SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠BAC=50°, AB is parallel to CQ, AB is parallel to DF, CQ∥DF, CD is perpendicular to FD. Find the measure of ∠DCA.
问题：如图所示，∠BAC=50°，AB平行于CQ，AB∥DF，CQ平行于DF，CD⊥FD。求∠DCA的大小。
Construction (CDL):
Shape(CA,AG)
Shape(BA,AC)
Shape(PC,CA)
Shape(AC,CQ)
Shape(DC,CP)
Shape(QC,CD)
Shape(ED,DC)
Shape(CD,DF)
Shape(DC,CA,CD)
Collinear(GAB)
Collinear(PCQ)
Collinear(EDF)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),50)
ParallelBetweenLine(AB,CQ)
ParallelBetweenLine(AB,DF)
ParallelBetweenLine(CQ,DF)
PerpendicularBetweenLine(CD,FD)
Goal: Value(MeasureOfAngle(DCA))
Answer: 140
Theorem sequence:
parallel_property_collinear_extend(2,QC,BA,P)
parallel_property_alternate_interior_angle(1,CP,BA)
parallel_property_collinear_extend(1,CQ,DF,P)
parallel_property_alternate_interior_angle(2,PC,DF)
angle_addition(1,DCP,PCA)
Problem: As shown in the diagram, DE=10, D bisects segment AC, E bisects segment BC. Find the length of line AB.
问题：如图所示，DE=10，D平分线段AC，E平分线段BC。求直线AB的长度。
Construction (CDL):
Shape(CD,DE,EC)
Shape(DA,AB,BE,ED)
Collinear(ADC)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(DE),10)
IsMidpointOfLine(D,AC)
IsMidpointOfLine(E,BC)
Goal: Value(LengthOfLine(AB))
Answer: 20
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,DE,CAB)
midsegment_of_triangle_property_parallel(1,DE,CAB)
parallel_property_corresponding_angle(2,BA,ED,C)
similar_triangle_judgment_aa(1,DEC,ABC)
line_addition(1,CD,DA)
similar_triangle_property_line_ratio(1,ECD,BCA)
similar_triangle_property_line_ratio(1,CDE,CAB)
Problem: As shown in the diagram, AC=1/2, AO=1, OB=8, OA is perpendicular to CA, OB⊥DB. Find the length of line DB.
问题：如图所示，AC=1/2，AO=1，OB=8，OA垂直于CA，OB⊥DB。求直线DB的长度。
Construction (CDL):
Shape(OA,AC,CO)
Shape(DO,OB,BD)
Collinear(AOB)
Collinear(COD)
Conditions (CDL):
Equal(LengthOfLine(AC),1/2)
Equal(LengthOfLine(AO),1)
Equal(LengthOfLine(OB),8)
PerpendicularBetweenLine(OA,CA)
PerpendicularBetweenLine(OB,DB)
Goal: Value(LengthOfLine(DB))
Answer: 4
Theorem sequence:
vertical_angle(1,COA,DOB)
similar_triangle_judgment_aa(1,COA,DOB)
similar_triangle_property_line_ratio(1,COA,DOB)
similar_triangle_property_line_ratio(1,OAC,OBD)
Problem: As shown in the diagram, ∠EBC=70°, O is the center of ⊙O, the diameter of ⊙O is BA. Find the measure of ∠EDB.
问题：如图所示，∠EBC=70°，⊙O的圆心为O，⊙O的直径为BA。求∠EDB的大小。
Construction (CDL):
Shape(OBC,CB)
Shape(BC,CE,EB)
Shape(OCA,AC)
Shape(CA,AO,OE,EC)
Shape(OAD,DE,EO,OA)
Shape(ODB,BD)
Shape(DB,BE,ED)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(EBC),70)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(BA,O)
Goal: Value(MeasureOfAngle(EDB))
Answer: 20
Theorem sequence:
diameter_of_circle_property_right_angle(1,BCA,O)
triangle_property_angle_sum(1,CAB)
arc_property_circumference_angle_external(1,OBC,A)
arc_property_circumference_angle_external(1,OBC,D)
Problem: As shown in the diagram, ∠DBC=30°, D bisects segment AB, E bisects segment AC, △ADE is mirror congruent to △PED. Find the measure of ∠QDB.
