SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, PA=5, the center of circle O is O, PA is the tangent to ⊙O, PB is the tangent to ⊙O. Find the length of line PB.
问题：如图所示，PA=5，圆O的圆心为O，PA是⊙O的切线，PB是⊙O的切线。求直线PB的长度。
Construction (CDL):
Shape(PA,OBA,BP)
Shape(O)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(PA),5)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(LengthOfLine(PB))
Answer: 5
Theorem sequence:
tangent_of_circle_property_length_equal(1,PA,PB,O)
Problem: As shown in the diagram, AC=BC, OA=5, OC=3, the center of ⊙O is O, AC is the tangent to circle O. Find the length of line AB.
问题：如图所示，AC=BC，OA=5，OC=3，⊙O的圆心为O，⊙O的切线为AC。求直线AB的长度。
Construction (CDL):
Shape(OM,OMC,CO)
Shape(OC,OCN,NO)
Shape(ON,ONM,MO)
Shape(AC,OMC,MA)
Shape(CB,BN,ONC)
Collinear(OMA)
Collinear(ONB)
Collinear(ACB)
Cocircular(O,MCN)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(BC))
Equal(LengthOfLine(OA),5)
Equal(LengthOfLine(OC),3)
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,AC,CB)
Problem: As shown in the diagram, ∠CAD=25°, the center of circle O is O, the tangent to circle O is DC. Find the measure of ∠ADC.
问题：如图所示，∠CAD=25°，O是⊙O的圆心，DC是圆O的切线。求∠ADC的大小。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(AO,OC,CA)
Shape(OCA,AC)
Shape(OB,OBC,CO)
Shape(BD,DC,OBC)
Collinear(DBOA)
Cocircular(O,BCA)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(ADC))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DC,O,O)
triangle_property_angle_sum(1,COD)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, ∠BPA=70°, the center of ⊙O is O, PA is the tangent to ⊙O, the tangent to ⊙O is PB. Find the measure of ∠BCA.
问题：如图所示，∠BPA=70°，O是⊙O的圆心，圆O的切线为PA，⊙O的切线为PB。求∠BCA的大小。
Construction (CDL):
Shape(PA,OCA,OBC,BP)
Shape(OCA,AC)
Shape(OBC,CB)
Shape(AO,OB,BC,CA)
Shape(AD,DB,BO,OA)
Shape(OAD,DA)
Shape(ODN,ND)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),70)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 125
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,BPAO)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_internal(1,OBA,C)
Problem: As shown in the diagram, ∠DAB=40°, O is the center of circle O, the tangent to circle O is BA. Find the measure of ∠BCA.
问题：如图所示，∠DAB=40°，⊙O的圆心为O，⊙O的切线为BA。求∠BCA的大小。
Construction (CDL):
Shape(OCA,AO,OC)
Shape(CO,OA,AD,DC)
Shape(ODC,CD)
Shape(OAD,DA)
Shape(OAD,AB,BD)
Collinear(COA)
Collinear(CDB)
Cocircular(O,CAD)
Conditions (CDL):
Equal(MeasureOfAngle(DAB),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BA,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BA,O,O)
angle_addition(1,CAD,DAB)
diameter_of_circle_judgment_pass_centre(1,COA,O)
diameter_of_circle_property_right_angle(1,ADC,O)
triangle_property_angle_sum(1,CAD)
Problem: As shown in the diagram, ∠ACO=40°, O is the center of ⊙O, CA is the tangent to circle O. Find the measure of ∠DBA.
问题：如图所示，∠ACO=40°，O是⊙O的圆心，圆O的切线为CA。求∠DBA的大小。
Construction (CDL):
Shape(OBA,AO,OB)
Shape(BO,OD,DB)
Shape(OA,OAD,DO)
Shape(BD,ODB)
Shape(OAD,AC,CD)
Collinear(BOA)
Collinear(ODC)
Cocircular(O,ADB)
Conditions (CDL):
Equal(MeasureOfAngle(ACO),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(DBA))
Answer: 25
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, ∠ABC=25°, O is the center of ⊙O, CA is the tangent to ⊙O. Find the measure of ∠BCA.
