SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BA=4*y-11, CA=2*y+3, DA=3*y+5, EA=2*x+4, DBEC is a parallelogram. Find the value of y.
问题：如图所示，BA=4*y-11，CA=2*y+3，DA=3*y+5，EA=2*x+4，DC和BE是▱DBEC的一组对边。求y的值。
Construction (CDL):
Shape(DB,BA,AD)
Shape(AB,BE,EA)
Shape(AE,EC,CA)
Shape(DA,AC,CD)
Collinear(DAE)
Collinear(BAC)
Conditions (CDL):
Equal(LengthOfLine(BA),4*y-11)
Equal(LengthOfLine(CA),2*y+3)
Equal(LengthOfLine(DA),3*y+5)
Equal(LengthOfLine(EA),2*x+4)
Parallelogram(DBEC)
Goal: Value(y)
Answer: 7
Theorem sequence:
parallelogram_property_diagonal_bisection(1,CDBE,A)
Problem: As shown in the diagram, AB=z, AC=x, AY=17, BC=y, CY=6, AY⊥BY, BA⊥CA. Find the value of y.
问题：如图所示，AB=z，AC=x，AY=17，BC=y，CY=6，AY垂直于BY，BA垂直于CA。求y的值。
Construction (CDL):
Shape(BA,AY,YB)
Shape(AC,CY,YA)
Collinear(BYC)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AY),17)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CY),6)
PerpendicularBetweenLine(AY,BY)
PerpendicularBetweenLine(BA,CA)
Goal: Value(y)
Answer: 325/6
Theorem sequence:
line_addition(1,BY,YC)
adjacent_complementary_angle(1,CYA,AYB)
right_triangle_judgment_angle(1,BAC)
right_triangle_judgment_angle(1,AYB)
right_triangle_judgment_angle(1,CYA)
right_triangle_property_pythagorean(1,CYA)
right_triangle_property_pythagorean(1,AYB)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, ∠VWX=67°, ∠XZY=65°, ∠ZYX=58°. Find the measure of ∠YXZ.
问题：如图所示，∠VWX=67°，∠XZY=65°，∠ZYX=58°。求∠YXZ的大小。
Construction (CDL):
Shape(VW,WX,XV)
Shape(YX,XZ,ZY)
Shape(XW,WZ,ZX)
Collinear(VXZ)
Collinear(YXW)
Conditions (CDL):
Equal(MeasureOfAngle(VWX),67)
Equal(MeasureOfAngle(XZY),65)
Equal(MeasureOfAngle(ZYX),58)
Goal: Value(MeasureOfAngle(YXZ))
Answer: 57
Theorem sequence:
triangle_property_angle_sum(1,YXZ)
Problem: As shown in the diagram, ∠WXY=4*x+15°, ∠ZXV=2*x+65°. Find the measure of ∠WXY.
问题：如图所示，∠WXY=4*x+15°，∠ZXV=2*x+65°。求∠WXY的大小。
Construction (CDL):
Shape(XV,XVZ,ZX)
Shape(XZ,XZY,YX)
Shape(XY,XYW,WX)
Shape(XW,XWV,VX)
Collinear(ZXW)
Collinear(YXV)
Cocircular(X,ZYWV)
Conditions (CDL):
Equal(MeasureOfAngle(WXY),4*x+15)
Equal(MeasureOfAngle(ZXV),2*x+65)
Goal: Value(MeasureOfAngle(WXY))
Answer: 115
Theorem sequence:
vertical_angle(1,ZXV,WXY)
Problem: As shown in the diagram, ∠GFJ=4*x+2°, ∠JHG=9*x-3°, the center of circle J is J, the diameter of circle J is FH. Find the value of x.
问题：如图所示，∠GFJ=4*x+2°，∠JHG=9*x-3°，圆J的圆心为J，⊙J的直径为FH。求x的值。
Construction (CDL):
Shape(HJ,JF,JFH)
Shape(GH,JHG)
Shape(FG,JGF)
Shape(FJ,JH,HG,GF)
Collinear(FJH)
Cocircular(J,FHG)
Conditions (CDL):
Equal(MeasureOfAngle(GFJ),4*x+2)
Equal(MeasureOfAngle(JHG),9*x-3)
IsCentreOfCircle(J,J)
IsDiameterOfCircle(FH,J)
Goal: Value(x)
Answer: 7
Theorem sequence:
diameter_of_circle_property_right_angle(1,HGF,J)
triangle_property_angle_sum(1,HGF)
Problem: As shown in the diagram, AB=y, AC=8*sqrt(3), BC=x, ∠ABC=60°, BC⊥AC. Find the value of y.
