SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AC=5, CD=9, ∠DCA=60°, DB and CA are opposite sides of the parallelogram DCAB, CA is perpendicular to DA. Find the perimeter of DCAB.
问题：如图所示，AC=5，CD=9，∠DCA=60°，DCAB是▱，CA垂直于DA。求四边形DCAB的周长。
Construction (CDL):
Shape(DC,CA,AD)
Shape(DA,AB,BD)
Conditions (CDL):
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(CD),9)
Equal(MeasureOfAngle(DCA),60)
Parallelogram(DCAB)
PerpendicularBetweenLine(CA,DA)
Goal: Value(PerimeterOfQuadrilateral(DCAB))
Answer: 28
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DCAB)
parallelogram_property_opposite_line_equal(1,CABD)
quadrilateral_perimeter_formula(1,DCAB)
Problem: As shown in the diagram, AB=3*y-8, AD=13, CB=2*x+7, CD=10, ∠ADF=59°, ∠BFC=49°, ∠FBA=20°, DA and CB are opposite sides of the ▱ ADCB. Find the measure of ∠AFB.
问题：如图所示，AB=3*y-8，AD=13，CB=2*x+7，CD=10，∠ADF=59°，∠BFC=49°，∠FBA=20°，四边形ADCB是▱。求∠AFB的大小。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(BF,FC,CB)
Shape(AF,FB,BA)
Collinear(AFC)
Collinear(BFD)
Conditions (CDL):
Equal(LengthOfLine(AB),3*y-8)
Equal(LengthOfLine(AD),13)
Equal(LengthOfLine(CB),2*x+7)
Equal(LengthOfLine(CD),10)
Equal(MeasureOfAngle(ADF),59)
Equal(MeasureOfAngle(BFC),49)
Equal(MeasureOfAngle(FBA),20)
Parallelogram(ADCB)
Goal: Value(MeasureOfAngle(AFB))
Answer: 131
Theorem sequence:
adjacent_complementary_angle(1,AFB,BFC)
Problem: As shown in the diagram, ∠ADF=18*x-8°, ∠FBA=70-4*x°, ADCB is a rectangle. Find the measure of ∠ADF.
问题：如图所示，∠ADF=18*x-8°，∠FBA=70-4*x°，四边形ADCB是矩形。求∠ADF的大小。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(FC,CB,BF)
Shape(AF,FB,BA)
Collinear(AFC)
Collinear(DFB)
Conditions (CDL):
Equal(MeasureOfAngle(ADF),18*x-8)
Equal(MeasureOfAngle(FBA),70-4*x)
Rectangle(ADCB)
Goal: Value(MeasureOfAngle(ADF))
Answer: 28
Theorem sequence:
triangle_property_angle_sum(1,ADB)
Problem: As shown in the diagram, AC=9, AY=18, PO=x, PQ=y, RO=14, YB=21, PORQ is similar to AYBC. Find the value of x.
问题：如图所示，AC=9，AY=18，PO=x，PQ=y，RO=14，YB=21，PORQ相似于AYBC。求x的值。
Construction (CDL):
Shape(QP,PO,OR,RQ)
Shape(AY,YB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),9)
Equal(LengthOfLine(AY),18)
Equal(LengthOfLine(PO),x)
Equal(LengthOfLine(PQ),y)
Equal(LengthOfLine(RO),14)
Equal(LengthOfLine(YB),21)
SimilarBetweenQuadrilateral(PORQ,AYBC)
Goal: Value(x)
Answer: 12
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,PORQ,AYBC)
similar_quadrilateral_property_line_ratio(1,ORQP,YBCA)
Problem: As shown in the diagram, ∠AGK=60°, ∠MJB=50°, HA∥CG, MH is parallel to EL, WD∥BT. Find the measure of ∠LNQ.
