construction (CDL):
Shape(MQ,QS,SM)
Shape(SR,RN,NS)
Collinear(MQL)
Collinear(NRP)
Cocircular(S,MLPN)
Conditions (CDL):
Equal(LengthOfLine(LM),16)
Equal(LengthOfLine(MS),LengthOfLine(SN))
Equal(LengthOfLine(PN),4*x)
Equal(LengthOfLine(QS),7)
Equal(LengthOfLine(RS),7)
Equal(RadiusOfCircle(S),LengthOfLine(MS))
Equal(RadiusOfCircle(S),LengthOfLine(SN))
IsCentreOfCircle(S,S)
PerpendicularBetweenLine(MQ,SQ)
PerpendicularBetweenLine(PR,SR)
Goal: Value(x)
Answer: 4
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,S,SQ,ML)
line_addition(1,MQ,QL)
right_triangle_judgment_angle(1,MQS)
right_triangle_property_pythagorean(1,MQS)
circle_property_chord_perpendicular_bisect_chord(1,S,SR,PN)
line_addition(1,PR,RN)
right_triangle_judgment_angle(1,SRN)
right_triangle_property_pythagorean(1,SRN)
Problem: As shown in the diagram, BC=24, CE=18, DF=21, EB=12, triangle AFD is similar to triangle BEC.. Find the perimeter of triangle AFD.
问题：如图所示，BC=24，CE=18，DF=21，EB=12，三角形AFD与三角形BEC是相似三角形。求△AFD的周长。
Construction (CDL):
Shape(AF,FD,DC,CE,EB,BA)
Shape(BE,EC,CB)
Collinear(ABCD)
Conditions (CDL):
Equal(LengthOfLine(BC),24)
Equal(LengthOfLine(CE),18)
Equal(LengthOfLine(DF),21)
Equal(LengthOfLine(EB),12)
SimilarBetweenTriangle(AFD,BEC)
Goal: Value(PerimeterOfTriangle(AFD))
Answer: 63
Theorem sequence:
triangle_perimeter_formula(1,BEC)
similar_triangle_property_line_ratio(1,AFD,BEC)
similar_triangle_property_perimeter_ratio(1,AFD,BEC)
Problem: As shown in the diagram, ∠EDF=39°, ∠FBA=48°, CF∥DE, AF⊥BF, DC⊥FC, FE is perpendicular to EE. Find the measure of ∠BAF.
问题：如图所示，∠EDF=39°，∠FBA=48°，CF∥DE，AF⊥BF，DC⊥FC，FE垂直于EE。求∠BAF的大小。
Construction (CDL):
Shape(AF,FB,BA)
Shape(FE,ED,DF)
Shape(FD,DC,CF)
Collinear(FBC)
Collinear(AFE)
Conditions (CDL):
Equal(MeasureOfAngle(EDF),39)
Equal(MeasureOfAngle(FBA),48)
ParallelBetweenLine(CF,DE)
PerpendicularBetweenLine(AF,BF)
PerpendicularBetweenLine(DC,FC)
PerpendicularBetweenLine(FE,ED)
Goal: Value(MeasureOfAngle(BAF))
Answer: 42
Theorem sequence:
triangle_property_angle_sum(1,AFB)
Problem: As shown in the diagram, ∠CGA=x°, ∠DGC=36°. Find the value of x.
问题：如图所示，∠CGA=x°，∠DGC=36°。求x的值。
Construction (CDL):
Shape(GA,GAE,EG)
Shape(GE,GED,DG)
Shape(GD,GDC,CG)
Shape(GC,GCA,AG)
Collinear(EGC)
Collinear(AGD)
Cocircular(G,AEDC)
Conditions (CDL):
Equal(MeasureOfAngle(CGA),x)
Equal(MeasureOfAngle(DGC),36)
Goal: Value(x)
Answer: 144
Theorem sequence:
flat_angle(1,DGA)
angle_addition(1,DGC,CGA)
Problem: As shown in the diagram, ∠BGC=4*x+10°, ∠IHA=110°, IH∥CG. Find the value of x.
