SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BC=32, BD=15, BE=z, CD=9, ∠BDE=∠EDC. Find the value of z.
问题：如图所示，BC=32，BD=15，BE=z，CD=9，∠BDE=∠EDC。求z的值。
Construction (CDL):
Shape(DC,CE,ED)
Shape(DE,EB,BD)
Collinear(CEB)
Conditions (CDL):
Equal(LengthOfLine(BC),32)
Equal(LengthOfLine(BD),15)
Equal(LengthOfLine(BE),z)
Equal(LengthOfLine(CD),9)
Equal(MeasureOfAngle(BDE),MeasureOfAngle(EDC))
Goal: Value(z)
Answer: 20
Theorem sequence:
bisector_of_angle_judgment_angle_equal(1,DE,BDC)
bisector_of_angle_property_line_ratio(1,DE,BDC)
line_addition(1,CE,EB)
Problem: As shown in the diagram, AB=AD, BT=5, CB=CD, TC=8. Find the length of line CD.
问题：如图所示，AB=AD，BT=5，CB=CD，TC=8。求直线CD的长度。
Construction (CDL):
Shape(CB,BT,TC)
Shape(BA,AT,TB)
Shape(CT,TD,DC)
Shape(DT,TA,AD)
Collinear(BTD)
Collinear(CTA)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AD))
Equal(LengthOfLine(BT),5)
Equal(LengthOfLine(CB),LengthOfLine(CD))
Equal(LengthOfLine(TC),8)
Goal: Value(LengthOfLine(CD))
Answer: sqrt(89)
Theorem sequence:
kite_judgment_equal_and_equal(1,CBAD)
kite_property_diagonal_perpendicular_bisection(1,CBAD,T)
right_triangle_judgment_angle(1,BTC)
right_triangle_property_pythagorean(1,BTC)
Problem: As shown in the diagram, ∠EAD=42°, A is the center of ⊙A. Find the measure of ⌒ACD.
问题：如图所示，∠EAD=42°，A是圆A的圆心。求弧ACD的角度。
Construction (CDL):
Shape(ABC,CA,AB)
Shape(ACD,DA,AC)
Shape(ADE,EA,AD)
Shape(AEB,BA,AE)
Collinear(BAD)
Collinear(CAE)
Cocircular(A,CDEB)
Conditions (CDL):
Equal(MeasureOfAngle(EAD),42)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfArc(ACD))
Answer: 138
Theorem sequence:
flat_angle(1,EAC)
angle_addition(1,EAD,DAC)
arc_property_center_angle(1,ACD,A)
Problem: As shown in the diagram, ∠CFB=80°, the measure of arc EAD is x, the measure of ⌒EBC is 110. Find the value of x.
问题：如图所示，∠CFB=80°，⌒EAD的角度为x，弧EBC的角度为110。求x的值。
Construction (CDL):
Shape(EBC,CE,EB)
Shape(ECA,AC)
Shape(EAD,DF,FA)
Shape(EDB,BD)
Shape(CA,AF,FC)
Shape(FD,DB,BF)
Shape(CF,FB,BE,EC)
Collinear(AFB)
Collinear(DFC)
Cocircular(E,CADB)
Conditions (CDL):
Equal(MeasureOfAngle(CFB),80)
Equal(MeasureOfArc(EAD),x)
Equal(MeasureOfArc(EBC),110)
Goal: Value(x)
Answer: 50
Theorem sequence:
flat_angle(1,CFD)
angle_addition(1,CFB,BFD)
arc_property_circumference_angle_external(1,EBC,D)
triangle_property_angle_sum(1,BFD)
arc_property_circumference_angle_external(1,EAD,B)
Problem: As shown in the diagram, ∠DCA=40°, ∠ECB=35°. Find the measure of ∠BCD.
问题：如图所示，∠DCA=40°，∠ECB=35°。求∠BCD的大小。
Construction (CDL):
Shape(CEA,AC,CE)
Shape(CAD,DC,CA)
Shape(CDB,BC,CD)
Shape(CBE,EC,CB)
Collinear(ACB)
Cocircular(C,ADBE)
Conditions (CDL):
Equal(MeasureOfAngle(DCA),40)
Equal(MeasureOfAngle(ECB),35)
Goal: Value(MeasureOfAngle(BCD))
Answer: 140
Theorem sequence:
flat_angle(1,BCA)
angle_addition(1,BCD,DCA)
Problem: As shown in the diagram, AB=c, CA=b, CB=a, a=8, b=15, c=17, BC⊥AC. Find the value of cos(AB).
