SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, CE=6, ∠OEC=30°, the center of circle O is O, the diameter of ⊙O is AB, the tangent to ⊙O is EC, OD is perpendicular to AD. Find the length of line OD.
问题：如图所示，CE=6，∠OEC=30°，O是圆O的圆心，⊙O的直径为AB，EC是圆O的切线，OD⊥AD。求直线OD的长度。
Construction (CDL):
Shape(OD,DA,AO)
Shape(CD,DO,OC)
Shape(CO,OB,OBC)
Shape(OBC,BE,EC)
Shape(OCA,AC)
Collinear(AOBE)
Collinear(CDA)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(CE),6)
Equal(MeasureOfAngle(OEC),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(EC,O)
PerpendicularBetweenLine(OD,AD)
Goal: Value(LengthOfLine(OD))
Answer: sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,COE)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
tangent_of_circle_property_perpendicular(2,EC,O,O)
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)
flat_angle(1,CDA)
angle_addition(1,CDO,ODA)
sine_theorem(1,OCD)
sine_theorem(1,COE)
Problem: As shown in the diagram, OA=1, ∠OBA=30°, O is the center of circle O, the tangent to circle O is BA. Find the length of line OB.
问题：如图所示，OA=1，∠OBA=30°，O是圆O的圆心，⊙O的切线为BA。求直线OB的长度。
Construction (CDL):
Shape(OC,OCA,AO)
Shape(CB,BA,OCA)
Collinear(OCB)
Cocircular(O,CA)
Conditions (CDL):
Equal(LengthOfLine(OA),1)
Equal(MeasureOfAngle(OBA),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BA,O)
Goal: Value(LengthOfLine(OB))
Answer: 2
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BA,O,O)
sine_theorem(1,OBA)
Problem: As shown in the diagram, OA=2, ∠OBA=45°, O is the center of ⊙O, BA is the tangent to circle O. Find the length of line OB minus the length of line OA.
问题：如图所示，OA=2，∠OBA=45°，圆O的圆心为O，BA是⊙O的切线。求直线OB的长度减去直线OA的长度。
Construction (CDL):
Shape(OC,OCA,AO)
Shape(CB,BA,OCA)
Collinear(OCB)
Cocircular(O,CA)
Conditions (CDL):
Equal(LengthOfLine(OA),2)
Equal(MeasureOfAngle(OBA),45)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BA,O)
Goal: Value(Sub(LengthOfLine(OB),LengthOfLine(OA)))
Answer: -2+2*sqrt(2)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BA,O,O)
sine_theorem(1,OBA)
Problem: As shown in the diagram, ∠OPA=50°, O is the center of ⊙O, the diameter of circle O is AB, PA is the tangent to circle O. Find the measure of ∠ABC.
问题：如图所示，∠OPA=50°，圆O的圆心为O，⊙O的直径为AB，PA是圆O的切线。求∠ABC的大小。
Construction (CDL):
Shape(PA,OCA,CP)
Shape(OC,OCA,AO)
Shape(CO,OB,BC)
Shape(OBC,CB)
Collinear(AOB)
Collinear(PCO)
Cocircular(O,CAB)
Conditions (CDL):
Equal(MeasureOfAngle(OPA),50)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(ABC))
Answer: 20
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
triangle_property_angle_sum(1,PAO)
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_external(1,OCA,B)
Problem: As shown in the diagram, ∠ACB=42°, O is the center of circle O, the diameter of circle O is AB, CB is the tangent to circle O. Find the measure of ∠DBA.
问题：如图所示，∠ACB=42°，O是圆O的圆心，AB是圆O的直径，⊙O的切线为CB。求∠DBA的大小。
Construction (CDL):
Shape(AD,DB,BO,OA)
Shape(OAD,DA)
Shape(ODB,BD)
Shape(DC,CB,ODB)
Collinear(AOB)
Collinear(ADC)
Cocircular(O,ADB)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),42)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(DBA))
Answer: 42
Theorem sequence:
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,D)
tangent_of_circle_property_perpendicular(2,CB,O,O)
triangle_property_angle_sum(1,ADB)
triangle_property_angle_sum(1,BAC)
flat_angle(1,AOB)
Problem: As shown in the diagram, ∠BCD=25°, the center of ⊙O is O, DC is the tangent to circle O. Find the measure of ∠CBO.
