SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠EAD=75°. Find the sum of the measure of ∠EDB and the measure of ∠CED.
问题：如图所示，∠EAD=75°。求∠EDB的大小与∠CED的大小之和。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CE,ED)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(MeasureOfAngle(EAD),75)
Goal: Value(Add(MeasureOfAngle(EDB),MeasureOfAngle(CED)))
Answer: 255
Theorem sequence:
adjacent_complementary_angle(1,ADE,EDB)
adjacent_complementary_angle(1,CED,DEA)
triangle_property_angle_sum(1,ADE)
Problem: As shown in the diagram, AC=8, BC=4, P bisects segment AC, Q bisects segment CB. Find the length of line PQ.
问题：如图所示，AC=8，BC=4，P平分线段AC，Q平分线段CB。求直线PQ的长度。
Construction (CDL):
Collinear(APCQB)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),4)
IsMidpointOfLine(P,AC)
IsMidpointOfLine(Q,CB)
Goal: Value(LengthOfLine(PQ))
Answer: 6
Theorem sequence:
line_addition(1,AP,PC)
line_addition(1,CQ,QB)
line_addition(1,PC,CQ)
Problem: As shown in the diagram, ∠GCF=36°, triangle FGC is mirror congruent to triangle FDG. Find the measure of ∠EFA minus the measure of ∠DGE.
问题：如图所示，∠GCF=36°，三角形FGC与三角形FDG是镜像全等三角形。求∠EFA的大小减去∠DGE的大小。
Construction (CDL):
Shape(AB,BE,EF,FA)
Shape(ED,DG,GE)
Shape(FE,EG,GF)
Shape(FG,GC,CF)
Collinear(AFC)
Collinear(BEGC)
Collinear(DEF)
Conditions (CDL):
Equal(MeasureOfAngle(GCF),36)
MirrorCongruentBetweenTriangle(FGC,FDG)
Goal: Value(Sub(MeasureOfAngle(EFA),MeasureOfAngle(DGE)))
Answer: 72
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,CFG,DGF)
adjacent_complementary_angle(1,CFD,DFA)
triangle_property_angle_sum(1,FEC)
adjacent_complementary_angle(1,FEC,CED)
triangle_property_angle_sum(1,EDG)
Problem: As shown in the diagram, BC=Mul(LengthOfLine(AB),1/2), DC=3, D is the midpoint of segment AC. Find the length of line AB.
问题：如图所示，BC=Mul(LengthOfLine(AB),1/2)，DC=3，D平分线段AC。求直线AB的长度。
Construction (CDL):
Collinear(ADBC)
Conditions (CDL):
Equal(LengthOfLine(BC),Mul(LengthOfLine(AB),1/2))
Equal(LengthOfLine(DC),3)
IsMidpointOfLine(D,AC)
Goal: Value(LengthOfLine(AB))
Answer: 4
Theorem sequence:
line_addition(1,AD,DC)
line_addition(1,AB,BC)
Problem: As shown in the diagram, ∠ACB=90°, ∠FCA=72°, △DCB is mirror congruent to △DFC. Find the measure of ∠ACD.
问题：如图所示，∠ACB=90°，∠FCA=72°，三角形DCB与三角形DFC是镜像全等三角形。求∠ACD的大小。
Construction (CDL):
Shape(AE,ED,DA)
Shape(EF,FC,CE)
Shape(EC,CD,DE)
Shape(DC,CB,BD)
Collinear(FED)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),90)
Equal(MeasureOfAngle(FCA),72)
MirrorCongruentBetweenTriangle(DCB,DFC)
Goal: Value(MeasureOfAngle(ACD))
Answer: 9
Theorem sequence:
angle_addition(1,FCA,ACB)
angle_addition(1,FCD,DCB)
mirror_congruent_triangle_property_angle_equal(1,CBD,CDF)
angle_addition(1,ACD,DCB)
Problem: As shown in the diagram, AC=8, EC=3, D bisects segment AB, E is the midpoint of segment BC. Find the length of line AD.
