SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠CAB=30°, the center of ⊙O is O, the tangent to ⊙O is BA. Find the measure of ∠ABC.
问题：如图所示，∠CAB=30°，⊙O的圆心为O，⊙O的切线为BA。求∠ABC的大小。
Construction (CDL):
Shape(AO,OC,OCA)
Shape(OA,AC,CO)
Shape(CA,OAC)
Shape(AB,BC,OAC)
Collinear(OCB)
Cocircular(O,AC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BA,O)
Goal: Value(MeasureOfAngle(ABC))
Answer: 30
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(1,BA,O,O)
angle_addition(1,OAC,CAB)
isosceles_triangle_judgment_line_equal(1,OAC)
isosceles_triangle_property_angle_equal(1,OAC)
triangle_property_angle_sum(1,OAC)
triangle_property_angle_sum(1,OAB)
Problem: As shown in the diagram, ∠DBO=20°, O is the center of ⊙O, CD is the tangent to ⊙O. Find the measure of ∠OCD.
问题：如图所示，∠DBO=20°，O是⊙O的圆心，圆O的切线为CD。求∠OCD的大小。
Construction (CDL):
Shape(DO,OA,OAD)
Shape(DB,BO,OD)
Shape(BD,ODB)
Shape(AO,OB,OBA)
Shape(AC,CD,OAD)
Collinear(BOAC)
Cocircular(O,BAD)
Conditions (CDL):
Equal(MeasureOfAngle(DBO),20)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CD,O)
Goal: Value(MeasureOfAngle(OCD))
Answer: 50
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CD,O,O)
arc_property_circumference_angle_external(1,OAD,B)
arc_property_center_angle(1,OAD,O)
triangle_property_angle_sum(1,CDO)
Problem: As shown in the diagram, ∠BCD=115°, the center of ⊙O is O, the tangent to circle O is PD. Find the measure of ∠ADP.
问题：如图所示，∠BCD=115°，O是圆O的圆心，圆O的切线为PD。求∠ADP的大小。
Construction (CDL):
Shape(DP,PA,OAD)
Shape(AD,OAD)
Shape(OD,DA,AO)
Shape(CD,DO,OB,BC)
Shape(CB,OBC)
Shape(DC,OCD)
Shape(BO,OA,OAB)
Collinear(PAOB)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),115)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PD,O)
Goal: Value(MeasureOfAngle(ADP))
Answer: 25
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
tangent_of_circle_property_perpendicular(1,PD,O,O)
arc_property_center_angle(1,OBD,O)
arc_property_center_angle(1,ODB,O)
arc_property_circumference_angle_external(1,OBD,A)
arc_property_circumference_angle_external(1,ODB,C)
round_angle(1,DOB,BOD)
isosceles_triangle_judgment_line_equal(1,ODA)
isosceles_triangle_property_angle_equal(1,ODA)
angle_addition(1,ODA,ADP)
Problem: As shown in the diagram, AB=6, AC=3, the center of ⊙O is O, AB is the diameter of ⊙O. Find the length of line BC.
问题：如图所示，AB=6，AC=3，⊙O的圆心为O，⊙O的直径为AB。求直线BC的长度。
Construction (CDL):
Shape(CA,AO,OB,BC)
Shape(AC,OCA)
Shape(BO,OA,OAB)
Shape(CB,OBC)
Collinear(AOB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AC),3)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
Goal: Value(LengthOfLine(BC))
Answer: 3*sqrt(3)
Theorem sequence:
diameter_of_circle_property_right_angle(1,BCA,O)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠OAC=25°, the center of circle O is O, DC is the tangent to circle O. Find the measure of ∠CDB.
问题：如图所示，∠OAC=25°，⊙O的圆心为O，⊙O的切线为DC。求∠CDB的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(AC,CO,OA)
Shape(OC,CB,BO)
Shape(BC,OCB)
Shape(CA,OAC)
Shape(CD,DB,OCB)
Collinear(AOBD)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(OAC),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(CDB))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DC,O,O)
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,A)
triangle_property_angle_sum(1,OCD)
Problem: As shown in the diagram, AB=4, OD=2, ∠FCB=30°, O is the center of circle O, the diameter of circle O is AB, the tangent to ⊙O is ED. Find the length of line DE.