问题：如图所示，∠BCD=25°，O是圆O的圆心，DC是圆O的切线。求∠CBO的大小。
Construction (CDL):
Shape(OC,CB,BO)
Shape(OCB,BC)
Shape(BC,CD)
Cocircular(O,CB)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(CBO))
Answer: 65
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DC,O,O)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OCB)
isosceles_triangle_property_angle_equal(1,OCB)
angle_addition(1,OCB,BCD)
Problem: As shown in the diagram, ∠ACD=50°, ∠OAC=90°, O is the center of circle O, AB is the diameter of ⊙O. Find the measure of ∠DOA.
问题：如图所示，∠ACD=50°，∠OAC=90°，O是圆O的圆心，AB是⊙O的直径。求∠DOA的大小。
Construction (CDL):
Shape(BO,OD,DB)
Shape(DO,OA,OAD)
Shape(ODB,BD)
Shape(OAD,AC,CD)
Collinear(BOA)
Collinear(BDC)
Cocircular(O,BAD)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),50)
Equal(MeasureOfAngle(OAC),90)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
Goal: Value(MeasureOfAngle(DOA))
Answer: 80
Theorem sequence:
arc_property_center_angle(1,OAD,O)
arc_property_circumference_angle_external(1,OAD,B)
triangle_property_angle_sum(1,BAC)
Problem: As shown in the diagram, AB=20, BC=20, CF=DF, ED=EA, the center of circle O is O. Find the perimeter of triangle BEF.
问题：如图所示，AB=20，BC=20，CF=DF，ED=EA，⊙O的圆心为O。求△BEF的周长。
Construction (CDL):
Shape(BE,ED,DF,FB)
Shape(DE,EA,ODA)
Shape(CF,FD,OCD)
Shape(ODA,OAC,OCD)
Collinear(EDF)
Collinear(AEB)
Collinear(BFC)
Cocircular(O,DAC)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(BC),20)
Equal(LengthOfLine(CF),LengthOfLine(DF))
Equal(LengthOfLine(ED),LengthOfLine(EA))
IsCentreOfCircle(O,O)
Goal: Value(PerimeterOfTriangle(BEF))
Answer: 40
Theorem sequence:
line_addition(1,ED,DF)
line_addition(1,AE,EB)
line_addition(1,BF,FC)
triangle_perimeter_formula(1,BEF)
Problem: As shown in the diagram, PO=4, ∠APO=30°, ∠OAP=90°, ∠OPB=30°, ∠PBO=90°. Find the length of line BO.
问题：如图所示，PO=4，∠APO=30°，∠OAP=90°，∠OPB=30°，∠PBO=90°。求直线BO的长度。
Construction (CDL):
Shape(PA,AO,OA)
Shape(OP,PB,BO)
Conditions (CDL):
Equal(LengthOfLine(PO),4)
Equal(MeasureOfAngle(APO),30)
Equal(MeasureOfAngle(OAP),90)
Equal(MeasureOfAngle(OPB),30)
Equal(MeasureOfAngle(PBO),90)
Goal: Value(LengthOfLine(BO))
Answer: 2
Theorem sequence:
sine_theorem(1,OPB)
Problem: As shown in the diagram, ∠APO=36°, O is the center of circle O, the diameter of ⊙O is AB, the tangent to circle O is PA. Find the measure of ∠CBA.
问题：如图所示，∠APO=36°，圆O的圆心为O，AB是⊙O的直径，圆O的切线为PA。求∠CBA的大小。
Construction (CDL):
Shape(AP,PC,OAC)
Shape(OA,OAC,CO)
Shape(OC,CB,BO)
Shape(OCB,BC)
Collinear(AOB)
Collinear(PCO)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(APO),36)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(CBA))
Answer: 27
Theorem sequence:
arc_property_center_angle(1,OAC,O)
arc_property_circumference_angle_external(1,OAC,B)
tangent_of_circle_property_perpendicular(1,PA,O,O)
triangle_property_angle_sum(1,APO)
Problem: As shown in the diagram, CD=8, CO=5, O is the center of circle O, DE⊥OE, OE is perpendicular to CE. Find the length of line AC.
