SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠ABD=36°, O is the center of circle O, the diameter of ⊙O is AC. Find the measure of ∠DAC.
问题：如图所示，∠ABD=36°，⊙O的圆心为O，AC是⊙O的直径。求∠DAC的大小。
Construction (CDL):
Shape(DA,AE,ED)
Shape(AB,BE,EA)
Shape(DE,EO,OC,CD)
Shape(AD,ODA)
Shape(BA,OAB)
Shape(DC,OCD)
Shape(CO,OE,EB,OBC)
Collinear(AEOC)
Collinear(DEB)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),36)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AC,O)
Goal: Value(MeasureOfAngle(DAC))
Answer: 54
Theorem sequence:
arc_property_circumference_angle_external(1,ODA,B)
arc_property_circumference_angle_external(1,ODA,C)
diameter_of_circle_property_right_angle(1,CDA,O)
triangle_property_angle_sum(1,CDA)
Problem: As shown in the diagram, AE=3, the perimeter of triangle ABD is 13, DE perpendicular bisects CA. Find the perimeter of △ABC.
问题：如图所示，AE=3，△ABD的周长为13，DE是CA的垂直平分线。求△ABC的周长。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DE,EA)
Shape(ED,DC,CE)
Collinear(AEC)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AE),3)
Equal(PerimeterOfTriangle(ABD),13)
IsPerpendicularBisectorOfLine(DE,CA)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 19
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,DE,CA)
line_addition(1,AE,EC)
triangle_perimeter_formula(1,ABD)
line_addition(1,BD,DC)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, ∠GBH=110°, DE∥FG. Find the measure of ∠DAC.
问题：如图所示，∠GBH=110°，DE平行于FG。求∠DAC的大小。
Construction (CDL):
Shape(DA,AC)
Shape(CA,AE)
Shape(BA,AD)
Shape(EA,AB)
Shape(FB,BA)
Shape(AB,BG)
Shape(HB,BF)
Shape(GB,BH)
Collinear(DAE)
Collinear(FBG)
Collinear(CABH)
Conditions (CDL):
Equal(MeasureOfAngle(GBH),110)
ParallelBetweenLine(DE,FG)
Goal: Value(MeasureOfAngle(DAC))
Answer: 110
Theorem sequence:
parallel_property_collinear_extend(3,DE,FG,A)
parallel_property_collinear_extend(3,GF,EA,B)
parallel_property_corresponding_angle(2,AE,BG,H)
vertical_angle(1,EAB,DAC)
Problem: As shown in the diagram, ∠OAB=28°, the center of circle O is O. Find the measure of ∠BCA.
问题：如图所示，∠OAB=28°，O是⊙O的圆心。求∠BCA的大小。
Construction (CDL):
Shape(CA,AO,OB,BC)
Shape(AC,OCA)
Shape(OA,AB,BO)
Shape(BA,OAB)
Shape(CB,OBC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(OAB),28)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 62
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_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
Problem: As shown in the diagram, BC=8, DE=5, F is the midpoint of segment BC, CD is perpendicular to BD, CE is perpendicular to BE. Find the perimeter of triangle DFE.
问题：如图所示，BC=8，DE=5，F是线段BC的中点，CD⊥BD，CE⊥BE。求三角形DFE的周长。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DH,HE,ED)
Shape(DG,GH,HD)
Shape(EH,HI,IE)
Shape(GF,FI,IH,HG)
Shape(DB,BG,GD)
Shape(GB,BF,FG)
Shape(IF,FC,CI)
Shape(EI,IC,CE)
Collinear(ADB)
Collinear(AEC)
Collinear(BFC)
Collinear(DGF)
Collinear(DHIC)
Collinear(EHGB)
Collinear(EIF)
Conditions (CDL):
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(DE),5)
IsMidpointOfLine(F,BC)
PerpendicularBetweenLine(CD,BD)
PerpendicularBetweenLine(CE,BE)
Goal: Value(PerimeterOfTriangle(DFE))
Answer: 13
Theorem sequence:
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,CEB)
median_of_triangle_judgment(1,DF,DBC)
median_of_triangle_judgment(1,EF,EBC)
right_triangle_property_length_of_median(1,CDB,F)
right_triangle_property_length_of_median(1,CEB,F)
triangle_perimeter_formula(1,DFE)
Problem: As shown in the diagram, AB=8, OC=3, the center of ⊙O is O, BC is perpendicular to OC. Find the radius of circle O.
