SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=30, CD=30, the measure of arc XCZ is 40, X is the center of ⊙X, AM⊥MM, DN is perpendicular to ZN. Find the length of line CN.
问题：如图所示，AB=30，CD=30，弧XCZ的角度为40，X是圆X的圆心，AM⊥MM，DN垂直于ZN。求直线CN的长度。
Construction (CDL):
Shape(AM,MY,XYA)
Shape(YM,MB,XBY)
Shape(CX,XM,MA,XAC)
Shape(BM,MX,XD,XDB)
Shape(XC,CN,NX)
Shape(XN,ND,DX)
Shape(ZN,NC,XCZ)
Shape(DN,NZ,XZD)
Collinear(YMX)
Collinear(XNZ)
Collinear(AMB)
Collinear(CND)
Cocircular(X,CZDBYA)
Conditions (CDL):
Equal(LengthOfLine(AB),30)
Equal(LengthOfLine(CD),30)
Equal(MeasureOfArc(XCZ),40)
IsCentreOfCircle(X,X)
PerpendicularBetweenLine(AM,MY)
PerpendicularBetweenLine(DN,ZN)
Goal: Value(LengthOfLine(CN))
Answer: 15
Theorem sequence:
line_addition(1,CN,ND)
radius_of_circle_property_length_equal(1,XC,X)
radius_of_circle_property_length_equal(1,XD,X)
isosceles_triangle_judgment_line_equal(1,XCD)
isosceles_triangle_property_angle_equal(1,XCD)
adjacent_complementary_angle(1,XND,DNZ)
adjacent_complementary_angle(1,CNX,XND)
mirror_congruent_triangle_judgment_aas(1,XCN,XND)
mirror_congruent_triangle_property_line_equal(1,XCN,XND)
Problem: As shown in the diagram, AS=4*z, AT=2*x-5, AU=3, DA=6, EA=y, FA=5, S bisects segment DE, T is the midpoint of segment EF, U is the midpoint of segment DF. Find the value of z.
问题：如图所示，AS=4*z，AT=2*x-5，AU=3，DA=6，EA=y，FA=5，S平分线段DE，T是线段EF的中点，U是线段DF的中点。求z的值。
Construction (CDL):
Shape(ES,SA,AE)
Shape(EA,AT,TE)
Shape(SD,DA,AS)
Shape(AD,DU,UA)
Shape(TA,AF,FT)
Shape(AU,UF,FA)
Collinear(ESD)
Collinear(ETF)
Collinear(DUF)
Collinear(SAF)
Collinear(EAU)
Collinear(DAT)
Conditions (CDL):
Equal(LengthOfLine(AS),4*z)
Equal(LengthOfLine(AT),2*x-5)
Equal(LengthOfLine(AU),3)
Equal(LengthOfLine(DA),6)
Equal(LengthOfLine(EA),y)
Equal(LengthOfLine(FA),5)
IsMidpointOfLine(S,DE)
IsMidpointOfLine(T,EF)
IsMidpointOfLine(U,DF)
Goal: Value(z)
Answer: 5/8
Theorem sequence:
median_of_triangle_judgment(1,DT,DFE)
median_of_triangle_judgment(1,FS,FED)
centroid_of_triangle_judgment_intersection(1,A,EDF,S,T)
centroid_of_triangle_property_line_ratio(1,A,FED,S)
Problem: As shown in the diagram, AB=6, A is the center of circle A. Find the area of the ⊙A.
问题：如图所示，AB=6，⊙A的圆心为A。求⊙A的面积。
Construction (CDL):
Shape(AB)
Cocircular(A,B)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
IsCentreOfCircle(A,A)
Goal: Value(AreaOfCircle(A))
Answer: 36*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,AB,A)
circle_area_formula(1,A)
Problem: As shown in the diagram, ∠XYZ=128°, ∠XWY and ∠WYX are the base angles of the isosceles triangle XWY, ∠YZX and ∠ZXY are the base angles of the isosceles triangle YZX. Find the measure of ∠YXW.
问题：如图所示，∠XYZ=128°，三角形XWY为等腰三角形，∠YZX和∠ZXY是等腰三角形YZX的底角。求∠YXW的大小。
Construction (CDL):
Shape(XW,WY,YX)
Shape(XY,YZ,ZX)
Collinear(WYZ)
Conditions (CDL):
Equal(MeasureOfAngle(XYZ),128)
IsoscelesTriangle(XWY)
IsoscelesTriangle(YZX)
Goal: Value(MeasureOfAngle(YXW))
Answer: 76
Theorem sequence:
flat_angle(1,WYZ)
angle_addition(1,WYX,XYZ)
isosceles_triangle_property_angle_equal(1,XWY)
triangle_property_angle_sum(1,XWY)
Problem: As shown in the diagram, ∠HEL=105°, LE is parallel to CG. Find the measure of ∠GEJ.