问题：如图所示，∠DBC=30°，D平分线段AB，E平分线段AC，三角形ADE镜像全等于三角形PED。求∠QDB的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DQ,QE)
Shape(PE,EQ,QP)
Shape(QD,DB,BC,CQ)
Collinear(AEQC)
Collinear(ADB)
Collinear(DQP)
Conditions (CDL):
Equal(MeasureOfAngle(DBC),30)
IsMidpointOfLine(D,AB)
IsMidpointOfLine(E,AC)
MirrorCongruentBetweenTriangle(ADE,PED)
Goal: Value(MeasureOfAngle(QDB))
Answer: 120
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,DE,ABC)
midsegment_of_triangle_property_parallel(1,DE,ABC)
parallel_property_corresponding_angle(1,DE,BC,A)
mirror_congruent_triangle_property_angle_equal(1,DEA,DPE)
angle_addition(1,ADE,EDQ)
flat_angle(1,ADB)
angle_addition(1,ADQ,QDB)
Problem: As shown in the diagram, AB=AD, AB=DC, AD=DC, ∠ABD=70°. Find the measure of ∠DCA.
问题：如图所示，AB=AD，AB=DC，AD=DC，∠ABD=70°。求∠DCA的大小。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AD))
Equal(LengthOfLine(AB),LengthOfLine(DC))
Equal(LengthOfLine(AD),LengthOfLine(DC))
Equal(MeasureOfAngle(ABD),70)
Goal: Value(MeasureOfAngle(DCA))
Answer: 35
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABD)
isosceles_triangle_property_angle_equal(1,ABD)
flat_angle(1,BDC)
angle_addition(1,BDA,ADC)
isosceles_triangle_judgment_line_equal(1,DCA)
isosceles_triangle_property_angle_equal(1,DCA)
triangle_property_angle_sum(1,ADC)
Problem: As shown in the diagram, ∠CAH=30°, ∠EBH=60°, HB is perpendicular to CB, quadrilateral DEFG is a rectangle, triangle ABC is a right triangle. Find the measure of ∠HCA.
问题：如图所示，∠CAH=30°，∠EBH=60°，HB⊥CB，四边形DEFG是长方形，△ABC为直角△。求∠HCA的大小。
Construction (CDL):
Shape(AH,HC,CA)
Shape(DE,EB,BH,HD)
Shape(HB,BC,CH)
Shape(CB,BF,FG,GC)
Collinear(DHCG)
Collinear(EBF)
Collinear(AHB)
Conditions (CDL):
Equal(MeasureOfAngle(CAH),30)
Equal(MeasureOfAngle(EBH),60)
PerpendicularBetweenLine(HB,CB)
Rectangle(DEFG)
RightTriangle(ABC)
Goal: Value(MeasureOfAngle(HCA))
Answer: 30
Theorem sequence:
angle_addition(1,EBH,HBC)
flat_angle(1,EBF)
angle_addition(1,EBC,CBF)
parallel_property_collinear_extend(3,FE,GD,B)
parallel_property_collinear_extend(3,DG,BF,C)
parallel_property_alternate_interior_angle(2,DC,BF)
triangle_property_angle_sum(1,ABC)
angle_addition(1,BCH,HCA)
Problem: As shown in the diagram, ∠AMP=60°, ∠BCA and ∠BCA are the base angles of the isosceles right △ BCA, quadrilateral DEFG is a rectangle. Find the measure of ∠QNP.
问题：如图所示，∠AMP=60°，∠ABC是等腰直角三角形BCA的直角，DEFG是长方形。求∠QNP的大小。
Construction (CDL):
Shape(AM,MP,PA)
Shape(MB,BQ,QN,NP,PM)
Shape(MD,DE,EQ,QB,BM)
Shape(GP,PN,NF,FG)
Shape(CN,NQ,QC)
Collinear(DMPG)
Collinear(EQNF)
Collinear(APNC)
Collinear(BQC)
Collinear(AMB)
Conditions (CDL):
Equal(MeasureOfAngle(AMP),60)
IsoscelesRightTriangle(BCA)
Rectangle(DEFG)
Goal: Value(MeasureOfAngle(QNP))
Answer: 75
Theorem sequence:
isosceles_triangle_property_angle_equal(1,BCA)
triangle_property_angle_sum(1,ABC)
triangle_property_angle_sum(1,AMP)
parallel_property_collinear_extend(3,FE,GD,N)
parallel_property_collinear_extend(3,DG,EN,P)
parallel_property_corresponding_angle(2,NE,PD,A)
Problem: As shown in the diagram, ∠APE=60°, ∠MQB=40°, PQ∥ND. Find the measure of ∠PNQ.