问题：如图所示，∠BGC=4*x+10°，∠IHA=110°，IH平行于CG。求x的值。
Construction (CDL):
Shape(AI,IH,HA)
Shape(HI,IC,CG,GH)
Shape(GC,CB,BG)
Collinear(AHGB)
Conditions (CDL):
Equal(MeasureOfAngle(BGC),4*x+10)
Equal(MeasureOfAngle(IHA),110)
ParallelBetweenLine(IH,CG)
Goal: Value(x)
Answer: 15
Theorem sequence:
parallel_property_corresponding_angle(1,GC,HI,B)
adjacent_complementary_angle(1,GHI,IHA)
Problem: As shown in the diagram, ∠BPY=3*Z°, the measure of arc PAC is 2*Z+15, the measure of ⌒PCY is 3*Z-3, the center of circle P is P. Find the measure of ⌒PCB.
问题：如图所示，∠BPY=3*Z°，弧PAC的角度为2*Z+15，弧PCY的角度为3*Z-3，P是圆P的圆心。求⌒PCB的角度。
Construction (CDL):
Shape(PY,PYB,BP)
Shape(PB,PBX,XP)
Shape(PX,PXA,AP)
Shape(PA,PAC,CP)
Shape(PC,PCY,YP)
Collinear(BPA)
Cocircular(P,YBXAC)
Conditions (CDL):
Equal(MeasureOfAngle(BPY),3*Z)
Equal(MeasureOfArc(PAC),2*Z+15)
Equal(MeasureOfArc(PCY),3*Z-3)
IsCentreOfCircle(P,P)
Goal: Value(MeasureOfArc(PCB))
Answer: 123
Theorem sequence:
arc_property_center_angle(1,PAC,P)
arc_property_center_angle(1,PCY,P)
flat_angle(1,BPA)
angle_addition(1,BPY,YPC)
angle_addition(1,BPC,CPA)
arc_property_center_angle(1,PCB,P)
Problem: As shown in the diagram, WT=x, XY=16, YW=20, ZT=10, ZT∥XY. Find the length of line TY.
问题：如图所示，WT=x，XY=16，YW=20，ZT=10，ZT∥XY。求直线TY的长度。
Construction (CDL):
Shape(WZ,ZT,TW)
Shape(ZX,XY,YT,TZ)
Collinear(WZX)
Collinear(WTY)
Conditions (CDL):
Equal(LengthOfLine(WT),x)
Equal(LengthOfLine(XY),16)
Equal(LengthOfLine(YW),20)
Equal(LengthOfLine(ZT),10)
ParallelBetweenLine(ZT,XY)
Goal: Value(LengthOfLine(TY))
Answer: 15/2
Theorem sequence:
parallel_property_corresponding_angle(1,ZT,XY,W)
similar_triangle_judgment_aa(1,TWZ,YWX)
similar_triangle_property_line_ratio(1,WZT,WXY)
similar_triangle_property_line_ratio(1,ZTW,XYW)
line_addition(1,WT,TY)
Problem: As shown in the diagram, ∠UZY=2*x+24°, ∠VZU=4*x°, ∠XZW=∠YZX, the center of ⊙Z is Z. Find the measure of arc ZXY.
