SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=16, AC=x-3, BE=20, CD=x+5, DE is parallel to CB. Find the value of x.
问题：如图所示，AB=16，AC=x-3，BE=20，CD=x+5，DE∥CB。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Shape(CB,BE,ED,DC)
Collinear(ABE)
Collinear(ACD)
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(x)
Answer: 35
Theorem sequence:
parallel_property_corresponding_angle(1,BC,ED,A)
similar_triangle_judgment_aa(1,CAB,DAE)
line_addition(1,AB,BE)
line_addition(1,AC,CD)
similar_triangle_property_line_ratio(1,CAB,DAE)
similar_triangle_property_line_ratio(1,BCA,EDA)
Problem: As shown in the diagram, DA=39, DC=39, ∠CBD=17°, BA⊥DA, DC⊥BC. Find the measure of ∠DBA.
问题：如图所示，DA=39，DC=39，∠CBD=17°，BA⊥DA，DC垂直于BC。求∠DBA的大小。
Construction (CDL):
Shape(AD,DB,BA)
Shape(DC,CB,BD)
Conditions (CDL):
Equal(LengthOfLine(DA),39)
Equal(LengthOfLine(DC),39)
Equal(MeasureOfAngle(CBD),17)
PerpendicularBetweenLine(BA,DA)
PerpendicularBetweenLine(DC,BC)
Goal: Value(MeasureOfAngle(DBA))
Answer: 17
Theorem sequence:
mirror_similar_triangle_judgment_hl(2,BAD,BDC)
mirror_similar_triangle_property_angle_equal(1,BAD,BDC)
Problem: As shown in the diagram, AC=x, BC=6*sqrt(2), ∠CAB=45°, AB⊥CB. Find the value of x.
问题：如图所示，AC=x，BC=6*sqrt(2)，∠CAB=45°，AB垂直于CB。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),6*sqrt(2))
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(AB,CB)
Goal: Value(x)
Answer: 12
Theorem sequence:
sine_theorem(1,CAB)
Problem: As shown in the diagram, AB=12, AC=x, BC=y, ∠BCA=60°, CA⊥BA. Find the value of x.
问题：如图所示，AB=12，AC=x，BC=y，∠BCA=60°，CA垂直于BA。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 4*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,CAB)
sine_theorem(1,CAB)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=14, AC=20, AD=x, BC=11, ∠DBC=∠ABD. Find the value of x.
问题：如图所示，AB=14，AC=20，AD=x，BC=11，∠DBC=∠ABD。求x的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AB),14)
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BC),11)
Equal(MeasureOfAngle(DBC),MeasureOfAngle(ABD))
Goal: Value(x)
Answer: 56/5
Theorem sequence:
bisector_of_angle_judgment_angle_equal(1,BD,ABC)
bisector_of_angle_property_line_ratio(1,BD,ABC)
line_addition(1,CD,DA)
Problem: As shown in the diagram, ∠VWX=67°, ∠XWZ=∠WZX, ∠XZY=65°, ∠ZYX=58°. Find the measure of ∠WXV.
问题：如图所示，∠VWX=67°，∠XWZ=∠WZX，∠XZY=65°，∠ZYX=58°。求∠WXV的大小。
Construction (CDL):
Shape(VW,WX,XV)
Shape(XW,WZ,ZX)
Shape(XZ,ZY,YX)
Collinear(VXZ)
Collinear(WXY)
Conditions (CDL):
Equal(MeasureOfAngle(VWX),67)
Equal(MeasureOfAngle(XWZ),MeasureOfAngle(WZX))
Equal(MeasureOfAngle(XZY),65)
Equal(MeasureOfAngle(ZYX),58)
Goal: Value(MeasureOfAngle(WXV))
Answer: 57
Theorem sequence:
adjacent_complementary_angle(1,ZXW,WXV)
adjacent_complementary_angle(1,YXZ,ZXW)
triangle_property_angle_sum(1,XZY)
Problem: As shown in the diagram, JM=PM, ML=PL, ∠JLP=34°. Find the measure of ∠MPJ.
问题：如图所示，JM=PM，ML=PL，∠JLP=34°。求∠MPJ的大小。
Construction (CDL):
Shape(PJ,JM,MP)
Shape(PM,ML,LP)
Collinear(JML)
Conditions (CDL):
Equal(LengthOfLine(JM),LengthOfLine(PM))
Equal(LengthOfLine(ML),LengthOfLine(PL))
Equal(MeasureOfAngle(JLP),34)
Goal: Value(MeasureOfAngle(MPJ))
Answer: 73/2
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,MPJ)
isosceles_triangle_property_angle_equal(1,MPJ)
isosceles_triangle_judgment_line_equal(1,LPM)
isosceles_triangle_property_angle_equal(1,LPM)
triangle_property_angle_sum(1,PJM)
triangle_property_angle_sum(1,PML)
adjacent_complementary_angle(1,JMP,PML)
Problem: As shown in the diagram, ∠AGB=30°, G is the center of circle G, CG is perpendicular to DG. Find the measure of arc GDB.