问题：如图所示，AC=8，EC=3，D是线段AB的中点，E是线段BC的中点。求直线AD的长度。
Construction (CDL):
Collinear(ADBEC)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(EC),3)
IsMidpointOfLine(D,AB)
IsMidpointOfLine(E,BC)
Goal: Value(LengthOfLine(AD))
Answer: 1
Theorem sequence:
line_addition(1,BE,EC)
line_addition(1,AB,BC)
line_addition(1,AD,DB)
Problem: As shown in the diagram, ∠AOE=60°, ∠BOD=20°, ∠COE=90°, ∠DOC=90°. Find the measure of ∠BOA.
问题：如图所示，∠AOE=60°，∠BOD=20°，∠COE=90°，∠DOC=90°。求∠BOA的大小。
Construction (CDL):
Shape(EO,OA)
Shape(AO,OC)
Shape(CO,OD)
Shape(BO,OD)
Shape(FO,OB)
Collinear(EOD)
Collinear(COF)
Conditions (CDL):
Equal(MeasureOfAngle(AOE),60)
Equal(MeasureOfAngle(BOD),20)
Equal(MeasureOfAngle(COE),90)
Equal(MeasureOfAngle(DOC),90)
Goal: Value(MeasureOfAngle(BOA))
Answer: 140
Theorem sequence:
angle_addition(1,COA,AOE)
angle_addition(1,BOD,DOC)
angle_addition(1,BOC,COA)
Problem: As shown in the diagram, ∠ABC=32°, AD bisects ∠EAC, AD∥BC. Find the measure of ∠BCA.
问题：如图所示，∠ABC=32°，AD平分∠EAC，AD平行于BC。求∠BCA的大小。
Construction (CDL):
Shape(EA,AD)
Shape(DA,AC)
Shape(AB,BC,CA)
Collinear(BAE)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),32)
IsBisectorOfAngle(AD,EAC)
ParallelBetweenLine(AD,BC)
Goal: Value(MeasureOfAngle(BCA))
Answer: 32
Theorem sequence:
parallel_property_corresponding_angle(1,AD,BC,E)
angle_addition(1,EAD,DAC)
adjacent_complementary_angle(1,EAC,CAB)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, AB=Add(LengthOfLine(AC),3), the perimeter of △ABD is 22, AD is the median of triangle ABC. Find the perimeter of △ADC.
问题：如图所示，AB=Add(LengthOfLine(AC),3)，三角形ABD的周长为22，AD是△ABC的中线。求△ADC的周长。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),Add(LengthOfLine(AC),3))
Equal(PerimeterOfTriangle(ABD),22)
IsMedianOfTriangle(AD,ABC)
Goal: Value(PerimeterOfTriangle(ADC))
Answer: 19
Theorem sequence:
triangle_perimeter_formula(1,ABD)
triangle_perimeter_formula(1,ADC)
Problem: As shown in the diagram, ∠BCD=90°. Find the sum of the measure of ∠BEF and the measure of ∠EFD.
问题：如图所示，∠BCD=90°。求∠BEF的大小与∠EFD的大小之和。
Construction (CDL):
Shape(AB,BE,EF,FD,DA)
Shape(EC,CF,FE)
Collinear(BEC)
Collinear(CFD)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),90)
Goal: Value(Add(MeasureOfAngle(BEF),MeasureOfAngle(EFD)))
Answer: 270
Theorem sequence:
triangle_property_angle_sum(1,ECF)
adjacent_complementary_angle(1,BEF,FEC)
adjacent_complementary_angle(1,CFE,EFD)
Problem: As shown in the diagram, AB=20, EB=3, C bisects segment AB, E is the midpoint of segment DB. Find the length of line CD.