问题：如图所示，AB=c，CA=b，CB=a，a=8，b=15，c=17，BC垂直于AC。求cos(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),c)
Equal(LengthOfLine(CA),b)
Equal(LengthOfLine(CB),a)
Equal(a,8)
Equal(b,15)
Equal(c,17)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Cos(MeasureOfAngle(ABC)))
Answer: 8/17
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, BC=DB, BC=DE, EC=6, EC=BC, CB⊥DB. Find the area of quadrilateral CBDE.
问题：如图所示，BC=DB，BC=DE，EC=6，EC=BC，CB垂直于DB。求CBDE的面积。
Construction (CDL):
Shape(CB,BD,DE,EC)
Conditions (CDL):
Equal(LengthOfLine(BC),LengthOfLine(DB))
Equal(LengthOfLine(BC),LengthOfLine(DE))
Equal(LengthOfLine(EC),6)
Equal(LengthOfLine(EC),LengthOfLine(BC))
PerpendicularBetweenLine(CB,DB)
Goal: Value(AreaOfQuadrilateral(CBDE))
Answer: 36
Theorem sequence:
kite_judgment_equal_and_equal(1,CBDE)
kite_area_formula_sine(1,CBDE)
Problem: As shown in the diagram, ∠AEC=63°, ∠DAB=47°, ∠DFB=136°, ∠EAD=69°. Find the measure of ∠CAE.
问题：如图所示，∠AEC=63°，∠DAB=47°，∠DFB=136°，∠EAD=69°。求∠CAE的大小。
Construction (CDL):
Shape(EC,CA,AE)
Shape(CG,GA,AC)
Shape(DA,AF,FD)
Shape(EA,AD)
Shape(EA,AF)
Shape(DF,FB)
Collinear(CAFB)
Collinear(EAG)
Conditions (CDL):
Equal(MeasureOfAngle(AEC),63)
Equal(MeasureOfAngle(DAB),47)
Equal(MeasureOfAngle(DFB),136)
Equal(MeasureOfAngle(EAD),69)
Goal: Value(MeasureOfAngle(CAE))
Answer: 64
Theorem sequence:
angle_addition(1,EAD,DAF)
angle_addition(1,CAE,EAF)
flat_angle(1,CAF)
Problem: As shown in the diagram, AB=4*z, AT=2*x-5, AU=2.9, DA=6, EA=y, FA=4.6, S is the midpoint of segment DE, T is the midpoint of segment EF, U is the midpoint of segment DF. Find the value of x.
问题：如图所示，AB=4*z，AT=2*x-5，AU=2.9，DA=6，EA=y，FA=4.6，S是线段DE的中点，T平分线段EF，U是线段DF的中点。求x的值。
Construction (CDL):
Shape(ES,SA,AE)
Shape(SD,DA,AS)
Shape(AD,DU,UA)
Shape(AU,UF,FA)
Shape(AF,FT,TA)
Shape(AT,TE,EA)
Collinear(ESD)
Collinear(DUF)
Collinear(FTE)
Collinear(EAU)
Collinear(SAF)
Collinear(DAT)
Conditions (CDL):
Equal(LengthOfLine(AB),4*z)
Equal(LengthOfLine(AT),2*x-5)
Equal(LengthOfLine(AU),2.9)
Equal(LengthOfLine(DA),6)
Equal(LengthOfLine(EA),y)
Equal(LengthOfLine(FA),4.6)
IsMidpointOfLine(S,DE)
IsMidpointOfLine(T,EF)
IsMidpointOfLine(U,DF)
Goal: Value(x)
Answer: 4
Theorem sequence:
median_of_triangle_judgment(1,DT,DFE)
median_of_triangle_judgment(1,EU,EDF)
centroid_of_triangle_judgment_intersection(1,A,FED,T,U)
centroid_of_triangle_property_line_ratio(1,A,DFE,T)
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 median of △ QPR. Find the length of line PR.
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是△QPR的中线。求直线PR的长度。
Construction (CDL):
Shape(QZ,ZC,CB,BQ)
Shape(ZP,PC,CZ)
Shape(CP,PY,YA,AC)
Shape(BC,CA,AB)
Shape(QB,BX,XQ)
Shape(XB,BA,AR,RX)
Shape(AY,YR,RA)
Collinear(QZP)
Collinear(PYR)
Collinear(RXQ)
Collinear(QBAY)
Collinear(PCBX)
Collinear(ZCAR)
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)
IsMedianOfTriangle(QY,QPR)
Goal: Value(LengthOfLine(PR))
Answer: 18
Theorem sequence:
line_addition(1,PY,YR)
Problem: As shown in the diagram, JO=12, KN=7, LN=x+3, LO=4*x-9, the tangent to circle R is JM, the tangent to circle R is JO, KM is the tangent to ⊙R, KN is the tangent to ⊙R, LN is the tangent to circle R, the tangent to ⊙R is LO. Find the value of x.