问题：如图所示，∠AGB=30°，⊙G的圆心为G，CG垂直于DG。求⌒GDB的角度。
Construction (CDL):
Shape(BG,GC,GCB)
Shape(AG,GB,GBA)
Shape(FG,GA,GAF)
Shape(DG,GF,GFD)
Shape(CG,GD,GDC)
Collinear(AGD)
Cocircular(G,AFDCB)
Conditions (CDL):
Equal(MeasureOfAngle(AGB),30)
IsCentreOfCircle(G,G)
PerpendicularBetweenLine(CG,DG)
Goal: Value(MeasureOfArc(GDB))
Answer: 150
Theorem sequence:
adjacent_complementary_angle(1,AGB,BGD)
arc_property_center_angle(1,GDB,G)
Problem: As shown in the diagram, AB=17, CE=9, ∠ACD=45°, AB and CD are opposite sides of the ▱ ACDB, AE⊥DE. Find the area of quadrilateral ACDB.
问题：如图所示，AB=17，CE=9，∠ACD=45°，CA和DB是▱ACDB的一组对边，AE⊥DE。求ACDB的面积。
Construction (CDL):
Shape(AC,CE,EA)
Shape(AE,ED,DB,BA)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AB),17)
Equal(LengthOfLine(CE),9)
Equal(MeasureOfAngle(ACD),45)
Parallelogram(ACDB)
PerpendicularBetweenLine(AE,DE)
Goal: Value(AreaOfQuadrilateral(ACDB))
Answer: 153
Theorem sequence:
triangle_property_angle_sum(1,ACE)
sine_theorem(1,EAC)
adjacent_complementary_angle(1,CEA,AED)
altitude_of_quadrilateral_judgment_left_vertex(1,AE,ACDB)
parallelogram_property_opposite_line_equal(1,CDBA)
parallelogram_area_formula_common(1,ACDB)
Problem: As shown in the diagram, the diameter of circle P is equal to the length of line CB, CB=9, AD⊥BD, BC is perpendicular to AC, CA is perpendicular to DA, DB⊥CB. Find the circumference of the ⊙P.
问题：如图所示，⊙P的直径与直线CB的长度相等，CB=9，AD⊥BD，BC垂直于AC，CA⊥DA，DB⊥CB。求圆P的周长。
Construction (CDL):
Shape(FC,CH,PFH)
Shape(HA,AG,PHG)
Shape(GD,DE,PGE)
Shape(EB,BF,PEF)
Collinear(CFB)
Collinear(CHA)
Collinear(AGD)
Collinear(DEB)
Cocircular(P,FHGE)
Conditions (CDL):
Equal(DiameterOfCircle(P),LengthOfLine(CB))
Equal(LengthOfLine(CB),9)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(CA,DA)
PerpendicularBetweenLine(DB,CB)
Goal: Value(PerimeterOfCircle(P))
Answer: 9*pi
Theorem sequence:
circle_property_length_of_radius_and_diameter(1,P)
circle_perimeter_formula(1,P)
Problem: As shown in the diagram, AB=10, the measure of arc OBA is 60, the radius of ⊙O is 10, the center of ⊙O is O, BX is perpendicular to OX. Find the length of line OX.
问题：如图所示，AB=10，弧OBA的角度为60，⊙O的半径为10，O是圆O的圆心，BX垂直于OX。求直线OX的长度。
Construction (CDL):
Shape(AX,XY,OYA)
Shape(YX,XB,OBY)
Shape(OA,OAB,BO)
Shape(OX,XA,AO)
Shape(XO,OB,BX)
Collinear(YXO)
Collinear(AXB)
Cocircular(O,BYA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(MeasureOfArc(OBA),60)
Equal(RadiusOfCircle(O),10)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BX,OX)
Goal: Value(LengthOfLine(OX))
Answer: 5*sqrt(3)
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OBA)
isosceles_triangle_property_angle_equal(1,OBA)
arc_property_center_angle(1,OBA,O)
triangle_property_angle_sum(1,OBA)
circle_property_chord_perpendicular_bisect_chord(1,O,OX,BA)
line_addition(1,AX,XB)
triangle_property_angle_sum(1,BXO)
sine_theorem(1,XOB)
Problem: As shown in the diagram, ∠ADB=11*x-1°, ∠BDF=7*y-4°, ∠DFH=7*x+9°, ∠GAD=2*y+5°, ∠HFC=z°, BD is parallel to GA, HF is parallel to BD. Find the value of z.