问题：如图所示，AB=8，OC=3，圆O的圆心为O，BC⊥OC。求圆O的半径。
Construction (CDL):
Shape(AC,CB,OBA)
Shape(AO,OC,CA)
Shape(BC,CO,OA,OAB)
Collinear(ACB)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(OC),3)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BC,OC)
Goal: Value(RadiusOfCircle(O))
Answer: 5
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OC,BA)
line_addition(1,AC,CB)
right_triangle_judgment_angle(1,OCA)
right_triangle_property_pythagorean(1,OCA)
radius_of_circle_property_length_equal(1,OA,O)
Problem: As shown in the diagram, ∠ABC=51°, O is the center of circle O. Find the measure of ∠AOC.
问题：如图所示，∠ABC=51°，O是圆O的圆心。求∠AOC的大小。
Construction (CDL):
Shape(AB,BC,CO,OA)
Shape(AO,OC,OCA)
Shape(BA,OAB)
Shape(CB,OBC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),51)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(AOC))
Answer: 102
Theorem sequence:
arc_property_circumference_angle_external(1,OCA,B)
arc_property_center_angle(1,OCA,O)
Problem: As shown in the diagram, CD=CB, ∠BAD=70°, the center of circle O is O, the diameter of circle O is AD, C bisects arc ODB. Find the measure of ∠ADC.
问题：如图所示，CD=CB，∠BAD=70°，O是⊙O的圆心，⊙O的直径为AD，C是⌒ODB的中点。求∠ADC的大小。
Construction (CDL):
Shape(BA,AO,OD,DB)
Shape(CB,BD,DC)
Shape(AB,OBA)
Shape(BC,OCB)
Shape(CD,ODC)
Collinear(AOD)
Cocircular(O,DCBA)
Conditions (CDL):
Equal(LengthOfLine(CD),LengthOfLine(CB))
Equal(MeasureOfAngle(BAD),70)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AD,O)
IsMidpointOfArc(C,ODB)
Goal: Value(MeasureOfAngle(ADC))
Answer: 55
Theorem sequence:
arc_property_circumference_angle_external(1,ODB,A)
arc_property_circumference_angle_internal(1,ODB,C)
isosceles_triangle_judgment_line_equal(1,CBD)
isosceles_triangle_property_angle_equal(1,CBD)
diameter_of_circle_property_right_angle(1,DBA,O)
triangle_property_angle_sum(1,CBD)
triangle_property_angle_sum(1,BAD)
angle_addition(1,ADB,BDC)
Problem: As shown in the diagram, triangle ACB is congruent to triangle ADE, ∠BAC=35°, ∠CAD=120°. Find the measure of ∠CAE.
问题：如图所示，三角形ACB与三角形ADE是全等三角形，∠BAC=35°，∠CAD=120°。求∠CAE的大小。
Construction (CDL):
Shape(BA,AC,CB)
Shape(EA,AD,DE)
Shape(CA,AE)
Conditions (CDL):
CongruentBetweenTriangle(ACB,ADE)
Equal(MeasureOfAngle(BAC),35)
Equal(MeasureOfAngle(CAD),120)
Goal: Value(MeasureOfAngle(CAE))
Answer: 85
Theorem sequence:
congruent_triangle_property_angle_equal(1,ACB,ADE)
angle_addition(1,CAE,EAD)
Problem: As shown in the diagram, ∠EDC=140°, BE bisects ∠ABC, DC∥AB. Find the measure of ∠BCD.