问题：如图所示，CD=8，CO=5，⊙O的圆心为O，DE垂直于OE，OE垂直于CE。求直线AC的长度。
Construction (CDL):
Shape(EC,CO,OE)
Shape(OC,CA,AO)
Shape(ODC,CE,ED)
Shape(OCA,AC)
Shape(DE,EO,OA,OAD)
Collinear(CED)
Collinear(AOE)
Cocircular(O,CAD)
Conditions (CDL):
Equal(LengthOfLine(CD),8)
Equal(LengthOfLine(CO),5)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(DE,OE)
PerpendicularBetweenLine(OE,CE)
Goal: Value(LengthOfLine(AC))
Answer: 4*sqrt(5)
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OE,DC)
line_addition(1,CE,ED)
line_addition(1,AO,OE)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OA,O)
right_triangle_judgment_angle(1,OEC)
right_triangle_judgment_angle(1,AEC)
right_triangle_property_pythagorean(1,OEC)
right_triangle_property_pythagorean(1,AEC)
Problem: As shown in the diagram, PC=3, ∠CAP=30°, △DOC is an equilateral △, the center of ⊙O is O, the diameter of ⊙O is AB, the tangent to circle O is PC, OE⊥AE. Find the length of line CD.
问题：如图所示，PC=3，∠CAP=30°，△DOC为等边△，O是圆O的圆心，⊙O的直径为AB，⊙O的切线为PC，OE⊥AE。求直线CD的长度。
Construction (CDL):
Shape(EA,AO,OE)
Shape(DE,EC,CD)
Shape(OCD,DC)
Shape(EO,OC,CE)
Shape(ED,ODA,AE)
Shape(CO,OB,OBC)
Shape(OBC,BP,PC)
Collinear(AOBP)
Collinear(DEO)
Collinear(AEC)
Cocircular(O,DABC)
Conditions (CDL):
Equal(LengthOfLine(PC),3)
Equal(MeasureOfAngle(CAP),30)
EquilateralTriangle(DOC)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PC,O)
PerpendicularBetweenLine(OE,AE)
Goal: Value(LengthOfLine(CD))
Answer: sqrt(3)
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)
angle_addition(1,PCO,OCE)
tangent_of_circle_property_perpendicular(2,PC,O,O)
triangle_property_angle_sum(1,AOC)
triangle_property_angle_sum(1,APC)
isosceles_triangle_judgment_angle_equal(1,CAP)
sine_theorem(1,CAO)
Problem: As shown in the diagram, AC=12, AO=13, ∠ACO=90°, O is the center of circle O. Find the length of line OC.
问题：如图所示，AC=12，AO=13，∠ACO=90°，⊙O的圆心为O。求直线OC的长度。
Construction (CDL):
Shape(OD,ODC,CO)
Shape(DA,AC,ODC)
Collinear(ADO)
Cocircular(O,DC)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(AO),13)
Equal(MeasureOfAngle(ACO),90)
IsCentreOfCircle(O,O)
Goal: Value(LengthOfLine(OC))
Answer: 5
Theorem sequence:
right_triangle_judgment_angle(1,ACO)
right_triangle_property_pythagorean(1,ACO)
Problem: As shown in the diagram, ∠BPO=28°, the center of ⊙O is O, PB is the tangent to circle O, BE is perpendicular to OE. Find the measure of ∠OBE.
问题：如图所示，∠BPO=28°，圆O的圆心为O，PB是⊙O的切线，BE⊥OE。求∠OBE的大小。
Construction (CDL):
Shape(EO,OB,BE)
Shape(DE,EB,BD)
Shape(OBD,DB)
Shape(OBD,BP,PD)
Collinear(OEDP)
Cocircular(O,BD)
Conditions (CDL):
Equal(MeasureOfAngle(BPO),28)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PB,O)
PerpendicularBetweenLine(BE,OE)
Goal: Value(MeasureOfAngle(OBE))
Answer: 28
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PB,O,O)
triangle_property_angle_sum(1,EOB)
triangle_property_angle_sum(1,OBP)
Problem: As shown in the diagram, AB=5, AC=3, the tangent to ⊙O is AC, the tangent to circle O is AP, the tangent to circle O is BD, the tangent to ⊙O is BP. Find the length of line BD.