问题：如图所示，∠ADB=11*x-1°，∠BDF=7*y-4°，∠DFH=7*x+9°，∠GAD=2*y+5°，∠HFC=z°，BD∥GA，HF平行于BD。求z的值。
Construction (CDL):
Shape(HF,FC)
Shape(DF,FH)
Shape(BD,DF)
Shape(AD,DB)
Shape(GA,AD)
Shape(EA,AG)
Shape(FD,DA)
Collinear(DFC)
Collinear(DAE)
Conditions (CDL):
Equal(MeasureOfAngle(ADB),11*x-1)
Equal(MeasureOfAngle(BDF),7*y-4)
Equal(MeasureOfAngle(DFH),7*x+9)
Equal(MeasureOfAngle(GAD),2*y+5)
Equal(MeasureOfAngle(HFC),z)
ParallelBetweenLine(BD,GA)
ParallelBetweenLine(HF,BD)
Goal: Value(z)
Answer: 73
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,DB,FH)
parallel_property_ipsilateral_internal_angle(1,AG,DB)
parallel_property_corresponding_angle(2,DB,FH,C)
Problem: As shown in the diagram, GK=14, MF=8, the measure of ⌒FKG is 142, F is the center of circle F, HJ is perpendicular to KJ. Find the length of line JK.
问题：如图所示，GK=14，MF=8，弧FKG的角度为142，圆F的圆心为F，HJ⊥KJ。求直线JK的长度。
Construction (CDL):
Shape(GJ,JH,FHG)
Shape(HJ,JK,FKH)
Shape(MF,FJ,JG,FGM)
Shape(KJ,JF,FM,FMK)
Collinear(GJK)
Collinear(HJFM)
Cocircular(F,GMKH)
Conditions (CDL):
Equal(LengthOfLine(GK),14)
Equal(LengthOfLine(MF),8)
Equal(MeasureOfArc(FKG),142)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(HJ,KJ)
Goal: Value(LengthOfLine(JK))
Answer: 7
Theorem sequence:
adjacent_complementary_angle(1,HJK,KJF)
circle_property_chord_perpendicular_bisect_chord(1,F,FJ,KG)
line_addition(1,GJ,JK)
Problem: As shown in the diagram, AD=7*y, CA=5*y+16, CE=8*x-18, CE=BE, EB=12-2*x, ∠BDA=∠EAC. Find the value of y.
问题：如图所示，AD=7*y，CA=5*y+16，CE=8*x-18，CE=BE，EB=12-2*x，∠BDA=∠EAC。求y的值。
Construction (CDL):
Shape(DA,AE,EB,BD)
Shape(AC,CE,EA)
Collinear(BEC)
Collinear(DAC)
Conditions (CDL):
Equal(LengthOfLine(AD),7*y)
Equal(LengthOfLine(CA),5*y+16)
Equal(LengthOfLine(CE),8*x-18)
Equal(LengthOfLine(CE),LengthOfLine(BE))
Equal(LengthOfLine(EB),12-2*x)
Equal(MeasureOfAngle(BDA),MeasureOfAngle(EAC))
Goal: Value(y)
Answer: 8
Theorem sequence:
similar_triangle_judgment_aa(1,EAC,BDC)
line_addition(1,BE,EC)
line_addition(1,DA,AC)
similar_triangle_property_line_ratio(1,ACE,DCB)
similar_triangle_property_line_ratio(1,EAC,BDC)
Problem: As shown in the diagram, ∠JQR=131°, QR is parallel to TG, TQ is parallel to GR. Find the measure of ∠QRH.
问题：如图所示，∠JQR=131°，QR平行于TG，TQ平行于GR。求∠QRH的大小。
Construction (CDL):
Shape(JQ,QR)
Shape(QR,RH)
Shape(GT,TC)
Shape(BG,GT)
Shape(QT,TG,GR,RQ)
Collinear(JQTC)
Collinear(HRGB)
Conditions (CDL):
Equal(MeasureOfAngle(JQR),131)
ParallelBetweenLine(QR,TG)
ParallelBetweenLine(TQ,GR)
Goal: Value(MeasureOfAngle(QRH))
Answer: 49
Theorem sequence:
parallel_property_collinear_extend(2,TQ,GR,J)
parallel_property_collinear_extend(1,RG,JQ,H)
parallel_property_ipsilateral_internal_angle(1,QJ,RH)
Problem: As shown in the diagram, ∠BDA=32°, ∠CAS=125°, ∠CBA=62°. Find the measure of ∠BAC.