问题：如图所示，∠EDC=140°，BE平分∠ABC，DC平行于AB。求∠BCD的大小。
Construction (CDL):
Shape(ED,DC)
Shape(CD,DB,BC)
Shape(AB,BD)
Collinear(EDB)
Conditions (CDL):
Equal(MeasureOfAngle(EDC),140)
IsBisectorOfAngle(BE,ABC)
ParallelBetweenLine(DC,AB)
Goal: Value(MeasureOfAngle(BCD))
Answer: 100
Theorem sequence:
adjacent_complementary_angle(1,EDC,CDB)
parallel_property_alternate_interior_angle(1,DC,AB)
triangle_property_angle_sum(1,BCD)
Problem: As shown in the diagram, ∠ACD=145°, ∠CAB=80°. Find the measure of ∠ABC.
问题：如图所示，∠ACD=145°，∠CAB=80°。求∠ABC的大小。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD)
Collinear(BCD)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),145)
Equal(MeasureOfAngle(CAB),80)
Goal: Value(MeasureOfAngle(ABC))
Answer: 65
Theorem sequence:
adjacent_complementary_angle(1,BCA,ACD)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠BPA=40°, O is the center of circle O, AC is the diameter of circle O, the tangent to ⊙O is PA, PB is the tangent to circle O. Find the measure of ∠BAC.
问题：如图所示，∠BPA=40°，O是⊙O的圆心，AC是⊙O的直径，PA是圆O的切线，圆O的切线为PB。求∠BAC的大小。
Construction (CDL):
Shape(CO,OA,OAC)
Shape(AO,OB,BA)
Shape(OC,CB,BO)
Shape(BC,OCB)
Shape(AB,OBA)
Shape(BP,PA,OBA)
Collinear(AOC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),40)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AC,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(BAC))
Answer: 20
Theorem sequence:
diameter_of_circle_property_right_angle(1,CBA,O)
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)
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, ∠DCB=35°, O is the center of circle O, the diameter of ⊙O is AB. Find the measure of ∠ABD.
问题：如图所示，∠DCB=35°，⊙O的圆心为O，⊙O的直径为AB。求∠ABD的大小。
Construction (CDL):
Shape(AD,ODA)
Shape(CE,EO,OA,OAC)
Shape(BC,OCB)
Shape(DB,OBD)
Shape(DA,AO,OE,ED)
Shape(DE,EB,BD)
Shape(EC,CB,BE)
Collinear(AOEB)
Collinear(DEC)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(MeasureOfAngle(DCB),35)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
Goal: Value(MeasureOfAngle(ABD))
Answer: 55
Theorem sequence:
arc_property_circumference_angle_external(1,OBD,C)
arc_property_circumference_angle_external(1,OBD,A)
diameter_of_circle_property_right_angle(1,BDA,O)
triangle_property_angle_sum(1,ABD)
Problem: As shown in the diagram, ∠AOB=150°, ∠AOC=90°, ∠DOB=90°. Find the measure of ∠DOC.
问题：如图所示，∠AOB=150°，∠AOC=90°，∠DOB=90°。求∠DOC的大小。
Construction (CDL):
Shape(AO,OD)
Shape(DO,OC)
Shape(CO,OB)
Conditions (CDL):
Equal(MeasureOfAngle(AOB),150)
Equal(MeasureOfAngle(AOC),90)
Equal(MeasureOfAngle(DOB),90)
Goal: Value(MeasureOfAngle(DOC))
Answer: 30
Theorem sequence:
angle_addition(1,AOC,COB)
angle_addition(1,DOC,COB)
Problem: As shown in the diagram, ∠ACB=50°, O is the center of circle O. Find the measure of ∠AOB.