问题：如图所示，AB=5，AC=3，⊙O的切线为AC，AP是⊙O的切线，⊙O的切线为BD，⊙O的切线为BP。求直线BD的长度。
Construction (CDL):
Shape(CA,AP,OCP)
Shape(PB,BD,OPD)
Shape(OCP,OPD,ODC)
Collinear(APB)
Cocircular(O,CPD)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),3)
IsTangentOfCircle(AC,O)
IsTangentOfCircle(AP,O)
IsTangentOfCircle(BD,O)
IsTangentOfCircle(BP,O)
Goal: Value(LengthOfLine(BD))
Answer: 2
Theorem sequence:
tangent_of_circle_property_length_equal(1,AP,AC,O)
tangent_of_circle_property_length_equal(1,BP,BD,O)
line_addition(1,AP,PB)
Problem: As shown in the diagram, AB=4, BF=5, CD=9, CF=AD, ∠CFB=90°, AD is the diameter of ⊙O, the tangent to circle O is BA, the tangent to circle O is BE, CD is the tangent to circle O, CE is the tangent to ⊙O. Find the radius of ⊙O.
问题：如图所示，AB=4，BF=5，CD=9，CF=AD，∠CFB=90°，AD是⊙O的直径，BA是⊙O的切线，圆O的切线为BE，CD是圆O的切线，圆O的切线为CE。求⊙O的半径。
Construction (CDL):
Shape(EB,BA,OEA)
Shape(ODE,DC,CE)
Shape(BE,EC,CF,FB)
Shape(OD,ODE,OEA,AO)
Collinear(AOD)
Collinear(BEC)
Collinear(ABF)
Cocircular(O,DEA)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(BF),5)
Equal(LengthOfLine(CD),9)
Equal(LengthOfLine(CF),LengthOfLine(AD))
Equal(MeasureOfAngle(CFB),90)
IsDiameterOfCircle(AD,O)
IsTangentOfCircle(BA,O)
IsTangentOfCircle(BE,O)
IsTangentOfCircle(CD,O)
IsTangentOfCircle(CE,O)
Goal: Value(RadiusOfCircle(O))
Answer: 6
Theorem sequence:
tangent_of_circle_property_length_equal(1,BA,BE,O)
tangent_of_circle_property_length_equal(1,CD,CE,O)
line_addition(1,BE,EC)
right_triangle_judgment_angle(1,CFB)
right_triangle_property_pythagorean(1,CFB)
circle_property_length_of_radius_and_diameter(1,O)
diameter_of_circle_property_length_equal(1,AD,O)
Problem: As shown in the diagram, OB=2, OD=2, ∠BAO=30°, the center of circle O is O, the tangent to circle O is AB. Find the length of line AD.
问题：如图所示，OB=2，OD=2，∠BAO=30°，O是⊙O的圆心，AB是⊙O的切线。求直线AD的长度。
Construction (CDL):
Shape(BA,AD,OBD)
Shape(OB,OBD,DO)
Collinear(ADO)
Cocircular(O,BD)
Conditions (CDL):
Equal(LengthOfLine(OB),2)
Equal(LengthOfLine(OD),2)
Equal(MeasureOfAngle(BAO),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(LengthOfLine(AD))
Answer: 2
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AB,O,O)
sine_theorem(1,OBA)
line_addition(1,AD,DO)
Problem: As shown in the diagram, ∠COB=120°, O is the center of ⊙O, AB is the diameter of ⊙O, PC is the tangent to circle O. Find the measure of ∠ACP.
问题：如图所示，∠COB=120°，圆O的圆心为O，AB是圆O的直径，PC是⊙O的切线。求∠ACP的大小。
Construction (CDL):
Shape(CP,PA,OCA)
Shape(OCA,AC)
Shape(CA,AO,OC)
Shape(OB,OBC,CO)
Collinear(PAOB)
Cocircular(O,CAB)
Conditions (CDL):
Equal(MeasureOfAngle(COB),120)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PC,O)
Goal: Value(MeasureOfAngle(ACP))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PC,O,O)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OA,O)
isosceles_triangle_judgment_line_equal(1,OCA)
isosceles_triangle_property_angle_equal(1,OCA)
triangle_property_angle_sum(1,CAO)
flat_angle(1,POB)
angle_addition(1,AOC,COB)
angle_addition(1,OCA,ACP)
Problem: As shown in the diagram, AE=1, AH=sqrt(3), A is the center of circle A, HE is the tangent to circle A. Find the length of line EH.