问题：如图所示，∠UZY=2*x+24°，∠VZU=4*x°，∠XZW=∠YZX，⊙Z的圆心为Z。求弧ZXY的角度。
Construction (CDL):
Shape(ZV,ZVW,WZ)
Shape(ZW,ZWX,XZ)
Shape(ZX,ZXY,YZ)
Shape(ZY,ZYU,UZ)
Shape(ZU,ZUV,VZ)
Collinear(WZU)
Collinear(VZY)
Cocircular(Z,VWXYU)
Conditions (CDL):
Equal(MeasureOfAngle(UZY),2*x+24)
Equal(MeasureOfAngle(VZU),4*x)
Equal(MeasureOfAngle(XZW),MeasureOfAngle(YZX))
IsCentreOfCircle(Z,Z)
Goal: Value(MeasureOfArc(ZXY))
Answer: 52
Theorem sequence:
flat_angle(1,VZY)
angle_addition(1,VZU,UZY)
vertical_angle(1,VZU,YZW)
angle_addition(1,YZX,XZW)
arc_property_center_angle(1,ZXY,Z)
Problem: As shown in the diagram, PY=2*c-1, YR=4*c-11, ZP=a+5, ZQ=3*a-11, ∠PRZ=4*b-17°, ∠QYR=7*b+6°, ∠RXP=2*a+10°, ∠ZRQ=3*b-4°, QY is the perpendicular bisector of PR. Find the value of b.
问题：如图所示，PY=2*c-1，YR=4*c-11，ZP=a+5，ZQ=3*a-11，∠PRZ=4*b-17°，∠QYR=7*b+6°，∠RXP=2*a+10°，∠ZRQ=3*b-4°，QY垂直平分AB。求b的值。
Construction (CDL):
Shape(PB,BZ,ZP)
Shape(ZB,BC,CQ,QZ)
Shape(CX,XQ,QC)
Shape(BP,PY,YA,AB)
Shape(CB,BA,AC)
Shape(CA,AR,RX,XC)
Shape(AY,YR,RA)
Collinear(PZQ)
Collinear(PBCX)
Collinear(PYR)
Collinear(QXR)
Collinear(QCAY)
Collinear(ZBAR)
Conditions (CDL):
Equal(LengthOfLine(PY),2*c-1)
Equal(LengthOfLine(YR),4*c-11)
Equal(LengthOfLine(ZP),a+5)
Equal(LengthOfLine(ZQ),3*a-11)
Equal(MeasureOfAngle(PRZ),4*b-17)
Equal(MeasureOfAngle(QYR),7*b+6)
Equal(MeasureOfAngle(RXP),2*a+10)
Equal(MeasureOfAngle(ZRQ),3*b-4)
IsPerpendicularBisectorOfLine(QY,PR)
Goal: Value(b)
Answer: 12
Theorem sequence:
Problem: As shown in the diagram, ∠ABD=28°, ∠DBE=x°. Find the value of x.
问题：如图所示，∠ABD=28°，∠DBE=x°。求x的值。
Construction (CDL):
Shape(BD,BDA,AB)
Shape(BA,BAC,CB)
Shape(BC,BCE,EB)
Shape(BE,BED,DB)
Collinear(DBC)
Collinear(ABE)
Cocircular(B,DACE)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),28)
Equal(MeasureOfAngle(DBE),x)
Goal: Value(x)
Answer: 152
Theorem sequence:
flat_angle(1,DBC)
vertical_angle(1,ABD,EBC)
angle_addition(1,DBE,EBC)
Problem: As shown in the diagram, EF=3*x+4, PQ=20, P bisects segment DE, Q is the midpoint of segment DF. Find the value of x.
问题：如图所示，EF=3*x+4，PQ=20，P是线段DE的中点，Q平分线段DF。求x的值。
Construction (CDL):
Shape(QD,DP,PQ)
Shape(FQ,QP,PE,EF)
Collinear(DQF)
Collinear(DPE)
Conditions (CDL):
Equal(LengthOfLine(EF),3*x+4)
Equal(LengthOfLine(PQ),20)
IsMidpointOfLine(P,DE)
IsMidpointOfLine(Q,DF)
Goal: Value(x)
Answer: 12
Theorem sequence:
line_addition(1,DP,PE)
line_addition(1,DQ,QF)
similar_triangle_judgment_sas(1,DPQ,DEF)
similar_triangle_property_line_ratio(1,PQD,EFD)
similar_triangle_property_line_ratio(1,DPQ,DEF)
Problem: As shown in the diagram, BA=BC, ∠CBA=60°. Find the measure of ∠BAC.
