SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=8, OC=5, O is the center of ⊙O, AD is perpendicular to OD. Find the length of line OD.
问题：如图所示，AB=8，OC=5，⊙O的圆心为O，AD⊥OD。求直线OD的长度。
Construction (CDL):
Shape(OA,AD,DO)
Shape(OAC,CD,DA)
Shape(OCB,BD,DC)
Shape(OBA,AO,OD,DB)
Collinear(ADB)
Collinear(ODC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(OC),5)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AD,OD)
Goal: Value(LengthOfLine(OD))
Answer: 3
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OD,AB)
line_addition(1,AD,DB)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
right_triangle_judgment_angle(1,ADO)
right_triangle_property_pythagorean(1,ADO)
Problem: As shown in the diagram, AB=10, OE=3, O is the center of circle O, DE⊥OE. Find the length of line CD.
问题：如图所示，AB=10，OE=3，O是圆O的圆心，DE垂直于OE。求直线CD的长度。
Construction (CDL):
Shape(OD,DE,EO)
Shape(OAD,DO,OA)
Shape(ODB,BE,ED)
Shape(OBC,CE,EB)
Shape(OCA,AO,OE,EC)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(OE),3)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(DE,OE)
Goal: Value(LengthOfLine(CD))
Answer: 8
Theorem sequence:
line_addition(1,AO,OB)
line_addition(1,CE,ED)
circle_property_chord_perpendicular_bisect_chord(1,O,OE,DC)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OB,O)
right_triangle_judgment_angle(1,DEO)
right_triangle_property_pythagorean(1,DEO)
Problem: As shown in the diagram, AD=1, ∠BCA=22.5°, O is the center of ⊙O, AD is perpendicular to OD. Find the length of line DB.
问题：如图所示，AD=1，∠BCA=22.5°，圆O的圆心为O，AD⊥OD。求直线DB的长度。
Construction (CDL):
Shape(OA,AD,DO)
Shape(DB,BC,CD)
Shape(OAB,BD,DA)
Shape(OBC,CB)
Shape(OCA,AO,OD,DC)
Collinear(ADC)
Collinear(ODB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(AD),1)
Equal(MeasureOfAngle(BCA),22.5)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AD,OD)
Goal: Value(LengthOfLine(DB))
Answer: -1+sqrt(2)
Theorem sequence:
line_addition(1,OD,DB)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
triangle_property_angle_sum(1,OAD)
isosceles_triangle_judgment_angle_equal(1,DOA)
right_triangle_judgment_angle(1,ADO)
right_triangle_property_pythagorean(1,ADO)
Problem: As shown in the diagram, ∠AOB=70°, O is the center of ⊙O, BM is perpendicular to OM. Find the measure of ∠CDA.
问题：如图所示，∠AOB=70°，O是⊙O的圆心，BM垂直于OM。求∠CDA的大小。
Construction (CDL):
Shape(OAC,CD,DA)
Shape(OAC,CM,MA)
Shape(OCD,DC)
Shape(ODB,BO,OM,MC,CD)
Shape(MO,OB,BM)
Shape(OBA,AM,MB)
Collinear(CMB)
Collinear(AMO)
Cocircular(O,CDBA)
Conditions (CDL):
Equal(MeasureOfAngle(AOB),70)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BM,OM)
Goal: Value(MeasureOfAngle(CDA))
Answer: 35
Theorem sequence:
circle_property_chord_perpendicular_bisect_arc(1,OBC,OMA)
congruent_arc_judgment_length_equal(1,OAC,OBA)
congruent_arc_property_measure_equal(1,OAC,OBA)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OAC,D)
Problem: As shown in the diagram, CE=4, the radius of ⊙O is 10, the center of ⊙O is O, BE is perpendicular to OE. Find the length of line AB.
问题：如图所示，CE=4，圆O的半径为10，⊙O的圆心为O，BE⊥OE。求直线AB的长度。
Construction (CDL):
Shape(OB,BE,EO)
Shape(OAB,BO,OE,EA)
Shape(OBC,CE,EB)
Shape(OCA,AE,EC)
Collinear(AEB)
Collinear(OEC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(CE),4)
Equal(RadiusOfCircle(O),10)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BE,OE)
Goal: Value(LengthOfLine(AB))
Answer: 16
Theorem sequence:
line_addition(1,AE,EB)
line_addition(1,OE,EC)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
circle_property_chord_perpendicular_bisect_chord(1,O,OE,BA)
right_triangle_judgment_angle(1,BEO)
right_triangle_property_pythagorean(1,BEO)
Problem: As shown in the diagram, AB=4, BC=6, DE=2, O is the center of ⊙O, CM⊥OM, quadrilateral ADHG is a rectangle. Find the length of line DF.