问题：如图所示，∠HEL=105°，LE∥CG。求∠GEJ的大小。
Construction (CDL):
Shape(HE,EL)
Shape(KJ,ID)
Shape(CG,GP)
Shape(FA,AI)
Shape(EJAG)
Collinear(HEGP)
Collinear(DJAF)
Collinear(LEJB)
Collinear(CGAI)
Conditions (CDL):
Equal(MeasureOfAngle(HEL),105)
ParallelBetweenLine(LE,CG)
Goal: Value(MeasureOfAngle(GEJ))
Answer: 105
Theorem sequence:
vertical_angle(1,HEL,GEJ)
Problem: As shown in the diagram, AB=15, CA=25, CB=20, CB⊥AB. Find the value of tan(AC).
问题：如图所示，AB=15，CA=25，CB=20，CB⊥AB。求tan(AC)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(CA),25)
Equal(LengthOfLine(CB),20)
PerpendicularBetweenLine(CB,AB)
Goal: Value(Tan(MeasureOfAngle(ACB)))
Answer: 3/4
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, AH=15, CA=x, CG=8, CH=18, ∠GAC=∠HAG. Find the value of x.
问题：如图所示，AH=15，CA=x，CG=8，CH=18，∠GAC=∠HAG。求x的值。
Construction (CDL):
Shape(AC,CG,GA)
Shape(AG,GH,HA)
Collinear(CGH)
Conditions (CDL):
Equal(LengthOfLine(AH),15)
Equal(LengthOfLine(CA),x)
Equal(LengthOfLine(CG),8)
Equal(LengthOfLine(CH),18)
Equal(MeasureOfAngle(GAC),MeasureOfAngle(HAG))
Goal: Value(x)
Answer: 12
Theorem sequence:
line_addition(1,CG,GH)
bisector_of_angle_judgment_angle_equal(1,AG,HAC)
bisector_of_angle_property_line_ratio(1,AG,HAC)
Problem: As shown in the diagram, ∠DAE=33°, the measure of ⌒FBC is x, the measure of arc FDE is 51. Find the value of x.
问题：如图所示，∠DAE=33°，⌒FBC的角度为x，⌒FDE的角度为51。求x的值。
Construction (CDL):
Shape(DA,AE,FDE)
Shape(FEB,BE)
Shape(FDE,EB,FBC,CD)
Shape(FCD,DC)
Collinear(ADC)
Collinear(AEB)
Cocircular(F,DEBC)
Conditions (CDL):
Equal(MeasureOfAngle(DAE),33)
Equal(MeasureOfArc(FBC),x)
Equal(MeasureOfArc(FDE),51)
Goal: Value(x)
Answer: 117
Theorem sequence:
circle_property_circular_power_segment_and_segment_angle(1,ADC,AEB,F)
Problem: As shown in the diagram, PQ=25, RS=10, RT=x, SQ=5, PQ∥TS, TR⊥SR. Find the length of line ST.
问题：如图所示，PQ=25，RS=10，RT=x，SQ=5，PQ平行于TS，TR垂直于SR。求直线ST的长度。
Construction (CDL):
Shape(TR,RS,ST)
Shape(PT,TS,SQ,QP)
Collinear(PTR)
Collinear(RSQ)
Conditions (CDL):
Equal(LengthOfLine(PQ),25)
Equal(LengthOfLine(RS),10)
Equal(LengthOfLine(RT),x)
Equal(LengthOfLine(SQ),5)
ParallelBetweenLine(PQ,TS)
PerpendicularBetweenLine(TR,SR)
Goal: Value(LengthOfLine(ST))
Answer: 50/3
Theorem sequence:
line_addition(1,RS,SQ)
parallel_property_corresponding_angle(1,ST,QP,R)
similar_triangle_judgment_aa(1,TRS,PRQ)
similar_triangle_property_line_ratio(1,TRS,PRQ)
similar_triangle_property_line_ratio(1,RST,RQP)
Problem: As shown in the diagram, ∠FJE=75°, FM∥BD, FM∥JE. Find the measure of ∠BFM.
问题：如图所示，∠FJE=75°，FM平行于BD，FM∥JE。求∠BFM的大小。
Construction (CDL):
Shape(LB,BD)
Shape(GB,BL)
Shape(DB,BF)
Shape(FB,BG)
Shape(BF,FM)
Shape(IF,FB)
Shape(MF,FJ)
Shape(JF,FI)
Shape(FJ,FE)
Shape(CJ,JF)
Shape(EJ,JT)
Shape(TJ,JC)
Collinear(LBFJT)
Collinear(DBG)
Collinear(MFI)
Collinear(EJC)
Conditions (CDL):
Equal(MeasureOfAngle(FJE),75)
ParallelBetweenLine(FM,BD)
ParallelBetweenLine(FM,JE)
Goal: Value(MeasureOfAngle(BFM))
Answer: 75
Theorem sequence:
parallel_property_corresponding_angle(1,FM,JE,B)
Problem: As shown in the diagram, AB=x+1, AC=sqrt(2), BC=x, BC⊥AC. Find the value of x.