问题：如图所示，AB=y，AC=8*sqrt(3)，BC=x，∠ABC=60°，BC⊥AC。求y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),8*sqrt(3))
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ABC),60)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 16
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, ∠ABC=x°, ∠BCD=2*x+18°, ∠CDA=x°, ∠DAB=4*x-26°. Find the measure of ∠DAB.
问题：如图所示，∠ABC=x°，∠BCD=2*x+18°，∠CDA=x°，∠DAB=4*x-26°。求∠DAB的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),x)
Equal(MeasureOfAngle(BCD),2*x+18)
Equal(MeasureOfAngle(CDA),x)
Equal(MeasureOfAngle(DAB),4*x-26)
Goal: Value(MeasureOfAngle(DAB))
Answer: 158
Theorem sequence:
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, AB=13, AC=10, BC=13, BD is perpendicular to CD. Find the area of triangle BAC.
问题：如图所示，AB=13，AC=10，BC=13，BD垂直于CD。求△BAC的面积。
Construction (CDL):
Shape(BA,AD,DB)
Shape(BD,DC,CB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(BC),13)
PerpendicularBetweenLine(BD,CD)
Goal: Value(AreaOfTriangle(BAC))
Answer: 60
Theorem sequence:
cosine_theorem(1,CBA)
triangle_area_formula_sine(1,CBA)
Problem: As shown in the diagram, CD=24, the radius of circle O is 13, the center of circle O is O, BX is perpendicular to DX, DX is perpendicular to OX. Find the length of line OX.
问题：如图所示，CD=24，⊙O的半径为13，O是⊙O的圆心，BX⊥DX，DX⊥OX。求直线OX的长度。
Construction (CDL):
Shape(OD,DX,XO)
Shape(OX,XC,CO)
Shape(OD,ODB,BO)
Shape(OB,OBC,CO)
Collinear(CXD)
Collinear(OXB)
Cocircular(O,BCD)
Conditions (CDL):
Equal(LengthOfLine(CD),24)
Equal(RadiusOfCircle(O),13)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BX,DX)
PerpendicularBetweenLine(DX,OX)
Goal: Value(LengthOfLine(OX))
Answer: 5
Theorem sequence:
radius_of_circle_property_length_equal(1,OD,O)
circle_property_chord_perpendicular_bisect_chord(1,O,OX,DC)
line_addition(1,DX,XC)
right_triangle_judgment_angle(1,DXO)
right_triangle_property_pythagorean(1,DXO)
Problem: As shown in the diagram, AB=y, AD=x, BD=32, BD=BC, ∠ACB=54°, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=y，AD=x，BD=32，BD=BC，∠ACB=54°，BA⊥CA。求x的值。
Construction (CDL):
Shape(BD,DA,AB)
Shape(BA,AC,CB)
Collinear(DAC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BD),32)
Equal(LengthOfLine(BD),LengthOfLine(BC))
Equal(MeasureOfAngle(ACB),54)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 8*sqrt(10-2*sqrt(5))
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,BDC)
isosceles_triangle_property_angle_equal(1,BDC)
adjacent_complementary_angle(1,DAB,BAC)
mirror_congruent_triangle_judgment_aas(1,BDA,BAC)
mirror_congruent_triangle_property_line_equal(1,BDA,BAC)
triangle_property_angle_sum(1,ACB)
sine_theorem(1,BAC)
sine_theorem(1,ACB)
Problem: As shown in the diagram, BM=DM, ∠MTB=28°, M is the center of ⊙M, AD is perpendicular to MD, TB is perpendicular to MB. Find the measure of ∠MAC.