问题：如图所示，BA=BC，∠CBA=60°。求∠BAC的大小。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(BA),LengthOfLine(BC))
Equal(MeasureOfAngle(CBA),60)
Goal: Value(MeasureOfAngle(BAC))
Answer: 60
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,BAC)
isosceles_triangle_property_angle_equal(1,BAC)
triangle_property_angle_sum(1,BAC)
Problem: As shown in the diagram, BC=14, ∠BCD=55°, ∠CDB is the right angle of the right triangle CDB. Find the perimeter of △CDB.
问题：如图所示，BC=14，∠BCD=55°，CD和DB是直角三角形CDB的两个直角边。求△CDB的周长。
Construction (CDL):
Shape(CD,DB,BC)
Conditions (CDL):
Equal(LengthOfLine(BC),14)
Equal(MeasureOfAngle(BCD),55)
RightTriangle(CDB)
Goal: Value(PerimeterOfTriangle(CDB))
Answer: 14+14*sqrt(2)*cos(pi/18)
Theorem sequence:
sine_theorem(1,BCD)
right_triangle_property_pythagorean(1,CDB)
triangle_perimeter_formula(1,CDB)
Problem: As shown in the diagram, AB=x+3, AD=6, BC=x, CD=4, ∠BAC=∠CAD. Find the value of x.
问题：如图所示，AB=x+3，AD=6，BC=x，CD=4，∠BAC=∠CAD。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Shape(DC,CA,AD)
Collinear(BCD)
Conditions (CDL):
Equal(LengthOfLine(AB),x+3)
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CD),4)
Equal(MeasureOfAngle(BAC),MeasureOfAngle(CAD))
Goal: Value(x)
Answer: 6
Theorem sequence:
bisector_of_angle_judgment_angle_equal(1,AC,BAD)
bisector_of_angle_property_line_ratio(1,AC,BAD)
Problem: As shown in the diagram, ∠ADG=36°, ∠AGF=104°, ∠EFC=40°, AB is perpendicular to BB. Find the measure of ∠GBF.
问题：如图所示，∠ADG=36°，∠AGF=104°，∠EFC=40°，AB垂直于BB。求∠GBF的大小。
Construction (CDL):
Shape(AD,DG,GA)
Shape(GB,BF,FG)
Shape(FB,BC,CF)
Shape(GF,FE)
Shape(EF,FC)
Shape(AG,GF)
Collinear(DGFC)
Collinear(AGB)
Collinear(BFE)
Conditions (CDL):
Equal(MeasureOfAngle(ADG),36)
Equal(MeasureOfAngle(AGF),104)
Equal(MeasureOfAngle(EFC),40)
PerpendicularBetweenLine(AB,BC)
Goal: Value(MeasureOfAngle(GBF))
Answer: 64
Theorem sequence:
vertical_angle(1,BFG,EFC)
adjacent_complementary_angle(1,AGF,FGB)
triangle_property_angle_sum(1,GBF)
Problem: As shown in the diagram, ST=6, WX=5, the perimeter of △SRT is 15, triangle WZX is similar to triangle SRT.. Find the perimeter of triangle WZX.
问题：如图所示，ST=6，WX=5，△SRT的周长为15，三角形WZX相似于三角形SRT。求△WZX的周长。
Construction (CDL):
Shape(WZ,ZX,XW)
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(ST),6)
Equal(LengthOfLine(WX),5)
Equal(PerimeterOfTriangle(SRT),15)
SimilarBetweenTriangle(WZX,SRT)
Goal: Value(PerimeterOfTriangle(WZX))
Answer: 25/2
Theorem sequence:
similar_triangle_property_line_ratio(1,ZXW,RTS)
similar_triangle_property_perimeter_ratio(1,WZX,SRT)
Problem: As shown in the diagram, VW=72, XV=75, XW=21, VW is perpendicular to XW. Find the value of sin(XV).