问题：如图所示，∠APE=60°，∠MQB=40°，PQ平行于ND。求∠PNQ的大小。
Construction (CDL):
Shape(PN,NQ,QP)
Shape(AP,PE)
Shape(NP,PA)
Shape(EP,PQ)
Shape(PQ,QM)
Shape(MQ,QB)
Shape(BQ,QN)
Shape(CN,NP)
Shape(FN,NC)
Shape(DN,NF)
Shape(QN,ND)
Collinear(APQB)
Collinear(CND)
Collinear(EPNF)
Collinear(MQN)
Conditions (CDL):
Equal(MeasureOfAngle(APE),60)
Equal(MeasureOfAngle(MQB),40)
ParallelBetweenLine(PQ,ND)
Goal: Value(MeasureOfAngle(PNQ))
Answer: 80
Theorem sequence:
vertical_angle(1,APE,QPN)
vertical_angle(1,MQB,NQA)
triangle_property_angle_sum(1,PNQ)
Problem: As shown in the diagram, ∠CQG=38°, OB is perpendicular to QB, quadrilateral DEFG is a rectangle, ∠ABC is the right angle of the right △ ABC. Find the measure of ∠EBP.
问题：如图所示，∠CQG=38°，OB垂直于QB，四边形DEFG是长方形，△ABC为直角△。求∠EBP的大小。
Construction (CDL):
Shape(CA,AP,PQ,QC)
Shape(PD,DE,EB,BP)
Shape(QP,PB,BQ)
Shape(GQ,QB,BF,FG)
Shape(CQ,QG)
Collinear(DPQG)
Collinear(EBF)
Collinear(APB)
Collinear(CQB)
Conditions (CDL):
Equal(MeasureOfAngle(CQG),38)
PerpendicularBetweenLine(OB,QB)
Rectangle(DEFG)
RightTriangle(ABC)
Goal: Value(MeasureOfAngle(EBP))
Answer: 52
Theorem sequence:
parallel_property_collinear_extend(3,FE,GD,B)
parallel_property_collinear_extend(3,DG,BF,Q)
parallel_property_corresponding_angle(1,QG,BF,C)
angle_addition(1,PBQ,QBF)
flat_angle(1,EBF)
angle_addition(1,EBP,PBF)
Problem: As shown in the diagram, ∠BPA=40°, the center of circle O is O, PA is the tangent to ⊙O, PB is the tangent to ⊙O. Find the measure of ∠OCB.
问题：如图所示，∠BPA=40°，圆O的圆心为O，⊙O的切线为PA，圆O的切线为PB。求∠OCB的大小。
Construction (CDL):
Shape(OBA,AB)
Shape(OBA,BP,PA)
Shape(OCB,BC)
Shape(BO,OC,CB)
Shape(BA,AO,OB)
Shape(OAC,CO,OA)
Collinear(AOC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(OCB))
Answer: 70
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
quadrilateral_property_angle_sum(1,AOBP)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,C)
Problem: As shown in the diagram, ∠BCE=20°, O is the center of ⊙O, the diameter of ⊙O is AB. Find the measure of ∠DBE.
问题：如图所示，∠BCE=20°，⊙O的圆心为O，AB是⊙O的直径。求∠DBE的大小。
Construction (CDL):
Shape(OBC,CB)
Shape(BC,CE,EB)
Shape(OCA,AO,OE,EC)
Shape(ODB,BD)
Shape(DB,BE,ED)
Shape(DE,EO,OA,AD)
Shape(OAD,DA)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(BCE),20)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
Goal: Value(MeasureOfAngle(DBE))
Answer: 70
Theorem sequence:
arc_property_circumference_angle_external(1,ODB,C)
arc_property_circumference_angle_external(1,ODB,A)
diameter_of_circle_property_right_angle(1,ADB,O)
triangle_property_angle_sum(1,ADB)
Problem: As shown in the diagram, ∠BDE=62°, CE is perpendicular to AE. Find the measure of ∠ACE.