问题：如图所示，BM=DM，∠MTB=28°，⊙M的圆心为M，AD⊥MD，TB⊥MB。求∠MAC的大小。
Construction (CDL):
Shape(MA,AD,DM)
Shape(MT,TB,BM)
Collinear(CDA)
Collinear(TBN)
Cocircular(M,CTNA)
Conditions (CDL):
Equal(LengthOfLine(BM),LengthOfLine(DM))
Equal(MeasureOfAngle(MTB),28)
IsCentreOfCircle(M,M)
PerpendicularBetweenLine(AD,MD)
PerpendicularBetweenLine(TB,MB)
Goal: Value(MeasureOfAngle(MAC))
Answer: 28
Theorem sequence:
radius_of_circle_property_length_equal(1,MA,M)
radius_of_circle_property_length_equal(1,MT,M)
congruent_triangle_judgment_hl(2,TBM,ADM)
congruent_triangle_property_angle_equal(1,TBM,ADM)
Problem: As shown in the diagram, ∠BAD=38°, ∠HCB=33°, BJ∥HC, DA is parallel to BJ, DF is parallel to HE. Find the measure of ∠ABC.
问题：如图所示，∠BAD=38°，∠HCB=33°，BJ∥HC，DA∥BJ，DF∥HE。求∠ABC的大小。
Construction (CDL):
Shape(AB,BJ,JA)
Shape(BC,CJ,JB)
Shape(HC,CB,BA,AD,DH)
Collinear(AJC)
Collinear(GAB)
Collinear(BCI)
Collinear(DAF)
Collinear(HCE)
Conditions (CDL):
Equal(MeasureOfAngle(BAD),38)
Equal(MeasureOfAngle(HCB),33)
ParallelBetweenLine(BJ,HC)
ParallelBetweenLine(DA,BJ)
ParallelBetweenLine(DF,HE)
Goal: Value(MeasureOfAngle(ABC))
Answer: 71
Theorem sequence:
parallel_property_alternate_interior_angle(2,DA,BJ)
parallel_property_alternate_interior_angle(1,BJ,HC)
angle_addition(1,ABJ,JBC)
Problem: As shown in the diagram, ∠EDC=79°, ∠FBD=x-1°, ∠GIB=2*x°, ∠HCI=x+10°. Find the value of x.
问题：如图所示，∠EDC=79°，∠FBD=x-1°，∠GIB=2*x°，∠HCI=x+10°。求x的值。
Construction (CDL):
Shape(ED,DC,CE)
Shape(FB,BD,DF)
Shape(HC,CI,IH)
Shape(BG,GI,IB)
Shape(DB,BI,IC,CD)
Collinear(EDB)
Collinear(CIG)
Collinear(FBI)
Collinear(DCH)
Conditions (CDL):
Equal(MeasureOfAngle(EDC),79)
Equal(MeasureOfAngle(FBD),x-1)
Equal(MeasureOfAngle(GIB),2*x)
Equal(MeasureOfAngle(HCI),x+10)
Goal: Value(x)
Answer: 68
Theorem sequence:
flat_angle(1,EDB)
flat_angle(1,FBI)
flat_angle(1,HCD)
flat_angle(1,GIC)
angle_addition(1,EDH,HDB)
angle_addition(1,FBD,DBI)
angle_addition(1,GIB,BIC)
angle_addition(1,HCI,ICD)
quadrilateral_property_angle_sum(1,DBIC)
Problem: As shown in the diagram, XM=2, XN=6, XY=10, NM∥ZY. Find the length of line NZ.
问题：如图所示，XM=2，XN=6，XY=10，NM平行于ZY。求直线NZ的长度。
Construction (CDL):
Shape(MX,XN,NM)
Shape(MN,NZ,ZY,YM)
Collinear(XMY)
Collinear(XNZ)
Conditions (CDL):
Equal(LengthOfLine(XM),2)
Equal(LengthOfLine(XN),6)
Equal(LengthOfLine(XY),10)
ParallelBetweenLine(NM,ZY)
Goal: Value(LengthOfLine(NZ))
Answer: 24
Theorem sequence:
parallel_property_corresponding_angle(1,NM,ZY,X)
similar_triangle_judgment_aa(1,MXN,YXZ)
similar_triangle_property_line_ratio(1,NMX,ZYX)
similar_triangle_property_line_ratio(1,MXN,YXZ)
line_addition(1,XN,NZ)
Problem: As shown in the diagram, AB=y, AC=21, BC=x, ∠ACB=60°, ∠CBA=30°, BA is perpendicular to CA. Find the value of y.
问题：如图所示，AB=y，AC=21，BC=x，∠ACB=60°，∠CBA=30°，BA垂直于CA。求y的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),21)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ACB),60)
Equal(MeasureOfAngle(CBA),30)
PerpendicularBetweenLine(BA,CA)
Goal: Value(y)
Answer: 21*sqrt(3)
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, AB=16, AC=x-3, BE=20, CD=x+5, DE∥CB. Find the length of line CD.