问题：如图所示，∠ACB=50°，圆O的圆心为O。求∠AOB的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(BC,OCB)
Shape(CA,OAC)
Shape(BO,OA,AC,CB)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),50)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(AOB))
Answer: 100
Theorem sequence:
arc_property_circumference_angle_external(1,OBA,C)
arc_property_center_angle(1,OBA,O)
Problem: As shown in the diagram, ∠CAB=30°, O is the center of ⊙O, the diameter of circle O is AB. Find the measure of ∠ADC.
问题：如图所示，∠CAB=30°，⊙O的圆心为O，AB是⊙O的直径。求∠ADC的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(DA,OAD)
Shape(BE,ED,ODB)
Shape(CB,OBC)
Shape(CA,AO,OE,EC)
Shape(CE,EB,BC)
Shape(AD,DE,EO,OA)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
Goal: Value(MeasureOfAngle(ADC))
Answer: 60
Theorem sequence:
diameter_of_circle_property_right_angle(1,BCA,O)
triangle_property_angle_sum(1,ABC)
arc_property_circumference_angle_external(1,OCA,B)
arc_property_circumference_angle_external(1,OCA,D)
Problem: As shown in the diagram, AB=130, BC=120, ∠ABC=20°, ∠CAB=70°. Find the length of line AC.
问题：如图所示，AB=130，BC=120，∠ABC=20°，∠CAB=70°。求直线AC的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),130)
Equal(LengthOfLine(BC),120)
Equal(MeasureOfAngle(ABC),20)
Equal(MeasureOfAngle(CAB),70)
Goal: Value(LengthOfLine(AC))
Answer: 50
Theorem sequence:
triangle_property_angle_sum(1,BCA)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, AC=60, ∠BAD=30°, ∠BCD=60°, ∠CEB=90°. Find the length of line BE.
问题：如图所示，AC=60，∠BAD=30°，∠BCD=60°，∠CEB=90°。求直线BE的长度。
Construction (CDL):
Shape(BA,AC,CB)
Shape(BC,CE,EB)
Shape(BE,ED)
Collinear(ACED)
Conditions (CDL):
Equal(LengthOfLine(AC),60)
Equal(MeasureOfAngle(BAD),30)
Equal(MeasureOfAngle(BCD),60)
Equal(MeasureOfAngle(CEB),90)
Goal: Value(LengthOfLine(BE))
Answer: 30*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,ACB,BCE)
triangle_property_angle_sum(1,ACB)
isosceles_triangle_judgment_angle_equal(1,CBA)
sine_theorem(1,BCE)
Problem: As shown in the diagram, AC=4, the perimeter of △ADC is 15, triangle EDA is mirror congruent to triangle EBD. Find the length of line BC.
问题：如图所示，AC=4，△ADC的周长为15，△EDA镜像全等于△EBD。求直线BC的长度。
Construction (CDL):
Shape(EB,BD,DE)
Shape(AE,ED,DA)
Shape(AD,DC,CA)
Collinear(BEA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(PerimeterOfTriangle(ADC),15)
MirrorCongruentBetweenTriangle(EDA,EBD)
Goal: Value(LengthOfLine(BC))
Answer: 11
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,EDA,EBD)
triangle_perimeter_formula(1,ADC)
line_addition(1,BD,DC)
Problem: As shown in the diagram, AD=3, the perimeter of triangle BCE is 15, ED perpendicular bisects AB. Find the perimeter of triangle ABC.
问题：如图所示，AD=3，△BCE的周长为15，ED是AB的垂直平分线。求△ABC的周长。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BE,ED)
Shape(EB,BC,CE)
Collinear(AEC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),3)
Equal(PerimeterOfTriangle(BCE),15)
IsPerpendicularBisectorOfLine(ED,AB)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 21
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,ED,AB)
line_addition(1,AD,DB)
triangle_perimeter_formula(1,EBC)
line_addition(1,AE,EC)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, AB=AD, AD=DC, ∠DAB=36°. Find the measure of ∠DCA.