问题：如图所示，JO=12，KN=7，LN=x+3，LO=4*x-9，JM是⊙O的切线，圆O的切线为JO，KM是⊙O的切线，KN是⊙O的切线，LN是圆O的切线，圆O的切线为LO。求x的值。
Construction (CDL):
Shape(RNM,RMO,RON)
Shape(RNM,NK,KM)
Shape(RMO,MJ,JO)
Shape(RON,OL,LN)
Collinear(JMK)
Collinear(JOL)
Collinear(KNL)
Cocircular(R,NMO)
Conditions (CDL):
Equal(LengthOfLine(JO),12)
Equal(LengthOfLine(KN),7)
Equal(LengthOfLine(LN),x+3)
Equal(LengthOfLine(LO),4*x-9)
IsTangentOfCircle(JM,R)
IsTangentOfCircle(JO,R)
IsTangentOfCircle(KM,R)
IsTangentOfCircle(KN,R)
IsTangentOfCircle(LN,R)
IsTangentOfCircle(LO,R)
Goal: Value(x)
Answer: 4
Theorem sequence:
tangent_of_circle_property_length_equal(1,LO,LN,R)
Problem: As shown in the diagram, EG=x+5, FD=3*x-7, DG is perpendicular to FG, DGFE is a rectangle. Find the length of line EG.
问题：如图所示，EG=x+5，FD=3*x-7，DG⊥FG，DGFE是长方形。求直线EG的长度。
Construction (CDL):
Shape(AD,DG,GA)
Shape(AG,GF,FA)
Shape(AF,FE,EA)
Shape(ED,DA,AE)
Collinear(DAF)
Collinear(GAE)
Conditions (CDL):
Equal(LengthOfLine(EG),x+5)
Equal(LengthOfLine(FD),3*x-7)
PerpendicularBetweenLine(DG,FG)
Rectangle(DGFE)
Goal: Value(LengthOfLine(EG))
Answer: 11
Theorem sequence:
rectangle_property_diagonal_equal(1,DGFE)
Problem: As shown in the diagram, BA=16, DA=9, DC=15, CA is perpendicular to DA. Find the area of △CBA.
问题：如图所示，BA=16，DA=9，DC=15，CA垂直于DA。求三角形CBA的面积。
Construction (CDL):
Shape(CB,BA,AC)
Shape(CA,AD,DC)
Collinear(BAD)
Conditions (CDL):
Equal(LengthOfLine(BA),16)
Equal(LengthOfLine(DA),9)
Equal(LengthOfLine(DC),15)
PerpendicularBetweenLine(CA,DA)
Goal: Value(AreaOfTriangle(CBA))
Answer: 96
Theorem sequence:
adjacent_complementary_angle(1,BAC,CAD)
right_triangle_judgment_angle(1,CAD)
right_triangle_property_pythagorean(1,CAD)
triangle_area_formula_sine(1,ACB)
Problem: As shown in the diagram, AB=x, AC=9*sqrt(2), BC=9*sqrt(2), ∠CBA=45°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=x，AC=9*sqrt(2)，BC=9*sqrt(2)，∠CBA=45°，AC⊥BC。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),9*sqrt(2))
Equal(LengthOfLine(BC),9*sqrt(2))
Equal(MeasureOfAngle(CBA),45)
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, AB=BI, AC=4, AC=ID, BE=8, CD=9, AC⊥EC, CE is perpendicular to FE, ED⊥ID. Find the sum of the area of triangle BAI and the area of ACDI.