问题：如图所示，∠BDA=32°，∠CAS=125°，∠CBA=62°。求∠BAC的大小。
Construction (CDL):
Shape(CB,BA,AC)
Shape(AB,BD,DA)
Shape(CA,AS)
Shape(SA,AD)
Collinear(BAS)
Collinear(CAD)
Conditions (CDL):
Equal(MeasureOfAngle(BDA),32)
Equal(MeasureOfAngle(CAS),125)
Equal(MeasureOfAngle(CBA),62)
Goal: Value(MeasureOfAngle(BAC))
Answer: 55
Theorem sequence:
adjacent_complementary_angle(1,BAC,CAS)
Problem: As shown in the diagram, QR=6, RT=x, UT=4, VW=8, WP=3, the tangent to circle Z is RQ, RS is the tangent to ⊙Z, the tangent to circle Z is TS, TU is the tangent to ⊙Z. Find the value of x.
问题：如图所示，QR=6，RT=x，UT=4，VW=8，WP=3，RQ是圆O的切线，RS是圆O的切线，TS是圆O的切线，TU是圆O的切线。求x的值。
Construction (CDL):
Shape(QP,PW,ZQW)
Shape(WV,VU,ZWU)
Shape(UT,TS,ZUS)
Shape(SR,RQ,ZSQ)
Collinear(PQR)
Collinear(PWV)
Collinear(VUT)
Collinear(RST)
Cocircular(Z,QWUS)
Conditions (CDL):
Equal(LengthOfLine(QR),6)
Equal(LengthOfLine(RT),x)
Equal(LengthOfLine(UT),4)
Equal(LengthOfLine(VW),8)
Equal(LengthOfLine(WP),3)
IsTangentOfCircle(RQ,Z)
IsTangentOfCircle(RS,Z)
IsTangentOfCircle(TS,Z)
IsTangentOfCircle(TU,Z)
Goal: Value(x)
Answer: 10
Theorem sequence:
tangent_of_circle_property_length_equal(1,RS,RQ,Z)
tangent_of_circle_property_length_equal(1,TU,TS,Z)
line_addition(1,RS,ST)
Problem: As shown in the diagram, PM=LM, the measure of arc OMP is x, the measure of ⌒OPL is 106. Find the value of x.
问题：如图所示，PM=LM，⌒OMP的角度为x，⌒OPL的角度为106。求x的值。
Construction (CDL):
Shape(LM,MP,PL)
Shape(ML,OLM)
Shape(PM,OMP)
Shape(LP,OOL)
Cocircular(O,LMP)
Conditions (CDL):
Equal(LengthOfLine(PM),LengthOfLine(LM))
Equal(MeasureOfArc(OMP),x)
Equal(MeasureOfArc(OPL),106)
Goal: Value(x)
Answer: 127
Theorem sequence:
arc_property_circumference_angle_external(1,OPL,M)
arc_property_circumference_angle_external(1,OMP,L)
isosceles_triangle_judgment_line_equal(1,MPL)
isosceles_triangle_property_angle_equal(1,MPL)
triangle_property_angle_sum(1,MPL)
Problem: As shown in the diagram, ∠AEJ=3*y+44°, ∠CAE=94°, ∠EJC=2*x°, ∠JCA=96°, AC is parallel to EJ. Find the value of x.
问题：如图所示，∠AEJ=3*y+44°，∠CAE=94°，∠EJC=2*x°，∠JCA=96°，AC平行于EJ。求x的值。
Construction (CDL):
Shape(CA,AE,EJ,JC)
Conditions (CDL):
Equal(MeasureOfAngle(AEJ),3*y+44)
Equal(MeasureOfAngle(CAE),94)
Equal(MeasureOfAngle(EJC),2*x)
Equal(MeasureOfAngle(JCA),96)
ParallelBetweenLine(AC,EJ)
Goal: Value(x)
Answer: 42
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,JE,CA)
Problem: As shown in the diagram, DA=7, the center of circle F is F, DA is the diameter of ⊙F. Find the length of line EF.