问题：如图所示，AB=20，EB=3，C平分线段AB，E是线段DB的中点。求直线CD的长度。
Construction (CDL):
Collinear(ACDEB)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(EB),3)
IsMidpointOfLine(C,AB)
IsMidpointOfLine(E,DB)
Goal: Value(LengthOfLine(CD))
Answer: 4
Theorem sequence:
line_addition(1,AC,CB)
line_addition(1,DE,EB)
line_addition(1,CD,DB)
Problem: As shown in the diagram, ∠ABC=90°, ∠BCD=Mul(MeasureOfAngle(ABC),1/2)°, ∠DAB=90°. Find the measure of ∠CDA.
问题：如图所示，∠ABC=90°，∠BCD=Mul(MeasureOfAngle(ABC),1/2)°，∠DAB=90°。求∠CDA的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),90)
Equal(MeasureOfAngle(BCD),Mul(MeasureOfAngle(ABC),1/2))
Equal(MeasureOfAngle(DAB),90)
Goal: Value(MeasureOfAngle(CDA))
Answer: 135
Theorem sequence:
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, ∠CAD=20°, AD is the angle bisector of ∠CAB, AE bisects ∠FAC. Find the measure of ∠EAC.
问题：如图所示，∠CAD=20°，AD平分∠CAB，AE平分∠FAC。求∠EAC的大小。
Construction (CDL):
Shape(FA,AE)
Shape(EA,AC)
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Shape(AC,CG)
Collinear(FAB)
Collinear(BDCG)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),20)
IsBisectorOfAngle(AD,CAB)
IsBisectorOfAngle(AE,FAC)
Goal: Value(MeasureOfAngle(EAC))
Answer: 70
Theorem sequence:
angle_addition(1,CAD,DAB)
adjacent_complementary_angle(1,FAC,CAB)
angle_addition(1,FAE,EAC)
Problem: As shown in the diagram, AB=11, CD=3, D bisects segment CB. Find the length of line AC.
问题：如图所示，AB=11，CD=3，D是线段CB的中点。求直线AC的长度。
Construction (CDL):
Collinear(ACDB)
Conditions (CDL):
Equal(LengthOfLine(AB),11)
Equal(LengthOfLine(CD),3)
IsMidpointOfLine(D,CB)
Goal: Value(LengthOfLine(AC))
Answer: 5
Theorem sequence:
line_addition(1,CD,DB)
line_addition(1,AC,CB)
Problem: As shown in the diagram, CB=4, DB=7, D is the midpoint of segment AC. Find the length of line AB.
问题：如图所示，CB=4，DB=7，D平分线段AC。求直线AB的长度。
Construction (CDL):
Collinear(ADCB)
Conditions (CDL):
Equal(LengthOfLine(CB),4)
Equal(LengthOfLine(DB),7)
IsMidpointOfLine(D,AC)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
line_addition(1,DC,CB)
line_addition(1,AD,DC)
line_addition(1,AC,CB)
Problem: As shown in the diagram, ∠CDA=111°, ∠DAB=93°, DE∥AB. Find the measure of ∠CDE.
问题：如图所示，∠CDA=111°，∠DAB=93°，DE∥AB。求∠CDE的大小。
Construction (CDL):
Shape(DA,AB,BE,ED)
Shape(DE,EC,CD)
Collinear(FDA)
Collinear(CEB)
Conditions (CDL):
Equal(MeasureOfAngle(CDA),111)
Equal(MeasureOfAngle(DAB),93)
ParallelBetweenLine(DE,AB)
Goal: Value(MeasureOfAngle(CDE))
Answer: 24
Theorem sequence:
parallel_property_corresponding_angle(1,DE,AB,F)
adjacent_complementary_angle(1,FDC,CDA)
angle_addition(1,FDC,CDE)
Problem: As shown in the diagram, ∠ABC=55°, ∠DCE=63°, BA is parallel to DE. Find the measure of ∠CED.