问题：如图所示，AB=16，AC=x-3，BE=20，CD=x+5，DE平行于CB。求直线CD的长度。
Construction (CDL):
Shape(CA,AB,BC)
Shape(CB,BE,ED,DC)
Collinear(ACD)
Collinear(ABE)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
Equal(LengthOfLine(AC),x-3)
Equal(LengthOfLine(BE),20)
Equal(LengthOfLine(CD),x+5)
ParallelBetweenLine(DE,CB)
Goal: Value(LengthOfLine(CD))
Answer: 40
Theorem sequence:
line_addition(1,AC,CD)
line_addition(1,AB,BE)
parallel_property_corresponding_angle(2,DE,CB,A)
similar_triangle_judgment_aa(1,BCA,EDA)
similar_triangle_property_line_ratio(1,CAB,DAE)
similar_triangle_property_line_ratio(1,BCA,EDA)
Problem: As shown in the diagram, ∠JMK=55°, ∠JMK=∠PMR, ∠MRP=70°, quadrilateral JMLK is a rectangle, MRPL is a rhombus. Find the measure of ∠KML.
问题：如图所示，∠JMK=55°，∠JMK=∠PMR，∠MRP=70°，四边形JMLK是矩形，四边形MRPL是菱形。求∠KML的大小。
Construction (CDL):
Shape(JM,MK,KJ)
Shape(ML,LK,KM)
Shape(MR,RP,PM)
Shape(MP,PL,LM)
Conditions (CDL):
Equal(MeasureOfAngle(JMK),55)
Equal(MeasureOfAngle(JMK),MeasureOfAngle(PMR))
Equal(MeasureOfAngle(MRP),70)
Rectangle(JMLK)
Rhombus(MRPL)
Goal: Value(MeasureOfAngle(KML))
Answer: 35
Theorem sequence:
angle_addition(1,JMK,KML)
Problem: As shown in the diagram, BC=6, ∠DAB=Mul(MeasureOfAngle(CDA),2)°, DABC is a rhombus. Find the measure of ∠ACD.
问题：如图所示，BC=6，∠DAB=Mul(MeasureOfAngle(CDA),2)°，四边形DABC是菱形。求∠ACD的大小。
Construction (CDL):
Shape(DA,AE,ED)
Shape(EA,AB,BE)
Shape(EB,BC,CE)
Shape(DE,EC,CD)
Collinear(DEB)
Collinear(CEA)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
Equal(MeasureOfAngle(DAB),Mul(MeasureOfAngle(CDA),2))
Rhombus(DABC)
Goal: Value(MeasureOfAngle(ACD))
Answer: 60
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,CDAB)
parallelogram_property_opposite_angle_equal(1,DABC)
quadrilateral_property_angle_sum(1,DABC)
kite_property_diagonal_perpendicular_bisection(1,CDAB,E)
perpendicular_bisector_property_bisector(1,CE,DB)
angle_addition(1,BCE,ECD)
Problem: As shown in the diagram, AB=12, AC=16, ED=5, CD∥BE. Find the length of line AE.
问题：如图所示，AB=12，AC=16，ED=5，CD平行于BE。求直线AE的长度。
Construction (CDL):
Shape(AE,EB,BA)
Shape(CB,BE,ED,DC)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),16)
Equal(LengthOfLine(ED),5)
ParallelBetweenLine(CD,BE)
Goal: Value(LengthOfLine(AE))
Answer: 15
Theorem sequence:
line_addition(1,AE,ED)
parallel_property_corresponding_angle(2,CD,BE,A)
similar_triangle_judgment_aa(1,EBA,DCA)
similar_triangle_property_line_ratio(1,EBA,DCA)
similar_triangle_property_line_ratio(1,BAE,CAD)
Problem: As shown in the diagram, MN=3*x-4, NQ=15.4, PN=2*y+5, PQ=11.1, RM=17.9, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, MN and RP are opposite sides of the ▱ MRPN. Find the measure of ∠PNR.