问题：如图所示，AB=AD，AD=DC，∠DAB=36°。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(AD),LengthOfLine(DC))
Equal(MeasureOfAngle(DAB),36)
Goal: Value(MeasureOfAngle(DCA))
Answer: 36
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABD)
isosceles_triangle_property_angle_equal(1,ABD)
triangle_property_angle_sum(1,ABD)
adjacent_complementary_angle(1,BDA,ADC)
isosceles_triangle_judgment_line_equal(1,DCA)
isosceles_triangle_property_angle_equal(1,DCA)
triangle_property_angle_sum(1,DCA)
Problem: As shown in the diagram, ∠DAB=132°, O is the center of ⊙O. Find the measure of ∠BOD.
问题：如图所示，∠DAB=132°，O是⊙O的圆心。求∠BOD的大小。
Construction (CDL):
Shape(AB,BO,OD,DA)
Shape(BC,CD,DO,OB)
Shape(BA,OAB)
Shape(CB,OBC)
Shape(DC,OCD)
Shape(AD,ODA)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(DAB),132)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BOD))
Answer: 96
Theorem sequence:
arc_property_circumference_angle_external(1,OBD,A)
arc_property_circumference_angle_internal(1,OBD,C)
arc_property_circumference_angle_external(1,ODB,C)
arc_property_center_angle(1,ODB,O)
Problem: As shown in the diagram, ∠GEF=90°, ∠HFC=50°, AE is parallel to CF. Find the measure of ∠BEG.
问题：如图所示，∠GEF=90°，∠HFC=50°，AE平行于CF。求∠BEG的大小。
Construction (CDL):
Shape(FE,EA)
Shape(BE,EG)
Shape(CF,FE)
Shape(EG,GD)
Shape(EF,FG,GE)
Shape(HF,FC)
Shape(DG,GI)
Collinear(AEB)
Collinear(CFGD)
Collinear(EFH)
Collinear(EGI)
Conditions (CDL):
Equal(MeasureOfAngle(GEF),90)
Equal(MeasureOfAngle(HFC),50)
ParallelBetweenLine(AE,CF)
Goal: Value(MeasureOfAngle(BEG))
Answer: 40
Theorem sequence:
parallel_property_corresponding_angle(1,FC,EA,H)
angle_addition(1,GEF,FEA)
adjacent_complementary_angle(1,BEG,GEA)
Problem: As shown in the diagram, AB=6, AC=10, ∠ABE=90°, triangle DAE is mirror congruent to triangle DEC. Find the perimeter of triangle ABE.
问题：如图所示，AB=6，AC=10，∠ABE=90°，三角形DAE与三角形DEC是镜像全等三角形。求△ABE的周长。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,ED,DA)
Shape(EC,CD,DE)
Collinear(ADC)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AC),10)
Equal(MeasureOfAngle(ABE),90)
MirrorCongruentBetweenTriangle(DAE,DEC)
Goal: Value(PerimeterOfTriangle(ABE))
Answer: 14
Theorem sequence:
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,ABC)
mirror_congruent_triangle_property_line_equal(1,DAE,DEC)
line_addition(1,BE,EC)
triangle_perimeter_formula(1,ABE)
Problem: As shown in the diagram, AB=5, DA=2, DC=3, DE=5, ∠ACE=90°. Find the length of line BE.
问题：如图所示，AB=5，DA=2，DC=3，DE=5，∠ACE=90°。求直线BE的长度。
Construction (CDL):
Shape(BF,FE,EB)
Shape(FD,DA,AF)
Shape(FA,AC,CE,EF)
Collinear(BFA)
Collinear(DFE)
Collinear(BEC)
Collinear(DAC)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(DA),2)
Equal(LengthOfLine(DC),3)
Equal(LengthOfLine(DE),5)
Equal(MeasureOfAngle(ACE),90)
Goal: Value(LengthOfLine(BE))
Answer: -4+2*sqrt(6)
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_judgment_angle(1,DCE)
line_addition(1,DA,AC)
right_triangle_property_pythagorean(1,DCE)
right_triangle_property_pythagorean(1,ACB)
line_addition(1,BE,EC)
Problem: As shown in the diagram, ∠DAF=30°, EF is parallel to BG, CA is perpendicular to BA. Find the measure of ∠BCA.