问题：如图所示，AB=x+1，AC=sqrt(2)，BC=x，BC⊥AC。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),x+1)
Equal(LengthOfLine(AC),sqrt(2))
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 1/2
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, MJ=2*x-5, NJ=x-2, NM=3*x-9, ∠MJN and ∠JNM are the base angles of the isosceles △ MJN. Find the value of x.
问题：如图所示，MJ=2*x-5，NJ=x-2，NM=3*x-9，∠MJN和∠JNM是等腰△MJN的底角。求x的值。
Construction (CDL):
Shape(MJ,JN,NM)
Conditions (CDL):
Equal(LengthOfLine(MJ),2*x-5)
Equal(LengthOfLine(NJ),x-2)
Equal(LengthOfLine(NM),3*x-9)
IsoscelesTriangle(MJN)
Goal: Value(x)
Answer: 4
Theorem sequence:
Problem: As shown in the diagram, CH=14, MD=6, OH=4, Sub(AreaOfQuadrilateral(MHCD)=AreaOfSector(OHM)), O is the center of circle O, the diameter of ⊙O is MH, CH is the tangent to ⊙O, the tangent to circle O is DM. Find the value of x.
问题：如图所示，CH=14，MD=6，OH=4，Sub(AreaOfQuadrilateral(MHCD)=AreaOfSector(OHM))，O是圆O的圆心，圆O的直径为MH，CH是圆O的切线，⊙O的切线为DM。求x的值。
Construction (CDL):
Shape(DM,MO,OH,HC,CD)
Shape(MO,OH,OHM)
Collinear(MOH)
Cocircular(O,HM)
Conditions (CDL):
Equal(LengthOfLine(CH),14)
Equal(LengthOfLine(MD),6)
Equal(LengthOfLine(OH),4)
Equal(Sub(AreaOfQuadrilateral(MHCD),AreaOfSector(OHM)),x)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(MH,O)
IsTangentOfCircle(CH,O)
IsTangentOfCircle(DM,O)
Goal: Value(x)
Answer: 80-8*pi
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DM,O,O)
tangent_of_circle_property_perpendicular(1,CH,O,O)
parallel_judgment_ipsilateral_internal_angle(1,MD,HC)
circle_property_length_of_radius_and_diameter(1,O)
flat_angle(1,MOH)
diameter_of_circle_property_length_equal(1,MH,O)
arc_property_center_angle(1,OHM,O)
trapezoid_judgment_parallel(1,MHCD)
radius_of_circle_property_length_equal(1,OH,O)
sector_area_formula(1,OHM)
right_trapezoid_judgment_right_angle(1,MHCD)
right_trapezoid_area_formular(1,MHCD)
Problem: As shown in the diagram, AB=4, AD=7, BX=5, FX=x. Find the value of x.
问题：如图所示，AB=4，AD=7，BX=5，FX=x。求x的值。
Construction (CDL):
Shape(CXF,FC)
Shape(CFD,DC,CA,AX,XF)
Shape(CDA,AD)
Shape(CAX,AB,BX)
Collinear(BXF)
Collinear(BACD)
Cocircular(C,XFDA)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AD),7)
Equal(LengthOfLine(BX),5)
Equal(LengthOfLine(FX),x)
Goal: Value(x)
Answer: 19/5
Theorem sequence:
line_addition(1,DA,AB)
line_addition(1,FX,XB)
circle_property_circular_power_segment_and_segment_line(1,BXF,BAD,C)
Problem: As shown in the diagram, DA=10, DC=4, DE=5, AB and DC are opposite sides of the parallelogram ADCB, DE is perpendicular to BE. Find the perimeter of quadrilateral ADCB.
问题：如图所示，DA=10，DC=4，DE=5，四边形ADCB是平行四边形，DE⊥BE。求ADCB的周长。
Construction (CDL):
Shape(AD,DE,EB,BA)
Shape(DC,CE,ED)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(DA),10)
Equal(LengthOfLine(DC),4)
Equal(LengthOfLine(DE),5)
Parallelogram(ADCB)
PerpendicularBetweenLine(DE,BE)
Goal: Value(PerimeterOfQuadrilateral(ADCB))
Answer: 28
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ADCB)
parallelogram_property_opposite_line_equal(1,DCBA)
quadrilateral_perimeter_formula(1,ADCB)
Problem: As shown in the diagram, BR=x, SB=2, SQ=4, R is the center of circle R, the tangent to circle R is SQ. Find the value of x.