问题：如图所示，∠AGK=60°，∠MJB=50°，HA平行于CG，MH平行于EL，WD平行于BT。求∠LNQ的大小。
Construction (CDL):
Shape(HC,CP,PN,NH)
Shape(SJ,JF,FO,OS)
Shape(RG,GA,AQ,QR)
Shape(WS)
Shape(BJ)
Shape(MJ)
Shape(EF)
Shape(IC)
Shape(NL)
Shape(CS,SO,OP,PC)
Shape(PO,OR,RP)
Shape(NP,PR,RQ,QN)
Shape(OF,FG,GR,RO)
Shape(QD)
Shape(AT)
Shape(AU)
Shape(GK)
Shape(MJ,JB)
Shape(EF,FJ)
Shape(AG,GK)
Shape(NQ,QD)
Shape(LN,NQ)
Collinear(WSORQD)
Collinear(BJFGAT)
Collinear(HCSJM)
Collinear(LNPOFE)
Collinear(ICPRGK)
Collinear(HNQAU)
Conditions (CDL):
Equal(MeasureOfAngle(AGK),60)
Equal(MeasureOfAngle(MJB),50)
ParallelBetweenLine(HA,CG)
ParallelBetweenLine(MH,EL)
ParallelBetweenLine(WD,BT)
Goal: Value(MeasureOfAngle(LNQ))
Answer: 110
Theorem sequence:
parallel_property_collinear_extend(3,WD,BT,O)
parallel_property_collinear_extend(3,OD,BT,R)
parallel_property_collinear_extend(3,TB,RO,F)
parallel_property_collinear_extend(3,TF,RO,G)
parallel_property_collinear_extend(3,MH,EL,S)
parallel_property_collinear_extend(3,MS,EL,J)
parallel_property_collinear_extend(3,LE,SJ,O)
parallel_property_collinear_extend(3,OE,SJ,F)
parallel_property_collinear_extend(3,HA,CG,N)
parallel_property_collinear_extend(3,GC,AN,P)
vertical_angle(1,AGK,FGR)
vertical_angle(1,MJB,SJF)
parallel_property_corresponding_angle(2,JS,FO,G)
parallel_property_corresponding_angle(2,GF,RO,P)
parallel_property_corresponding_angle(1,OR,FG,P)
triangle_property_angle_sum(1,POR)
parallel_property_corresponding_angle(2,NA,PG,O)
adjacent_complementary_angle(1,LNQ,QNP)
Problem: As shown in the diagram, ∠AEP=3*x-50°, ∠CDG=2*x-5°, BP∥GF. Find the value of x.
问题：如图所示，∠AEP=3*x-50°，∠CDG=2*x-5°，BP平行于GF。求x的值。
Construction (CDL):
Shape(AE)
Shape(ED)
Shape(DC)
Shape(PE)
Shape(EB)
Shape(FD)
Shape(DG)
Shape(AE,EP)
Shape(PE,ED)
Shape(ED,DF)
Shape(FD,DC)
Shape(BE,EA)
Shape(DE,EB)
Shape(GD,DE)
Shape(CD,DG)
Collinear(AEDC)
Collinear(PEB)
Collinear(FDG)
Conditions (CDL):
Equal(MeasureOfAngle(AEP),3*x-50)
Equal(MeasureOfAngle(CDG),2*x-5)
ParallelBetweenLine(BP,GF)
Goal: Value(x)
Answer: 45
Theorem sequence:
parallel_property_collinear_extend(3,BP,GF,E)
parallel_property_collinear_extend(3,FG,PE,D)
parallel_property_corresponding_angle(1,EP,DF,A)
vertical_angle(1,EDF,CDG)
Problem: As shown in the diagram, AB=x, AC=8, ∠ACB=67°, ∠CBA=73°. Find the value of x.
问题：如图所示，AB=x，AC=8，∠ACB=67°，∠CBA=73°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),8)
Equal(MeasureOfAngle(ACB),67)
Equal(MeasureOfAngle(CBA),73)
Goal: Value(x)
Answer: 8*sin(67*pi/180)/sin(73*pi/180)
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, RA=14*x-34, TA=12*x-8, QR and TS are opposite sides of the ▱ QTSR. Find the value of x.
问题：如图所示，RA=14*x-34，TA=12*x-8，TQ和SR是平行四边形QTSR的一组对边。求x的值。
Construction (CDL):
Shape(QT,TA,AQ)
Shape(AT,TS,SA)
Shape(AS,SR,RA)
Shape(QA,AR,RQ)
Collinear(QAS)
Collinear(TAR)
Conditions (CDL):
Equal(LengthOfLine(RA),14*x-34)
Equal(LengthOfLine(TA),12*x-8)
Parallelogram(QTSR)
Goal: Value(x)
Answer: 13
Theorem sequence:
parallelogram_property_diagonal_bisection(1,TSRQ,A)
Problem: As shown in the diagram, AC=2*x, CD=8, DY=YF, EF=9, EY=x, XD=DA, ∠AXC=∠FDE, ∠CAX=∠EFD. Find the value of x.