问题：如图所示，VW=72，XV=75，XW=21，VW⊥XW。求sin(XV)的值。
Construction (CDL):
Shape(VW,WX,XV)
Conditions (CDL):
Equal(LengthOfLine(VW),72)
Equal(LengthOfLine(XV),75)
Equal(LengthOfLine(XW),21)
PerpendicularBetweenLine(VW,XW)
Goal: Value(Sin(MeasureOfAngle(XVW)))
Answer: 7/25
Theorem sequence:
sine_theorem(1,XVW)
Problem: As shown in the diagram, ∠AGC=60°, DG⊥AG. Find the measure of ∠CGB.
问题：如图所示，∠AGC=60°，DG⊥AG。求∠CGB的大小。
Construction (CDL):
Shape(GB,GBC,CG)
Shape(GC,GCA,AG)
Shape(GA,GAD,DG)
Shape(GD,GDE,EG)
Shape(GE,GEB,BG)
Collinear(AGB)
Collinear(CGE)
Cocircular(G,CADEB)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),60)
PerpendicularBetweenLine(DG,AG)
Goal: Value(MeasureOfAngle(CGB))
Answer: 120
Theorem sequence:
flat_angle(1,EGC)
angle_addition(1,DGA,AGC)
angle_addition(1,EGD,DGC)
angle_addition(1,EGD,DGA)
vertical_angle(1,CGB,EGA)
Problem: As shown in the diagram, WX=8, YW=8, ∠WXY=62°, ∠XYW=4*x-2°. Find the value of x.
问题：如图所示，WX=8，YW=8，∠WXY=62°，∠XYW=4*x-2°。求x的值。
Construction (CDL):
Shape(WX,XY,YW)
Conditions (CDL):
Equal(LengthOfLine(WX),8)
Equal(LengthOfLine(YW),8)
Equal(MeasureOfAngle(WXY),62)
Equal(MeasureOfAngle(XYW),4*x-2)
Goal: Value(x)
Answer: 16
Theorem sequence:
sine_theorem(1,WXY)
Problem: As shown in the diagram, JN=14-x, KL=3*x+2*y, MK=6*x, LK and JM are opposite sides of the parallelogram LJMK, quadrilateral PNML is a rectangle. Find the value of x.
问题：如图所示，JN=14-x，KL=3*x+2*y，MK=6*x，JL和MK是▱LJMK的一组对边，四边形PNML是矩形。求x的值。
Construction (CDL):
Shape(PN,NJ,JP)
Shape(NM,MJ,JN)
Shape(JM,ML,LJ)
Shape(PJ,JL,LP)
Shape(LM,MK,KL)
Collinear(PJM)
Collinear(LJN)
Conditions (CDL):
Equal(LengthOfLine(JN),14-x)
Equal(LengthOfLine(KL),3*x+2*y)
Equal(LengthOfLine(MK),6*x)
Parallelogram(LJMK)
Rectangle(PNML)
Goal: Value(x)
Answer: 2
Theorem sequence:
parallelogram_property_diagonal_bisection(1,NMLP,J)
parallelogram_property_opposite_line_equal(1,LJMK)
Problem: As shown in the diagram, ∠UEN=64°, HD∥EU, IL∥BN. Find the measure of ∠BHD.