问题：如图所示，∠BDE=62°，CE垂直于AE。求∠ACE的大小。
Construction (CDL):
Shape(OAC,CA)
Shape(AC,CE,EA)
Shape(ODA,AE,ED)
Shape(OCB,BE,EC)
Shape(OBD,DB)
Shape(EB,BD,DE)
Collinear(AEB)
Collinear(CED)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(MeasureOfAngle(BDE),62)
PerpendicularBetweenLine(CE,AE)
Goal: Value(MeasureOfAngle(ACE))
Answer: 28
Theorem sequence:
vertical_angle(1,CEA,DEB)
triangle_property_angle_sum(1,DEB)
arc_property_circumference_angle_external(1,ODA,B)
arc_property_circumference_angle_external(1,ODA,C)
Problem: As shown in the diagram, ∠ADE=155°, ∠DAB=90°, DE∥BC, DA⊥BA. Find the measure of ∠ABC.
问题：如图所示，∠ADE=155°，∠DAB=90°，DE∥BC，DA⊥BA。求∠ABC的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(AD,DE)
Shape(ED,DC)
Collinear(ADC)
Conditions (CDL):
Equal(MeasureOfAngle(ADE),155)
Equal(MeasureOfAngle(DAB),90)
ParallelBetweenLine(DE,BC)
PerpendicularBetweenLine(DA,BA)
Goal: Value(MeasureOfAngle(ABC))
Answer: 65
Theorem sequence:
flat_angle(1,ADC)
angle_addition(1,ADE,EDC)
parallel_property_alternate_interior_angle(1,DE,BC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠FEA=40°, AB∥CD. Find the measure of ∠EFD.
问题：如图所示，∠FEA=40°，AB平行于CD。求∠EFD的大小。
Construction (CDL):
Shape(AE,EM)
Shape(ME,EB)
Shape(FE,EA)
Shape(BE,EF)
Shape(CF,FE)
Shape(EF,FD)
Shape(NF,FC)
Shape(DF,FN)
Collinear(AEB)
Collinear(CFD)
Collinear(MEFN)
Conditions (CDL):
Equal(MeasureOfAngle(FEA),40)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(EFD))
Answer: 40
Theorem sequence:
parallel_property_collinear_extend(3,DC,BA,F)
parallel_property_collinear_extend(3,AB,FD,E)
parallel_property_alternate_interior_angle(2,AE,FD)
Problem: As shown in the diagram, ∠OAB=35°, the center of ⊙O is O. Find the measure of ∠BCA.
问题：如图所示，∠OAB=35°，O是⊙O的圆心。求∠BCA的大小。
Construction (CDL):
Shape(OCA,AC)
Shape(OAB,BA)
Shape(OBC,CB)
Shape(OB,BC,CA,AO)
Shape(OA,AB,BO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(OAB),35)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 55
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,OAB)
isosceles_triangle_property_angle_equal(1,OAB)
triangle_property_angle_sum(1,OAB)
arc_property_circumference_angle_external(1,OAB,C)
arc_property_center_angle(1,OAB,O)
Problem: As shown in the diagram, ∠BCN=30°, quadrilateral DEFG is a rectangle, ∠BCA is the right angle of the right triangle BCA. Find the measure of ∠GHC.
问题：如图所示，∠BCN=30°，四边形DEFG是矩形，∠BCA是直角三角形BCA的直角。求∠GHC的大小。
Construction (CDL):
Shape(AM,MH,HA)
Shape(MD,DE,EN,NM)
Shape(HM,MN,NC,CH)
Shape(HC,CF,FG,GH)
Shape(CN,NB,BC)
Collinear(DMHG)
Collinear(ENCF)
Collinear(AMNB)
Collinear(AHC)
Conditions (CDL):
Equal(MeasureOfAngle(BCN),30)
Rectangle(DEFG)
RightTriangle(BCA)
Goal: Value(MeasureOfAngle(GHC))
Answer: 60
Theorem sequence:
angle_addition(1,BCN,NCH)
parallel_property_collinear_extend(3,FE,GD,C)
parallel_property_collinear_extend(3,DG,EC,H)
parallel_property_alternate_interior_angle(1,HG,EC)
Problem: As shown in the diagram, ∠DBC=40°, ∠DEA=60°, DE∥BC. Find the measure of ∠EAD.