问题：如图所示，AC=2*x，CD=8，DY=YF，EF=9，EY=x，XD=DA，∠AXC=∠FDE，∠CAX=∠EFD。求x的值。
Construction (CDL):
Shape(XC,CD,DX)
Shape(DC,CA,AD)
Shape(DE,EY,YD)
Shape(YE,EF,FY)
Collinear(DYF)
Collinear(XDA)
Conditions (CDL):
Equal(LengthOfLine(AC),2*x)
Equal(LengthOfLine(CD),8)
Equal(LengthOfLine(DY),LengthOfLine(YF))
Equal(LengthOfLine(EF),9)
Equal(LengthOfLine(EY),x)
Equal(LengthOfLine(XD),LengthOfLine(DA))
Equal(MeasureOfAngle(AXC),MeasureOfAngle(FDE))
Equal(MeasureOfAngle(CAX),MeasureOfAngle(EFD))
Goal: Value(x)
Answer: 6
Theorem sequence:
similar_triangle_judgment_aa(1,EFD,CAX)
similar_triangle_property_line_ratio(1,DEF,XCA)
similar_triangle_property_line_ratio(1,EFD,CAX)
line_addition(1,AD,DX)
line_addition(1,DY,YF)
similar_triangle_judgment_sas(1,FYE,ADC)
similar_triangle_property_line_ratio(1,FYE,ADC)
similar_triangle_property_line_ratio(1,YEF,DCA)
Problem: As shown in the diagram, ∠DOB=∠AOF, ∠DOB=∠EOD, ∠DOB=∠FOE, the center of ⊙O is O, CO⊥AO. Find the measure of arc OBE.
问题：如图所示，∠DOB=∠AOF，∠DOB=∠EOD，∠DOB=∠FOE，圆O的圆心为O，CO⊥AO。求⌒OBE的角度。
Construction (CDL):
Shape(ODE,EO,OD)
Shape(OEF,FO,OE)
Shape(OFA,AO,OF)
Shape(OAC,CO,OA)
Shape(OCB,BO,OC)
Shape(OBD,DO,OB)
Collinear(AOB)
Collinear(EOC)
Cocircular(O,ACBDEF)
Conditions (CDL):
Equal(MeasureOfAngle(DOB),MeasureOfAngle(AOF))
Equal(MeasureOfAngle(DOB),MeasureOfAngle(EOD))
Equal(MeasureOfAngle(DOB),MeasureOfAngle(FOE))
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CO,AO)
Goal: Value(MeasureOfArc(OBE))
Answer: 90
Theorem sequence:
vertical_angle(1,EOB,COA)
arc_property_center_angle(1,OBE,O)
Problem: As shown in the diagram, TY=15, XY=WZ, ZX=20, WX is parallel to ZY. Find the length of line WT.
问题：如图所示，TY=15，XY=WZ，ZX=20，WX∥ZY。求直线WT的长度。
Construction (CDL):
Shape(WT,TX,XW)
Shape(WZ,ZT,TW)
Shape(TZ,ZY,YT)
Shape(XT,TY,YX)
Collinear(WTY)
Collinear(XTZ)
Conditions (CDL):
Equal(LengthOfLine(TY),15)
Equal(LengthOfLine(XY),LengthOfLine(WZ))
Equal(LengthOfLine(ZX),20)
ParallelBetweenLine(WX,ZY)
Goal: Value(LengthOfLine(WT))
Answer: 5
Theorem sequence:
line_addition(1,WT,TY)
trapezoid_judgment_parallel(1,WZYX)
isosceles_trapezoid_judgment_line_equal(1,WZYX)
isosceles_trapezoid_property_diagonal_equal(1,WZYX)
Problem: As shown in the diagram, BA=7, ED=7, the measure of arc OBA is 105, the measure of arc OED is 5*x. Find the value of x.
问题：如图所示，BA=7，ED=7，⌒OBA的角度为105，⌒OED的角度为5*x。求x的值。
Construction (CDL):
Shape(OBA,AB)
Shape(BA,OAE,ED,ODB)
Shape(DE,OED)
Cocircular(O,AEDB)
Conditions (CDL):
Equal(LengthOfLine(BA),7)
Equal(LengthOfLine(ED),7)
Equal(MeasureOfArc(OBA),105)
Equal(MeasureOfArc(OED),5*x)
Goal: Value(x)
Answer: 21
Theorem sequence:
congruent_arc_judgment_chord_equal(1,OBA,OED)
congruent_arc_property_measure_equal(1,OBA,OED)
Problem: As shown in the diagram, PQ=2*x+1, PT=10, RQ=6, ST=4, QT is parallel to RS, SR is parallel to TQ. Find the length of line PQ.