问题：如图所示，MN=3*x-4，NQ=15.4，PN=2*y+5，PQ=11.1，RM=17.9，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，RM和PN是▱MRPN的一组对边。求∠PNR的大小。
Construction (CDL):
Shape(MR,RQ,QM)
Shape(QR,RP,PQ)
Shape(QP,PN,NQ)
Shape(QN,NM,MQ)
Collinear(MQP)
Collinear(RQN)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(NQ),15.4)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),11.1)
Equal(LengthOfLine(RM),17.9)
Equal(LengthOfLine(RP),20)
Equal(LengthOfLine(RQ),3*z-3)
Equal(MeasureOfAngle(MRQ),38)
Equal(MeasureOfAngle(NQP),83)
Equal(MeasureOfAngle(QNM),33)
Parallelogram(MRPN)
Goal: Value(MeasureOfAngle(PNR))
Answer: 38
Theorem sequence:
parallel_property_alternate_interior_angle(1,NP,MR)
Problem: As shown in the diagram, ∠ABC=25*y-20°, ∠BAD=10*x°, ∠EAB=8*y+2°, AE is parallel to BC. Find the value of y.
问题：如图所示，∠ABC=25*y-20°，∠BAD=10*x°，∠EAB=8*y+2°，AE平行于BC。求y的值。
Construction (CDL):
Shape(HA,AE)
Shape(AB,BC)
Shape(BA,AD)
Shape(FB,BG)
Shape(EA,AB)
Collinear(HABF)
Collinear(DAE)
Collinear(GBC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),25*y-20)
Equal(MeasureOfAngle(BAD),10*x)
Equal(MeasureOfAngle(EAB),8*y+2)
ParallelBetweenLine(AE,BC)
Goal: Value(y)
Answer: 6
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AE,BC)
Problem: As shown in the diagram, ∠RPS=130°, ∠TPQ=112°, the radius of ⊙P is 2, P is the center of circle P. Find the length of ⌒PSR.
问题：如图所示，∠RPS=130°，∠TPQ=112°，⊙P的半径为2，圆P的圆心为P。求弧PSR的长度。
Construction (CDL):
Shape(PRQ,QP,PR)
Shape(PQT,TP,PQ)
Shape(PTS,SP,PT)
Shape(PSR,RP,PS)
Cocircular(P,RQST)
Conditions (CDL):
Equal(MeasureOfAngle(RPS),130)
Equal(MeasureOfAngle(TPQ),112)
Equal(RadiusOfCircle(P),2)
IsCentreOfCircle(P,P)
Goal: Value(LengthOfArc(PSR))
Answer: 13*pi/9
Theorem sequence:
arc_length_formula(1,PSR)
arc_property_center_angle(1,PSR,P)
Problem: As shown in the diagram, BA=6, DA=9, FC=6.86, ∠BAF=32°, ∠CBF=40.1°, ∠FAD=20°, CD and BA are opposite sides of the parallelogram DCBA. Find the measure of ∠ADC.
问题：如图所示，BA=6，DA=9，FC=6.86，∠BAF=32°，∠CBF=40.1°，∠FAD=20°，CD和BA是▱DCBA的一组对边。求∠ADC的大小。
Construction (CDL):
Shape(DF,FA,AD)
Shape(AF,FB,BA)
Shape(FC,CB,BF)
Shape(DC,CF,FD)
Collinear(DFB)
Collinear(CFA)
Conditions (CDL):
Equal(LengthOfLine(BA),6)
Equal(LengthOfLine(DA),9)
Equal(LengthOfLine(FC),6.86)
Equal(MeasureOfAngle(BAF),32)
Equal(MeasureOfAngle(CBF),40.1)
Equal(MeasureOfAngle(FAD),20)
Parallelogram(DCBA)
Goal: Value(MeasureOfAngle(ADC))
Answer: 128
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AB,DC)
angle_addition(1,BAF,FAD)
Problem: As shown in the diagram, AC=6, CF=8, E is the center of circle E, FA is the diameter of circle E, FGAC is a rectangle. Find the circumference of the circle E.