问题：如图所示，AB=4，BC=6，DE=2，O是圆O的圆心，CM⊥OM，四边形ADHG是矩形。求直线DF的长度。
Construction (CDL):
Shape(OCB,BC)
Shape(OF,OFC,CM,MO)
Shape(OM,MB,OBE,EO)
Shape(OEF,FO,OE)
Shape(AD,DE,OBE,BA)
Shape(OFC,FH,HG,GC)
Collinear(ABMCG)
Collinear(DEOFH)
Cocircular(O,CBEF)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(DE),2)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CM,OM)
Rectangle(ADHG)
Goal: Value(LengthOfLine(DF))
Answer: 12
Theorem sequence:
adjacent_complementary_angle(1,CMO,OMA)
parallel_judgment_ipsilateral_internal_angle(1,AM,DO)
parallel_judgment_ipsilateral_internal_angle(1,MO,AD)
line_addition(1,AB,BM)
line_addition(1,AB,BC)
line_addition(1,AM,MC)
line_addition(1,DE,EO)
line_addition(1,DO,OF)
circle_property_chord_perpendicular_bisect_chord(1,O,OM,CB)
parallelogram_judgment_parallel_and_parallel(1,OMAD)
parallelogram_property_opposite_line_equal(1,MADO)
radius_of_circle_property_length_equal(1,OE,O)
radius_of_circle_property_length_equal(1,OF,O)
Problem: As shown in the diagram, AB=8, AC=6, AD=4, the center of circle O is O, AD is perpendicular to CD. Find the diameter of ⊙O.
问题：如图所示，AB=8，AC=6，AD=4，⊙O的圆心为O，AD垂直于CD。求圆O的直径。
Construction (CDL):
Shape(OAB,BA)
Shape(OBC,CD,DB)
Shape(OCA,AC)
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Shape(AB,BE,EO,OA)
Collinear(BDC)
Collinear(AOE)
Cocircular(O,CABE)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(AD),4)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AD,CD)
Goal: Value(DiameterOfCircle(O))
Answer: 12
Theorem sequence:
line_addition(1,AO,OE)
diameter_of_circle_judgment_pass_centre(1,AOE,O)
diameter_of_circle_property_right_angle(1,ABE,O)
arc_property_circumference_angle_external(1,OAB,C)
arc_property_circumference_angle_external(1,OAB,E)
similar_triangle_judgment_aa(1,ADC,ABE)
similar_triangle_property_line_ratio(1,DCA,BEA)
similar_triangle_property_line_ratio(1,CAD,EAB)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OE,O)
circle_property_length_of_radius_and_diameter(1,O)
Problem: As shown in the diagram, ∠BOA=100°, the center of circle O is O. Find the measure of ∠BCA.
问题：如图所示，∠BOA=100°，O是⊙O的圆心。求∠BCA的大小。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(OBC,CB)
Shape(OCA,AC)
Shape(OB,BC,CA,AO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BOA),100)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 50
Theorem sequence:
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
Problem: As shown in the diagram, ∠OCA=20°, the center of circle O is O. Find the measure of ∠COB.
问题：如图所示，∠OCA=20°，O是圆O的圆心。求∠COB的大小。
Construction (CDL):
Shape(OCA,AC)
Shape(OC,CA,AO)
Shape(OAB,BO,OA)
Shape(OBC,CO,OB)
Collinear(AOB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(OCA),20)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(COB))
Answer: 40
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OCA)
isosceles_triangle_property_angle_equal(1,OCA)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, ∠ABD=58°, O is the center of circle O. Find the measure of ∠DCB.