问题：如图所示，PQ=2*x+1，PT=10，RQ=6，ST=4，QT平行于RS，SR平行于TQ。求直线PQ的长度。
Construction (CDL):
Shape(RS,ST,TQ,QR)
Shape(QT,TP,PQ)
Collinear(PTS)
Collinear(PQR)
Conditions (CDL):
Equal(LengthOfLine(PQ),2*x+1)
Equal(LengthOfLine(PT),10)
Equal(LengthOfLine(RQ),6)
Equal(LengthOfLine(ST),4)
ParallelBetweenLine(QT,RS)
ParallelBetweenLine(SR,TQ)
Goal: Value(LengthOfLine(PQ))
Answer: 15
Theorem sequence:
parallel_property_corresponding_angle(1,QT,RS,P)
parallel_property_corresponding_angle(2,SR,TQ,P)
similar_triangle_judgment_aa(1,PRS,PQT)
line_addition(1,RQ,QP)
line_addition(1,ST,TP)
similar_triangle_property_line_ratio(1,SPR,TPQ)
similar_triangle_property_line_ratio(1,RSP,QTP)
Problem: As shown in the diagram, CA=10, CI=15, ∠ACI=45°, BI and AC are opposite sides of the ▱ IBAC. Find the perimeter of quadrilateral IBAC.
问题：如图所示，CA=10，CI=15，∠ACI=45°，BI和AC是平行四边形IBAC的一组对边。求IBAC的周长。
Construction (CDL):
Shape(IB,BA,AC,CI)
Conditions (CDL):
Equal(LengthOfLine(CA),10)
Equal(LengthOfLine(CI),15)
Equal(MeasureOfAngle(ACI),45)
Parallelogram(IBAC)
Goal: Value(PerimeterOfQuadrilateral(IBAC))
Answer: 50
Theorem sequence:
parallelogram_property_opposite_line_equal(1,IBAC)
parallelogram_property_opposite_line_equal(1,BACI)
quadrilateral_perimeter_formula(1,IBAC)
Problem: As shown in the diagram, AB=x, AD=3*sqrt(2), BC=y, BD=sqrt(2), CD=z, AB⊥CB, BD⊥AD. Find the value of x.
问题：如图所示，AB=x，AD=3*sqrt(2)，BC=y，BD=sqrt(2)，CD=z，AB垂直于CB，BD垂直于AD。求x的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),3*sqrt(2))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),sqrt(2))
Equal(LengthOfLine(CD),z)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(BD,AD)
Goal: Value(x)
Answer: 2*sqrt(5)
Theorem sequence:
right_triangle_judgment_angle(1,BDA)
right_triangle_property_pythagorean(1,BDA)
Problem: As shown in the diagram, BC=12/5, ∠CBA=161°, the center of ⊙B is B. Find the area of the sector BCA.
问题：如图所示，BC=12/5，∠CBA=161°，B是⊙B的圆心。求扇形BCA的面积。
Construction (CDL):
Shape(BCA,AB,BC)
Shape(BA,BAC,CB)
Cocircular(B,AC)
Conditions (CDL):
Equal(LengthOfLine(BC),12/5)
Equal(MeasureOfAngle(CBA),161)
IsCentreOfCircle(B,B)
Goal: Value(AreaOfSector(BCA))
Answer: 398*pi/125
Theorem sequence:
round_angle(1,ABC,CBA)
radius_of_circle_property_length_equal(1,BC,B)
arc_property_center_angle(1,BCA,B)
sector_area_formula(1,BCA)
Problem: As shown in the diagram, AB=z, AC=5, BC=15, BA⊥CA. Find the value of z.
问题：如图所示，AB=z，AC=5，BC=15，BA⊥CA。求z的值。
Construction (CDL):
Shape(BA,AY,YB)
Shape(AC,CY,YA)
Collinear(BYC)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),15)
PerpendicularBetweenLine(BA,CA)
Goal: Value(z)
Answer: 10*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, GF=x, HF=32, ∠FHG=29°, GF is perpendicular to HF. Find the value of x.
问题：如图所示，GF=x，HF=32，∠FHG=29°，GF⊥HF。求x的值。
Construction (CDL):
Shape(HG,GF,FH)
Conditions (CDL):
Equal(LengthOfLine(GF),x)
Equal(LengthOfLine(HF),32)
Equal(MeasureOfAngle(FHG),29)
PerpendicularBetweenLine(GF,HF)
Goal: Value(x)
Answer: 32*tan(29*pi/180)
Theorem sequence:
triangle_property_angle_sum(1,HGF)
sine_theorem(1,FHG)
Problem: As shown in the diagram, AB=x, AC=z, AD=y, BD=12, CD=4, AD⊥CD, BD is perpendicular to AD, CA⊥BA. Find the value of z.