问题：如图所示，DA=7，F是圆F的圆心，DA是圆F的直径。求直线EF的长度。
Construction (CDL):
Shape(CF,FD,FDC)
Shape(BF,FC,FCB)
Shape(AF,FB,FBA)
Shape(FA,AE,EF)
Shape(DF,FE,ED)
Shape(EA,FAE)
Shape(DE,FED)
Collinear(BFE)
Collinear(AFD)
Cocircular(F,AEDCB)
Conditions (CDL):
Equal(LengthOfLine(DA),7)
IsCentreOfCircle(F,F)
IsDiameterOfCircle(DA,F)
Goal: Value(LengthOfLine(EF))
Answer: 7/2
Theorem sequence:
diameter_of_circle_property_length_equal(1,DA,F)
radius_of_circle_property_length_equal(1,FE,F)
circle_property_length_of_radius_and_diameter(1,F)
Problem: As shown in the diagram, VX=75, WV=72, WX=21, VW⊥XW. Find the value of tan(XV).
问题：如图所示，VX=75，WV=72，WX=21，VW垂直于XW。求tan(XV)的值。
Construction (CDL):
Shape(VW,WX,XV)
Conditions (CDL):
Equal(LengthOfLine(VX),75)
Equal(LengthOfLine(WV),72)
Equal(LengthOfLine(WX),21)
PerpendicularBetweenLine(VW,XW)
Goal: Value(Tan(MeasureOfAngle(XVW)))
Answer: 7/24
Theorem sequence:
cosine_theorem(1,VWX)
Problem: As shown in the diagram, QS=14, RS=3, PS⊥QS, QP⊥RP. Find the length of line PS.
问题：如图所示，QS=14，RS=3，PS垂直于QS，QP⊥RP。求直线PS的长度。
Construction (CDL):
Shape(PR,RS,SP)
Shape(PS,SQ,QP)
Collinear(RSQ)
Conditions (CDL):
Equal(LengthOfLine(QS),14)
Equal(LengthOfLine(RS),3)
PerpendicularBetweenLine(PS,QS)
PerpendicularBetweenLine(QP,RP)
Goal: Value(LengthOfLine(PS))
Answer: sqrt(42)
Theorem sequence:
adjacent_complementary_angle(1,RSP,PSQ)
mirror_similar_triangle_judgment_aa(1,PRS,QPR)
mirror_similar_triangle_property_line_ratio(1,PRS,QPR)
mirror_similar_triangle_property_line_ratio(1,SPR,PRQ)
line_addition(1,RS,SQ)
right_triangle_judgment_angle(1,RSP)
right_triangle_property_pythagorean(1,RSP)
Problem: As shown in the diagram, PQ=6-x, QS=3+x, RQ=6+x, TQ=3, TP∥RS. Find the value of x.
问题：如图所示，PQ=6-x，QS=3+x，RQ=6+x，TQ=3，TP∥RS。求x的值。
Construction (CDL):
Shape(PT,TQ,QP)
Shape(QR,RS,SQ)
Collinear(PQR)
Collinear(TQS)
Conditions (CDL):
Equal(LengthOfLine(PQ),6-x)
Equal(LengthOfLine(QS),3+x)
Equal(LengthOfLine(RQ),6+x)
Equal(LengthOfLine(TQ),3)
ParallelBetweenLine(TP,RS)
Goal: Value(x)
Answer: 0
Theorem sequence:
parallel_property_alternate_interior_angle(1,TP,RS)
parallel_property_alternate_interior_angle(2,TP,RS)
similar_triangle_judgment_aa(1,QPT,QRS)
similar_triangle_property_line_ratio(1,TQP,SQR)
similar_triangle_property_line_ratio(1,PTQ,RSQ)
Problem: As shown in the diagram, AB=y, AC=x, BC=15, ∠BAC=60°, AC⊥BC. Find the value of y.
问题：如图所示，AB=y，AC=x，BC=15，∠BAC=60°，AC⊥BC。求y的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),15)
Equal(MeasureOfAngle(BAC),60)
PerpendicularBetweenLine(AC,BC)
Goal: Value(y)
Answer: 10*sqrt(3)
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, RS=TR, RT=3*x+3, SR=4*x-4, ST=12+x. Find the value of x.
问题：如图所示，RS=TR，RT=3*x+3，SR=4*x-4，ST=12+x。求x的值。
Construction (CDL):
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(RS),LengthOfLine(TR))
Equal(LengthOfLine(RT),3*x+3)
Equal(LengthOfLine(SR),4*x-4)
Equal(LengthOfLine(ST),12+x)
Goal: Value(x)
Answer: 7
Theorem sequence:
Problem: As shown in the diagram, HP=5*x-16, PJ=3*x+8, ∠AJP=4*y+23°, ∠GMA=4*z+14°, ∠MJA=6*y-3°, JN is the angle bisector of ∠GJH. Find the measure of ∠GJH.