问题：如图所示，BR=x，SB=2，SQ=4，圆R的圆心为R，圆O的切线为SQ。求x的值。
Construction (CDL):
Shape(QS,SB,RQB)
Shape(RQ,RQB,BR)
Shape(RBQ,QR,RB)
Collinear(SBR)
Cocircular(R,QB)
Conditions (CDL):
Equal(LengthOfLine(BR),x)
Equal(LengthOfLine(SB),2)
Equal(LengthOfLine(SQ),4)
IsCentreOfCircle(R,R)
IsTangentOfCircle(SQ,R)
Goal: Value(x)
Answer: 3
Theorem sequence:
line_addition(1,SB,BR)
tangent_of_circle_property_perpendicular(1,SQ,R,R)
right_triangle_judgment_angle(1,RQS)
radius_of_circle_property_length_equal(1,RQ,R)
radius_of_circle_property_length_equal(1,RB,R)
right_triangle_property_pythagorean(1,RQS)
Problem: As shown in the diagram, DB=15, EB=6, Sub(AreaOfQuadrilateral(ADBE)=AreaOfCircle(C)), C is the center of circle C, FG is the diameter of circle C, the tangent to circle C is AF, DG is the tangent to circle C, quadrilateral ADBE is a rectangle. Find the value of x.
问题：如图所示，DB=15，EB=6，Sub(AreaOfQuadrilateral(ADBE)=AreaOfCircle(C))，C是圆C的圆心，圆C的直径为FG，圆O的切线为AF，圆O的切线为DG，四边形ADBE是长方形。求x的值。
Construction (CDL):
Shape(AD,DG,GC,CF,FA)
Shape(FC,CG,GB,BE,EF)
Shape(CGF)
Collinear(EFA)
Collinear(DGB)
Collinear(FCG)
Cocircular(C,FG)
Conditions (CDL):
Equal(LengthOfLine(DB),15)
Equal(LengthOfLine(EB),6)
Equal(Sub(AreaOfQuadrilateral(ADBE),AreaOfCircle(C)),x)
IsCentreOfCircle(C,C)
IsDiameterOfCircle(FG,C)
IsTangentOfCircle(AF,C)
IsTangentOfCircle(DG,C)
Rectangle(ADBE)
Goal: Value(x)
Answer: 90-9*pi
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AF,C,C)
parallel_judgment_ipsilateral_internal_angle(1,FG,AD)
parallel_property_collinear_extend(3,AE,DB,F)
parallel_property_collinear_extend(3,BD,FA,G)
parallelogram_judgment_parallel_and_parallel(1,ADGF)
parallelogram_property_opposite_line_equal(1,ADBE)
parallelogram_property_opposite_line_equal(1,ADGF)
diameter_of_circle_property_length_equal(1,FG,C)
circle_property_length_of_radius_and_diameter(1,C)
circle_area_formula(1,C)
parallelogram_area_formula_sine(1,ADBE)
Problem: As shown in the diagram, ∠BXE=5/2*x+15°, ∠CAD=3*x-9°, FD∥EG. Find the value of x.
问题：如图所示，∠BXE=5/2*x+15°，∠CAD=3*x-9°，FD平行于EG。求x的值。
Construction (CDL):
Shape(CA,AD)
Shape(FA,AC)
Shape(DA,AX)
Shape(XA,AF)
Shape(AX,XG)
Shape(EX,XA)
Shape(GX,XB)
Shape(BX,XE)
Collinear(CAXB)
Collinear(DAF)
Collinear(GXE)
Conditions (CDL):
Equal(MeasureOfAngle(BXE),5/2*x+15)
Equal(MeasureOfAngle(CAD),3*x-9)
ParallelBetweenLine(FD,EG)
Goal: Value(x)
Answer: 48
Theorem sequence:
parallel_property_collinear_extend(3,FD,EG,A)
parallel_property_collinear_extend(3,GE,AF,X)
parallel_property_corresponding_angle(1,XE,AF,B)
vertical_angle(1,CAD,XAF)
Problem: As shown in the diagram, BD=13, BI=12, ID=5, BI⊥DI, BDCG is a square, quadrilateral EFIB is a square, IAHD is a square. Find the sum of the area of △BID, the area of quadrilateral IAHD, the area of EFIB and the area of BDCG.