问题：如图所示，AB=BI，AC=4，AC=ID，BE=8，CD=9，AC垂直于EC，CE垂直于FE，ED垂直于ID。求三角形BAI的面积与ACDI的面积之和。
Construction (CDL):
Shape(BA,AF,FB)
Shape(BF,FI,IB)
Shape(FE,ED,DI,IF)
Shape(AC,CE,EF,FA)
Collinear(AFI)
Collinear(BFE)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(BI))
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(AC),LengthOfLine(ID))
Equal(LengthOfLine(BE),8)
Equal(LengthOfLine(CD),9)
PerpendicularBetweenLine(AC,EC)
PerpendicularBetweenLine(CE,FE)
PerpendicularBetweenLine(ED,ID)
Goal: Value(Add(AreaOfTriangle(BAI),AreaOfQuadrilateral(ACDI)))
Answer: 54
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,CA,EF)
parallel_judgment_ipsilateral_internal_angle(1,CA,DI)
parallelogram_judgment_parallel_and_equal(1,DIAC)
parallel_property_collinear_extend(3,DC,IA,E)
parallel_property_collinear_extend(3,AI,CE,F)
parallel_property_corresponding_angle(2,EC,FA,B)
parallelogram_judgment_parallel_and_parallel(1,ACEF)
parallelogram_property_opposite_line_equal(1,IACD)
parallelogram_property_opposite_line_equal(1,ACEF)
line_addition(1,BF,FE)
altitude_of_triangle_judgment(1,BF,BAI)
triangle_area_formula_common(1,BAI)
parallelogram_area_formula_sine(1,ACDI)
Problem: As shown in the diagram, AD=x-3, AF=y+1, DB=12, FB=AF, LM=8, LO=10, LO=NO, MN=16, ∠AFB=∠NOL, ∠DAF=∠MNO, BD⊥AD, FB⊥DB, LM is perpendicular to NM, OL is perpendicular to ML, quadrilateral BDAF is similar to quadrilateral LMNO. Find the value of y.
问题：如图所示，AD=x-3，AF=y+1，DB=12，FB=AF，LM=8，LO=10，LO=NO，MN=16，∠AFB=∠NOL，∠DAF=∠MNO，BD⊥AD，FB垂直于DB，LM垂直于NM，OL垂直于ML，四边形BDAF与四边形LMNO相似。求y的值。
Construction (CDL):
Shape(BD,DA,AF,FB)
Shape(LM,MN,NO,OL)
Conditions (CDL):
Equal(LengthOfLine(AD),x-3)
Equal(LengthOfLine(AF),y+1)
Equal(LengthOfLine(DB),12)
Equal(LengthOfLine(FB),LengthOfLine(AF))
Equal(LengthOfLine(LM),8)
Equal(LengthOfLine(LO),10)
Equal(LengthOfLine(LO),LengthOfLine(NO))
Equal(LengthOfLine(MN),16)
Equal(MeasureOfAngle(AFB),MeasureOfAngle(NOL))
Equal(MeasureOfAngle(DAF),MeasureOfAngle(MNO))
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(FB,DB)
PerpendicularBetweenLine(LM,NM)
PerpendicularBetweenLine(OL,ML)
SimilarBetweenQuadrilateral(BDAF,LMNO)
Goal: Value(y)
Answer: 14
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,BDAF,LMNO)
similar_quadrilateral_property_line_ratio(1,AFBD,NOLM)
Problem: As shown in the diagram, WZ=8, XY=20, ∠XYZ=70°, WZ and XY are the parallel sides of trapezoid WXYZ. Find the measure of ∠YZW.
问题：如图所示，WZ=8，XY=20，∠XYZ=70°，WXYZ是梯形。求∠YZW的大小。
Construction (CDL):
Shape(WX,XY,YZ,ZW)
Conditions (CDL):
Equal(LengthOfLine(WZ),8)
Equal(LengthOfLine(XY),20)
Equal(MeasureOfAngle(XYZ),70)
Trapezoid(WXYZ)
Goal: Value(MeasureOfAngle(YZW))
Answer: 110
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,YX,ZW)
Problem: As shown in the diagram, WT=3, WX is perpendicular to YX, XY is perpendicular to ZY, YZ⊥WZ, ZW is perpendicular to XW, WXYZ is a square. Find the length of line ZX.
问题：如图所示，WT=3，WX⊥YX，XY垂直于ZY，YZ⊥WZ，ZW垂直于XW，WXYZ是正方形。求直线ZX的长度。
Construction (CDL):
Shape(XY,YT,TX)
Shape(TY,YZ,ZT)
Shape(TZ,ZW,WT)
Shape(XT,TW,WX)
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(LengthOfLine(ZX))
Answer: 6
Theorem sequence:
line_addition(1,YT,TW)
parallelogram_property_diagonal_bisection(1,WXYZ,T)
rectangle_property_diagonal_equal(1,WXYZ)
Problem: As shown in the diagram, ∠ACB=80°, ∠CBA=∠GHF, ∠FGH=2*x°, BA is perpendicular to CA, HF is perpendicular to GF. Find the value of x.