问题：如图所示，AE=1，AH=sqrt(3)，A是圆A的圆心，HE是圆O的切线。求直线EH的长度。
Construction (CDL):
Shape(AE,AED,DA)
Shape(AED,EH,HD)
Collinear(ADH)
Cocircular(A,ED)
Conditions (CDL):
Equal(LengthOfLine(AE),1)
Equal(LengthOfLine(AH),sqrt(3))
IsCentreOfCircle(A,A)
IsTangentOfCircle(HE,A)
Goal: Value(LengthOfLine(EH))
Answer: sqrt(2)
Theorem sequence:
tangent_of_circle_property_perpendicular(1,HE,A,A)
right_triangle_judgment_angle(1,AEH)
right_triangle_property_pythagorean(1,AEH)
Problem: As shown in the diagram, EA=AF, ∠OEF=40°, the center of ⊙O is O, EA is the tangent to circle O, the tangent to ⊙O is FA. Find the measure of ∠DBC.
问题：如图所示，EA=AF，∠OEF=40°，O是圆O的圆心，EA是⊙O的切线，圆O的切线为FA。求∠DBC的大小。
Construction (CDL):
Shape(OC,OCA,AO)
Shape(OA,OAD,DO)
Shape(CE,EA,OCA)
Shape(OAD,AF,FD)
Shape(OBC,CB)
Shape(ODB,BD)
Shape(BC,CO,OD,DB)
Collinear(EAF)
Collinear(ODF)
Collinear(OCE)
Cocircular(O,BCAD)
Conditions (CDL):
Equal(LengthOfLine(EA),LengthOfLine(AF))
Equal(MeasureOfAngle(OEF),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(EA,O)
IsTangentOfCircle(FA,O)
Goal: Value(MeasureOfAngle(DBC))
Answer: 50
Theorem sequence:
tangent_of_circle_property_perpendicular(2,EA,O,O)
arc_property_center_angle(1,OCD,O)
arc_property_circumference_angle_external(1,OCD,B)
perpendicular_bisector_judgment_per_and_mid(1,OA,EF)
perpendicular_bisector_property_distance_equal(1,OA,EF)
isosceles_triangle_judgment_line_equal(1,OEF)
isosceles_triangle_property_angle_equal(1,OEF)
triangle_property_angle_sum(1,EFO)
Problem: As shown in the diagram, AB=7, ∠BPA=60°, O is the center of circle O, the tangent to circle O is PA, the tangent to circle O is PB. Find the length of line PA.
问题：如图所示，AB=7，∠BPA=60°，O是圆O的圆心，⊙O的切线为PA，圆O的切线为PB。求直线PA的长度。
Construction (CDL):
Shape(PA,OBA,BP)
Shape(OBA,AB)
Cocircular(O,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),7)
Equal(MeasureOfAngle(BPA),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(LengthOfLine(PA))
Answer: 7
Theorem sequence:
tangent_of_circle_property_length_equal(1,PA,PB,O)
isosceles_triangle_judgment_line_equal(1,PAB)
isosceles_triangle_property_angle_equal(1,PAB)
triangle_property_angle_sum(1,BPA)
isosceles_triangle_judgment_angle_equal(1,ABP)
Problem: As shown in the diagram, ∠CPA=36°, O is the center of ⊙O, the diameter of ⊙O is AB, the tangent to ⊙O is PA. Find the measure of ∠OBC.
问题：如图所示，∠CPA=36°，O是圆O的圆心，AB是⊙O的直径，PA是⊙O的切线。求∠OBC的大小。
Construction (CDL):
Shape(PA,OCA,CP)
Shape(AO,OC,OCA)
Shape(CO,OB,BC)
Shape(OBC,CB)
Shape(OA,OAB,BO)
Collinear(AOB)
Collinear(OCP)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CPA),36)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(OBC))
Answer: 27
Theorem sequence:
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_external(1,OCA,B)
tangent_of_circle_property_perpendicular(2,PA,O,O)
triangle_property_angle_sum(1,PAO)
Problem: As shown in the diagram, ∠ABM=140°, the center of circle O is O, the tangent to ⊙O is MB. Find the measure of ∠BCA.