问题：如图所示，∠ABD=58°，O是圆O的圆心。求∠DCB的大小。
Construction (CDL):
Shape(ODA,AD)
Shape(OAC,CM,MO,OA)
Shape(OCB,BC)
Shape(OBD,DB)
Shape(AO,OM,MD,DA)
Shape(DM,MB,BD)
Shape(MC,CB,BM)
Collinear(AOMB)
Collinear(DMC)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),58)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(DCB))
Answer: 32
Theorem sequence:
diameter_of_circle_judgment_pass_centre(1,AOB,O)
diameter_of_circle_property_right_angle(1,BDA,O)
triangle_property_angle_sum(1,DAB)
arc_property_circumference_angle_external(1,OBD,A)
arc_property_circumference_angle_external(1,OBD,C)
Problem: As shown in the diagram, ∠ADC=35°, the center of ⊙O is O. Find the measure of ∠CAO.
问题：如图所示，∠ADC=35°，O是⊙O的圆心。求∠CAO的大小。
Construction (CDL):
Shape(OAB,BA)
Shape(OBD,DM,MO,OB)
Shape(ODC,CD)
Shape(OCA,AC)
Shape(OA,AB,BO)
Shape(AO,OM,MA)
Shape(AM,MC,CA)
Shape(MD,DC,CM)
Collinear(BOMC)
Collinear(AMD)
Cocircular(O,ABDC)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),35)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(CAO))
Answer: 55
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
arc_property_circumference_angle_external(1,OCA,B)
arc_property_circumference_angle_external(1,OCA,D)
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_judgment_line_equal(1,OCA)
isosceles_triangle_property_angle_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OCA)
triangle_property_angle_sum(1,OAB)
Problem: As shown in the diagram, CD=DO, the radius of circle O is 3, O is the center of circle O, AD is perpendicular to OD. Find the length of line AB.
问题：如图所示，CD=DO，⊙O的半径为3，O是⊙O的圆心，AD垂直于OD。求直线AB的长度。
Construction (CDL):
Shape(OAC,CD,DA)
Shape(OCB,BD,DC)
Shape(OBA,AO,OB)
Shape(OA,AD,DO)
Shape(OD,DB,BO)
Collinear(ADB)
Collinear(CDO)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(CD),LengthOfLine(DO))
Equal(RadiusOfCircle(O),3)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AD,OD)
Goal: Value(LengthOfLine(AB))
Answer: 3*sqrt(3)
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OB,O)
adjacent_complementary_angle(1,ADO,ODB)
right_triangle_judgment_angle(1,ADO)
right_triangle_judgment_angle(1,ODB)
right_triangle_property_pythagorean(1,ADO)
right_triangle_property_pythagorean(1,ODB)
line_addition(1,AD,DB)
line_addition(1,CD,DO)
Problem: As shown in the diagram, AM=2, DE=1, EF=8, O is the center of ⊙O, FH is perpendicular to OH, NP is perpendicular to OP, quadrilateral ADCB is a rectangle. Find the length of line MN.
问题：如图所示，AM=2，DE=1，EF=8，圆O的圆心为O，FH垂直于OH，NP⊥OP，ADCB是长方形。求直线MN的长度。
Construction (CDL):
Shape(ONM,MN)
Shape(OE,OEF,FH,HO)
Shape(PM,OME,EH,HP)
Shape(PH,HF,OFN,NP)
Shape(AD,DE,OME,MA)
Shape(CB,BN,OFN,FC)
Shape(HE,EO,OH)
Collinear(AMPNB)
Collinear(DEHFC)
Collinear(OHP)
Cocircular(O,MEFN)
Conditions (CDL):
Equal(LengthOfLine(AM),2)
Equal(LengthOfLine(DE),1)
Equal(LengthOfLine(EF),8)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(FH,OH)
PerpendicularBetweenLine(NP,OP)
Rectangle(ADCB)
Goal: Value(LengthOfLine(MN))
Answer: 6
Theorem sequence:
adjacent_complementary_angle(1,NPH,HPA)
parallel_judgment_ipsilateral_internal_angle(1,AP,DH)
parallel_judgment_ipsilateral_internal_angle(1,PH,AD)
line_addition(1,AM,MP)
line_addition(1,MP,PN)
line_addition(1,DE,EH)
line_addition(1,DE,EF)
line_addition(1,DH,HF)
circle_property_chord_perpendicular_bisect_chord(1,O,OP,NM)
circle_property_chord_perpendicular_bisect_chord(1,O,OH,FE)
parallelogram_judgment_parallel_and_parallel(1,HPAD)
parallelogram_property_opposite_line_equal(1,PADH)
Problem: As shown in the diagram, AB=3, △DCQ is an equilateral △, △PCD is an equilateral △, quadrilateral ABCD is a square. Find the length of line PQ.