问题：如图所示，AB=x，AC=z，AD=y，BD=12，CD=4，AD垂直于CD，BD⊥AD，CA垂直于BA。求z的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),z)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BD),12)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(z)
Answer: 8
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,ADC)
line_addition(1,BD,DC)
right_triangle_property_pythagorean(1,CAB)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,ADC)
Problem: As shown in the diagram, AC=23, BC=x, ∠ABC=21°, ∠CAB=128°. Find the value of x.
问题：如图所示，AC=23，BC=x，∠ABC=21°，∠CAB=128°。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AC),23)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ABC),21)
Equal(MeasureOfAngle(CAB),128)
Goal: Value(x)
Answer: 368*sin(13*pi/45)/(-sqrt(30)-sqrt(3)*sqrt(sqrt(5)+5)-sqrt(sqrt(5)+5)-sqrt(6)+sqrt(2)+sqrt(10)+sqrt(5)*sqrt(sqrt(5)+5)+sqrt(15)*sqrt(sqrt(5)+5))
Theorem sequence:
sine_theorem(1,CAB)
Problem: As shown in the diagram, the measure of ⌒AST is 48, A is the center of circle A. Find the measure of ∠AST.
问题：如图所示，弧AST的角度为48，⊙A的圆心为A。求∠AST的大小。
Construction (CDL):
Shape(AR,ARS,SA)
Shape(AS,ST,TA)
Shape(ST,ATS)
Shape(RA,AT,ATR)
Collinear(RAS)
Cocircular(A,RST)
Conditions (CDL):
Equal(MeasureOfArc(AST),48)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfAngle(AST))
Answer: 66
Theorem sequence:
radius_of_circle_property_length_equal(1,AS,A)
radius_of_circle_property_length_equal(1,AT,A)
arc_property_center_angle(1,AST,A)
sine_theorem(1,AST)
triangle_property_angle_sum(1,AST)
Problem: As shown in the diagram, ∠FDL=4*x-9°, ∠KHG=65°, BL is parallel to EK. Find the value of x.
问题：如图所示，∠FDL=4*x-9°，∠KHG=65°，BL平行于EK。求x的值。
Construction (CDL):
Shape(FD)
Shape(DH)
Shape(HG)
Shape(BD)
Shape(DL)
Shape(EH)
Shape(HK)
Shape(FD,DL)
Shape(LD,DH)
Shape(DH,HK)
Shape(KH,HG)
Shape(BD,DF)
Shape(HD,DB)
Shape(EH,HD)
Shape(GH,HE)
Collinear(FDHG)
Collinear(BDL)
Collinear(EHK)
Conditions (CDL):
Equal(MeasureOfAngle(FDL),4*x-9)
Equal(MeasureOfAngle(KHG),65)
ParallelBetweenLine(BL,EK)
Goal: Value(x)
Answer: 31
Theorem sequence:
parallel_property_collinear_extend(3,BL,EK,D)
parallel_property_collinear_extend(3,KE,LD,H)
parallel_property_corresponding_angle(2,DL,HK,G)
adjacent_complementary_angle(1,FDL,LDH)
Problem: As shown in the diagram, FH=FL, FK=17, HF=8, FL⊥KL, JH⊥FH. Find the length of line LK.
问题：如图所示，FH=FL，FK=17，HF=8，FL垂直于KL，JH垂直于FH。求直线LK的长度。
Construction (CDL):
Shape(FK,FKJ,JH,HF)
Shape(HJ,FJG,GH)
Shape(FH,HG,FGM,ML,LF)
Shape(KF,FL,LK)
Shape(KL,LM,FMK)
Collinear(JHG)
Collinear(HFL)
Collinear(KLM)
Cocircular(F,JGMK)
Conditions (CDL):
Equal(LengthOfLine(FH),LengthOfLine(FL))
Equal(LengthOfLine(FK),17)
Equal(LengthOfLine(HF),8)
PerpendicularBetweenLine(FL,KL)
PerpendicularBetweenLine(JH,FH)
Goal: Value(LengthOfLine(LK))
Answer: 15
Theorem sequence:
right_triangle_judgment_angle(1,FLK)
right_triangle_property_pythagorean(1,FLK)
Problem: As shown in the diagram, LN=12, J is the center of circle J, the diameter of ⊙J is LN. Find the length of line JM.