问题：如图所示，HP=5*x-16，PJ=3*x+8，∠AJP=4*y+23°，∠GMA=4*z+14°，∠MJA=6*y-3°，JN平分∠GJH。求∠GJH的大小。
Construction (CDL):
Shape(NG,GB,BN)
Shape(BG,GM,MA,AB)
Shape(AM,MJ,JA)
Shape(AJ,JP,PZ,ZA)
Shape(ZP,PH,HZ)
Shape(ZH,HN,NB,BZ)
Shape(BA,AZ,ZB)
Collinear(GNH)
Collinear(GMJ)
Collinear(HPJ)
Collinear(NBAJ)
Collinear(MAZH)
Collinear(GBZP)
Conditions (CDL):
Equal(LengthOfLine(HP),5*x-16)
Equal(LengthOfLine(PJ),3*x+8)
Equal(MeasureOfAngle(AJP),4*y+23)
Equal(MeasureOfAngle(GMA),4*z+14)
Equal(MeasureOfAngle(MJA),6*y-3)
IsBisectorOfAngle(JN,GJH)
Goal: Value(MeasureOfAngle(GJH))
Answer: 150
Theorem sequence:
angle_addition(1,GJN,NJH)
Problem: As shown in the diagram, ∠JMK=55°, ∠JMK=∠PMR, ∠MRP=70°, JMLK is a rectangle, MRPL is a rhombus. Find the measure of ∠RPM.
问题：如图所示，∠JMK=55°，∠JMK=∠PMR，∠MRP=70°，JMLK是矩形，四边形MRPL是菱形。求∠RPM的大小。
Construction (CDL):
Shape(JM,MK,KJ)
Shape(ML,LK,KM)
Shape(MP,PL,LM)
Shape(MR,RP,PM)
Conditions (CDL):
Equal(MeasureOfAngle(JMK),55)
Equal(MeasureOfAngle(JMK),MeasureOfAngle(PMR))
Equal(MeasureOfAngle(MRP),70)
Rectangle(JMLK)
Rhombus(MRPL)
Goal: Value(MeasureOfAngle(RPM))
Answer: 55
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,RPM)
isosceles_triangle_property_angle_equal(1,RPM)
Problem: As shown in the diagram, the area of ADCB is 750, BA=35, CD=25, DF⊥EF, AD and CB are the non-parallel sides (legs) of trapezoid ADCB. Find the height of ADCB.
问题：如图所示，ADCB的面积为750，BA=35，CD=25，DF⊥EF，ADCB是梯形。求四边形ADCB的高。
Construction (CDL):
Shape(AD,DF,FE,EA)
Shape(EF,FC,CB,BE)
Collinear(AEB)
Collinear(DFC)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ADCB),750)
Equal(LengthOfLine(BA),35)
Equal(LengthOfLine(CD),25)
PerpendicularBetweenLine(DF,EF)
Trapezoid(ADCB)
Goal: Value(HeightOfQuadrilateral(ADCB))
Answer: 25
Theorem sequence:
trapezoid_area_formula(1,ADCB)
Problem: As shown in the diagram, FH=8, FH=FL, FK=17, F is the center of ⊙F, FL is perpendicular to KL, JH is perpendicular to FH. Find the length of line JH.
问题：如图所示，FH=8，FH=FL，FK=17，圆F的圆心为F，FL垂直于KL，JH垂直于FH。求直线JH的长度。
Construction (CDL):
Shape(FJG,GH,HJ)
Shape(FMK,KL,LM)
Shape(FKJ,JF,FK)
Shape(FJ,JH,HF)
Shape(KF,FL,LK)
Shape(FGM,ML,LF,FH,HG)
Collinear(JHG)
Collinear(KLM)
Collinear(HFL)
Cocircular(F,KJGM)
Conditions (CDL):
Equal(LengthOfLine(FH),8)
Equal(LengthOfLine(FH),LengthOfLine(FL))
Equal(LengthOfLine(FK),17)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(FL,KL)
PerpendicularBetweenLine(JH,FH)
Goal: Value(LengthOfLine(JH))
Answer: 15
Theorem sequence:
radius_of_circle_property_length_equal(1,FK,F)
radius_of_circle_property_length_equal(1,FJ,F)
right_triangle_judgment_angle(1,JHF)
right_triangle_property_pythagorean(1,JHF)
Problem: As shown in the diagram, EK=y, JQ=2*x+6, KM=3/5*y+2, RJ=20-5*x, RJ=JQ, JK∥QM, RE∥JK, RE is parallel to QM. Find the value of y.