问题：如图所示，AB=3，三角形DCQ为等边三角形，△PCD为等边△，ABCD是正方形。求直线PQ的长度。
Construction (CDL):
Shape(AB,BC,CE,EP,PD,DA)
Shape(DP,PE,ED)
Shape(PC,CE,EP)
Shape(EC,CQ,QE)
Shape(DE,EQ,QD)
Collinear(PEQ)
Collinear(DEC)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
EquilateralTriangle(DCQ)
EquilateralTriangle(PCD)
Square(ABCD)
Goal: Value(LengthOfLine(PQ))
Answer: 3*sqrt(3)
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DPQ)
isosceles_triangle_judgment_line_equal(1,CQP)
isosceles_triangle_property_angle_equal(1,PCD)
isosceles_triangle_property_angle_equal(1,CDP)
isosceles_triangle_property_angle_equal(1,DCQ)
isosceles_triangle_property_angle_equal(1,CQD)
isosceles_triangle_property_angle_equal(1,DPQ)
isosceles_triangle_property_angle_equal(1,CQP)
triangle_property_angle_sum(1,DPE)
triangle_property_angle_sum(1,PCE)
triangle_property_angle_sum(1,ECQ)
triangle_property_angle_sum(1,DEQ)
triangle_property_angle_sum(1,DPC)
triangle_property_angle_sum(1,CQD)
mirror_congruent_triangle_judgment_sas(1,DPE,DEQ)
mirror_congruent_triangle_property_line_equal(1,DPE,DEQ)
adjacent_complementary_angle(1,PED,DEQ)
adjacent_complementary_angle(1,QEC,CEP)
right_triangle_judgment_angle(1,PED)
right_triangle_judgment_angle(1,CEP)
right_triangle_property_pythagorean(1,PED)
right_triangle_property_pythagorean(1,CEP)
line_addition(1,PE,EQ)
line_addition(1,DE,EC)
Problem: As shown in the diagram, AE=3, CF=1, BE is perpendicular to AE, CF⊥EF, ABCD is a square. Find the length of line AB.
问题：如图所示，AE=3，CF=1，BE⊥AE，CF⊥EF，ABCD是正方形。求直线AB的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(BC,CF,FE,EB)
Shape(AE,EF,FG,GD,DA)
Shape(FC,CG,GF)
Collinear(BEFG)
Collinear(CGD)
Conditions (CDL):
Equal(LengthOfLine(AE),3)
Equal(LengthOfLine(CF),1)
PerpendicularBetweenLine(BE,AE)
PerpendicularBetweenLine(CF,EF)
Square(ABCD)
Goal: Value(LengthOfLine(AB))
Answer: sqrt(10)
Theorem sequence:
triangle_property_angle_sum(1,BCF)
angle_addition(1,ABE,EBC)
congruent_triangle_judgment_aas(1,ABE,BCF)
congruent_triangle_property_line_equal(1,ABE,BCF)
right_triangle_judgment_angle(1,BEA)
right_triangle_property_pythagorean(1,BEA)
Problem: As shown in the diagram, △BCE is congruent to △DCF, ∠CEB=70°. Find the measure of ∠EFD.
问题：如图所示，三角形BCE与三角形DCF是全等三角形，∠CEB=70°。求∠EFD的大小。
Construction (CDL):
Shape(AB,BE,ED,DA)
Shape(BC,CE,EB)
Shape(EC,CF,FE)
Shape(DE,EF,FD)
Collinear(BCF)
Collinear(DEC)
Conditions (CDL):
CongruentBetweenTriangle(BCE,DCF)
Equal(MeasureOfAngle(CEB),70)
Goal: Value(MeasureOfAngle(EFD))
Answer: 25
Theorem sequence:
congruent_triangle_property_line_equal(1,BCE,DCF)
congruent_triangle_property_angle_equal(1,CEB,CFD)
congruent_triangle_property_angle_equal(1,EBC,FDC)
adjacent_complementary_angle(1,BCE,ECF)
angle_addition(1,CFE,EFD)
isosceles_triangle_judgment_line_equal(1,CFE)
isosceles_triangle_property_angle_equal(1,CFE)
triangle_property_angle_sum(1,CFE)
Problem: As shown in the diagram, ∠DEB=126°, ABCD is a square. Find the measure of ∠EDA.