问题：如图所示，AB=4，OD=2，∠FCB=30°，O是圆O的圆心，AB是圆O的直径，ED是圆O的切线。求直线DE的长度。
Construction (CDL):
Shape(AO,OD,ODA)
Shape(DO,OF,FD)
Shape(DF,FB,OBD)
Shape(BE,ED,OBD)
Shape(BF,FC,CB)
Shape(BC,OCB)
Shape(CF,FO,OA,OAC)
Collinear(AOFBE)
Collinear(DFC)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(OD),2)
Equal(MeasureOfAngle(FCB),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(ED,O)
Goal: Value(LengthOfLine(DE))
Answer: 2*sqrt(3)
Theorem sequence:
arc_property_center_angle(1,OBD,O)
arc_property_circumference_angle_external(1,OBD,C)
tangent_of_circle_property_perpendicular(2,ED,O,O)
triangle_property_angle_sum(1,DOE)
sine_theorem(1,DOE)
Problem: As shown in the diagram, ∠BPA=50°, the center of ⊙O is O, the tangent to ⊙O is PA, the tangent to circle O is PB. Find the measure of ∠OCB.
问题：如图所示，∠BPA=50°，⊙O的圆心为O，圆O的切线为PA，PB是圆O的切线。求∠OCB的大小。
Construction (CDL):
Shape(CO,OA,OAC)
Shape(AB,OBA)
Shape(AO,OB,BA)
Shape(CB,BO,OC)
Shape(BC,OCB)
Shape(BP,PA,OBA)
Collinear(AOC)
Cocircular(O,CBA)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),50)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(OCB))
Answer: 65
Theorem sequence:
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OB,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)
flat_angle(1,AOC)
angle_addition(1,AOB,BOC)
isosceles_triangle_judgment_line_equal(1,OCB)
isosceles_triangle_property_angle_equal(1,OCB)
triangle_property_angle_sum(1,OCB)
Problem: As shown in the diagram, ∠CDB=100°, the center of circle O is O, the tangent to circle O is DB, the tangent to ⊙O is DC. Find the measure of ∠OAC.
问题：如图所示，∠CDB=100°，圆O的圆心为O，DB是⊙O的切线，DC是圆O的切线。求∠OAC的大小。
Construction (CDL):
Shape(BC,OCB)
Shape(CD,DB,OCB)
Shape(BO,OC,CB)
Shape(OA,AC,CO)
Shape(AO,OB,OBA)
Collinear(AOB)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(CDB),100)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DB,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(OAC))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DB,O,O)
tangent_of_circle_property_perpendicular(1,DC,O,O)
quadrilateral_property_angle_sum(1,BOCD)
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,A)
Problem: As shown in the diagram, ∠DAB=34°, the center of ⊙O is O, the tangent to ⊙O is AB. Find the measure of ∠BCO.
问题：如图所示，∠DAB=34°，圆O的圆心为O，圆O的切线为AB。求∠BCO的大小。
Construction (CDL):
Shape(CB,OBC)
Shape(CO,OB,BC)
Shape(BO,OD,ODB)
Shape(DO,OC,OCD)
Shape(DA,AB,ODB)
Collinear(CODA)
Cocircular(O,DBC)
Conditions (CDL):
Equal(MeasureOfAngle(DAB),34)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(MeasureOfAngle(BCO))
Answer: 28
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
triangle_property_angle_sum(1,OAB)
arc_property_center_angle(1,ODB,O)
arc_property_circumference_angle_external(1,ODB,C)
Problem: As shown in the diagram, ∠BPA=62°, the center of ⊙O is O, PA is the tangent to ⊙O, PB is the tangent to circle O. Find the measure of ∠BOC.
问题：如图所示，∠BPA=62°，⊙O的圆心为O，圆O的切线为PA，PB是⊙O的切线。求∠BOC的大小。
Construction (CDL):
Shape(CO,OA,OAC)
Shape(AO,OB,BA)
Shape(AB,OBA)
Shape(BP,PA,OBA)
Shape(BO,OC,OCB)
Collinear(AOC)
Cocircular(O,CBA)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),62)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(BOC))
Answer: 62
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
quadrilateral_property_angle_sum(1,PAOB)
flat_angle(1,AOC)
angle_addition(1,AOB,BOC)
Problem: As shown in the diagram, ∠BPA=80°, the center of ⊙O is O, PA is the tangent to ⊙O, PB is the tangent to ⊙O. Find the measure of ∠ACB.