问题：如图所示，∠DBC=40°，∠DEA=60°，DE平行于BC。求∠EAD的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CE,ED)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(MeasureOfAngle(DBC),40)
Equal(MeasureOfAngle(DEA),60)
ParallelBetweenLine(DE,BC)
Goal: Value(MeasureOfAngle(EAD))
Answer: 80
Theorem sequence:
parallel_property_corresponding_angle(1,DE,BC,A)
triangle_property_angle_sum(1,ADE)
Problem: As shown in the diagram, ∠BCA=60°, the center of circle O is O. Find the measure of ∠ABO.
问题：如图所示，∠BCA=60°，O是⊙O的圆心。求∠ABO的大小。
Construction (CDL):
Shape(OCA,AC)
Shape(OAB,BA)
Shape(OBC,CB)
Shape(OB,BC,CA,AO)
Shape(OA,AB,BO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),60)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(ABO))
Answer: 30
Theorem sequence:
arc_property_circumference_angle_external(1,OAB,C)
arc_property_center_angle(1,OAB,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OAB)
triangle_property_angle_sum(1,OAB)
Problem: As shown in the diagram, ∠DAB=90°, ∠EDC=35°, DE is parallel to BC. Find the measure of ∠ABC.
问题：如图所示，∠DAB=90°，∠EDC=35°，DE平行于BC。求∠ABC的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(AD,DE)
Shape(ED,DC)
Collinear(ADC)
Conditions (CDL):
Equal(MeasureOfAngle(DAB),90)
Equal(MeasureOfAngle(EDC),35)
ParallelBetweenLine(DE,BC)
Goal: Value(MeasureOfAngle(ABC))
Answer: 55
Theorem sequence:
parallel_property_alternate_interior_angle(1,DE,BC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠EFG=50°, EG bisects ∠BEF, AB∥CD. Find the measure of ∠FGE.
问题：如图所示，∠EFG=50°，EG是∠BEF的角平分线，AB∥CD。求∠FGE的大小。
Construction (CDL):
Shape(EF,FG,GE)
Shape(FE,EA)
Shape(BE,EG)
Shape(BE,EF)
Shape(CF,FE)
Shape(EG,GD)
Collinear(AEB)
Collinear(CFGD)
Conditions (CDL):
Equal(MeasureOfAngle(EFG),50)
IsBisectorOfAngle(EG,BEF)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(FGE))
Answer: 65
Theorem sequence:
parallel_property_collinear_extend(3,DC,BA,F)
parallel_property_collinear_extend(3,AB,FD,E)
parallel_property_ipsilateral_internal_angle(1,EB,FD)
angle_addition(1,BEG,GEF)
triangle_property_angle_sum(1,EFG)
Problem: As shown in the diagram, ∠EBA=70°, PQ∥EF, BA⊥CA. Find the measure of ∠CAP.
问题：如图所示，∠EBA=70°，PQ平行于EF，BA⊥CA。求∠CAP的大小。
Construction (CDL):
Shape(CA,AP)
Shape(CA,AM)
Shape(PA,AM)
Shape(MA,AD)
Shape(BA,AC)
Shape(QA,AB)
Shape(DA,AQ)
Shape(EB,BA)
Shape(AB,BF)
Shape(NB,BE)
Shape(FB,BN)
Collinear(PAQ)
Collinear(CAD)
Collinear(EBF)
Collinear(MABN)
Conditions (CDL):
Equal(MeasureOfAngle(EBA),70)
ParallelBetweenLine(PQ,EF)
PerpendicularBetweenLine(BA,CA)
Goal: Value(MeasureOfAngle(CAP))
Answer: 20
Theorem sequence:
parallel_property_collinear_extend(3,FE,QP,B)
parallel_property_collinear_extend(3,PQ,EB,A)
parallel_property_corresponding_angle(2,BE,AP,M)
adjacent_complementary_angle(1,BAC,CAM)
angle_addition(1,CAP,PAM)
Problem: As shown in the diagram, ∠BOD=50°, the center of ⊙O is O, OB is parallel to AC. Find the measure of ∠OAD.