问题：如图所示，BD=13，BI=12，ID=5，BI垂直于DI，BDCG是正方形，EFIB是正方形，IAHD是正方形。求三角形BID的面积、IAHD的面积、四边形EFIB的面积与BDCG的面积之和。
Construction (CDL):
Shape(BE,EF,FJ,JI,IB)
Shape(BI,ID,DB)
Shape(GB,BD,DC,CG)
Shape(DI,IA,AH,HD)
Collinear(FJID)
Collinear(BIA)
Conditions (CDL):
Equal(LengthOfLine(BD),13)
Equal(LengthOfLine(BI),12)
Equal(LengthOfLine(ID),5)
PerpendicularBetweenLine(BI,DI)
Square(BDCG)
Square(EFIB)
Square(IAHD)
Goal: Value(Add(AreaOfTriangle(BID),AreaOfQuadrilateral(IAHD),AreaOfQuadrilateral(EFIB),AreaOfQuadrilateral(BDCG)))
Answer: 368
Theorem sequence:
triangle_area_formula_sine(1,IDB)
parallelogram_area_formula_sine(1,EFIB)
parallelogram_area_formula_sine(1,IAHD)
parallelogram_area_formula_sine(1,BDCG)
Problem: As shown in the diagram, the measure of ⌒DEA is 100, the center of circle D is D, the tangent to circle D is CA. Find the measure of ∠EAC.
问题：如图所示，⌒DEA的角度为100，D是⊙D的圆心，⊙O的切线为CA。求∠EAC的大小。
Construction (CDL):
Shape(BE)
Shape(AC)
Shape(DEA,AE)
Shape(EA,AC)
Shape(DA,AC)
Shape(DE,EA,AD)
Shape(DAE,EA)
Collinear(BEA)
Cocircular(D,EA)
Conditions (CDL):
Equal(MeasureOfArc(DEA),100)
IsCentreOfCircle(D,D)
IsTangentOfCircle(CA,D)
Goal: Value(MeasureOfAngle(EAC))
Answer: 130
Theorem sequence:
arc_property_center_angle(1,DEA,D)
triangle_property_angle_sum(1,DEA)
radius_of_circle_property_length_equal(1,DA,D)
radius_of_circle_property_length_equal(1,DE,D)
isosceles_triangle_judgment_line_equal(1,DEA)
isosceles_triangle_property_angle_equal(1,DEA)
tangent_of_circle_property_perpendicular(1,CA,D,D)
angle_addition(1,EAD,DAC)
Problem: As shown in the diagram, AE=10, BE=11, CE=y-7, DE=x+6, AD and BC are opposite sides of the ▱ ABCD. Find the value of x.
问题：如图所示，AE=10，BE=11，CE=y-7，DE=x+6，ABCD是▱。求x的值。
Construction (CDL):
Shape(EA,AB,BE)
Shape(EB,BC,CE)
Shape(EC,CD,DE)
Shape(ED,DA,AE)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(AE),10)
Equal(LengthOfLine(BE),11)
Equal(LengthOfLine(CE),y-7)
Equal(LengthOfLine(DE),x+6)
Parallelogram(ABCD)
Goal: Value(x)
Answer: 5
Theorem sequence:
parallelogram_property_diagonal_bisection(1,BCDA,E)
Problem: As shown in the diagram, ∠ANE=30°, ∠CEK=110°, ∠ENJ=130°, ∠NCF=∠CFN. Find the measure of ∠EAN.
问题：如图所示，∠ANE=30°，∠CEK=110°，∠ENJ=130°，∠NCF=∠CFN。求∠EAN的大小。
Construction (CDL):
Shape(AG)
Shape(AD)
Shape(EK)
Shape(NJ)
Shape(NH)
Shape(CA,AE,EC)
Shape(EA,AN,NE)
Shape(AF,FN,NA)
Shape(CE,EK)
Collinear(GANH)
Collinear(CEN)
Collinear(DAEK)
Conditions (CDL):
Equal(MeasureOfAngle(ANE),30)
Equal(MeasureOfAngle(CEK),110)
Equal(MeasureOfAngle(ENJ),130)
Equal(MeasureOfAngle(NCF),MeasureOfAngle(CFN))
Goal: Value(MeasureOfAngle(EAN))
Answer: 40
Theorem sequence:
vertical_angle(1,CEK,NEA)
triangle_property_angle_sum(1,EAN)
Problem: As shown in the diagram, ∠GKJ=x+21°, ∠HGK=4*y-11°, ∠JHG=2*x°, ∠KJH=3*y+9°. Find the measure of ∠JHG.