问题：如图所示，∠BPA=80°，⊙O的圆心为O，PA是⊙O的切线，⊙O的切线为PB。求∠ACB的大小。
Construction (CDL):
Shape(BP,PA,OBA)
Shape(CA,OAC)
Shape(BC,OCB)
Shape(AC,CB,BO,OA)
Shape(AO,OB,OBA)
Cocircular(O,CBA)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),80)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(ACB))
Answer: 50
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
quadrilateral_property_angle_sum(1,PAOB)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,C)
Problem: As shown in the diagram, ∠BPA=60°, O is the center of ⊙O, the tangent to circle O is PA, the tangent to ⊙O is PB. Find the measure of ∠AOB.
问题：如图所示，∠BPA=60°，圆O的圆心为O，PA是⊙O的切线，PB是圆O的切线。求∠AOB的大小。
Construction (CDL):
Shape(BP,PA,OBA)
Shape(AO,OB,OBA)
Shape(BO,OA,OAB)
Cocircular(O,BA)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(AOB))
Answer: 120
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
quadrilateral_property_angle_sum(1,PAOB)
Problem: As shown in the diagram, AC=16, OB=12, OC=12, O is the center of circle O, AC is the tangent to circle O. Find the length of line AB.
问题：如图所示，AC=16，OB=12，OC=12，圆O的圆心为O，圆O的切线为AC。求直线AB的长度。
Construction (CDL):
Shape(BA,AC,OBC)
Shape(CO,OB,OBC)
Shape(BO,OC,OCB)
Collinear(ABO)
Cocircular(O,BC)
Conditions (CDL):
Equal(LengthOfLine(AC),16)
Equal(LengthOfLine(OB),12)
Equal(LengthOfLine(OC),12)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AC,O)
Goal: Value(LengthOfLine(AB))
Answer: 8
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AC,O,O)
right_triangle_judgment_angle(1,ACO)
right_triangle_property_pythagorean(1,ACO)
line_addition(1,AB,BO)
Problem: As shown in the diagram, ∠OAC=20°, the center of circle O is O, DC is the tangent to ⊙O. Find the measure of ∠CDB.
问题：如图所示，∠OAC=20°，O是⊙O的圆心，圆O的切线为DC。求∠CDB的大小。
Construction (CDL):
Shape(CD,DB,OCB)
Shape(AC,CO,OA)
Shape(CA,OAC)
Shape(BO,OC,OCB)
Shape(AO,OB,OBA)
Collinear(AOBD)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(OAC),20)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(CDB))
Answer: 50
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DC,O,O)
arc_property_circumference_angle_external(1,OCB,A)
arc_property_center_angle(1,OCB,O)
triangle_property_angle_sum(1,OCD)
Problem: As shown in the diagram, AD=3, BC=6, the center of ⊙O is O, the tangent to circle O is AE, the tangent to circle O is AH, BE is the tangent to circle O, the tangent to ⊙O is BF, CF is the tangent to ⊙O, CG is the tangent to circle O, DG is the tangent to circle O, the tangent to ⊙O is DH. Find the sum of the length of line AB and the length of line CD.