问题：如图所示，∠ACB=80°，∠CBA=∠GHF，∠FGH=2*x°，BA⊥CA，HF垂直于GF。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Shape(GH,HF,FG)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),80)
Equal(MeasureOfAngle(CBA),MeasureOfAngle(GHF))
Equal(MeasureOfAngle(FGH),2*x)
PerpendicularBetweenLine(BA,CA)
PerpendicularBetweenLine(HF,GF)
Goal: Value(x)
Answer: 40
Theorem sequence:
similar_triangle_judgment_aa(1,CBA,GHF)
similar_triangle_property_angle_equal(1,CBA,GHF)
Problem: As shown in the diagram, GF=2, ∠FGH=99°, G is the center of ⊙G. Find the area of the sector GHF.
问题：如图所示，GF=2，∠FGH=99°，圆G的圆心为G。求扇形GHF的面积。
Construction (CDL):
Shape(GFH,HG,GF)
Shape(FG,GH,GHF)
Cocircular(G,FH)
Conditions (CDL):
Equal(LengthOfLine(GF),2)
Equal(MeasureOfAngle(FGH),99)
IsCentreOfCircle(G,G)
Goal: Value(AreaOfSector(GHF))
Answer: 11*pi/10
Theorem sequence:
arc_property_center_angle(1,GHF,G)
radius_of_circle_property_length_equal(1,GF,G)
sector_area_formula(1,GHF)
Problem: As shown in the diagram, UT=12, VS=9, S bisects segment RT, V bisects segment QU, QR and UT are the parallel sides of trapezoid QUTR. Find the length of line QR.
问题：如图所示，UT=12，VS=9，S是线段RT的中点，V是线段QU的中点，四边形QUTR是梯形。求直线QR的长度。
Construction (CDL):
Shape(QV,VS,SR,RQ)
Shape(VU,UT,TS,SV)
Collinear(QVU)
Collinear(RST)
Conditions (CDL):
Equal(LengthOfLine(UT),12)
Equal(LengthOfLine(VS),9)
IsMidpointOfLine(S,RT)
IsMidpointOfLine(V,QU)
Trapezoid(QUTR)
Goal: Value(LengthOfLine(QR))
Answer: 6
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,VS,QUTR)
midsegment_of_quadrilateral_property_length(1,VS,QUTR)
Problem: As shown in the diagram, AC=AB, BA=5*sqrt(2), BC=y, BE=x, AE⊥BE, BC is perpendicular to DC, CA is perpendicular to BA, CD is perpendicular to AD, EB is perpendicular to CB. Find the value of x.
问题：如图所示，AC=AB，BA=5*sqrt(2)，BC=y，BE=x，AE垂直于BE，BC⊥DC，CA⊥BA，CD垂直于AD，EB⊥CB。求x的值。
Construction (CDL):
Shape(CD,DA,AC)
Shape(AE,EB,BA)
Shape(CA,AB,BC)
Collinear(DAE)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(AB))
Equal(LengthOfLine(BA),5*sqrt(2))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BE),x)
PerpendicularBetweenLine(AE,BE)
PerpendicularBetweenLine(BC,DC)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(CD,AD)
PerpendicularBetweenLine(EB,CB)
Goal: Value(x)
Answer: 5
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,AEB)
triangle_property_angle_sum(1,CAB)
angle_addition(1,EBA,ABC)
sine_theorem(1,AEB)
sine_theorem(1,EBA)
Problem: As shown in the diagram, the diameter of circle A is 8, the diameter of ⊙B is 18, the diameter of ⊙C is 11, the center of circle A is A, B is the center of circle B, C is the center of ⊙C, the diameter of circle B is AG. Find the length of line FG.
问题：如图所示，圆A的直径为8，圆B的直径为18，⊙C的直径为11，⊙A的圆心为A，⊙B的圆心为B，C是⊙C的圆心，AG是⊙B的直径。求直线FG的长度。
Construction (CDL):
Shape(BAM,AMN,BNA)
Shape(AFM,BMA,AF)
Shape(FA,BAN,ANF)
Shape(BXM,AMF,AFN,BNY,CYE,CEX)
Shape(CXE,EG,BGX)
Shape(CEY,BYG,GE)
Shape(CYX,BXG,BGY)
Collinear(AFBEGC)
Cocircular(A,NFM)
Cocircular(B,MANYGX)
Cocircular(C,XEY)
Conditions (CDL):
Equal(DiameterOfCircle(A),8)
Equal(DiameterOfCircle(B),18)
Equal(DiameterOfCircle(C),11)
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
IsDiameterOfCircle(AG,B)
Goal: Value(LengthOfLine(FG))
Answer: 14
Theorem sequence:
diameter_of_circle_property_length_equal(1,AG,B)
circle_property_length_of_radius_and_diameter(1,A)
radius_of_circle_property_length_equal(1,AF,A)
line_addition(1,AF,FG)
Problem: As shown in the diagram, ∠EDH=x°, ∠HDA=3*x°. Find the value of x.