问题：如图所示，∠BOD=50°，O是⊙O的圆心，OB平行于AC。求∠OAD的大小。
Construction (CDL):
Shape(OBA,AO,OB)
Shape(OAC,CA)
Shape(OCB,BD,DC)
Shape(OA,AD,DO)
Shape(OD,DB,BO)
Shape(DA,AC,CD)
Collinear(ADB)
Collinear(ODC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BOD),50)
IsCentreOfCircle(O,O)
ParallelBetweenLine(OB,AC)
Goal: Value(MeasureOfAngle(OAD))
Answer: 25
Theorem sequence:
parallel_property_alternate_interior_angle(1,OB,AC)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OAC)
isosceles_triangle_property_angle_equal(1,OAC)
triangle_property_angle_sum(1,OAC)
angle_addition(1,BOD,DOA)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OAB)
triangle_property_angle_sum(1,OAB)
Problem: As shown in the diagram, ∠ABC=40°, ∠BCA=30°. Find the measure of ∠DAC.
问题：如图所示，∠ABC=40°，∠BCA=30°。求∠DAC的大小。
Construction (CDL):
Shape(CA,AB,BC)
Shape(DA,AC)
Collinear(DAB)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),40)
Equal(MeasureOfAngle(BCA),30)
Goal: Value(MeasureOfAngle(DAC))
Answer: 70
Theorem sequence:
triangle_property_angle_sum(1,ABC)
flat_angle(1,DAB)
angle_addition(1,DAC,CAB)
Problem: As shown in the diagram, ∠DAO=42°, ∠OBC=23°, BA and CD are opposite sides of the ▱ ABCD. Find the measure of ∠DOC.
问题：如图所示，∠DAO=42°，∠OBC=23°，AD和BC是▱ABCD的一组对边。求∠DOC的大小。
Construction (CDL):
Shape(AB,BO,OA)
Shape(DA,AO,OD)
Shape(OB,BC,CO)
Shape(DO,OC,CD)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(MeasureOfAngle(DAO),42)
Equal(MeasureOfAngle(OBC),23)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(DOC))
Answer: 65
Theorem sequence:
parallel_property_alternate_interior_angle(2,AD,BC)
triangle_property_angle_sum(1,AOD)
flat_angle(1,AOC)
angle_addition(1,AOD,DOC)
Problem: As shown in the diagram, ∠BMN=70°, ∠EPM=58°, ∠PQN=58°. Find the measure of ∠HNQ.
问题：如图所示，∠BMN=70°，∠EPM=58°，∠PQN=58°。求∠HNQ的大小。
Construction (CDL):
Shape(PQ,QN,NM,MP)
Shape(AP,PE)
Shape(EP,PM)
Shape(PM,MG)
Shape(GM,MB)
Shape(QP,PA)
Shape(BM,MN)
Shape(CQ,QP)
Shape(MN,ND)
Shape(FQ,QC)
Shape(NQ,QF)
Shape(HN,NQ)
Shape(DN,NH)
Collinear(APMB)
Collinear(CQND)
Collinear(EPQF)
Collinear(GMNH)
Conditions (CDL):
Equal(MeasureOfAngle(BMN),70)
Equal(MeasureOfAngle(EPM),58)
Equal(MeasureOfAngle(PQN),58)
Goal: Value(MeasureOfAngle(HNQ))
Answer: 110
Theorem sequence:
parallel_judgment_corresponding_angle(1,PM,QN,E)
parallel_property_collinear_extend(1,NQ,MP,D)
parallel_property_collinear_extend(2,PM,ND,B)
parallel_property_corresponding_angle(2,MB,ND,H)
flat_angle(1,DNQ)
angle_addition(1,DNH,HNQ)
Problem: As shown in the diagram, AB=AC, AC=AD, ∠EAB=44°, ∠EBC=Mul(MeasureOfAngle(CDE),2)°, A is the center of circle A. Find the measure of ∠DAE.
问题：如图所示，AB=AC，AC=AD，∠EAB=44°，∠EBC=Mul(MeasureOfAngle(CDE),2)°，圆A的圆心为A。求∠DAE的大小。
Construction (CDL):
Shape(ADB,BA,AD)
Shape(ABC,CB)
Shape(ACD,DC)
Shape(AB,BE,EA)
Shape(AE,ED,DA)
Shape(EB,BC,CE)
Shape(EC,CD,DE)
Collinear(AEC)
Collinear(BED)
Cocircular(A,BCD)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(AC),LengthOfLine(AD))
Equal(MeasureOfAngle(EAB),44)
Equal(MeasureOfAngle(EBC),Mul(MeasureOfAngle(CDE),2))
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfAngle(DAE))
Answer: 88
Theorem sequence:
arc_property_circumference_angle_external(1,ABC,D)
arc_property_circumference_angle_external(1,ACD,B)
arc_property_center_angle(1,ABC,A)
arc_property_center_angle(1,ACD,A)
Problem: As shown in the diagram, ∠BMN=158°, ∠HKQ=50°, ∠MNH=42°, AB∥CD, AB is parallel to PQ, CD∥EF, EF∥PQ. Find the measure of ∠KHN.