问题：如图所示，AD=3，BC=6，O是⊙O的圆心，AE是⊙O的切线，⊙O的切线为AH，圆O的切线为BE，BF是⊙O的切线，⊙O的切线为CF，圆O的切线为CG，圆O的切线为DG，⊙O的切线为DH。求直线AB的长度与直线CD的长度之和。
Construction (CDL):
Shape(HA,AE,OHE)
Shape(EB,BF,OEF)
Shape(FC,CG,OFG)
Shape(GD,DH,OGH)
Collinear(AEB)
Collinear(BFC)
Collinear(CGD)
Collinear(AHD)
Cocircular(O,EFGH)
Conditions (CDL):
Equal(LengthOfLine(AD),3)
Equal(LengthOfLine(BC),6)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AE,O)
IsTangentOfCircle(AH,O)
IsTangentOfCircle(BE,O)
IsTangentOfCircle(BF,O)
IsTangentOfCircle(CF,O)
IsTangentOfCircle(CG,O)
IsTangentOfCircle(DG,O)
IsTangentOfCircle(DH,O)
Goal: Value(Add(LengthOfLine(AB),LengthOfLine(CD)))
Answer: 9
Theorem sequence:
tangent_of_circle_property_length_equal(1,AE,AH,O)
tangent_of_circle_property_length_equal(1,BF,BE,O)
tangent_of_circle_property_length_equal(1,CG,CF,O)
tangent_of_circle_property_length_equal(1,DH,DG,O)
line_addition(1,AE,EB)
line_addition(1,BF,FC)
line_addition(1,CG,GD)
line_addition(1,AH,HD)
Problem: As shown in the diagram, AC=6, the radius of ⊙O is 4, O is the center of circle O, BC is the diameter of ⊙O, the tangent to circle O is AC. Find the length of line AB.
问题：如图所示，AC=6，圆O的半径为4，O是圆O的圆心，BC是圆O的直径，AC是圆O的切线。求直线AB的长度。
Construction (CDL):
Shape(BD,ODB)
Shape(CA,AD,OCD)
Shape(DB,BO,OC,OCD)
Shape(CO,OB,OBC)
Collinear(BOC)
Collinear(BDA)
Cocircular(O,CDB)
Conditions (CDL):
Equal(LengthOfLine(AC),6)
Equal(RadiusOfCircle(O),4)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(BC,O)
IsTangentOfCircle(AC,O)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
line_addition(1,BO,OC)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OB,O)
tangent_of_circle_property_perpendicular(1,AC,O,O)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠ADC=30°, O is the center of ⊙O, BA is the tangent to circle O. Find the measure of ∠CBA.
问题：如图所示，∠ADC=30°，O是⊙O的圆心，BA是⊙O的切线。求∠CBA的大小。
Construction (CDL):
Shape(CB,BA,OCA)
Shape(AO,OC,OCA)
Shape(AD,DC,CO,OA)
Shape(CD,ODC)
Shape(DA,OAD)
Collinear(OCB)
Cocircular(O,ADC)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BA,O)
Goal: Value(MeasureOfAngle(CBA))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BA,O,O)
arc_property_circumference_angle_external(1,OCA,D)
arc_property_center_angle(1,OCA,O)
triangle_property_angle_sum(1,AOB)
Problem: As shown in the diagram, ∠ABC=65°, the center of ⊙O is O, CB is the tangent to circle O. Find the measure of ∠BAO.
问题：如图所示，∠ABC=65°，O是圆O的圆心，CB是圆O的切线。求∠BAO的大小。
Construction (CDL):
Shape(OBA,BC)
Shape(BA,AO,OB)
Shape(AB,OBA)
Shape(BO,OA,OAB)
Cocircular(O,AB)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),65)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(BAO))
Answer: 25
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
tangent_of_circle_property_perpendicular(1,CB,O,O)
angle_addition(1,OBA,ABC)
isosceles_triangle_judgment_line_equal(1,OBA)
isosceles_triangle_property_angle_equal(1,OBA)
Problem: As shown in the diagram, ∠BPA=60°, O is the center of ⊙O, PA is the tangent to circle O, the tangent to ⊙O is PB. Find the measure of ∠AOB.
问题：如图所示，∠BPA=60°，O是⊙O的圆心，PA是圆O的切线，PB是圆O的切线。求∠AOB的大小。
Construction (CDL):
Shape(BP,PA,OBA)
Shape(AO,OB,OBA)
Shape(BO,OA,OAB)
Cocircular(O,BA)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(AOB))
Answer: 120
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
quadrilateral_property_angle_sum(1,PAOB)
Problem: As shown in the diagram, OA=2, OB=3, OC=2, O is the center of ⊙O, BA is the tangent to ⊙O. Find the length of line AB.