问题：如图所示，∠ABC=55°，∠DCE=63°，BA∥DE。求∠CED的大小。
Construction (CDL):
Shape(AB,BD,DE,EA)
Shape(ED,DC,CE)
Collinear(AEC)
Collinear(BDC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),55)
Equal(MeasureOfAngle(DCE),63)
ParallelBetweenLine(BA,DE)
Goal: Value(MeasureOfAngle(CED))
Answer: 62
Theorem sequence:
parallel_property_corresponding_angle(2,BA,DE,C)
triangle_property_angle_sum(1,EDC)
Problem: As shown in the diagram, AN=2, AN=Mul(LengthOfLine(NM),1/2), M is the midpoint of segment AB. Find the length of line AB.
问题：如图所示，AN=2，AN=Mul(LengthOfLine(NM),1/2)，M是线段AB的中点。求直线AB的长度。
Construction (CDL):
Collinear(ANMB)
Conditions (CDL):
Equal(LengthOfLine(AN),2)
Equal(LengthOfLine(AN),Mul(LengthOfLine(NM),1/2))
IsMidpointOfLine(M,AB)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
line_addition(1,AN,NM)
line_addition(1,AM,MB)
Problem: As shown in the diagram, ∠COB=130°, BO is the angle bisector of ∠ABC, CO is the angle bisector of ∠BCA. Find the measure of ∠CAB.
问题：如图所示，∠COB=130°，BO平分∠ABC，CO平分∠BCA。求∠CAB的大小。
Construction (CDL):
Shape(AB,BO,OC,CA)
Shape(OB,BC,CO)
Conditions (CDL):
Equal(MeasureOfAngle(COB),130)
IsBisectorOfAngle(BO,ABC)
IsBisectorOfAngle(CO,BCA)
Goal: Value(MeasureOfAngle(CAB))
Answer: 80
Theorem sequence:
angle_addition(1,ABO,OBC)
angle_addition(1,BCO,OCA)
triangle_property_angle_sum(1,OBC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠CAD=30°, ∠EDA=60°. Find the measure of ∠DCA.
问题：如图所示，∠CAD=30°，∠EDA=60°。求∠DCA的大小。
Construction (CDL):
Shape(AD,DC,CA)
Shape(ED,DA)
Shape(DB,BC,CD)
Collinear(ADB)
Collinear(EDC)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),30)
Equal(MeasureOfAngle(EDA),60)
Goal: Value(MeasureOfAngle(DCA))
Answer: 30
Theorem sequence:
adjacent_complementary_angle(1,EDA,ADC)
triangle_property_angle_sum(1,ADC)
Problem: As shown in the diagram, CD=8, OP=3, O is the center of ⊙O, the diameter of ⊙O is AB, OP⊥CP. Find the radius of ⊙O.
问题：如图所示，CD=8，OP=3，⊙O的圆心为O，AB是⊙O的直径，OP⊥CP。求⊙O的半径。
Construction (CDL):
Shape(AO,OC,OCA)
Shape(OP,PC,CO)
Shape(CP,PB,OBC)
Shape(DP,PO,OA,OAD)
Shape(BP,PD,ODB)
Collinear(AOPB)
Collinear(CPD)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(LengthOfLine(CD),8)
Equal(LengthOfLine(OP),3)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
PerpendicularBetweenLine(OP,CP)
Goal: Value(RadiusOfCircle(O))
Answer: 5
Theorem sequence:
adjacent_complementary_angle(1,DPO,OPC)
circle_property_chord_perpendicular_bisect_chord(1,O,OP,DC)
line_addition(1,DP,PC)
right_triangle_judgment_angle(1,OPC)
right_triangle_property_pythagorean(1,OPC)
radius_of_circle_property_length_equal(1,OC,O)
Problem: As shown in the diagram, AB=6, OA=5, ∠ADO=90°, O is the center of circle O. Find the length of line CD.