问题：如图所示，∠BMN=158°，∠HKQ=50°，∠MNH=42°，AB平行于CD，AB平行于PQ，CD∥EF，EF∥PQ。求∠KHN的大小。
Construction (CDL):
Shape(NM,MA)
Shape(BM,MN)
Shape(CN,NM)
Shape(MN,ND)
Shape(HN,NC)
Shape(DN,NH)
Shape(EH,HN)
Shape(NH,HF)
Shape(KH,HE)
Shape(FH,HK)
Shape(PK,KH)
Shape(HK,KQ)
Shape(MN,NH)
Shape(KH,HN)
Collinear(AMB)
Collinear(CND)
Collinear(EHF)
Collinear(PKQ)
Conditions (CDL):
Equal(MeasureOfAngle(BMN),158)
Equal(MeasureOfAngle(HKQ),50)
Equal(MeasureOfAngle(MNH),42)
ParallelBetweenLine(AB,CD)
ParallelBetweenLine(AB,PQ)
ParallelBetweenLine(CD,EF)
ParallelBetweenLine(EF,PQ)
Goal: Value(MeasureOfAngle(KHN))
Answer: 70
Theorem sequence:
parallel_property_collinear_extend(3,DC,BA,N)
parallel_property_collinear_extend(3,FE,DC,H)
parallel_property_collinear_extend(3,QP,FE,K)
parallel_property_collinear_extend(3,AB,ND,M)
parallel_property_collinear_extend(3,CD,EH,N)
parallel_property_collinear_extend(3,EF,KQ,H)
parallel_property_ipsilateral_internal_angle(1,MB,ND)
angle_addition(1,MND,DNH)
angle_addition(1,KHE,EHN)
parallel_property_alternate_interior_angle(1,ND,EH)
parallel_property_alternate_interior_angle(2,EH,KQ)
Problem: As shown in the diagram, ∠ABC=40°, CD∥AB, BC is perpendicular to AC. Find the measure of ∠ECD.
问题：如图所示，∠ABC=40°，CD平行于AB，BC垂直于AC。求∠ECD的大小。
Construction (CDL):
Shape(CA,AB,BC)
Shape(EC,CD)
Shape(DC,CB)
Shape(EC,CB)
Collinear(ECA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),40)
ParallelBetweenLine(CD,AB)
PerpendicularBetweenLine(BC,AC)
Goal: Value(MeasureOfAngle(ECD))
Answer: 50
Theorem sequence:
parallel_property_alternate_interior_angle(1,CD,AB)
angle_addition(1,ECD,DCB)
adjacent_complementary_angle(1,ECB,BCA)
Problem: As shown in the diagram, ∠DAB=70°, O is the center of circle O. Find the measure of ∠BCD.
问题：如图所示，∠DAB=70°，圆O的圆心为O。求∠BCD的大小。
Construction (CDL):
Shape(OBC,CB)
Shape(OCD,DC)
Shape(CD,DO,OB,BC)
Shape(ODA,AD)
Shape(OAB,BA)
Shape(OD,DA,AB,BO)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(DAB),70)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCD))
Answer: 110
Theorem sequence:
arc_property_circumference_angle_external(1,OBD,A)
arc_property_center_angle(1,OBD,O)
round_angle(1,DOB,BOD)
arc_property_circumference_angle_external(1,ODB,C)
arc_property_center_angle(1,ODB,O)
Problem: As shown in the diagram, AB=AC, ∠AEB=35°, BD bisects ∠ABC, DB∥AE. Find the measure of ∠DAB.
问题：如图所示，AB=AC，∠AEB=35°，BD是∠ABC的角平分线，DB平行于AE。求∠DAB的大小。
Construction (CDL):
Shape(AE,EB,BA)
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Collinear(EBC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(AEB),35)
IsBisectorOfAngle(BD,ABC)
ParallelBetweenLine(DB,AE)
Goal: Value(MeasureOfAngle(DAB))
Answer: 40
Theorem sequence:
parallel_property_corresponding_angle(2,EA,BD,C)
angle_addition(1,ABD,DBC)
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, AC=4, the perimeter of △NBC is 7, NM perpendicular bisects AB. Find the length of line BC.