问题：如图所示，∠GKJ=x+21°，∠HGK=4*y-11°，∠JHG=2*x°，∠KJH=3*y+9°。求∠JHG的大小。
Construction (CDL):
Shape(GJ,JH,HG)
Shape(JG,GK,KJ)
Shape(OJH,HJ)
Shape(OHG,GH)
Shape(OGK,KG)
Shape(OKJ,JK)
Cocircular(O,JHGK)
Conditions (CDL):
Equal(MeasureOfAngle(GKJ),x+21)
Equal(MeasureOfAngle(HGK),4*y-11)
Equal(MeasureOfAngle(JHG),2*x)
Equal(MeasureOfAngle(KJH),3*y+9)
Goal: Value(MeasureOfAngle(JHG))
Answer: 106
Theorem sequence:
arc_property_circumference_angle_external(1,OJG,K)
arc_property_circumference_angle_internal(1,OJG,H)
Problem: As shown in the diagram, AB=26, AD=12, DG=9/2, EF=8, GF=14, ∠AGF=108°, quadrilateral AGFE is similar to quadrilateral ADCB. Find the length of line BC.
问题：如图所示，AB=26，AD=12，DG=9/2，EF=8，GF=14，∠AGF=108°，AGFE相似于ADCB。求直线BC的长度。
Construction (CDL):
Shape(AG,GF,FE,EA)
Shape(GD,DC,CB,BE,EF,FG)
Collinear(AGD)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DG),9/2)
Equal(LengthOfLine(EF),8)
Equal(LengthOfLine(GF),14)
Equal(MeasureOfAngle(AGF),108)
SimilarBetweenQuadrilateral(AGFE,ADCB)
Goal: Value(LengthOfLine(BC))
Answer: 64/5
Theorem sequence:
line_addition(1,AG,GD)
similar_quadrilateral_property_line_ratio(1,AGFE,ADCB)
similar_quadrilateral_property_line_ratio(1,FEAG,CBAD)
Problem: As shown in the diagram, the length of ⌒BCD is 3/2, BD=1/2, the center of circle B is B. Find the measure of ⌒BCD.
问题：如图所示，⌒BCD的长度为3/2，BD=1/2，B是⊙B的圆心。求⌒BCD的角度。
Construction (CDL):
Shape(BCD,DB,BC)
Shape(BDC,DB,BD)
Cocircular(B,CD)
Conditions (CDL):
Equal(LengthOfArc(BCD),3/2)
Equal(LengthOfLine(BD),1/2)
IsCentreOfCircle(B,B)
Goal: Value(MeasureOfArc(BCD))
Answer: 540/pi
Theorem sequence:
radius_of_circle_property_length_equal(1,BD,B)
arc_length_formula(1,BCD)
Problem: As shown in the diagram, XY=2*x+3, XZ=8*x-4, YZ=4*x-1, triangleYXZ is an isosceles triangle. Find the length of line YZ.
问题：如图所示，XY=2*x+3，XZ=8*x-4，YZ=4*x-1，∠YXZ和∠XZY是等腰△YXZ的底角。求直线YZ的长度。
Construction (CDL):
Shape(YX,XZ,ZY)
Conditions (CDL):
Equal(LengthOfLine(XY),2*x+3)
Equal(LengthOfLine(XZ),8*x-4)
Equal(LengthOfLine(YZ),4*x-1)
IsoscelesTriangle(YXZ)
Goal: Value(LengthOfLine(YZ))
Answer: 7
Theorem sequence:
Problem: As shown in the diagram, BA=x+2, CB=6, DA=8, DE=5, EB is parallel to DC. Find the length of line AC.
问题：如图所示，BA=x+2，CB=6，DA=8，DE=5，EB∥DC。求直线AC的长度。
Construction (CDL):
Shape(AE,EB,BA)
Shape(ED,DC,CB,BE)
Collinear(AED)
Collinear(ABC)
Conditions (CDL):
Equal(LengthOfLine(BA),x+2)
Equal(LengthOfLine(CB),6)
Equal(LengthOfLine(DA),8)
Equal(LengthOfLine(DE),5)
ParallelBetweenLine(EB,DC)
Goal: Value(LengthOfLine(AC))
Answer: 48/5
Theorem sequence:
line_addition(1,AE,ED)
line_addition(1,AB,BC)
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
similar_triangle_property_line_ratio(1,BAE,CAD)
similar_triangle_property_line_ratio(1,EBA,DCA)
Problem: As shown in the diagram, CB=4, ∠CBO=45°, CA and BO are opposite sides of the ▱ CBOA, BO⊥CO. Find the perimeter of CBOA.