问题：如图所示，∠ABC=25°，圆O的圆心为O，⊙O的切线为CA。求∠BCA的大小。
Construction (CDL):
Shape(OAB,BA)
Shape(AB,BO,OA)
Shape(AO,OM,OMA)
Shape(OMA,MC,CA)
Shape(MO,OB,BM)
Collinear(BOMC)
Cocircular(O,ABM)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 40
Theorem sequence:
arc_property_center_angle(1,OMA,O)
arc_property_circumference_angle_external(1,OMA,B)
tangent_of_circle_property_perpendicular(2,CA,O,O)
triangle_property_angle_sum(1,AOC)
Problem: As shown in the diagram, ∠CAB=30°, the center of circle O is O, AB is the tangent to circle O. Find the measure of ∠BCA.
问题：如图所示，∠CAB=30°，O是圆O的圆心，圆O的切线为AB。求∠BCA的大小。
Construction (CDL):
Shape(DA,AB,ODB)
Shape(OD,ODB,BO)
Shape(CO,OB,BC)
Shape(CB,OBC)
Shape(OCD,DO,OC)
Collinear(ADOC)
Cocircular(O,DBC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
triangle_property_angle_sum(1,ABO)
arc_property_center_angle(1,ODB,O)
arc_property_circumference_angle_external(1,ODB,C)
Problem: As shown in the diagram, OC=5, ∠DAC=30°, the center of ⊙O is O, the tangent to ⊙O is DC. Find the length of line DB.
问题：如图所示，OC=5，∠DAC=30°，O是⊙O的圆心，⊙O的切线为DC。求直线DB的长度。
Construction (CDL):
Shape(DB,OCB,CD)
Shape(OC,OCB,BO)
Shape(OA,AC,CO)
Shape(OAC,CA)
Shape(AO,OB,OBA)
Collinear(AOBD)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(OC),5)
Equal(MeasureOfAngle(DAC),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(LengthOfLine(DB))
Answer: 5
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)
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,A)
triangle_property_angle_sum(1,CDO)
sine_theorem(1,OCD)
line_addition(1,OB,BD)
Problem: As shown in the diagram, AB=sqrt(2), ∠CAB=45°, O is the center of circle O, AB is the tangent to ⊙O. Find the length of line AC.
问题：如图所示，AB=sqrt(2)，∠CAB=45°，O是圆O的圆心，圆O的切线为AB。求直线AC的长度。
Construction (CDL):
Shape(AB,OMB,MA)
Shape(OM,OMB,BO)
Shape(OB,OBC,CO)
Shape(MO,OC,OCM)
Collinear(COMA)
Cocircular(O,MBC)
Conditions (CDL):
Equal(LengthOfLine(AB),sqrt(2))
Equal(MeasureOfAngle(CAB),45)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(LengthOfLine(AC))
Answer: sqrt(2)+2
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
triangle_property_angle_sum(1,ABO)
isosceles_triangle_judgment_angle_equal(1,BOA)
right_triangle_judgment_angle(1,ABO)
right_triangle_property_pythagorean(1,ABO)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
line_addition(1,CO,OA)
Problem: As shown in the diagram, ∠BPA=60°, O is the center of circle O, PA is the tangent to ⊙O, PB is the tangent to circle O. Find the measure of ∠ACB.
问题：如图所示，∠BPA=60°，O是⊙O的圆心，PA是⊙O的切线，⊙O的切线为PB。求∠ACB的大小。
Construction (CDL):
Shape(PA,OBA,BP)
Shape(OBA,AO,OB)
Shape(AC,CB,BO,OA)
Shape(OAC,CA)
Shape(OCB,BC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(ACB))
Answer: 60
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,BPAO)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,C)
Problem: As shown in the diagram, ∠ACD=20°, O is the center of circle O, CD is the tangent to circle O. Find the measure of ∠DAC.