问题：如图所示，AC=6，CF=8，圆E的圆心为E，FA是⊙E的直径，四边形FGAC是长方形。求圆E的周长。
Construction (CDL):
Shape(ECF,FC)
Shape(EFG,GF)
Shape(EGA,AG)
Shape(EAC,CA)
Shape(FG,GA,AC,CF)
Shape(FE,EA,AC,CF)
Collinear(FEA)
Cocircular(E,FGCA)
Conditions (CDL):
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(CF),8)
IsCentreOfCircle(E,E)
IsDiameterOfCircle(FA,E)
Rectangle(FGAC)
Goal: Value(PerimeterOfCircle(E))
Answer: 10*pi
Theorem sequence:
right_triangle_judgment_angle(1,ACF)
right_triangle_property_pythagorean(1,ACF)
diameter_of_circle_property_length_equal(1,FA,E)
circle_property_length_of_radius_and_diameter(1,E)
circle_perimeter_formula(1,E)
Problem: As shown in the diagram, BA=45, BC=15, ED=x, EF=3, ∠FDE=∠CAB, BC is perpendicular to AC, EF⊥DF. Find the length of line DE.
问题：如图所示，BA=45，BC=15，ED=x，EF=3，∠FDE=∠CAB，BC垂直于AC，EF垂直于DF。求直线DE的长度。
Construction (CDL):
Shape(FD,DE,EF)
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(BA),45)
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(ED),x)
Equal(LengthOfLine(EF),3)
Equal(MeasureOfAngle(FDE),MeasureOfAngle(CAB))
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(EF,DF)
Goal: Value(LengthOfLine(DE))
Answer: 9
Theorem sequence:
similar_triangle_judgment_aa(1,EFD,BCA)
similar_triangle_property_line_ratio(1,DEF,ABC)
similar_triangle_property_line_ratio(1,FDE,CAB)
Problem: As shown in the diagram, CB=12, DB=20, ∠DBC=60°, BD and CA are opposite sides of the ▱ DBCA, DE⊥CE. Find the perimeter of quadrilateral DBCA.
问题：如图所示，CB=12，DB=20，∠DBC=60°，DA和BC是平行四边形DBCA的一组对边，DE⊥CE。求四边形DBCA的周长。
Construction (CDL):
Shape(DB,BE,ED)
Shape(DE,EC,CA,AD)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(CB),12)
Equal(LengthOfLine(DB),20)
Equal(MeasureOfAngle(DBC),60)
Parallelogram(DBCA)
PerpendicularBetweenLine(DE,CE)
Goal: Value(PerimeterOfQuadrilateral(DBCA))
Answer: 64
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DBCA)
parallelogram_property_opposite_line_equal(1,BCAD)
quadrilateral_perimeter_formula(1,DBCA)
Problem: As shown in the diagram, AD=14, ∠AOB=80°, O is the center of circle O, the diameter of circle O is DA. Find the length of arc OBA.
问题：如图所示，AD=14，∠AOB=80°，O是⊙O的圆心，DA是圆O的直径。求⌒OBA的长度。
Construction (CDL):
Shape(OAD,DO,OA)
Shape(ODB,BO,OD)
Shape(OBA,AO,OB)
Collinear(DOA)
Cocircular(O,DBA)
Conditions (CDL):
Equal(LengthOfLine(AD),14)
Equal(MeasureOfAngle(AOB),80)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(DA,O)
Goal: Value(LengthOfArc(OBA))
Answer: 28*pi/9
Theorem sequence:
diameter_of_circle_property_length_equal(1,DA,O)
circle_property_length_of_radius_and_diameter(1,O)
arc_length_formula(1,OBA)
arc_property_center_angle(1,OBA,O)
Problem: As shown in the diagram, AB=2*x-7, AC=x-3, BA=BC, BC=4*x-21, BA and BC are the legs of the isosceles △ BAC. Find the length of line BA.
问题：如图所示，AB=2*x-7，AC=x-3，BA=BC，BC=4*x-21，△BAC为等腰△。求直线BA的长度。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x-7)
Equal(LengthOfLine(AC),x-3)
Equal(LengthOfLine(BA),LengthOfLine(BC))
Equal(LengthOfLine(BC),4*x-21)
IsoscelesTriangle(BAC)
Goal: Value(LengthOfLine(BA))
Answer: 7
Theorem sequence:
Problem: As shown in the diagram, ∠AGC=60°, DG⊥AG. Find the measure of ∠BGE.
问题：如图所示，∠AGC=60°，DG垂直于AG。求∠BGE的大小。
Construction (CDL):
Shape(GCA,AG,GC)
Shape(GAD,DG,GE)
Shape(GDE,EG,GD)
Shape(GEB,BG,GE)
Shape(GBC,CG,GB)
Collinear(AGB)
Collinear(CGE)
Cocircular(G,ADEBC)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),60)
PerpendicularBetweenLine(DG,AG)
Goal: Value(MeasureOfAngle(BGE))
Answer: 60
Theorem sequence:
vertical_angle(1,AGC,BGE)
Problem: As shown in the diagram, AB=73, AC=x, ∠ABC=60°, ∠CAB=55°. Find the value of x.