问题：如图所示，AB=6，OA=5，∠ADO=90°，圆O的圆心为O。求直线CD的长度。
Construction (CDL):
Shape(AO,OD,DB,OBA)
Shape(OA,AD,DO)
Shape(CD,DA,OAC)
Shape(BD,DC,OCB)
Collinear(ADB)
Collinear(ODC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(OA),5)
Equal(MeasureOfAngle(ADO),90)
IsCentreOfCircle(O,O)
Goal: Value(LengthOfLine(CD))
Answer: 1
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OD,AB)
line_addition(1,AD,DB)
right_triangle_judgment_angle(1,ADO)
right_triangle_property_pythagorean(1,ADO)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
line_addition(1,OD,DC)
Problem: As shown in the diagram, ∠BOC=40°, the center of circle O is O, AB is the diameter of ⊙O, E bisects segment CD. Find the measure of ∠ABD.
问题：如图所示，∠BOC=40°，O是⊙O的圆心，AB是⊙O的直径，E是线段CD的中点。求∠ABD的大小。
Construction (CDL):
Shape(CO,OA,OAC)
Shape(OC,CE,EO)
Shape(BE,EC,OCB)
Shape(AO,OE,ED,ODA)
Shape(EB,BD,DE)
Shape(DB,OBD)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(MeasureOfAngle(BOC),40)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsMidpointOfLine(E,CD)
Goal: Value(MeasureOfAngle(ABD))
Answer: 70
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(2,O,OE,CD)
adjacent_complementary_angle(1,OED,DEB)
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,D)
triangle_property_angle_sum(1,EBD)
Problem: As shown in the diagram, OP=6, ∠OPA=30°, ∠PCO=90°, the radius of circle O is 4, O is the center of ⊙O. Find the length of line AB.
问题：如图所示，OP=6，∠OPA=30°，∠PCO=90°，圆O的半径为4，O是圆O的圆心。求直线AB的长度。
Construction (CDL):
Shape(AO,OC,CA)
Shape(BC,CO,OD,ODB)
Shape(AC,CB,OBA)
Shape(DO,OA,OAD)
Shape(DP,PB,ODB)
Collinear(ACBP)
Collinear(ODP)
Cocircular(O,ADB)
Conditions (CDL):
Equal(LengthOfLine(OP),6)
Equal(MeasureOfAngle(OPA),30)
Equal(MeasureOfAngle(PCO),90)
Equal(RadiusOfCircle(O),4)
IsCentreOfCircle(O,O)
Goal: Value(LengthOfLine(AB))
Answer: 2*sqrt(7)
Theorem sequence:
sine_theorem(1,OPC)
adjacent_complementary_angle(1,PCO,OCA)
right_triangle_judgment_angle(1,OCA)
radius_of_circle_property_length_equal(1,OA,O)
right_triangle_property_pythagorean(1,OCA)
circle_property_chord_perpendicular_bisect_chord(1,O,OC,BA)
line_addition(1,AC,CB)
Problem: As shown in the diagram, ∠BOA=90°, the radius of circle O is 2, the center of circle O is O, AE⊥OE, CD is perpendicular to OD. Find the length of line DE.
问题：如图所示，∠BOA=90°，⊙O的半径为2，O是圆O的圆心，AE垂直于OE，CD⊥OD。求直线DE的长度。
Construction (CDL):
Shape(BO,OF,FB)
Shape(BF,FD,DB)
Shape(DF,FG,GE,ED)
Shape(FO,OG,GF)
Shape(EG,GA,AE)
Shape(GO,OA,AG)
Shape(DE,EC,CD)
Shape(BD,DC,OCB)
Shape(CE,EA,OAC)
Collinear(BDC)
Collinear(CEA)
Collinear(BFGA)
Collinear(OFD)
Collinear(OGE)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BOA),90)
Equal(RadiusOfCircle(O),2)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AE,OE)
PerpendicularBetweenLine(CD,OD)
Goal: Value(LengthOfLine(DE))
Answer: sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,BOA)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
right_triangle_property_pythagorean(1,BOA)
circle_property_chord_perpendicular_bisect_chord(1,O,OD,CB)
circle_property_chord_perpendicular_bisect_chord(1,O,OE,AC)
line_addition(1,BD,DC)
line_addition(1,CE,EA)
similar_triangle_judgment_sas(1,CDE,CBA)
similar_triangle_property_line_ratio(1,ECD,ACB)
similar_triangle_property_line_ratio(1,CDE,CBA)
Problem: As shown in the diagram, AB=12, ∠BOC=60°, O is the center of circle O, the diameter of circle O is AB, CE is perpendicular to AE. Find the length of line BE.