问题：如图所示，∠UEN=64°，HD平行于EU，IL∥BN。求∠BHD的大小。
Construction (CDL):
Shape(DH,HE,EU,UD)
Shape(WD,DU,UA,AW)
Shape(HJ,JK,KE,EH)
Shape(UE,EN,NL,LU)
Shape(DI,IB,BH,HD)
Shape(WI,ID,DW)
Shape(HB,BJ,JH)
Shape(AU,UL,LA)
Shape(EK,KN,NE)
Collinear(IDUQTSL)
Collinear(BHEN)
Collinear(JHDW)
Collinear(KEUGA)
Conditions (CDL):
Equal(MeasureOfAngle(UEN),64)
ParallelBetweenLine(HD,EU)
ParallelBetweenLine(IL,BN)
Goal: Value(MeasureOfAngle(BHD))
Answer: 116
Theorem sequence:
parallel_property_corresponding_angle(2,HD,EU,N)
flat_angle(1,BHE)
angle_addition(1,BHD,DHE)
Problem: As shown in the diagram, ∠CFD=12*a+45°, CF⊥DF. Find the value of a.
问题：如图所示，∠CFD=12*a+45°，CF垂直于DF。求a的值。
Construction (CDL):
Shape(CF,FD,DC)
Shape(DF,FE,ED)
Shape(EF,FA,AE)
Shape(FB,BA,AF)
Shape(BF,FC,CB)
Collinear(BFE)
Collinear(AFD)
Conditions (CDL):
Equal(MeasureOfAngle(CFD),12*a+45)
PerpendicularBetweenLine(CF,DF)
Goal: Value(a)
Answer: 15/4
Theorem sequence:
Problem: As shown in the diagram, ∠BNC=20°, the center of circle N is N. Find the measure of arc NCB.
问题：如图所示，∠BNC=20°，N是⊙N的圆心。求⌒NCB的角度。
Construction (CDL):
Shape(NC,NCB,BN)
Shape(NB,NBA,AN)
Shape(NA,NAE,ED,DN)
Shape(ND,NDC,CN)
Collinear(AND)
Cocircular(N,CBAED)
Conditions (CDL):
Equal(MeasureOfAngle(BNC),20)
IsCentreOfCircle(N,N)
Goal: Value(MeasureOfArc(NCB))
Answer: 20
Theorem sequence:
arc_property_center_angle(1,NCB,N)
Problem: As shown in the diagram, ∠CFT=5*x-20°, FB⊥NB, TF is perpendicular to BF. Find the value of x.
问题：如图所示，∠CFT=5*x-20°，FB⊥NB，TF⊥BF。求x的值。
Construction (CDL):
Shape(CF,FT,TC)
Shape(AF,FC,CA)
Shape(FB,BN,NT,TF)
Shape(AD,DB,BF,FA)
Shape(BE,EN,NB)
Shape(DE,EB,BD)
Collinear(AGFT)
Collinear(DBN)
Collinear(CFBE)
Conditions (CDL):
Equal(MeasureOfAngle(CFT),5*x-20)
PerpendicularBetweenLine(FB,NB)
PerpendicularBetweenLine(TF,BF)
Goal: Value(x)
Answer: 22
Theorem sequence:
flat_angle(1,CFB)
angle_addition(1,CFT,TFB)
Problem: As shown in the diagram, AE=x, ED=2, ∠AED=15°, ∠CAD=70°, DC is perpendicular to AC. Find the value of x.
问题：如图所示，AE=x，ED=2，∠AED=15°，∠CAD=70°，DC⊥AC。求x的值。
Construction (CDL):
Shape(AE,ED,DA)
Shape(AD,DC,CA)
Collinear(EDC)
Conditions (CDL):
Equal(LengthOfLine(AE),x)
Equal(LengthOfLine(ED),2)
Equal(MeasureOfAngle(AED),15)
Equal(MeasureOfAngle(CAD),70)
PerpendicularBetweenLine(DC,AC)
Goal: Value(x)
Answer: 2*sin(pi/9)/sin(pi/36)
Theorem sequence:
triangle_property_angle_sum(1,CAE)
angle_addition(1,CAD,DAE)
triangle_property_angle_sum(1,EDA)
sine_theorem(1,EDA)