问题：如图所示，LN=12，⊙J的圆心为J，LN是圆J的直径。求直线JM的长度。
Construction (CDL):
Shape(KP,JPK)
Shape(PK,JKL,LJ,JN,JNP)
Shape(JL,JLM,MJ)
Shape(JM,JMN,NJ)
Collinear(LJN)
Cocircular(J,KLMNP)
Conditions (CDL):
Equal(LengthOfLine(LN),12)
IsCentreOfCircle(J,J)
IsDiameterOfCircle(LN,J)
Goal: Value(LengthOfLine(JM))
Answer: 6
Theorem sequence:
diameter_of_circle_property_length_equal(1,LN,J)
radius_of_circle_property_length_equal(1,JM,J)
circle_property_length_of_radius_and_diameter(1,J)
Problem: As shown in the diagram, the area of quadrilateral DBCA is 78, AD=6, CB=13, DE=x, DE is perpendicular to CE, DB and CA are the non-parallel sides (legs) of trapezoid DBCA. Find the value of x.
问题：如图所示，四边形DBCA的面积为78，AD=6，CB=13，DE=x，DE垂直于CE，四边形DBCA是梯形。求x的值。
Construction (CDL):
Shape(DB,BE,ED)
Shape(AD,DE,EC,CA)
Collinear(BEC)
Conditions (CDL):
Equal(AreaOfQuadrilateral(DBCA),78)
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(CB),13)
Equal(LengthOfLine(DE),x)
PerpendicularBetweenLine(DE,CE)
Trapezoid(DBCA)
Goal: Value(x)
Answer: 156/19
Theorem sequence:
adjacent_complementary_angle(1,BED,DEC)
altitude_of_quadrilateral_judgment_left_vertex(2,DE,DBCA)
trapezoid_area_formula(1,DBCA)
Problem: As shown in the diagram, ∠EAD=42°, A is the center of ⊙A. Find the measure of arc ABE.
问题：如图所示，∠EAD=42°，A是圆A的圆心。求弧ABE的角度。
Construction (CDL):
Shape(ABC,CA,AB)
Shape(AC,ACD,DA)
Shape(AD,ADE,EA)
Shape(BA,AE,AEB)
Collinear(BAD)
Collinear(CAE)
Cocircular(A,BCDE)
Conditions (CDL):
Equal(MeasureOfAngle(EAD),42)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfArc(ABE))
Answer: 222
Theorem sequence:
adjacent_complementary_angle(1,BAE,EAD)
round_angle(1,BAE,EAB)
arc_property_center_angle(1,ABE,A)
Problem: As shown in the diagram, AB=x, BC=5, ∠BAC=45°, AC⊥BC. Find the value of x.
问题：如图所示，AB=x，BC=5，∠BAC=45°，AC垂直于BC。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),5)
Equal(MeasureOfAngle(BAC),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 5*sqrt(2)
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, ST=a+15, VT=3*b+5, VW=3*a+11, WS=b+11, ST and WV are opposite sides of the ▱ TSWV. Find the value of a.
问题：如图所示，ST=a+15，VT=3*b+5，VW=3*a+11，WS=b+11，TSWV是平行四边形。求a的值。
Construction (CDL):
Shape(TS,SW,WV,VT)
Conditions (CDL):
Equal(LengthOfLine(ST),a+15)
Equal(LengthOfLine(VT),3*b+5)
Equal(LengthOfLine(VW),3*a+11)
Equal(LengthOfLine(WS),b+11)
Parallelogram(TSWV)
Goal: Value(a)
Answer: 2
Theorem sequence:
parallelogram_property_opposite_line_equal(1,TSWV)
Problem: As shown in the diagram, ∠CBA=55°, the measure of arc DAC is x, the center of circle D is D, BA is the tangent to circle D, the tangent to circle D is BC. Find the value of x.
问题：如图所示，∠CBA=55°，弧DAC的角度为x，圆D的圆心为D，BA是⊙O的切线，BC是⊙O的切线。求x的值。
Construction (CDL):
Shape(DAC,CD,DA)
Shape(AD,DC,DCA)
Shape(BA,DCA,CB)
Cocircular(D,AC)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),55)
Equal(MeasureOfArc(DAC),x)
IsCentreOfCircle(D,D)
IsTangentOfCircle(BA,D)
IsTangentOfCircle(BC,D)
Goal: Value(x)
Answer: 235
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BA,D,D)
tangent_of_circle_property_perpendicular(1,BC,D,D)
quadrilateral_property_angle_sum(1,ADCB)
round_angle(1,CDA,ADC)
arc_property_center_angle(1,DAC,D)
Problem: As shown in the diagram, JM=LengthOfLine(P,M), ML=PL, ∠MLP=58°. Find the measure of ∠PJL.