问题：如图所示，AB=12，∠BOC=60°，O是圆O的圆心，⊙O的直径为AB，CE垂直于AE。求直线BE的长度。
Construction (CDL):
Shape(CO,OA,OAC)
Shape(OC,CE,EO)
Shape(BE,EC,OCB)
Shape(DE,EB,OBD)
Shape(AO,OE,ED,ODA)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,CBDA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(MeasureOfAngle(BOC),60)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
PerpendicularBetweenLine(CE,AE)
Goal: Value(LengthOfLine(BE))
Answer: 3
Theorem sequence:
diameter_of_circle_property_length_equal(1,AB,O)
circle_property_length_of_radius_and_diameter(1,O)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OB,O)
triangle_property_angle_sum(1,OCE)
sine_theorem(1,OCE)
line_addition(1,OE,EB)
Problem: As shown in the diagram, MN=3, O is the center of circle O, AM⊥OM, CN⊥ON. Find the length of line BC.
问题：如图所示，MN=3，圆O的圆心为O，AM⊥OM，CN垂直于ON。求直线BC的长度。
Construction (CDL):
Shape(CN,NO,OM,MB,BC)
Shape(NM,MO,ON)
Shape(NA,AM,MN)
Shape(AN,NC,OCA)
Shape(BM,MA,OAB)
Shape(CB,OBC)
Collinear(ANC)
Collinear(AMB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(MN),3)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AM,OM)
PerpendicularBetweenLine(CN,ON)
Goal: Value(LengthOfLine(BC))
Answer: 6
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OM,AB)
circle_property_chord_perpendicular_bisect_chord(1,O,ON,CA)
line_addition(1,AN,NC)
line_addition(1,AM,MB)
similar_triangle_judgment_sas(1,ABC,AMN)
similar_triangle_property_line_ratio(1,BCA,MNA)
similar_triangle_property_line_ratio(1,ABC,AMN)
Problem: As shown in the diagram, AB=10, CD=8, O is the center of ⊙O, the diameter of ⊙O is AB, CE⊥OE. Find the length of line OE.
问题：如图所示，AB=10，CD=8，O是⊙O的圆心，AB是圆O的直径，CE垂直于OE。求直线OE的长度。
Construction (CDL):
Shape(CO,OB,OCB)
Shape(CE,EO,OC)
Shape(AE,EC,OCA)
Shape(BO,OE,ED,ODB)
Shape(DE,EA,OAD)
Collinear(CED)
Collinear(BOEA)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(CD),8)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
PerpendicularBetweenLine(CE,OE)
Goal: Value(LengthOfLine(OE))
Answer: 3
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OE,CD)
line_addition(1,CE,ED)
radius_of_circle_property_length_equal(1,OC,O)
diameter_of_circle_property_length_equal(1,AB,O)
circle_property_length_of_radius_and_diameter(1,O)
right_triangle_judgment_angle(1,CEO)
right_triangle_property_pythagorean(1,CEO)
Problem: As shown in the diagram, EF=7/2, the center of circle O is O, AE is perpendicular to OE, CF⊥OF. Find the length of line BC.