问题：如图所示，∠DAF=30°，EF∥BG，CA垂直于BA。求∠BCA的大小。
Construction (CDL):
Shape(DA,AF)
Shape(EA,AD)
Shape(BA,AE)
Shape(FA,AC)
Shape(AC,CG)
Shape(AB,BC,CA)
Collinear(EAF)
Collinear(DAB)
Collinear(BCG)
Conditions (CDL):
Equal(MeasureOfAngle(DAF),30)
ParallelBetweenLine(EF,BG)
PerpendicularBetweenLine(CA,BA)
Goal: Value(MeasureOfAngle(BCA))
Answer: 60
Theorem sequence:
vertical_angle(1,DAF,BAE)
parallel_property_collinear_extend(3,EF,BG,A)
parallel_property_alternate_interior_angle(2,EA,BG)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠EDA=125°, DA is parallel to CB. Find the measure of ∠CBD.
问题：如图所示，∠EDA=125°，DA平行于CB。求∠CBD的大小。
Construction (CDL):
Shape(ED,DA)
Shape(AD,DB)
Shape(CB,BD)
Shape(FB,BC)
Collinear(EDBF)
Conditions (CDL):
Equal(MeasureOfAngle(EDA),125)
ParallelBetweenLine(DA,CB)
Goal: Value(MeasureOfAngle(CBD))
Answer: 55
Theorem sequence:
adjacent_complementary_angle(1,EDA,ADB)
parallel_property_alternate_interior_angle(1,DA,CB)
Problem: As shown in the diagram, triangle ABC is congruent to triangle EBD, ∠ABC=40°, ∠DBC=100°. Find the measure of ∠FBA.
问题：如图所示，三角形ABC全等于三角形EBD，∠ABC=40°，∠DBC=100°。求∠FBA的大小。
Construction (CDL):
Shape(DE,EF,FD)
Shape(EB,BF,FE)
Shape(AF,FB,BA)
Shape(AB,BC,CA)
Collinear(DFB)
Collinear(EFA)
Conditions (CDL):
CongruentBetweenTriangle(ABC,EBD)
Equal(MeasureOfAngle(ABC),40)
Equal(MeasureOfAngle(DBC),100)
Goal: Value(MeasureOfAngle(FBA))
Answer: 60
Theorem sequence:
angle_addition(1,FBC,ABC)
Problem: As shown in the diagram, ∠IEG=58°, FG is the angle bisector of ∠EFD, AB∥CD. Find the measure of ∠BGF.
问题：如图所示，∠IEG=58°，FG平分∠EFD，AB∥CD。求∠BGF的大小。
Construction (CDL):
Shape(AE,EI)
Shape(IE,EG)
Shape(FE,EA)
Shape(BG,GF)
Shape(CF,FE)
Shape(GF,FD)
Shape(EF,FG,GE)
Shape(JF,FC)
Shape(DF,FJ)
Collinear(AEGB)
Collinear(CFD)
Collinear(IEFJ)
Conditions (CDL):
Equal(MeasureOfAngle(IEG),58)
IsBisectorOfAngle(FG,EFD)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(BGF))
Answer: 151
Theorem sequence:
parallel_property_collinear_extend(3,AB,CD,E)
parallel_property_collinear_extend(3,DC,BE,F)
parallel_property_corresponding_angle(1,EB,FD,I)
angle_addition(1,EFG,GFD)
adjacent_complementary_angle(1,IEG,GEF)
triangle_property_angle_sum(1,EFG)
adjacent_complementary_angle(1,BGF,FGE)
Problem: As shown in the diagram, ∠BOA=60°, the center of circle O is O. Find the measure of ∠BCA.