问题：如图所示，JM=LengthOfLine(P,M)，ML=PL，∠MLP=58°。求∠PJL的大小。
Construction (CDL):
Shape(PJ,JM,MP)
Shape(PM,ML,LP)
Collinear(JML)
Conditions (CDL):
Equal(LengthOfLine(JM),LengthOfLine(P,M))
Equal(LengthOfLine(ML),LengthOfLine(PL))
Equal(MeasureOfAngle(MLP),58)
Goal: Value(MeasureOfAngle(PJL))
Answer: 61/2
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,MPJ)
isosceles_triangle_judgment_line_equal(1,LPM)
triangle_property_angle_sum(1,LPM)
isosceles_triangle_property_angle_equal(1,LPM)
adjacent_complementary_angle(1,JMP,PML)
triangle_property_angle_sum(1,MPJ)
isosceles_triangle_property_angle_equal(1,MPJ)
Problem: As shown in the diagram, AB=x, AC=4, BC=6, AC⊥BC. Find the value of x.
问题：如图所示，AB=x，AC=4，BC=6，AC⊥BC。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BC),6)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 2*sqrt(13)
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, EF=12, GF=x, RT=15, SR=20, ST=10, ∠EFG=110°, ∠GEF=27°, ∠RTS=110°, ∠SRT=27°, ∠TSR=43°, △SRT is similar to △GEF.. Find the length of line EG.
问题：如图所示，EF=12，GF=x，RT=15，SR=20，ST=10，∠EFG=110°，∠GEF=27°，∠RTS=110°，∠SRT=27°，∠TSR=43°，△SRT相似于△GEF。求直线EG的长度。
Construction (CDL):
Shape(SR,RT,TS)
Shape(GE,EF,FG)
Conditions (CDL):
Equal(LengthOfLine(EF),12)
Equal(LengthOfLine(GF),x)
Equal(LengthOfLine(RT),15)
Equal(LengthOfLine(SR),20)
Equal(LengthOfLine(ST),10)
Equal(MeasureOfAngle(EFG),110)
Equal(MeasureOfAngle(GEF),27)
Equal(MeasureOfAngle(RTS),110)
Equal(MeasureOfAngle(SRT),27)
Equal(MeasureOfAngle(TSR),43)
SimilarBetweenTriangle(SRT,GEF)
Goal: Value(LengthOfLine(EG))
Answer: 32*sqrt(2)*cos(pi/9)/(-sqrt(5)+1+sqrt(2)*sqrt(sqrt(5)+5))
Theorem sequence:
similar_triangle_property_line_ratio(1,SRT,GEF)
similar_triangle_property_line_ratio(1,RTS,EFG)
sine_theorem(1,GEF)
Problem: As shown in the diagram, BC=5*x-8, EB=3*x, the tangent to ⊙R is BC, the tangent to ⊙R is BE. Find the value of x.
问题：如图所示，BC=5*x-8，EB=3*x，圆O的切线为BC，⊙O的切线为BE。求x的值。
Construction (CDL):
Shape(REC,RCE)
Shape(BE,RCE,CB)
Cocircular(R,EC)
Conditions (CDL):
Equal(LengthOfLine(BC),5*x-8)
Equal(LengthOfLine(EB),3*x)
IsTangentOfCircle(BC,R)
IsTangentOfCircle(BE,R)
Goal: Value(x)
Answer: 4
Theorem sequence:
tangent_of_circle_property_length_equal(1,BE,BC,R)
Problem: As shown in the diagram, ∠ABC=35°, the measure of arc DAC is x, D is the center of circle D, BA is the tangent to ⊙D, BC is the tangent to circle D. Find the value of x.
问题：如图所示，∠ABC=35°，弧DAC的角度为x，D是⊙D的圆心，圆O的切线为BA，BC是⊙O的切线。求x的值。
Construction (CDL):
Shape(DC,DCA,AD)
Shape(CD,DA,DAC)
Shape(BC,DAC,AB)
Cocircular(D,CA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),35)
Equal(MeasureOfArc(DAC),x)
IsCentreOfCircle(D,D)
IsTangentOfCircle(BA,D)
IsTangentOfCircle(BC,D)
Goal: Value(x)
Answer: 145
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BA,D,D)
tangent_of_circle_property_perpendicular(2,BC,D,D)
quadrilateral_property_angle_sum(1,CDAB)
arc_property_center_angle(1,DAC,D)
Problem: As shown in the diagram, ST=UT, SY=YZ, UJ=9, VJ=3, ZT=18, ZV=UV. Find the length of line ZJ.