问题：如图所示，EF=7/2，O是圆O的圆心，AE⊥OE，CF⊥OF。求直线BC的长度。
Construction (CDL):
Shape(BC,CF,FO,OE,EB)
Shape(OF,FE,EO)
Shape(EF,FA,AE)
Shape(AF,FC,OCA)
Shape(BE,EA,OAB)
Shape(CB,OBC)
Collinear(BEA)
Collinear(CFA)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(EF),7/2)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AE,OE)
PerpendicularBetweenLine(CF,OF)
Goal: Value(LengthOfLine(BC))
Answer: 7
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OF,CA)
circle_property_chord_perpendicular_bisect_chord(1,O,OE,AB)
line_addition(1,BE,EA)
line_addition(1,CF,FA)
similar_triangle_judgment_sas(1,ABC,AEF)
similar_triangle_property_line_ratio(1,BCA,EFA)
similar_triangle_property_line_ratio(1,ABC,AEF)
Problem: As shown in the diagram, AB=8, OE=3, O is the center of ⊙O, AE⊥OE. Find the radius of circle O.
问题：如图所示，AB=8，OE=3，⊙O的圆心为O，AE⊥OE。求⊙O的半径。
Construction (CDL):
Shape(OE,EB,BO)
Shape(AE,EO,OB,OBA)
Shape(BE,EA,OAB)
Collinear(AEB)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(OE),3)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AE,OE)
Goal: Value(RadiusOfCircle(O))
Answer: 5
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OE,AB)
line_addition(1,AE,EB)
right_triangle_judgment_angle(1,OEB)
right_triangle_property_pythagorean(1,OEB)
radius_of_circle_property_length_equal(1,OB,O)
Problem: As shown in the diagram, the radius of circle O is 2, O is the center of circle O, OAPB is a ▱. Find the length of line AB.
问题：如图所示，⊙O的半径为2，O是圆O的圆心，OAPB是▱。求直线AB的长度。
Construction (CDL):
Shape(OA,AC,CO)
Shape(OC,CB,BO)
Shape(AP,PC,CA)
Shape(CP,PB,BC)
Shape(AO,OB,OBA)
Shape(PA,OAP)
Shape(BP,OPB)
Collinear(ACB)
Collinear(OCP)
Cocircular(O,APB)
Conditions (CDL):
Equal(RadiusOfCircle(O),2)
IsCentreOfCircle(O,O)
Parallelogram(OAPB)
Goal: Value(LengthOfLine(AB))
Answer: 2*sqrt(3)
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,OP,O)
parallelogram_property_opposite_line_equal(1,OAPB)
parallelogram_property_opposite_line_equal(1,APBO)
kite_judgment_equal_and_equal(1,OAPB)
parallelogram_property_diagonal_bisection(1,OAPB,C)
line_addition(1,OC,CP)
kite_property_diagonal_perpendicular_bisection(1,OAPB,C)
right_triangle_judgment_angle(1,ACO)
right_triangle_property_pythagorean(1,ACO)
line_addition(1,AC,CB)
Problem: As shown in the diagram, CD=2, OB=10, O is the center of circle O, BD is perpendicular to OD. Find the length of line AB.
问题：如图所示，CD=2，OB=10，O是⊙O的圆心，BD⊥OD。求直线AB的长度。
Construction (CDL):
Shape(AD,DC,OCA)
Shape(CD,DB,OBC)
Shape(DO,OB,BD)
Shape(BO,OD,DA,OAB)
Collinear(ADB)
Collinear(CDO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(CD),2)
Equal(LengthOfLine(OB),10)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BD,OD)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OD,BA)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
line_addition(1,CD,DO)
right_triangle_judgment_angle(1,BDO)
right_triangle_property_pythagorean(1,BDO)
line_addition(1,AD,DB)
Problem: As shown in the diagram, CD=8, O is the center of circle O, AB is the diameter of ⊙O, DE⊥OE. Find the length of line DE.