问题：如图所示，EK=y，JQ=2*x+6，KM=3/5*y+2，RJ=20-5*x，RJ=JQ，JK平行于QM，RE平行于JK，RE∥QM。求y的值。
Construction (CDL):
Shape(RJ,JK,KE,ER)
Shape(JQ,QM,MK,KJ)
Shape(GR,RB)
Shape(BR,RE)
Shape(JR,RG)
Shape(RE,EA)
Shape(AE,EO)
Shape(OE,EK)
Shape(XQ,QJ)
Shape(IQ,QX)
Shape(MQ,QI)
Shape(KM,MH)
Shape(HM,MY)
Shape(YM,MQ)
Shape(PJ,JR)
Shape(QJ,JP)
Shape(EK,KP)
Shape(PK,KM)
Collinear(BRJQI)
Collinear(AEKMY)
Collinear(GREO)
Collinear(PJKP)
Collinear(XQMH)
Conditions (CDL):
Equal(LengthOfLine(EK),y)
Equal(LengthOfLine(JQ),2*x+6)
Equal(LengthOfLine(KM),3/5*y+2)
Equal(LengthOfLine(RJ),20-5*x)
Equal(LengthOfLine(RJ),LengthOfLine(JQ))
ParallelBetweenLine(JK,QM)
ParallelBetweenLine(RE,JK)
ParallelBetweenLine(RE,QM)
Goal: Value(y)
Answer: 5
Theorem sequence:
trapezoid_judgment_parallel(1,RQME)
midsegment_of_quadrilateral_judgment_parallel(1,JK,RQME)
Problem: As shown in the diagram, △RVS is congruent to △TSV, RS=2*y-1, ST=24.5, VT=24, VS is perpendicular to RS. Find the value of y.
问题：如图所示，三角形RVS全等于三角形TSV，RS=2*y-1，ST=24.5，VT=24，VS垂直于RS。求y的值。
Construction (CDL):
Shape(RV,VS,SR)
Shape(SV,VT,TS)
Conditions (CDL):
CongruentBetweenTriangle(RVS,TSV)
Equal(LengthOfLine(RS),2*y-1)
Equal(LengthOfLine(ST),24.5)
Equal(LengthOfLine(VT),24)
PerpendicularBetweenLine(VS,RS)
Goal: Value(y)
Answer: 25/2
Theorem sequence:
congruent_triangle_property_line_equal(1,VSR,SVT)
Problem: As shown in the diagram, BK=8, KC=12, KF=16. Find the length of line JK.
问题：如图所示，BK=8，KC=12，KF=16。求直线JK的长度。
Construction (CDL):
Shape(DAB,BA)
Shape(DBJ,JK,KB)
Shape(BK,KM,MA,AB)
Shape(AM,MF,DFA)
Shape(DFA,FE,EA)
Shape(DJC,CK,KJ)
Shape(MK,KC,CD,DM)
Shape(MD,DF,FM)
Shape(DC,DCH,HD)
Shape(DH,DHF,FD)
Shape(DHF,HG,GF)
Collinear(JKMF)
Collinear(BKC)
Collinear(AMDC)
Collinear(DFE)
Cocircular(D,ABJCHF)
Conditions (CDL):
Equal(LengthOfLine(BK),8)
Equal(LengthOfLine(KC),12)
Equal(LengthOfLine(KF),16)
Goal: Value(LengthOfLine(JK))
Answer: 6
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,BKC,JKF,D)
Problem: As shown in the diagram, AB=6*y+1/2*x, AD=21, BC=3*x+3*y, DC=2*x+4*y, DABC is a parallelogram. Find the value of x.
问题：如图所示，AB=6*y+1/2*x，AD=21，BC=3*x+3*y，DC=2*x+4*y，DC和AB是平行四边形DABC的一组对边。求x的值。
Construction (CDL):
Shape(DA,AB,BC,CD)
Conditions (CDL):
Equal(LengthOfLine(AB),6*y+1/2*x)
Equal(LengthOfLine(AD),21)
Equal(LengthOfLine(BC),3*x+3*y)
Equal(LengthOfLine(DC),2*x+4*y)
Parallelogram(DABC)
Goal: Value(x)
Answer: 4
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DABC)
parallelogram_property_opposite_line_equal(1,ABCD)
Problem: As shown in the diagram, JH=4*x+7, JK=25, KH=6*x-2, NQ=12, PN=8, PQ=20, ∠JHK=∠QNP, ∠KJH=∠PQN. Find the length of line HJ.