问题：如图所示，∠ACD=20°，⊙O的圆心为O，圆O的切线为CD。求∠DAC的大小。
Construction (CDL):
Shape(CD,OBD,BC)
Shape(OBD,DO,OB)
Shape(OD,DA,AO)
Shape(ODA,AD)
Shape(BO,OA,OAB)
Collinear(AOBC)
Cocircular(O,ABD)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),20)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CD,O)
Goal: Value(MeasureOfAngle(DAC))
Answer: 35
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CD,O,O)
triangle_property_angle_sum(1,CDO)
arc_property_center_angle(1,OBD,O)
arc_property_circumference_angle_external(1,OBD,A)
Problem: As shown in the diagram, AB=2, AC=CB, OA=OB, OC=2, O is the center of ⊙O, AC is the tangent to circle O, BC is the tangent to circle O. Find the length of line OA.
问题：如图所示，AB=2，AC=CB，OA=OB，OC=2，⊙O的圆心为O，圆O的切线为AC，BC是⊙O的切线。求直线OA的长度。
Construction (CDL):
Shape(AC,OMC,MA)
Shape(CB,BN,OCN)
Shape(OM,OMC,CO)
Shape(OC,OCN,NO)
Shape(MO,ON,ONM)
Collinear(OMA)
Collinear(ONB)
Collinear(ACB)
Cocircular(O,MCN)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(AC),LengthOfLine(CB))
Equal(LengthOfLine(OA),LengthOfLine(OB))
Equal(LengthOfLine(OC),2)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AC,O)
IsTangentOfCircle(BC,O)
Goal: Value(LengthOfLine(OA))
Answer: sqrt(5)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AC,O,O)
line_addition(1,AC,CB)
right_triangle_judgment_angle(1,ACO)
right_triangle_property_pythagorean(1,ACO)
Problem: As shown in the diagram, ∠BPC=50°, O is the center of ⊙O, the diameter of ⊙O is AB, the tangent to circle O is PC. Find the measure of ∠CAB.
问题：如图所示，∠BPC=50°，圆O的圆心为O，AB是圆O的直径，PC是⊙O的切线。求∠CAB的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(CA,AO,OC)
Shape(CO,OB,OBC)
Shape(BP,PC,OBC)
Shape(BO,OA,OAB)
Collinear(PBOA)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BPC),50)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PC,O))
Goal: Value(MeasureOfAngle(CAB))
Answer: 20
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PC,O,O)
triangle_property_angle_sum(1,COP)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, DC=2, the radius of ⊙O is 5, the center of ⊙O is O, OD is the perpendicular bisector of AB. Find the length of line AB.
问题：如图所示，DC=2，⊙O的半径为5，O是⊙O的圆心，OD垂直平分AB。求直线AB的长度。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OA,AD,DO)
Shape(OD,DB,BO)
Shape(CD,DA,OAC)
Shape(BD,DC,OCB)
Collinear(ADB)
Collinear(ODC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(DC),2)
Equal(RadiusOfCircle(O),5)
IsCentreOfCircle(O,O)
IsPerpendicularBisectorOfLine(OD,AB)
Goal: Value(LengthOfLine(AB))
Answer: 8
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
right_triangle_judgment_angle(1,ADO)
line_addition(1,AD,DB)
line_addition(1,OD,DC)
right_triangle_property_pythagorean(1,ADO)
Problem: As shown in the diagram, ∠APD=30°, the radius of ⊙O is 1, O is the center of ⊙O, C is the midpoint of segment BA, the tangent to ⊙O is PA. Find the length of line CP.
问题：如图所示，∠APD=30°，圆O的半径为1，圆O的圆心为O，C平分线段BA，圆O的切线为PA。求直线CP的长度。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(BO,OC,CB)
Shape(OA,AC,CO)
Shape(BC,CD,ODB)
Shape(DC,CA,OAD)
Shape(AP,PD,OAD)
Collinear(OCDP)
Collinear(BCA)
Cocircular(O,ADB)
Conditions (CDL):
Equal(MeasureOfAngle(APD),30)
Equal(RadiusOfCircle(O),1)
IsCentreOfCircle(O,O)
IsMidpointOfLine(C,BA)
IsTangentOfCircle(PA,O)
Goal: Value(LengthOfLine(CP))
Answer: 3/2
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,PA,O,O)
triangle_property_angle_sum(1,BOC)
triangle_property_angle_sum(1,OAP)
sine_theorem(1,OAP)
line_addition(1,OC,CP)
isosceles_triangle_judgment_line_equal(1,OAB)
median_of_triangle_judgment(1,OC,OAB)
isosceles_triangle_property_line_coincidence(2,OAB,C)
sine_theorem(1,OCB)
Problem: As shown in the diagram, ∠OAB=36°, O is the center of ⊙O, the tangent to circle O is AB. Find the measure of ∠BCO.