问题：如图所示，∠DEB=126°，四边形ABCD是正方形。求∠EDA的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BC,CE)
Shape(EC,CD,DE)
Shape(ED,DA,AE)
Collinear(AEC)
Conditions (CDL):
Equal(MeasureOfAngle(DEB),126)
Square(ABCD)
Goal: Value(MeasureOfAngle(EDA))
Answer: 18
Theorem sequence:
angle_addition(1,BCE,ECD)
angle_addition(1,DEC,CEB)
angle_addition(1,CDE,EDA)
triangle_property_angle_sum(1,ECD)
triangle_property_angle_sum(1,ABC)
isosceles_triangle_judgment_line_equal(1,BCA)
isosceles_triangle_property_angle_equal(1,BCA)
mirror_congruent_triangle_judgment_sas(1,CEB,CDE)
mirror_congruent_triangle_property_angle_equal(1,EBC,ECD)
Problem: As shown in the diagram, Div(LengthOfLine(OM)=LengthOfLine(OA)), AB=15, the center of circle O is O, DM⊥OM. Find the length of line CD.
问题：如图所示，Div(LengthOfLine(OM)=LengthOfLine(OA))，AB=15，O是圆O的圆心，DM⊥OM。求直线CD的长度。
Construction (CDL):
Shape(OD,DM,MO)
Shape(OCA,AO,OM,MC)
Shape(OAD,DO,OA)
Shape(ODB,BM,MD)
Shape(OBC,CM,MB)
Collinear(AOMB)
Collinear(CMD)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(Div(LengthOfLine(OM),LengthOfLine(OA)),3/5)
Equal(LengthOfLine(AB),15)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(DM,OM)
Goal: Value(LengthOfLine(CD))
Answer: 12
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OB,O)
line_addition(1,AO,OB)
line_addition(1,CM,MD)
circle_property_chord_perpendicular_bisect_chord(1,O,OM,DC)
right_triangle_judgment_angle(1,DMO)
right_triangle_property_pythagorean(1,DMO)
Problem: As shown in the diagram, EB=1, EC=2, ABCD is a square. Find the area of quadrilateral ABCD.
问题：如图所示，EB=1，EC=2，ABCD是正方形。求ABCD的面积。
Construction (CDL):
Shape(AE,EC,CD,DA)
Shape(EB,BC,CE)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(EB),1)
Equal(LengthOfLine(EC),2)
Square(ABCD)
Goal: Value(AreaOfQuadrilateral(ABCD))
Answer: 3
Theorem sequence:
right_triangle_judgment_angle(1,EBC)
right_triangle_property_pythagorean(1,EBC)
parallelogram_area_formula_sine(1,ABCD)
Problem: As shown in the diagram, AB=8, the radius of ⊙O is 5, the center of ⊙O is O, AC is perpendicular to OC. Find the length of line OC.
问题：如图所示，AB=8，圆O的半径为5，O是圆O的圆心，AC垂直于OC。求直线OC的长度。
Construction (CDL):
Shape(OA,AC,CO)
Shape(OC,CB,BO)
Shape(OAB,BA)
Shape(OBA,AO,OB)
Collinear(ACB)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(RadiusOfCircle(O),5)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AC,OC)
Goal: Value(LengthOfLine(OC))
Answer: 3
Theorem sequence:
line_addition(1,AC,CB)
radius_of_circle_property_length_equal(1,OA,O)
circle_property_chord_perpendicular_bisect_chord(1,O,OC,AB)
right_triangle_judgment_angle(1,ACO)
right_triangle_property_pythagorean(1,ACO)
Problem: As shown in the diagram, ∠FAE=50°, the center of ⊙O is O, M bisects ⌒OAB, N bisects ⌒OCA. Find the measure of ∠MON.