问题：如图所示，AC=4，△NBC的周长为7，NM垂直平分AB。求直线BC的长度。
Construction (CDL):
Shape(AM,MN,NA)
Shape(NM,MB,BN)
Shape(CN,NB,BC)
Collinear(ANC)
Collinear(AMB)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(PerimeterOfTriangle(NBC),7)
IsPerpendicularBisectorOfLine(NM,AB)
Goal: Value(LengthOfLine(BC))
Answer: 3
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,NM,AB)
line_addition(1,AN,NC)
triangle_perimeter_formula(1,NBC)
Problem: As shown in the diagram, quadrilateral OABC is congruent to quadrilateral ODEF, ∠FON=40°, ∠MOA=40°, OABC is a square, quadrilateral ODEF is a square. Find the measure of ∠NFO.
问题：如图所示，四边形OABC全等于四边形ODEF，∠FON=40°，∠MOA=40°，OABC是正方形，ODEF是正方形。求∠NFO的大小。
Construction (CDL):
Shape(OA,AM,MO)
Shape(OM,MN,NO)
Shape(ON,NF,FO)
Shape(AB,BP,PD,DM,MA)
Shape(MD,DP,PC,CN,NM)
Shape(NC,CP,PE,EF,FN)
Collinear(AMNF)
Collinear(OMD)
Collinear(ONC)
Collinear(DPE)
Collinear(BPC)
Conditions (CDL):
CongruentBetweenQuadrilateral(OABC,ODEF)
Equal(MeasureOfAngle(FON),40)
Equal(MeasureOfAngle(MOA),40)
Square(OABC)
Square(ODEF)
Goal: Value(MeasureOfAngle(NFO))
Answer: 25
Theorem sequence:
angle_addition(1,FON,NOA)
congruent_quadrilateral_property_line_equal(1,OABC,ODEF)
isosceles_triangle_judgment_line_equal(1,OAF)
isosceles_triangle_property_angle_equal(1,OAF)
triangle_property_angle_sum(1,OAF)
Problem: As shown in the diagram, ∠CDH=26°, O is the center of circle O, BE is perpendicular to OE. Find the measure of ∠EOB.
问题：如图所示，∠CDH=26°，O是圆O的圆心，BE垂直于OE。求∠EOB的大小。
Construction (CDL):
Shape(OB,BE,EO)
Shape(OBA,AE,EB)
Shape(EA,AF,FE)
Shape(EF,FH,HO,OE)
Shape(HF,FC,CH)
Shape(CD,DH,HC)
Shape(OAC,CF,FA)
Shape(OCD,DC)
Shape(ODB,BO,OH,HD)
Collinear(BEFC)
Collinear(OEA)
Collinear(DHFA)
Collinear(OHC)
Cocircular(O,ACDB)
Conditions (CDL):
Equal(MeasureOfAngle(CDH),26)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BE,OE)
Goal: Value(MeasureOfAngle(EOB))
Answer: 52
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OE,BC)
perpendicular_bisector_property_bisector(1,OE,BC)
arc_property_circumference_angle_external(1,OAC,D)
arc_property_center_angle(1,OAC,O)
Problem: As shown in the diagram, ∠FNE=63°, EN is the angle bisector of ∠BEF, AB∥CD. Find the measure of ∠EFN.
问题：如图所示，∠FNE=63°，EN平分∠BEF，AB平行于CD。求∠EFN的大小。
Construction (CDL):
Shape(EF,FN,NE)
Shape(FE,EA)
Shape(BE,EN)
Shape(CF,FE)
Shape(EN,ND)
Collinear(AEB)
Collinear(CFND)
Conditions (CDL):
Equal(MeasureOfAngle(FNE),63)
IsBisectorOfAngle(EN,BEF)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(EFN))
Answer: 54
Theorem sequence:
parallel_property_collinear_extend(3,DC,BA,F)
parallel_property_collinear_extend(1,FC,BA,N)
parallel_property_collinear_extend(3,AB,FN,E)
parallel_property_alternate_interior_angle(1,EB,FN)
triangle_property_angle_sum(1,EFN)