问题：如图所示，CD=8，O是⊙O的圆心，圆O的直径为AB，DE⊥OE。求直线DE的长度。
Construction (CDL):
Shape(CE,EA,OAC)
Shape(AE,ED,ODA)
Shape(BO,OE,EC,OCB)
Shape(DE,EO,OB,OBD)
Collinear(CED)
Collinear(AEOB)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(LengthOfLine(CD),8)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
PerpendicularBetweenLine(DE,OE)
Goal: Value(LengthOfLine(DE))
Answer: 4
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OE,DC)
line_addition(1,CE,ED)
Problem: As shown in the diagram, the radius of ⊙O is 6, O is the center of ⊙O, D is the midpoint of segment OC, BD⊥OD. Find the length of line AB.
问题：如图所示，圆O的半径为6，O是⊙O的圆心，D平分线段OC，BD垂直于OD。求直线AB的长度。
Construction (CDL):
Shape(AO,OD,DA)
Shape(OB,BD,DO)
Shape(CD,DB,OBC)
Shape(AD,DC,OCA)
Shape(BO,OA,OAB)
Collinear(ODC)
Collinear(ADB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(RadiusOfCircle(O),6)
IsCentreOfCircle(O,O)
IsMidpointOfLine(D,OC)
PerpendicularBetweenLine(BD,OD)
Goal: Value(LengthOfLine(AB))
Answer: 6*sqrt(3)
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OD,BA)
radius_of_circle_property_length_equal(1,OC,O)
line_addition(1,OD,DC)
radius_of_circle_property_length_equal(1,OB,O)
right_triangle_judgment_angle(1,BDO)
right_triangle_property_pythagorean(1,BDO)
line_addition(1,BD,DA)
Problem: As shown in the diagram, AB=16, OC=6, the center of ⊙O is O, OC is perpendicular to AC. Find the radius of ⊙O.
问题：如图所示，AB=16，OC=6，O是圆O的圆心，OC垂直于AC。求⊙O的半径。
Construction (CDL):
Shape(OC,CA,AO)
Shape(AC,CB,OBA)
Shape(BC,CO,OA,OAB)
Collinear(ACB)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
Equal(LengthOfLine(OC),6)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(OC,AC)
Goal: Value(RadiusOfCircle(O))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,BCO,OCA)
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, DE=3, the center of circle O is O, AD⊥OD, CE is perpendicular to OE. Find the length of line BC.
问题：如图所示，DE=3，O是⊙O的圆心，AD垂直于OD，CE⊥OE。求直线BC的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DO,OE,ED)
Shape(DB,BC,CE,EO,OD)
Shape(CB,OBC)
Shape(AE,EC,OCA)
Shape(BD,DA,OAB)
Collinear(ADB)
Collinear(AEC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(DE),3)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AD,OD)
PerpendicularBetweenLine(CE,OE)
Goal: Value(LengthOfLine(BC))
Answer: 6
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OD,AB)
circle_property_chord_perpendicular_bisect_chord(1,O,OE,CA)
line_addition(1,AD,DB)
line_addition(1,AE,EC)
similar_triangle_judgment_sas(1,ADE,ABC)
similar_triangle_property_line_ratio(1,DEA,BCA)
similar_triangle_property_line_ratio(1,ADE,ABC)
Problem: As shown in the diagram, AB=8, OB=5, the center of circle O is O, AD⊥OD. Find the length of line OD.
问题：如图所示，AB=8，OB=5，O是圆O的圆心，AD垂直于OD。求直线OD的长度。
Construction (CDL):
Shape(OD,DB,BO)
Shape(AD,DO,OB,OBA)
Shape(CD,DA,OAC)
Shape(BD,DC,OCB)
Collinear(ADB)
Collinear(ODC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(OB),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)
right_triangle_judgment_angle(1,ODB)
right_triangle_property_pythagorean(1,ODB)