问题：如图所示，AB=73，AC=x，∠ABC=60°，∠CAB=55°。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),73)
Equal(LengthOfLine(AC),x)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),55)
Goal: Value(x)
Answer: 73*sqrt(3)/(2*sin(13*pi/36))
Theorem sequence:
sine_theorem(1,ABC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, AB=5, AC=3*sqrt(2), BC=x, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=5，AC=3*sqrt(2)，BC=x，AC垂直于BC。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),3*sqrt(2))
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: sqrt(7)
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, CB=14, ED=7, ∠EDB=60°, CADB is a parallelogram, AE is perpendicular to BE. Find the area of quadrilateral CADB.
问题：如图所示，CB=14，ED=7，∠EDB=60°，CB和AD是▱CADB的一组对边，AE垂直于BE。求CADB的面积。
Construction (CDL):
Shape(CA,AE,EB,BC)
Shape(BE,ED,DB)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(CB),14)
Equal(LengthOfLine(ED),7)
Equal(MeasureOfAngle(EDB),60)
Parallelogram(CADB)
PerpendicularBetweenLine(AE,BE)
Goal: Value(AreaOfQuadrilateral(CADB))
Answer: 98*sqrt(3)
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BCAD)
parallelogram_area_formula_sine(1,ADBC)
adjacent_complementary_angle(1,AEB,BED)
sine_theorem(1,DBE)
triangle_property_angle_sum(1,BED)
Problem: As shown in the diagram, XY=18+2*y, YW=ZW, ZX=5*y-6, ZW is perpendicular to XW. Find the length of line XZ.
问题：如图所示，XY=18+2*y，YW=ZW，ZX=5*y-6，ZW垂直于XW。求直线XZ的长度。
Construction (CDL):
Shape(XZ,ZW,WX)
Shape(XW,WY,YX)
Collinear(ZWY)
Conditions (CDL):
Equal(LengthOfLine(XY),18+2*y)
Equal(LengthOfLine(YW),LengthOfLine(ZW))
Equal(LengthOfLine(ZX),5*y-6)
PerpendicularBetweenLine(ZW,XW)
Goal: Value(LengthOfLine(XZ))
Answer: 34
Theorem sequence:
flat_angle(1,ZWY)
angle_addition(1,ZWX,XWY)
mirror_congruent_triangle_judgment_sas(1,WXZ,WYX)
mirror_congruent_triangle_property_line_equal(1,WXZ,WYX)
Problem: As shown in the diagram, AE=9, EB=16, AD⊥BD, BE is perpendicular to AE, DB⊥EB, EA⊥DA. Find the area of quadrilateral DBEA.
问题：如图所示，AE=9，EB=16，AD垂直于BD，BE垂直于AE，DB垂直于EB，EA⊥DA。求四边形DBEA的面积。
Construction (CDL):
Shape(DB,BE,EA,AD)
Conditions (CDL):
Equal(LengthOfLine(AE),9)
Equal(LengthOfLine(EB),16)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(BE,AE)
PerpendicularBetweenLine(DB,EB)
PerpendicularBetweenLine(EA,DA)
Goal: Value(AreaOfQuadrilateral(DBEA))
Answer: 144
Theorem sequence:
parallelogram_judgment_angle_and_angle(1,DBEA)
parallelogram_area_formula_sine(1,BEAD)
Problem: As shown in the diagram, AB=x, AC=y, BC=17, ∠ACB=60°, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=x，AC=y，BC=17，∠ACB=60°，BA⊥CA。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(BC),17)
Equal(MeasureOfAngle(ACB),60)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 17*sqrt(3)/2
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, ∠ABC=40°, ∠BCA=x°, ∠CAB=25°. Find the value of x.
问题：如图所示，∠ABC=40°，∠BCA=x°，∠CAB=25°。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),40)
Equal(MeasureOfAngle(BCA),x)
Equal(MeasureOfAngle(CAB),25)
Goal: Value(x)
Answer: 115
Theorem sequence:
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠WXY=4*x+15°, ∠ZXV=2*x+65°. Find the measure of ∠VXW.