问题：如图所示，CB=4，∠CBO=45°，四边形CBOA是平行四边形，BO垂直于CO。求四边形CBOA的周长。
Construction (CDL):
Shape(CB,BO,OC)
Shape(CO,OA,AC)
Conditions (CDL):
Equal(LengthOfLine(CB),4)
Equal(MeasureOfAngle(CBO),45)
Parallelogram(CBOA)
PerpendicularBetweenLine(BO,CO)
Goal: Value(PerimeterOfQuadrilateral(CBOA))
Answer: 4*sqrt(2)+8
Theorem sequence:
right_triangle_judgment_angle(1,BOC)
triangle_property_angle_sum(1,BOC)
isosceles_triangle_judgment_angle_equal(1,OCB)
right_triangle_property_pythagorean(1,BOC)
parallelogram_property_opposite_line_equal(1,CBOA)
parallelogram_property_opposite_line_equal(1,BOAC)
quadrilateral_perimeter_formula(1,CBOA)
Problem: As shown in the diagram, ∠WZY=a+2°, ∠XWZ=1/2*a+8°, ∠YXW=a°, ∠ZYX=a-28°. Find the measure of ∠XWZ.
问题：如图所示，∠WZY=a+2°，∠XWZ=1/2*a+8°，∠YXW=a°，∠ZYX=a-28°。求∠XWZ的大小。
Construction (CDL):
Shape(XW,WZ,ZY,YX)
Conditions (CDL):
Equal(MeasureOfAngle(WZY),a+2)
Equal(MeasureOfAngle(XWZ),1/2*a+8)
Equal(MeasureOfAngle(YXW),a)
Equal(MeasureOfAngle(ZYX),a-28)
Goal: Value(MeasureOfAngle(XWZ))
Answer: 62
Theorem sequence:
quadrilateral_property_angle_sum(1,XWZY)
Problem: As shown in the diagram, KN=9, LN=16, PM=Mul(LengthOfLine(KP),2), LN is parallel to RQ, KN is perpendicular to QN, RM is perpendicular to PM. Find the length of line KM.
问题：如图所示，KN=9，LN=16，PM=Mul(LengthOfLine(KP),2)，LN∥RQ，KN⊥QN，RM⊥PM。求直线KM的长度。
Construction (CDL):
Shape(LR,RQ,QN,NL)
Shape(KN,NQ,QP,PK)
Shape(RM,MQ,QR)
Shape(QM,MP,PQ)
Collinear(LNK)
Collinear(RQP)
Collinear(LRM)
Collinear(KPM)
Collinear(NQM)
Conditions (CDL):
Equal(LengthOfLine(KN),9)
Equal(LengthOfLine(LN),16)
Equal(LengthOfLine(PM),Mul(LengthOfLine(KP),2))
ParallelBetweenLine(LN,RQ)
PerpendicularBetweenLine(KN,QN)
PerpendicularBetweenLine(RM,PM)
Goal: Value(LengthOfLine(KM))
Answer: 15
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,MKN,LMK)
mirror_similar_triangle_property_angle_equal(1,MKN,LMK)
adjacent_complementary_angle(1,KNQ,QNL)
similar_triangle_judgment_aa(1,KNM,MNL)
similar_triangle_property_line_ratio(1,KNM,MNL)
similar_triangle_property_line_ratio(1,MKN,LMN)
right_triangle_judgment_angle(1,KNM)
right_triangle_property_pythagorean(1,KNM)
Problem: As shown in the diagram, DB=12, ∠DBC=18°, CD⊥BD. Find the perimeter of △BCD.
问题：如图所示，DB=12，∠DBC=18°，CD垂直于BD。求三角形BCD的周长。
Construction (CDL):
Shape(BC,CD,DB)
Conditions (CDL):
Equal(LengthOfLine(DB),12)
Equal(MeasureOfAngle(DBC),18)
PerpendicularBetweenLine(CD,BD)
Goal: Value(PerimeterOfTriangle(BCD))
Answer: 6*(sqrt(10)+3*sqrt(2)+2*sqrt(sqrt(5)+5))/sqrt(sqrt(5)+5)
Theorem sequence:
triangle_property_angle_sum(1,BCD)
sine_theorem(1,BCD)
sine_theorem(1,CDB)
triangle_perimeter_formula(1,BCD)
Problem: As shown in the diagram, ML=3*x+1, MN=4*x-2, ∠LPM=5*y°, triangle LNM is an equilateral triangle, MP is the median of triangle MLN. Find the value of x.
问题：如图所示，ML=3*x+1，MN=4*x-2，∠LPM=5*y°，三角形LNM为等边三角形，MP是三角形MLN的中线。求x的值。
Construction (CDL):
Shape(ML,LP,PM)
Shape(MP,PN,NM)
Collinear(LPN)
Conditions (CDL):
Equal(LengthOfLine(ML),3*x+1)
Equal(LengthOfLine(MN),4*x-2)
Equal(MeasureOfAngle(LPM),5*y)
EquilateralTriangle(LNM)
IsMedianOfTriangle(MP,MLN)
Goal: Value(x)
Answer: 3
Theorem sequence:
Problem: As shown in the diagram, the area of quadrilateral ABCX is 92, AC=x, XB=22, CB and CX are one pair of adjacent sides of the kite ABCX. Find the value of x.