问题：如图所示，∠BOA=60°，⊙O的圆心为O。求∠BCA的大小。
Construction (CDL):
Shape(OA,AD,DO)
Shape(DB,BC,CD)
Shape(AO,OD,DC,OCA)
Shape(BD,DA,OAB)
Shape(CB,OBC)
Collinear(ODB)
Collinear(ADC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BOA),60)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 30
Theorem sequence:
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
Problem: As shown in the diagram, AD=9, CD=2, DE=3, ∠BAD=90°, ∠CDE=90°. Find the length of line BA.
问题：如图所示，AD=9，CD=2，DE=3，∠BAD=90°，∠CDE=90°。求直线BA的长度。
Construction (CDL):
Shape(BA,AD,DC,CB)
Shape(CD,DE,EC)
Collinear(BCE)
Collinear(ADE)
Conditions (CDL):
Equal(LengthOfLine(AD),9)
Equal(LengthOfLine(CD),2)
Equal(LengthOfLine(DE),3)
Equal(MeasureOfAngle(BAD),90)
Equal(MeasureOfAngle(CDE),90)
Goal: Value(LengthOfLine(BA))
Answer: 8
Theorem sequence:
similar_triangle_judgment_aa(1,BAE,CDE)
line_addition(1,AD,DE)
similar_triangle_property_line_ratio(1,BAE,CDE)
similar_triangle_property_line_ratio(1,EBA,ECD)
Problem: As shown in the diagram, triangle ABC is congruent to triangle EFC, ∠BCM=35°, ∠DCE=35°, ∠EDC=90°. Find the measure of ∠DAN.
问题：如图所示，△ABC全等于△EFC，∠BCM=35°，∠DCE=35°，∠EDC=90°。求∠DAN的大小。
Construction (CDL):
Shape(AN,ND,DA)
Shape(FM,MN,NF)
Shape(ED,DC,CE)
Shape(DN,NM,MC,CD)
Shape(MB,BC,CM)
Collinear(FNDE)
Collinear(BMNA)
Collinear(FMC)
Collinear(ADC)
Conditions (CDL):
CongruentBetweenTriangle(ABC,EFC)
Equal(MeasureOfAngle(BCM),35)
Equal(MeasureOfAngle(DCE),35)
Equal(MeasureOfAngle(EDC),90)
Goal: Value(MeasureOfAngle(DAN))
Answer: 55
Theorem sequence:
triangle_property_angle_sum(1,EDC)
congruent_triangle_property_angle_equal(1,ABC,EFC)
Problem: As shown in the diagram, BC=8, EF=5, BE is the altitude of triangle BCA, CF is the altitude of triangle CAB, M bisects segment BC. Find the perimeter of triangle EFM.
问题：如图所示，BC=8，EF=5，BE是三角形BCA的高，CF是三角形CAB的高，M平分线段BC。求△EFM的周长。
Construction (CDL):
Shape(AF,FE,EA)
Shape(FB,BP,PF)
Shape(FP,PQ,QF)
Shape(FQ,QE,EF)
Shape(BM,MP,PB)
Shape(PM,MN,NQ,QP)
Shape(EQ,QN,NE)
Shape(MC,CN,NM)
Shape(EN,NC,CE)
Collinear(AFB)
Collinear(AEC)
Collinear(BMC)
Collinear(BPQE)
Collinear(FPM)
Collinear(FQNC)
Collinear(MNE)
Conditions (CDL):
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(EF),5)
IsAltitudeOfTriangle(BE,BCA)
IsAltitudeOfTriangle(CF,CAB)
IsMidpointOfLine(M,BC)
Goal: Value(PerimeterOfTriangle(EFM))
Answer: 13
Theorem sequence:
right_triangle_judgment_angle(1,CFB)
right_triangle_judgment_angle(1,CEB)
median_of_triangle_judgment(1,EM,EBC)
median_of_triangle_judgment(1,FM,FBC)
right_triangle_property_length_of_median(1,CEB,M)
right_triangle_property_length_of_median(1,CFB,M)
triangle_perimeter_formula(1,EFM)
Problem: As shown in the diagram, ∠CPD=40°, the center of ⊙O is O, the tangent to ⊙O is CE, the tangent to circle O is DE, the tangent to circle O is PA, PB is the tangent to circle O. Find the sum of the measure of ∠EAC and the measure of ∠DBE.