问题：如图所示，∠FAE=50°，O是圆O的圆心，M是弧OAB的中点，N平分弧OCA。求∠MON的大小。
Construction (CDL):
Shape(OCN,NF,FC)
Shape(ONA,AF,FN)
Shape(OAM,ME,EA)
Shape(OMB,BE,EM)
Shape(OBC,CF,FO,OE,EB)
Shape(OF,FA,AE,EO)
Shape(OC,CN,NO)
Shape(ON,NA,AO)
Shape(OM,MB,BO)
Shape(OA,AM,MO)
Shape(OB,OBC,CO)
Collinear(NFO)
Collinear(AFC)
Collinear(AEB)
Collinear(OEM)
Cocircular(O,AMBCN)
Conditions (CDL):
Equal(MeasureOfAngle(FAE),50)
IsCentreOfCircle(O,O)
IsMidpointOfArc(M,OAB)
IsMidpointOfArc(N,OCA)
Goal: Value(MeasureOfAngle(MON))
Answer: 130
Theorem sequence:
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
congruent_arc_judgment_length_equal(1,OAM,OMB)
congruent_arc_judgment_length_equal(1,ONA,OCN)
congruent_arc_property_chord_equal(1,OAM,OMB)
congruent_arc_property_chord_equal(1,ONA,OCN)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OM,O)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,ON,O)
congruent_triangle_judgment_sss(1,OCN,ONA)
congruent_triangle_judgment_sss(1,OMB,OAM)
congruent_triangle_property_angle_equal(1,OCN,ONA)
congruent_triangle_property_angle_equal(1,OMB,OAM)
angle_addition(1,AON,NOC)
angle_addition(1,BOM,MOA)
angle_addition(1,BOA,AOC)
angle_addition(1,MOA,AON)
round_angle(1,BOC,COB)
Problem: As shown in the diagram, OD=1, the center of ⊙O is O, CD is perpendicular to OD. Find the length of line BC.
问题：如图所示，OD=1，O是⊙O的圆心，CD垂直于OD。求直线BC的长度。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(OBC,CB)
Shape(OCA,AD,DC)
Shape(AO,OD,DA)
Shape(DO,OB,BC,CD)
Collinear(ADC)
Collinear(AOB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(OD),1)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CD,OD)
Goal: Value(LengthOfLine(BC))
Answer: 2
Theorem sequence:
adjacent_complementary_angle(1,CDO,ODA)
line_addition(1,AO,OB)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
diameter_of_circle_judgment_pass_centre(1,AOB,O)
diameter_of_circle_property_right_angle(1,BCA,O)
similar_triangle_judgment_aa(1,BCA,ODA)
similar_triangle_property_line_ratio(1,CAB,DAO)
similar_triangle_property_line_ratio(1,ABC,AOD)
Problem: As shown in the diagram, AP=5, CP is the angle bisector of ∠OCD, O is the center of ⊙O, CM is perpendicular to BM. Find the length of line BP.
问题：如图所示，AP=5，CP是∠OCD的角平分线，⊙O的圆心为O，CM垂直于BM。求直线BP的长度。
Construction (CDL):
Shape(OCA,AM,MC)
Shape(OAD,DM,MA)
Shape(CM,MN,NC)
Shape(ODP,PN,NM,MD)
Shape(OC,CN,NE,EO)
Shape(NP,PE,EN)
Shape(OBC,CO,OE,EB)
Shape(OPB,BE,EP)
Shape(OAD,ODP,PA)
Shape(OPB,BP)
Collinear(AMNEB)
Collinear(CMD)
Collinear(CNP)
Collinear(OEP)
Cocircular(O,ADPBC)
Conditions (CDL):
Equal(LengthOfLine(AP),5)
IsBisectorOfAngle(CP,OCD)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CM,BM)
Goal: Value(LengthOfLine(BP))
Answer: 5
Theorem sequence:
radius_of_circle_property_length_equal(1,OP,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OCP)
isosceles_triangle_property_angle_equal(1,OCP)
parallel_judgment_alternate_interior_angle(2,EP,CM)
parallel_property_collinear_extend(1,EP,CM,O)
parallel_property_corresponding_angle(2,MC,EO,B)
adjacent_complementary_angle(1,AEO,OEB)
circle_property_chord_perpendicular_bisect_arc(1,OAB,OEP)
congruent_arc_judgment_length_equal(1,OAP,OPB)
congruent_arc_property_chord_equal(1,OAP,OPB)
Problem: As shown in the diagram, PB=1, PC=3, the center of circle O is O, PC is the tangent to ⊙O. Find the radius of ⊙O.