问题：如图所示，ABCX的面积为92，AC=x，XB=22，ABCX是风筝形。求x的值。
Construction (CDL):
Shape(AB,BC,CX,XA)
Shape(BX)
Shape(AC)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ABCX),92)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(XB),22)
Kite(ABCX)
Goal: Value(x)
Answer: 92/11
Theorem sequence:
kite_area_formula_diagonal(1,ABCX)
Problem: As shown in the diagram, AZ=2*b-6, WA=10, XA=a-7, YA=2, WX and YZ are opposite sides of the parallelogram WYZX. Find the value of b.
问题：如图所示，AZ=2*b-6，WA=10，XA=a-7，YA=2，WX和YZ是平行四边形WYZX的一组对边。求b的值。
Construction (CDL):
Shape(AW,WY,YA)
Shape(AY,YZ,ZA)
Shape(AZ,ZX,XA)
Shape(WA,AX,XW)
Collinear(WAZ)
Collinear(XAY)
Conditions (CDL):
Equal(LengthOfLine(AZ),2*b-6)
Equal(LengthOfLine(WA),10)
Equal(LengthOfLine(XA),a-7)
Equal(LengthOfLine(YA),2)
Parallelogram(WYZX)
Goal: Value(b)
Answer: 8
Theorem sequence:
parallelogram_property_diagonal_bisection(1,WYZX,A)
Problem: As shown in the diagram, FH=FL, FK=17, FL=8, the center of ⊙F is F, FL⊥KL, JH is perpendicular to FH. Find the length of line KM.
问题：如图所示，FH=FL，FK=17，FL=8，圆F的圆心为F，FL垂直于KL，JH⊥FH。求直线KM的长度。
Construction (CDL):
Shape(FJG,GJ)
Shape(FMK,KL,LM)
Shape(FKJ,JH,HF,FK)
Shape(FGM,MF,FH,HG)
Shape(KF,FL,LK)
Shape(LF,FM,ML)
Collinear(HFL)
Collinear(JHG)
Collinear(KLM)
Cocircular(F,JGMK)
Conditions (CDL):
Equal(LengthOfLine(FH),LengthOfLine(FL))
Equal(LengthOfLine(FK),17)
Equal(LengthOfLine(FL),8)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(FL,KL)
PerpendicularBetweenLine(JH,FH)
Goal: Value(LengthOfLine(KM))
Answer: 30
Theorem sequence:
adjacent_complementary_angle(1,MLF,FLK)
right_triangle_judgment_angle(1,FLK)
right_triangle_property_pythagorean(1,FLK)
circle_property_chord_perpendicular_bisect_chord(1,F,FL,MK)
line_addition(1,KL,LM)
Problem: As shown in the diagram, BD=3, BD=GD, GDBE is a parallelogram, GD⊥BD. Find the perimeter of GDBE.
问题：如图所示，BD=3，BD=GD，GDBE是平行四边形，GD垂直于BD。求GDBE的周长。
Construction (CDL):
Shape(GD,DB,BE,EG)
Conditions (CDL):
Equal(LengthOfLine(BD),3)
Equal(LengthOfLine(BD),LengthOfLine(GD))
Parallelogram(GDBE)
PerpendicularBetweenLine(GD,BD)
Goal: Value(PerimeterOfQuadrilateral(GDBE))
Answer: 12
Theorem sequence:
parallelogram_property_opposite_line_equal(1,GDBE)
parallelogram_property_opposite_line_equal(1,DBEG)
quadrilateral_perimeter_formula(1,GDBE)
Problem: As shown in the diagram, ∠AFE=x-4°, ∠BAF=x+2°, ∠CBA=x-8°, ∠DCB=x+7°, ∠EDC=x-3°, ∠FED=x+6°. Find the measure of ∠AFE.
问题：如图所示，∠AFE=x-4°，∠BAF=x+2°，∠CBA=x-8°，∠DCB=x+7°，∠EDC=x-3°，∠FED=x+6°。求∠AFE的大小。
Construction (CDL):
Shape(AF,FE,ED,DA)
Shape(BA,AD,DC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(AFE),x-4)
Equal(MeasureOfAngle(BAF),x+2)
Equal(MeasureOfAngle(CBA),x-8)
Equal(MeasureOfAngle(DCB),x+7)
Equal(MeasureOfAngle(EDC),x-3)
Equal(MeasureOfAngle(FED),x+6)
Goal: Value(MeasureOfAngle(AFE))
Answer: 116
Theorem sequence:
angle_addition(1,BAD,DAF)
angle_addition(1,EDA,ADC)
quadrilateral_property_angle_sum(1,AFED)
quadrilateral_property_angle_sum(1,DCBA)