问题：如图所示，ST=UT，SY=YZ，UJ=9，VJ=3，ZT=18，ZV=UV。求直线ZJ的长度。
Construction (CDL):
Shape(SY,YJ,JS)
Shape(YZ,ZJ,JY)
Shape(JZ,ZV,VJ)
Shape(JV,VU,UJ)
Shape(TJ,JU,UT)
Shape(SJ,JT,TS)
Collinear(SJV)
Collinear(YJU)
Collinear(ZJT)
Collinear(SYZ)
Collinear(ZVU)
Collinear(UTS)
Conditions (CDL):
Equal(LengthOfLine(ST),LengthOfLine(UT))
Equal(LengthOfLine(SY),LengthOfLine(YZ))
Equal(LengthOfLine(UJ),9)
Equal(LengthOfLine(VJ),3)
Equal(LengthOfLine(ZT),18)
Equal(LengthOfLine(ZV),LengthOfLine(UV))
Goal: Value(LengthOfLine(ZJ))
Answer: 12
Theorem sequence:
median_of_triangle_judgment(1,UY,USZ)
median_of_triangle_judgment(1,SV,SZU)
centroid_of_triangle_judgment_intersection(1,J,ZUS,V,Y)
centroid_of_triangle_property_line_ratio(1,J,ZUS,T)
line_addition(1,ZJ,JT)
Problem: As shown in the diagram, DE=EC, the measure of arc DFC is 60, D is the center of ⊙D, DG is perpendicular to BG. Find the measure of arc DAF.
问题：如图所示，DE=EC，弧DFC的角度为60，D是⊙D的圆心，DG垂直于BG。求⌒DAF的角度。
Construction (CDL):
Shape(AB,DBA)
Shape(AD,DG,GB,BA)
Shape(BG,GE,DEB)
Shape(EG,GC,CE)
Shape(EC,DCE)
Shape(GD,DC,CG)
Shape(DA,AF,FD)
Shape(CD,DF,FC)
Shape(DAF,FD,DA)
Shape(CF,DFC)
Collinear(BGC)
Collinear(ADC)
Collinear(DGE)
Cocircular(D,AFCEB)
Conditions (CDL):
Equal(LengthOfLine(DE),LengthOfLine(EC))
Equal(MeasureOfArc(DFC),60)
IsCentreOfCircle(D,D)
PerpendicularBetweenLine(DG,BG)
Goal: Value(MeasureOfArc(DAF))
Answer: 120
Theorem sequence:
arc_property_center_angle(1,DAF,D)
arc_property_center_angle(1,DFC,D)
adjacent_complementary_angle(1,CDF,FDA)
Problem: As shown in the diagram, the measure of ⌒OBD is 6*a, the measure of arc OCB is 4*a, the measure of arc ODE is 3*a, the measure of ⌒OEC is 5*a, the center of circle O is O. Find the measure of ∠DAB.
问题：如图所示，⌒OBD的角度为6*a，弧OCB的角度为4*a，弧ODE的角度为3*a，⌒OEC的角度为5*a，圆O的圆心为O。求∠DAB的大小。
Construction (CDL):
Shape(AE,OEC,CA)
Shape(AD,ODE,EA)
Shape(OA,AC,CO)
Shape(OD,DA,AO)
Shape(OC,OCB,BO)
Shape(OB,OBD,DO)
Collinear(EAOB)
Collinear(CAD)
Cocircular(O,ECBD)
Conditions (CDL):
Equal(MeasureOfArc(OBD),6*a)
Equal(MeasureOfArc(OCB),4*a)
Equal(MeasureOfArc(ODE),3*a)
Equal(MeasureOfArc(OEC),5*a)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(DAB))
Answer: 110
Theorem sequence:
arc_property_center_angle(1,ODE,O)
arc_property_center_angle(1,OEC,O)
arc_property_center_angle(1,OCB,O)
arc_property_center_angle(1,OBD,O)
angle_addition(1,COE,EOD)
angle_addition(1,COD,DOB)
round_angle(1,COB,BOC)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,ODC)
isosceles_triangle_property_angle_equal(1,ODC)
triangle_property_angle_sum(1,ODA)
triangle_property_angle_sum(1,COD)