问题：如图所示，OA=2，OB=3，OC=2，⊙O的圆心为O，圆O的切线为BA。求直线AB的长度。
Construction (CDL):
Shape(AB,BC,OAC)
Shape(CO,OA,OAC)
Shape(AO,OC,OCA)
Collinear(BCO)
Cocircular(O,AC)
Conditions (CDL):
Equal(LengthOfLine(OA),2)
Equal(LengthOfLine(OB),3)
Equal(LengthOfLine(OC),2)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BA,O)
Goal: Value(LengthOfLine(AB))
Answer: sqrt(5)
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BA,O,O)
right_triangle_judgment_angle(1,OAB)
right_triangle_property_pythagorean(1,OAB)
Problem: As shown in the diagram, OA=2, OB=3, OC=2, the center of circle O is O, the tangent to ⊙O is BA. Find the length of line AB.
问题：如图所示，OA=2，OB=3，OC=2，⊙O的圆心为O，⊙O的切线为BA。求直线AB的长度。
Construction (CDL):
Shape(AB,BC,OAC)
Shape(CO,OA,OAC)
Shape(AO,OC,OCA)
Collinear(BCO)
Cocircular(O,AC)
Conditions (CDL):
Equal(LengthOfLine(OA),2)
Equal(LengthOfLine(OB),3)
Equal(LengthOfLine(OC),2)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BA,O)
Goal: Value(LengthOfLine(AB))
Answer: sqrt(5)
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BA,O,O)
right_triangle_judgment_angle(1,OAB)
right_triangle_property_pythagorean(1,OAB)
Problem: As shown in the diagram, ∠FDB=20°, the center of ⊙O is O, EC is the tangent to ⊙O. Find the measure of ∠BEC.
问题：如图所示，∠FDB=20°，O是圆O的圆心，EC是圆O的切线。求∠BEC的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(CO,OF,FC)
Shape(CF,FB,OBC)
Shape(BE,EC,OBC)
Shape(AO,OC,CA)
Shape(FD,DB,BF)
Shape(BD,ODB)
Shape(DF,FO,OA,OAD)
Collinear(CFD)
Collinear(AOFBE)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(FDB),20)
IsCentreOfCircle(O,O)
IsTangentOfCircle(EC,O)
Goal: Value(MeasureOfAngle(BEC))
Answer: 50
Theorem sequence:
arc_property_circumference_angle_external(1,OBC,D)
arc_property_center_angle(1,OBC,O)
tangent_of_circle_property_perpendicular(2,EC,O,O)
triangle_property_angle_sum(1,COE)
Problem: As shown in the diagram, PA=15, the center of ⊙O is O, the tangent to circle O is CA, the tangent to circle O is CE, the tangent to circle O is DB, the tangent to ⊙O is DE, PA is the tangent to circle O, PB is the tangent to ⊙O. Find the sum of the length of line PC, the length of line CD and the length of line DP.
问题：如图所示，PA=15，O是圆O的圆心，CA是⊙O的切线，⊙O的切线为CE，DB是⊙O的切线，DE是圆O的切线，PA是⊙O的切线，⊙O的切线为PB。求直线PC的长度、直线CD的长度与直线DP的长度之和。
Construction (CDL):
Shape(PC,CE,ED,DP)
Shape(BD,DE,OBE)
Shape(EC,CA,OEA)
Collinear(BDP)
Collinear(ACP)
Collinear(CED)
Cocircular(O,BEA)
Conditions (CDL):
Equal(LengthOfLine(PA),15)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
IsTangentOfCircle(CE,O)
IsTangentOfCircle(DB,O)
IsTangentOfCircle(DE,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(Add(LengthOfLine(PC),LengthOfLine(CD),LengthOfLine(DP)))
Answer: 30
Theorem sequence:
tangent_of_circle_property_length_equal(1,CE,CA,O)
tangent_of_circle_property_length_equal(1,DE,DB,O)
tangent_of_circle_property_length_equal(1,PA,PB,O)
line_addition(1,BD,DP)
line_addition(1,AC,CP)
line_addition(1,CE,ED)
Problem: As shown in the diagram, ∠MCB=38°, the center of circle O is O, MC is the tangent to circle O. Find the measure of ∠OBC.