问题：如图所示，JH=4*x+7，JK=25，KH=6*x-2，NQ=12，PN=8，PQ=20，∠JHK=∠QNP，∠KJH=∠PQN。求直线HJ的长度。
Construction (CDL):
Shape(JH,HK,KJ)
Shape(PQ,QN,NP)
Conditions (CDL):
Equal(LengthOfLine(JH),4*x+7)
Equal(LengthOfLine(JK),25)
Equal(LengthOfLine(KH),6*x-2)
Equal(LengthOfLine(NQ),12)
Equal(LengthOfLine(PN),8)
Equal(LengthOfLine(PQ),20)
Equal(MeasureOfAngle(JHK),MeasureOfAngle(QNP))
Equal(MeasureOfAngle(KJH),MeasureOfAngle(PQN))
Goal: Value(LengthOfLine(HJ))
Answer: 15
Theorem sequence:
similar_triangle_judgment_aa(1,KJH,PQN)
similar_triangle_property_line_ratio(1,KJH,PQN)
similar_triangle_property_line_ratio(1,HKJ,NPQ)
Problem: As shown in the diagram, PR=8, QR=17, RQ=15, QR⊥PR. Find the value of cos(PQ).
问题：如图所示，PR=8，QR=17，RQ=15，QR⊥PR。求cos(PQ)的值。
Construction (CDL):
Shape(QR,RP,PQ)
Conditions (CDL):
Equal(LengthOfLine(PR),8)
Equal(LengthOfLine(QP),17)
Equal(LengthOfLine(RQ),15)
PerpendicularBetweenLine(QR,PR)
Goal: Value(Cos(MeasureOfAngle(PQR)))
Answer: 15/17
Theorem sequence:
cosine_theorem(1,QRP)
Problem: As shown in the diagram, WP=6, XP=8, YP=24, ZP=8, WZYX is a kite. Find the length of line ZY.
问题：如图所示，WP=6，XP=8，YP=24，ZP=8，YZ和YX是风筝形WZYX的一组临边。求直线ZY的长度。
Construction (CDL):
Shape(XW,WP,PX)
Shape(WZ,ZP,PW)
Shape(XP,PY,YX)
Shape(PZ,ZY,YP)
Collinear(XPZ)
Collinear(WPY)
Conditions (CDL):
Equal(LengthOfLine(WP),6)
Equal(LengthOfLine(XP),8)
Equal(LengthOfLine(YP),24)
Equal(LengthOfLine(ZP),8)
Kite(WZYX)
Goal: Value(LengthOfLine(ZY))
Answer: 8*sqrt(10)
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,YXWZ,P)
right_triangle_judgment_angle(1,YPZ)
right_triangle_property_pythagorean(1,YPZ)
Problem: As shown in the diagram, CA=x, CB=24, ∠CAD=32°, AC is perpendicular to CC, BC⊥AC. Find the value of x.
问题：如图所示，CA=x，CB=24，∠CAD=32°，AC⊥CC，BC垂直于AC。求x的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(CA),x)
Equal(LengthOfLine(CB),24)
Equal(MeasureOfAngle(CAD),32)
PerpendicularBetweenLine(AC,CD)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 24*sqrt(1-sin(8*pi/45)**2)/sin(8*pi/45)
Theorem sequence:
sine_theorem(1,BCA)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, LM=7*x+1, LP=10*x-5, PN=NM, LN is perpendicular to MN. Find the length of line LP.
问题：如图所示，LM=7*x+1，LP=10*x-5，PN=NM，LN⊥MN。求直线LP的长度。
Construction (CDL):
Shape(LP,PN,NL)
Shape(LN,NM,ML)
Collinear(PNM)
Conditions (CDL):
Equal(LengthOfLine(LM),7*x+1)
Equal(LengthOfLine(LP),10*x-5)
Equal(LengthOfLine(PN),LengthOfLine(NM))
PerpendicularBetweenLine(LN,MN)
Goal: Value(LengthOfLine(LP))
Answer: 15
Theorem sequence:
adjacent_complementary_angle(1,PNL,LNM)
perpendicular_bisector_judgment_per_and_mid(1,LN,PM)
perpendicular_bisector_property_distance_equal(1,LN,PM)
Problem: As shown in the diagram, ∠TWV=3*x-4°, ∠UTW=x°, ∠VUT=3*x-4°, ∠WVU=x°. Find the measure of ∠WVU.
问题：如图所示，∠TWV=3*x-4°，∠UTW=x°，∠VUT=3*x-4°，∠WVU=x°。求∠WVU的大小。
Construction (CDL):
Shape(UT,TW,WV,VU)
Conditions (CDL):
Equal(MeasureOfAngle(TWV),3*x-4)
Equal(MeasureOfAngle(UTW),x)
Equal(MeasureOfAngle(VUT),3*x-4)
Equal(MeasureOfAngle(WVU),x)
Goal: Value(MeasureOfAngle(WVU))
Answer: 46
Theorem sequence:
quadrilateral_property_angle_sum(1,UTWV)