问题：如图所示，PB=1，PC=3，⊙O的圆心为O，圆O的切线为PC。求⊙O的半径。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(OBC,OCA,AO,OB)
Shape(PC,OBC,BP)
Collinear(AOBP)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(PB),1)
Equal(LengthOfLine(PC),3)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PC,O)
Goal: Value(RadiusOfCircle(O))
Answer: 4
Theorem sequence:
circle_property_circular_power_tangent_and_segment_line(1,PC,PBA,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
line_addition(1,AO,OB)
line_addition(1,AB,BP)
Problem: As shown in the diagram, ∠DAP=28°, ∠PBC=26°, the center of circle O is O, the tangent to ⊙O is BC. Find the measure of ∠CDP.
问题：如图所示，∠DAP=28°，∠PBC=26°，圆O的圆心为O，BC是圆O的切线。求∠CDP的大小。
Construction (CDL):
Shape(OPC,CO,OP)
Shape(OCD,DC)
Shape(ODE,EP,PO,OD)
Shape(DA,AE,ODE)
Shape(OEP,PE)
Shape(OC,CD,DO)
Shape(PB,BC,OPC)
Collinear(BPEA)
Collinear(CDA)
Collinear(POD)
Cocircular(O,PCDE)
Conditions (CDL):
Equal(MeasureOfAngle(DAP),28)
Equal(MeasureOfAngle(PBC),26)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BC,O)
Goal: Value(MeasureOfAngle(CDP))
Answer: 36
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BC,O,O)
triangle_property_angle_sum(1,ABC)
angle_addition(1,BCO,OCD)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,OCD)
isosceles_triangle_property_angle_equal(1,OCD)
Problem: As shown in the diagram, ∠AOB=60°, the center of circle O is O, the tangent to ⊙O is MA. Find the measure of ∠MAB.
问题：如图所示，∠AOB=60°，O是圆O的圆心，MA是⊙O的切线。求∠MAB的大小。
Construction (CDL):
Shape(OB,BA,AO)
Shape(OBA,AB)
Shape(OAB,BO,OA)
Shape(MA,AN)
Shape(MA,AB,BM)
Collinear(MAN)
Cocircular(O,BA)
Conditions (CDL):
Equal(MeasureOfAngle(AOB),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(MA,O)
Goal: Value(MeasureOfAngle(MAB))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(2,MA,O,O)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OA,O)
isosceles_triangle_judgment_line_equal(1,OBA)
isosceles_triangle_property_angle_equal(1,OBA)
triangle_property_angle_sum(1,OBA)
angle_addition(1,MAB,BAO)
Problem: As shown in the diagram, ∠PCB=29°, the center of ⊙O is O, the diameter of circle O is AB, PC is the tangent to ⊙O. Find the measure of ∠CDA.
问题：如图所示，∠PCB=29°，⊙O的圆心为O，AB是⊙O的直径，PC是圆O的切线。求∠CDA的大小。
Construction (CDL):
Shape(OC,CD,DA,AO)
Shape(OCD,DC)
Shape(ODA,AD)
Shape(CO,OB,BC)
Shape(OBC,CB)
Shape(OBC,BP,PC)
Shape(OA,OAB,BO)
Collinear(AOBP)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(PCB),29)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PC,O)
Goal: Value(MeasureOfAngle(CDA))
Answer: 119
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PC,O,O)
angle_addition(1,AOC,COB)
angle_addition(1,PCB,BCO)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OBC)
triangle_property_angle_sum(1,COB)
flat_angle(1,AOB)
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_internal(1,OCA,D)
Problem: As shown in the diagram, ∠ABC=40°, O is the center of ⊙O, AB is the diameter of circle O, the tangent to ⊙O is DA. Find the measure of ∠DAC.
问题：如图所示，∠ABC=40°，O是圆O的圆心，AB是⊙O的直径，DA是⊙O的切线。求∠DAC的大小。
Construction (CDL):
Shape(AO,OB,BC,CA)
Shape(DA,AC)
Collinear(AOB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),40)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(DA,O)
Goal: Value(MeasureOfAngle(DAC))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DA,O,O)
angle_addition(1,DAC,CAB)
diameter_of_circle_property_right_angle(1,BCA,O)
triangle_property_angle_sum(1,ABC)