问题：如图所示，∠MCB=38°，圆O的圆心为O，MC是⊙O的切线。求∠OBC的大小。
Construction (CDL):
Shape(MC,OBC)
Shape(BO,OA,OAB)
Shape(AD,ODA)
Shape(OB,BC,CO)
Shape(OC,CD,DA,AO)
Shape(DC,OCD)
Shape(CB,OBC)
Collinear(BOA)
Cocircular(O,BCDA)
Conditions (CDL):
Equal(MeasureOfAngle(MCB),38)
IsCentreOfCircle(O,O)
IsTangentOfCircle(MC,O)
Goal: Value(MeasureOfAngle(OBC))
Answer: 52
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
tangent_of_circle_property_perpendicular(2,MC,O,O)
angle_addition(1,MCB,BCO)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OBC)
Problem: As shown in the diagram, AB=5, AO=4, BO=3, O is the center of ⊙O, AC is the tangent to circle O, the tangent to ⊙O is BC. Find the length of line OC.
问题：如图所示，AB=5，AO=4，BO=3，O是圆O的圆心，AC是圆O的切线，圆O的切线为BC。求直线OC的长度。
Construction (CDL):
Shape(DA,AC,ODC)
Shape(CB,BE,OCE)
Shape(CO,OD,ODC)
Shape(EO,OC,OCE)
Shape(DO,OE,OED)
Collinear(ACB)
Collinear(BEO)
Collinear(ADO)
Cocircular(O,DCE)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AO),4)
Equal(LengthOfLine(BO),3)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AC,O)
IsTangentOfCircle(BC,O)
Goal: Value(LengthOfLine(OC))
Answer: 12/5
Theorem sequence:
right_triangle_judgment_pythagorean_inverse(1,BOA)
tangent_of_circle_property_perpendicular(1,BC,O,O)
mirror_similar_triangle_judgment_aa(1,OCB,ABO)
mirror_similar_triangle_property_line_ratio(1,CBO,OAB)
mirror_similar_triangle_property_line_ratio(1,BOC,BOA)
Problem: As shown in the diagram, ∠ABC=70°, O is the center of ⊙O, CB is the tangent to ⊙O. Find the measure of ∠BAO.
问题：如图所示，∠ABC=70°，圆O的圆心为O，CB是⊙O的切线。求∠BAO的大小。
Construction (CDL):
Shape(OB,BA,AO)
Shape(OBA,AB)
Shape(OAB,BO,OA)
Shape(OB,BC)
Shape(AB,BC)
Cocircular(O,BA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),70)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(BAO))
Answer: 20
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CB,O,O)
angle_addition(1,OBA,ABC)
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, ∠DCA=40°, O is the center of circle O, the tangent to ⊙O is CA. Find the measure of ∠ABD.
问题：如图所示，∠DCA=40°，圆O的圆心为O，CA是圆O的切线。求∠ABD的大小。
Construction (CDL):
Shape(OAB,BA)
Shape(OBD,DO,OB)
Shape(AB,BO,OA)
Shape(CA,ODA,DC)
Shape(AO,OD,ODA)
Collinear(BODC)
Cocircular(O,BDA)
Conditions (CDL):
Equal(MeasureOfAngle(DCA),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(ABD))
Answer: 25
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CA,O,O)
triangle_property_angle_sum(1,CAO)
arc_property_circumference_angle_external(1,ODA,B)
arc_property_center_angle(1,ODA,O)
Problem: As shown in the diagram, PA=8, ∠APB=60°, the center of ⊙O is O, the tangent to ⊙O is PA, PB is the tangent to ⊙O. Find the length of line OA.
问题：如图所示，PA=8，∠APB=60°，O是圆O的圆心，圆O的切线为PA，圆O的切线为PB。求直线OA的长度。
Construction (CDL):
Shape(AP,PD,DA)
Shape(OA,AD,DO)
Shape(OAE,ED,DA)
Shape(DP,PB,BD)
Shape(OEB,BD,DE)
Shape(OD,DB,BO)
Shape(OBA,AO,OB)
Collinear(PEDO)
Cocircular(O,AEB)
Conditions (CDL):
Equal(LengthOfLine(PA),8)
Equal(MeasureOfAngle(APB),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(LengthOfLine(OA))
Answer: 8*sqrt(3)/3
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
tangent_of_circle_property_perpendicular(2,PB,O,O)
tangent_of_circle_property_length_equal(1,PA,PB,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
mirror_congruent_triangle_judgment_sas(1,APO,BOP)
angle_addition(1,APO,OPB)
mirror_congruent_triangle_property_angle_equal(1,POA,PBO)
triangle_property_angle_sum(1,APO)
sine_theorem(1,APO)
Problem: As shown in the diagram, ∠BCD=40°, O is the center of ⊙O, DC is the tangent to ⊙O. Find the measure of ∠BAC.