问题：如图所示，∠EDH=x°，∠HDA=3*x°。求x的值。
Construction (CDL):
Shape(DA)
Shape(ED)
Shape(DH)
Shape(HD,DA)
Shape(ED,DH)
Collinear(ADE)
Conditions (CDL):
Equal(MeasureOfAngle(EDH),x)
Equal(MeasureOfAngle(HDA),3*x)
Goal: Value(x)
Answer: 45
Theorem sequence:
adjacent_complementary_angle(1,EDH,HDA)
Problem: As shown in the diagram, the length of ⌒ANX is x, XA=3, ∠XAN=78°, A is the center of circle A. Find the value of x.
问题：如图所示，弧ANX的长度为x，XA=3，∠XAN=78°，⊙A的圆心为A。求x的值。
Construction (CDL):
Shape(AN,ANX,XA)
Shape(AX,AXN,NA)
Cocircular(A,XN)
Conditions (CDL):
Equal(LengthOfArc(ANX),x)
Equal(LengthOfLine(XA),3)
Equal(MeasureOfAngle(XAN),78)
IsCentreOfCircle(A,A)
Goal: Value(x)
Answer: 13*pi/10
Theorem sequence:
radius_of_circle_property_length_equal(1,AX,A)
arc_property_center_angle(1,ANX,A)
arc_length_formula(1,ANX)
Problem: As shown in the diagram, ∠JQR=131°, GT is parallel to RQ, HR∥TC. Find the measure of ∠RQT.
问题：如图所示，∠JQR=131°，GT∥RQ，HR∥TC。求∠RQT的大小。
Construction (CDL):
Shape(JQ,QR)
Shape(QR,RH)
Shape(BG,GT)
Shape(GT,TC)
Shape(QT,TG,GR,RQ)
Collinear(JQTC)
Collinear(HRGB)
Conditions (CDL):
Equal(MeasureOfAngle(JQR),131)
ParallelBetweenLine(GT,RQ)
ParallelBetweenLine(HR,TC)
Goal: Value(MeasureOfAngle(RQT))
Answer: 49
Theorem sequence:
adjacent_complementary_angle(1,JQR,RQT)
Problem: As shown in the diagram, AF=4, GA=3, CG and CD are one pair of adjacent sides of the kite CGFD. Find the length of line GF.
问题：如图所示，AF=4，GA=3，FG和FD是风筝形CGFD的一组临边。求直线GF的长度。
Construction (CDL):
Shape(CG,GA,AC)
Shape(CA,AD,DC)
Shape(AG,GF,FA)
Shape(AF,FD,DA)
Collinear(GAD)
Collinear(CAF)
Conditions (CDL):
Equal(LengthOfLine(AF),4)
Equal(LengthOfLine(GA),3)
Kite(CGFD)
Goal: Value(LengthOfLine(GF))
Answer: 5
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,FDCG,A)
right_triangle_judgment_angle(1,FAG)
right_triangle_property_pythagorean(1,FAG)
Problem: As shown in the diagram, the area of triangle BAC is 875, the area of triangle DEF is 315, BM=x, DN=21, AM is perpendicular to BM, EN⊥DN, △BAC is similar to △DEF.. Find the value of x.
问题：如图所示，△BAC的面积为875，△DEF的面积为315，BM=x，DN=21，AM垂直于BM，EN垂直于DN，三角形BAC与三角形DEF是相似三角形。求x的值。
Construction (CDL):
Shape(BA,AM,MB)
Shape(BM,MC,CB)
Shape(DE,EN,ND)
Shape(DN,NF,FD)
Collinear(AMC)
Collinear(ENF)
Conditions (CDL):
Equal(AreaOfTriangle(BAC),875)
Equal(AreaOfTriangle(DEF),315)
Equal(LengthOfLine(BM),x)
Equal(LengthOfLine(DN),21)
PerpendicularBetweenLine(AM,BM)
PerpendicularBetweenLine(EN,DN)
SimilarBetweenTriangle(BAC,DEF)
Goal: Value(x)
Answer: 35
Theorem sequence:
altitude_of_triangle_judgment(1,BM,BAC)
altitude_of_triangle_judgment(1,DN,DEF)
triangle_area_formula_common(1,BAC)
triangle_area_formula_common(1,DEF)
similar_triangle_property_line_ratio(1,BAC,DEF)
similar_triangle_property_area_square_ratio(1,BAC,DEF)
Problem: As shown in the diagram, BD=5, DF=4, EB=DF, EF=DB, BE and DF are opposite sides of the parallelogram EBDF, EB is perpendicular to DB. Find the area of quadrilateral EBDF.