问题：如图所示，∠ABM=140°，O是⊙O的圆心，⊙O的切线为MB。求∠BCA的大小。
Construction (CDL):
Shape(OA,AB,BO)
Shape(OB,BC,CA,AO)
Shape(OAB,BA)
Shape(OCA,AC)
Shape(OBC,CB)
Shape(OB,BM)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABM),140)
IsCentreOfCircle(O,O)
IsTangentOfCircle(MB,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(1,MB,O,O)
angle_addition(1,ABO,OBM)
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, DO=4, OC=4, ∠OAD=30°, O is the center of circle O, EC is the diameter of circle O, AD is the tangent to circle O, the tangent to ⊙O is BC, BD is the tangent to circle O. Find the length of line BC.
问题：如图所示，DO=4，OC=4，∠OAD=30°，圆O的圆心为O，EC是⊙O的直径，AD是圆O的切线，圆O的切线为BC，BD是圆O的切线。求直线BC的长度。
Construction (CDL):
Shape(AD,OED,EA)
Shape(OED,DO,OE)
Shape(OD,ODC,CO)
Shape(DB,BC,ODC)
Collinear(AEOC)
Collinear(ADB)
Cocircular(O,EDC)
Conditions (CDL):
Equal(LengthOfLine(DO),4)
Equal(LengthOfLine(OC),4)
Equal(MeasureOfAngle(OAD),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(EC,O)
IsTangentOfCircle(AD,O)
IsTangentOfCircle(BC,O)
IsTangentOfCircle(BD,O)
Goal: Value(LengthOfLine(BC))
Answer: 4*sqrt(3)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AD,O,O)
tangent_of_circle_property_perpendicular(2,BC,O,O)
triangle_property_angle_sum(1,ABC)
line_addition(1,AO,OC)
sine_theorem(1,OAD)
sine_theorem(1,CAB)
Problem: As shown in the diagram, ∠CBA=140°, the center of circle O is O, the tangent to circle O is CB. Find the measure of ∠BAO.
问题：如图所示，∠CBA=140°，O是圆O的圆心，圆O的切线为CB。求∠BAO的大小。
Construction (CDL):
Shape(AO,OB,BA)
Shape(OBA,AB)
Shape(CB,BO)
Cocircular(O,BA)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),140)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(BAO))
Answer: 50
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CB,O,O)
angle_addition(1,CBO,OBA)
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)
Problem: As shown in the diagram, PA=8, O is the center of circle O, CA is the tangent to ⊙O, the tangent to circle O is CQ, DB is the tangent to circle O, DQ is the tangent to circle O, the tangent to ⊙O is PA, the tangent to circle O is PB. Find the perimeter of triangle PCD.
问题：如图所示，PA=8，圆O的圆心为O，CA是⊙O的切线，⊙O的切线为CQ，⊙O的切线为DB，DQ是⊙O的切线，PA是⊙O的切线，圆O的切线为PB。求三角形PCD的周长。
Construction (CDL):
Shape(PC,CQ,QD,DP)
Shape(QC,CA,OQA)
Shape(BD,DQ,OBQ)
Shape(OQA,OAB,OBQ)
Collinear(ACP)
Collinear(BDP)
Collinear(CQD)
Cocircular(O,ABQ)
Conditions (CDL):
Equal(LengthOfLine(PA),8)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
IsTangentOfCircle(CQ,O)
IsTangentOfCircle(DB,O)
IsTangentOfCircle(DQ,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(PerimeterOfTriangle(PCD))
Answer: 16
Theorem sequence:
tangent_of_circle_property_length_equal(1,PA,PB,O)
tangent_of_circle_property_length_equal(1,DQ,DB,O)
tangent_of_circle_property_length_equal(1,CA,CQ,O)
line_addition(1,AC,CP)
line_addition(1,BD,DP)
line_addition(1,CQ,QD)
triangle_perimeter_formula(1,PCD)
Problem: As shown in the diagram, ∠BAC=25°, O is the center of circle O, PA is the tangent to circle O, the tangent to circle O is PB. Find the measure of ∠BPA.
问题：如图所示，∠BAC=25°，圆O的圆心为O，⊙O的切线为PA，PB是圆O的切线。求∠BPA的大小。
Construction (CDL):
Shape(CO,OA,OAC)
Shape(BA,AC,OCB)
Shape(AB,OBA)
Shape(BP,PA,OBA)
Collinear(AOC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(BPA))
Answer: 50
Theorem sequence:
angle_addition(1,PAB,BAC)
circle_property_angle_of_osculation(1,OBA,P)
circle_property_angle_of_osculation(2,OBA,P)
tangent_of_circle_property_perpendicular(2,PA,O,O)
triangle_property_angle_sum(1,ABP)
Problem: As shown in the diagram, ∠BPA=50°, the center of circle O is O, AC is the diameter of circle O, the tangent to circle O is PA, the tangent to circle O is PB. Find the measure of ∠ACB.