问题：如图所示，∠WXY=4*x+15°，∠ZXV=2*x+65°。求∠VXW的大小。
Construction (CDL):
Shape(XZY.YX,XZ)
Shape(XYW,WX,XY)
Shape(XWV,VX,XW)
Shape(XVZ,ZX,XV)
Collinear(ZXW)
Collinear(YXV)
Cocircular(X,ZYWV)
Conditions (CDL):
Equal(MeasureOfAngle(WXY),4*x+15)
Equal(MeasureOfAngle(ZXV),2*x+65)
Goal: Value(MeasureOfAngle(VXW))
Answer: 65
Theorem sequence:
vertical_angle(1,ZXV,WXY)
adjacent_complementary_angle(1,ZXV,VXW)
Problem: As shown in the diagram, RW=x+6, SW=8, TR=2*x+6, TV=10, ∠SWR=∠VTR. Find the length of line WR.
问题：如图所示，RW=x+6，SW=8，TR=2*x+6，TV=10，∠SWR=∠VTR。求直线WR的长度。
Construction (CDL):
Shape(RS,SW,WR)
Shape(TR,RV,VT)
Collinear(WRT)
Collinear(SRV)
Conditions (CDL):
Equal(LengthOfLine(RW),x+6)
Equal(LengthOfLine(SW),8)
Equal(LengthOfLine(TR),2*x+6)
Equal(LengthOfLine(TV),10)
Equal(MeasureOfAngle(SWR),MeasureOfAngle(VTR))
Goal: Value(LengthOfLine(WR))
Answer: 8
Theorem sequence:
vertical_angle(1,WRS,TRV)
similar_triangle_judgment_aa(1,SWR,VTR)
similar_triangle_property_line_ratio(1,RSW,RVT)
similar_triangle_property_line_ratio(1,SWR,VTR)
Problem: As shown in the diagram, TR=SR, ∠TRS=70°. Find the measure of ∠RST.
问题：如图所示，TR=SR，∠TRS=70°。求∠RST的大小。
Construction (CDL):
Shape(RS,ST,TR)
Conditions (CDL):
Equal(LengthOfLine(TR),LengthOfLine(SR))
Equal(MeasureOfAngle(TRS),70)
Goal: Value(MeasureOfAngle(RST))
Answer: 55
Theorem sequence:
triangle_property_angle_sum(1,RST)
isosceles_triangle_judgment_line_equal(1,RST)
isosceles_triangle_property_angle_equal(1,RST)
Problem: As shown in the diagram, AB=26, AD=12, DG=4.5, EF=8, GF=14, ∠AGF=108°, ADCB is similar to AGFE. Find the measure of ∠ADC.
问题：如图所示，AB=26，AD=12，DG=4.5，EF=8，GF=14，∠AGF=108°，ADCB相似于AGFE。求∠ADC的大小。
Construction (CDL):
Shape(AG,GF,FE,EA)
Shape(GD,DC,CB,BE,EF,FG)
Collinear(AEB)
Collinear(AGD)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DG),4.5)
Equal(LengthOfLine(EF),8)
Equal(LengthOfLine(GF),14)
Equal(MeasureOfAngle(AGF),108)
SimilarBetweenQuadrilateral(ADCB,AGFE)
Goal: Value(MeasureOfAngle(ADC))
Answer: 108
Theorem sequence:
similar_quadrilateral_property_angle_equal(1,DCBA,GFEA)
Problem: As shown in the diagram, ∠HEL=55°, ∠JHF=110°, JC is parallel to HF. Find the measure of ∠DJC.
问题：如图所示，∠HEL=55°，∠JHF=110°，JC∥HF。求∠DJC的大小。
Construction (CDL):
Shape(DJ,JC)
Shape(KA,AJ)
Shape(LE,EG)
Shape(FH,HB)
Shape(HE,EL)
Shape(JH,HF)
Shape(JA,AE,EH,GJ)
Collinear(DJHB)
Collinear(JAEL)
Collinear(KAJC)
Collinear(GEHF)
Conditions (CDL):
Equal(MeasureOfAngle(HEL),55)
Equal(MeasureOfAngle(JHF),110)
ParallelBetweenLine(JC,HF)
Goal: Value(MeasureOfAngle(DJC))
Answer: 110
Theorem sequence:
parallel_property_corresponding_angle(1,JC,HF,D)