问题：如图所示，BD=5，DF=4，EB=DF，EF=DB，EBDF是平行四边形，EB垂直于DB。求EBDF的面积。
Construction (CDL):
Shape(EB,BD,DF,FE)
Conditions (CDL):
Equal(LengthOfLine(BD),5)
Equal(LengthOfLine(DF),4)
Equal(LengthOfLine(EB),LengthOfLine(DF))
Equal(LengthOfLine(EF),LengthOfLine(DB))
Parallelogram(EBDF)
PerpendicularBetweenLine(EB,DB)
Goal: Value(AreaOfQuadrilateral(EBDF))
Answer: 20
Theorem sequence:
parallelogram_area_formula_sine(1,EBDF)
Problem: As shown in the diagram, AB=3, AC=x, BC=y, ∠ACB=45°, BA⊥CA. Find the value of x.
问题：如图所示，AB=3，AC=x，BC=y，∠ACB=45°，BA垂直于CA。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(ACB),45)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 3
Theorem sequence:
triangle_property_angle_sum(1,ACB)
isosceles_triangle_judgment_angle_equal(1,ACB)
Problem: As shown in the diagram, HF=4*x-22, JO=17+5*y, OP=13+6*y, RH=3*x-9, RH=HF. Find the value of x.
问题：如图所示，HF=4*x-22，JO=17+5*y，OP=13+6*y，RH=3*x-9，RH=HF。求x的值。
Construction (CDL):
Shape(NJ,JC)
Shape(CJ,JR)
Shape(JR,RL)
Shape(LR,RD)
Shape(DR,RH)
Shape(RH,HK)
Shape(KH,HF)
Shape(HF,FA)
Shape(AF,FY)
Shape(YF,FP)
Shape(FP,PV)
Shape(VP,PM)
Shape(MP,PO)
Shape(PO,OE)
Shape(EO,OJ)
Shape(OJ,JN)
Shape(JO,OE,ER,RJ)
Shape(OP,PF,FH,HO)
Collinear(NJRD)
Collinear(EOHK)
Collinear(MPFA)
Collinear(CJOPN)
Collinear(LRHFY)
Conditions (CDL):
Equal(LengthOfLine(HF),4*x-22)
Equal(LengthOfLine(JO),17+5*y)
Equal(LengthOfLine(OP),13+6*y)
Equal(LengthOfLine(RH),3*x-9)
Equal(LengthOfLine(RH),LengthOfLine(HF))
Goal: Value(x)
Answer: 13
Theorem sequence:
Problem: As shown in the diagram, PS=3, RY=5, WX=10, WY=8, XY=6, RP∥XW, BQ is perpendicular to SQ, SY⊥BY, YS⊥PS. Find the length of line SY.
问题：如图所示，PS=3，RY=5，WX=10，WY=8，XY=6，RP∥XW，BQ⊥SQ，SY垂直于BY，YS垂直于PS。求直线SY的长度。
Construction (CDL):
Shape(YR,RB,BY)
Shape(YB,BQ,QS,SY)
Shape(PY,YS,SP)
Shape(BX,XQ,QB)
Shape(SQ,QW,WS)
Collinear(RYP)
Collinear(YSW)
Collinear(RBQ)
Collinear(XQW)
Collinear(YBX)
Collinear(PSQ)
Conditions (CDL):
Equal(LengthOfLine(PS),3)
Equal(LengthOfLine(RY),5)
Equal(LengthOfLine(WX),10)
Equal(LengthOfLine(WY),8)
Equal(LengthOfLine(XY),6)
ParallelBetweenLine(RP,XW)
PerpendicularBetweenLine(BQ,SQ)
PerpendicularBetweenLine(SY,BY)
PerpendicularBetweenLine(YS,PS)
Goal: Value(LengthOfLine(SY))
Answer: 4
Theorem sequence:
cosine_theorem(1,XWY)
parallel_property_collinear_extend(3,RP,XW,Y)
parallel_property_ipsilateral_internal_angle(1,YP,XW)
angle_addition(1,PYS,SYB)
triangle_property_angle_sum(1,PYS)
sine_theorem(1,SPY)
Problem: As shown in the diagram, HQ=y, JQ=2*z, KQ=7, LQ=3, QM=4, QP=2*x-6, J is the midpoint of segment HK, L is the midpoint of segment KM, P is the midpoint of segment MH. Find the value of y.