问题：如图所示，∠BCD=40°，O是圆O的圆心，DC是圆O的切线。求∠BAC的大小。
Construction (CDL):
Shape(OC,CB,BO)
Shape(AC,CB,BA)
Shape(OCB,BC)
Shape(OBA,AB)
Shape(OAC,CA)
Shape(BC,CD)
Cocircular(O,CBA)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(BAC))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DC,O,O)
angle_addition(1,OCB,BCD)
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)
triangle_property_angle_sum(1,OCB)
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,A)
Problem: As shown in the diagram, ∠EDC=30°, the radius of ⊙O is 2, the center of circle O is O, BE is the tangent to ⊙O. Find the length of line CB.
问题：如图所示，∠EDC=30°，圆O的半径为2，O是⊙O的圆心，圆O的切线为BE。求直线CB的长度。
Construction (CDL):
Shape(OEA,AE)
Shape(OAD,DO,OA)
Shape(EA,AO,OE)
Shape(OCE,EO,OC)
Shape(ODC,CD)
Shape(CO,OD,DC)
Shape(BE,OCE,CB)
Collinear(AOCB)
Collinear(EOD)
Cocircular(O,ADCE)
Conditions (CDL):
Equal(MeasureOfAngle(EDC),30)
Equal(RadiusOfCircle(O),2)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BE,O)
Goal: Value(LengthOfLine(CB))
Answer: 2
Theorem sequence:
arc_property_circumference_angle_external(1,OCE,D)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OE,O)
arc_property_center_angle(1,OCE,O)
triangle_property_angle_sum(1,EOB)
tangent_of_circle_property_perpendicular(2,BE,O,O)
sine_theorem(1,OBE)
line_addition(1,OC,CB)
Problem: As shown in the diagram, ∠BAC=30°, the center of ⊙O is O, the tangent to ⊙O is AB. Find the measure of ∠BCO.
问题：如图所示，∠BAC=30°，O是⊙O的圆心，AB是⊙O的切线。求∠BCO的大小。
Construction (CDL):
Shape(OBC,CB)
Shape(OB,BC,CO)
Shape(BA,AC,OBC)
Shape(OCB,BO,OC)
Collinear(ACO)
Cocircular(O,BC)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(MeasureOfAngle(BCO))
Answer: 60
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
triangle_property_angle_sum(1,OBC)
triangle_property_angle_sum(1,OBA)
tangent_of_circle_property_perpendicular(1,AB,O,O)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OBC)
Problem: As shown in the diagram, ∠DCB=38°, the center of ⊙O is O, AB is the diameter of circle O, CB is the tangent to ⊙O. Find the measure of ∠DEA.
问题：如图所示，∠DCB=38°，O是⊙O的圆心，圆O的直径为AB，圆O的切线为CB。求∠DEA的大小。
Construction (CDL):
Shape(AD,DB,BO,OA)
Shape(ODB,BD)
Shape(OBE,EA,AO,OB)
Shape(OEA,AE)
Shape(OAD,DA)
Shape(AD,DE,EA)
Shape(DC,CB,ODB)
Collinear(AOB)
Collinear(CDA)
Cocircular(O,DBEA)
Conditions (CDL):
Equal(MeasureOfAngle(DCB),38)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(DEA))
Answer: 38
Theorem sequence:
diameter_of_circle_property_right_angle(1,ADB,O)
tangent_of_circle_property_perpendicular(2,CB,O,O)
triangle_property_angle_sum(1,ADB)
triangle_property_angle_sum(1,BAC)
arc_property_circumference_angle_external(1,OAD,B)
arc_property_circumference_angle_external(1,OAD,E)