问题：如图所示，∠BPA=50°，O是圆O的圆心，AC是圆O的直径，PA是⊙O的切线，PB是⊙O的切线。求∠ACB的大小。
Construction (CDL):
Shape(CO,OA,OAC)
Shape(AO,OC,CB,BA)
Shape(BC,OCB)
Shape(AB,OBA)
Shape(BP,PA,OBA)
Collinear(AOC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),50)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AC,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(ACB))
Answer: 65
Theorem sequence:
angle_addition(1,PAB,BAC)
tangent_of_circle_property_length_equal(1,PB,PA,O)
isosceles_triangle_judgment_line_equal(1,PAB)
isosceles_triangle_property_angle_equal(1,PAB)
diameter_of_circle_property_right_angle(1,CBA,O)
tangent_of_circle_property_perpendicular(2,PA,O,O)
triangle_property_angle_sum(1,ACB)
triangle_property_angle_sum(1,ABP)
Problem: As shown in the diagram, ∠ABC=31°, the center of ⊙O is O, DC is the tangent to ⊙O. Find the measure of ∠CDO.
问题：如图所示，∠ABC=31°，⊙O的圆心为O，圆O的切线为DC。求∠CDO的大小。
Construction (CDL):
Shape(BA,OBA)
Shape(CB,OBC)
Shape(OA,AB,BC,CO)
Shape(AO,OC,OCA)
Shape(CD,DA,OCA)
Collinear(OAD)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),31)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(CDO))
Answer: 28
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DC,O,O)
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_external(1,OCA,B)
triangle_property_angle_sum(1,OCD)
Problem: As shown in the diagram, ∠ACD=42°, the center of ⊙O is O, the diameter of ⊙O is AB, the tangent to circle O is CA. Find the measure of ∠DBA.
问题：如图所示，∠ACD=42°，⊙O的圆心为O，⊙O的直径为AB，CA是圆O的切线。求∠DBA的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(DO,OA,OAD)
Shape(AC,CD,OAD)
Shape(BO,OD,DB)
Shape(BD,ODB)
Collinear(AOB)
Collinear(ODC)
Cocircular(O,ADB)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),42)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(DBA))
Answer: 24
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CA,O,O)
triangle_property_angle_sum(1,OAC)
arc_property_center_angle(1,OAD,O)
arc_property_circumference_angle_external(1,OAD,B)
Problem: As shown in the diagram, ∠MNB=52°, O is the center of ⊙O, MN is the tangent to circle O. Find the measure of ∠AON.
问题：如图所示，∠MNB=52°，圆O的圆心为O，圆O的切线为MN。求∠AON的大小。
Construction (CDL):
Shape(AO,ON,ONA)
Shape(BO,OA,OAB)
Shape(NO,OB,BN)
Shape(NB,OBN)
Shape(MN,NB)
Collinear(AOB)
Cocircular(O,ABN)
Conditions (CDL):
Equal(MeasureOfAngle(MNB),52)
IsCentreOfCircle(O,O)
IsTangentOfCircle(MN,O)
Goal: Value(MeasureOfAngle(AON))
Answer: 76
Theorem sequence:
circle_property_angle_of_osculation(2,OBN,M)
arc_property_center_angle(1,OBN,O)
adjacent_complementary_angle(1,AON,NOB)
Problem: As shown in the diagram, ∠CAD=26°, the center of ⊙O is O, the tangent to circle O is DC. Find the measure of ∠ADC.
问题：如图所示，∠CAD=26°，⊙O的圆心为O，DC是⊙O的切线。求∠ADC的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(CA,AO,OC)
Shape(BO,OA,OBA)
Shape(BD,DC,OBC)
Shape(CO,OB,OBC)
Shape(BD,DC,OBC)
Collinear(AOBD)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),26)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(ADC))
Answer: 38
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
tangent_of_circle_property_perpendicular(2,DC,O,O)
isosceles_triangle_judgment_line_equal(1,OCA)
isosceles_triangle_property_angle_equal(1,OCA)
angle_addition(1,DCO,OCA)
triangle_property_angle_sum(1,CAD)