问题：如图所示，HQ=y，JQ=2*z，KQ=7，LQ=3，QM=4，QP=2*x-6，J是线段HK的中点，L是线段KM的中点，P是线段MH的中点。求y的值。
Construction (CDL):
Shape(HJ,JQ,QH)
Shape(JK,KQ,QJ)
Shape(QK,KL,LQ)
Shape(QL,LM,MQ)
Shape(QM,MP,PQ)
Shape(QP,PH,HQ)
Collinear(HJK)
Collinear(KLM)
Collinear(MPH)
Collinear(HQL)
Collinear(JQM)
Collinear(KQP)
Conditions (CDL):
Equal(LengthOfLine(HQ),y)
Equal(LengthOfLine(JQ),2*z)
Equal(LengthOfLine(KQ),7)
Equal(LengthOfLine(LQ),3)
Equal(LengthOfLine(QM),4)
Equal(LengthOfLine(QP),2*x-6)
IsMidpointOfLine(J,HK)
IsMidpointOfLine(L,KM)
IsMidpointOfLine(P,MH)
Goal: Value(y)
Answer: 6
Theorem sequence:
median_of_triangle_judgment(1,KP,KMH)
median_of_triangle_judgment(1,HL,HKM)
centroid_of_triangle_judgment_intersection(1,Q,MHK,P,L)
centroid_of_triangle_property_line_ratio(1,Q,HKM,L)
Problem: As shown in the diagram, FJ=y+12, FK=3*x+7, GK=4*x-1, HJ=2*y-5, HJ=FJ, FK⊥JK, KG is perpendicular to HG. Find the value of y.
问题：如图所示，FJ=y+12，FK=3*x+7，GK=4*x-1，HJ=2*y-5，HJ=FJ，FK⊥JK，KG垂直于HG。求y的值。
Construction (CDL):
Shape(JF,FK,KJ)
Shape(JK,KG,GH,HJ)
Collinear(FKG)
Collinear(FJH)
Conditions (CDL):
Equal(LengthOfLine(FJ),y+12)
Equal(LengthOfLine(FK),3*x+7)
Equal(LengthOfLine(GK),4*x-1)
Equal(LengthOfLine(HJ),2*y-5)
Equal(LengthOfLine(HJ),LengthOfLine(FJ))
PerpendicularBetweenLine(FK,JK)
PerpendicularBetweenLine(KG,HG)
Goal: Value(y)
Answer: 17
Theorem sequence:
Problem: As shown in the diagram, AB=y, AC=15, BC=x, ∠CBA=60°, AC⊥BC. Find the value of x.
问题：如图所示，AB=y，AC=15，BC=x，∠CBA=60°，AC垂直于BC。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),15)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(CBA),60)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 5*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,CBA)
sine_theorem(1,CBA)
Problem: As shown in the diagram, the measure of arc AJL is 92. Find the measure of ∠LKJ.
问题：如图所示，弧AJL的角度为92。求∠LKJ的大小。
Construction (CDL):
Shape(AKJ,JK)
Shape(AJL,LK,KJ)
Shape(ALK,KL)
Cocircular(A,KJL)
Conditions (CDL):
Equal(MeasureOfArc(AJL),92)
Goal: Value(MeasureOfAngle(LKJ))
Answer: 46
Theorem sequence:
arc_property_circumference_angle_external(1,AJL,K)
Problem: As shown in the diagram, ∠WZY=35°, ∠XWZ=13*x+14°, ∠YXW=13*x+24°, quadrilateral XWZY is a kite. Find the measure of ∠ZYX.
问题：如图所示，∠WZY=35°，∠XWZ=13*x+14°，∠YXW=13*x+24°，XWZY是风筝形。求∠ZYX的大小。
Construction (CDL):
Shape(XW,WU,UX)
Shape(UW,WZ,ZU)
Shape(UZ,ZY,YU)
Shape(UY,YX,XU)
Collinear(ZUX)
Collinear(WUY)
Conditions (CDL):
Equal(MeasureOfAngle(WZY),35)
Equal(MeasureOfAngle(XWZ),13*x+14)
Equal(MeasureOfAngle(YXW),13*x+24)
Kite(XWZY)
Goal: Value(MeasureOfAngle(ZYX))
Answer: 105
Theorem sequence:
kite_property_opposite_angle_equal(1,XWZY)
quadrilateral_property_angle_sum(1,XWZY)