问题：如图所示，∠CPD=40°，⊙O的圆心为O，CE是⊙O的切线，圆O的切线为DE，PA是⊙O的切线，⊙O的切线为PB。求∠EAC的大小与∠DBE的大小之和。
Construction (CDL):
Shape(OBA,AO,OB)
Shape(OAE,EA)
Shape(OEB,BE)
Shape(OA,AE,EB,BO)
Shape(OAE,AC,CE)
Shape(OEB,ED,DB)
Shape(CP,PD,DE,EC)
Collinear(PCA)
Collinear(PDB)
Collinear(CED)
Cocircular(O,AEB)
Conditions (CDL):
Equal(MeasureOfAngle(CPD),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CE,O)
IsTangentOfCircle(DE,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(Add(MeasureOfAngle(EAC),MeasureOfAngle(DBE)))
Answer: 70
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
tangent_of_circle_judgment_perpendicular(1,CA,O,O)
tangent_of_circle_judgment_perpendicular(1,DB,O,O)
tangent_of_circle_property_length_equal(1,CE,CA,O)
tangent_of_circle_property_length_equal(1,DE,DB,O)
isosceles_triangle_judgment_line_equal(1,CEA)
isosceles_triangle_judgment_line_equal(1,DBE)
isosceles_triangle_property_angle_equal(1,CEA)
isosceles_triangle_property_angle_equal(1,DBE)
adjacent_complementary_angle(1,ACE,ECP)
adjacent_complementary_angle(1,PDE,EDB)
triangle_property_angle_sum(1,PDC)
triangle_property_angle_sum(1,ACE)
triangle_property_angle_sum(1,EDB)
Problem: As shown in the diagram, ∠NAM=110°, PM is the perpendicular bisector of AB, QN perpendicular bisects AC. Find the measure of ∠QAP.
问题：如图所示，∠NAM=110°，PM垂直平分AB，QN垂直平分AB。求∠QAP的大小。
Construction (CDL):
Shape(AM,MP,PA)
Shape(MB,BP,PM)
Shape(AP,PQ,QA)
Shape(AQ,QN,NA)
Shape(NQ,QC,CN)
Collinear(AMB)
Collinear(ANC)
Collinear(BPQC)
Conditions (CDL):
Equal(MeasureOfAngle(NAM),110)
IsPerpendicularBisectorOfLine(PM,AB)
IsPerpendicularBisectorOfLine(QN,AC)
Goal: Value(MeasureOfAngle(QAP))
Answer: 40
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,PM,AB)
perpendicular_bisector_property_distance_equal(1,QN,AC)
isosceles_triangle_judgment_line_equal(1,PAB)
isosceles_triangle_judgment_line_equal(1,QCA)
isosceles_triangle_property_angle_equal(1,PAB)
isosceles_triangle_property_angle_equal(1,QCA)
triangle_property_angle_sum(1,ABC)
angle_addition(1,QAP,PAM)
angle_addition(1,NAQ,QAM)
Problem: As shown in the diagram, ∠DOA=50°. Find the measure of ∠COB.
问题：如图所示，∠DOA=50°。求∠COB的大小。
Construction (CDL):
Shape(DO,OA)
Shape(BO,OD)
Shape(AO,OC)
Shape(CO,OB)
Collinear(AOB)
Collinear(COD)
Conditions (CDL):
Equal(MeasureOfAngle(DOA),50)
Goal: Value(MeasureOfAngle(COB))
Answer: 50
Theorem sequence:
vertical_angle(1,COB,DOA)