问题：如图所示，∠OAB=36°，⊙O的圆心为O，AB是⊙O的切线。求∠BCO的大小。
Construction (CDL):
Shape(DO,OC,OCD)
Shape(DA,AB,ODB)
Shape(BO,OD,ODB)
Shape(CO,OB,BC)
Shape(CB,OBC)
Collinear(CODA)
Cocircular(O,BCD)
Conditions (CDL):
Equal(MeasureOfAngle(OAB),36)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(MeasureOfAngle(BCO))
Answer: 27
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
triangle_property_angle_sum(1,ABO)
arc_property_center_angle(1,ODB,O)
arc_property_circumference_angle_external(1,ODB,C)
Problem: As shown in the diagram, ∠OPA=40°, the center of circle O is O, the diameter of ⊙O is AB, the tangent to circle O is PA. Find the measure of ∠ABC.
问题：如图所示，∠OPA=40°，O是⊙O的圆心，AB是⊙O的直径，⊙O的切线为PA。求∠ABC的大小。
Construction (CDL):
Shape(BO,OA,OAB)
Shape(AO,OC,OCA)
Shape(OB,BC,CO)
Shape(CB,OBC)
Shape(CP,PA,OCA)
Collinear(AOB)
Collinear(OCP)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(OPA),40)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(ABC))
Answer: 25
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
triangle_property_angle_sum(1,AOP)
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_external(1,OCA,B)
Problem: As shown in the diagram, PO=2, ∠APO=30°, O is the center of ⊙O, PA is the tangent to ⊙O. Find the length of line OA.
问题：如图所示，PO=2，∠APO=30°，圆O的圆心为O，⊙O的切线为PA。求直线OA的长度。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(BO,OA,OAB)
Shape(AP,PB,OAB)
Collinear(OBP)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(PO),2)
Equal(MeasureOfAngle(APO),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(LengthOfLine(OA))
Answer: 1
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
sine_theorem(1,OAP)
Problem: As shown in the diagram, ∠CDB=25°, the center of ⊙O is O, EC is the tangent to ⊙O. Find the measure of ∠BEC.
问题：如图所示，∠CDB=25°，圆O的圆心为O，EC是⊙O的切线。求∠BEC的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(CA,AO,OC)
Shape(CO,OF,FC)
Shape(DF,FO,OA,OAD)
Shape(CF,FB,OBC)
Shape(FD,DB,BF)
Shape(BD,ODB)
Shape(BE,EC,OBC)
Collinear(AOFBE)
Collinear(CFD)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(CDB),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(EC,O)
Goal: Value(MeasureOfAngle(BEC))
Answer: 40
Theorem sequence:
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,D)
tangent_of_circle_property_perpendicular(2,EC,O,O)
triangle_property_angle_sum(1,COE)
Problem: As shown in the diagram, ∠CAB=35°, O is the center of ⊙O, the tangent to circle O is DC. Find the measure of ∠BDC.
问题：如图所示，∠CAB=35°，⊙O的圆心为O，DC是⊙O的切线。求∠BDC的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(CA,AO,OC)
Shape(CO,OB,OBC)
Shape(BO,OA,OAB)
Shape(BD,DC,OBC)
Collinear(AOBD)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),35)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(BDC))
Answer: 20
Theorem sequence:
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
tangent_of_circle_property_perpendicular(2,DC,O,O)
triangle_property_angle_sum(1,COD)
Problem: As shown in the diagram, ∠OAB=20°, the center of circle O is O, the tangent to circle O is AB. Find the measure of ∠ODC.
问题：如图所示，∠OAB=20°，O是圆O的圆心，AB是圆O的切线。求∠ODC的大小。
Construction (CDL):
Shape(DO,OB,OBD)
Shape(OD,DC,CO)
Shape(CD,ODC)
Shape(BO,OE,OEB)
Shape(EO,OC,OCE)
Shape(EA,AB,OEB)
Collinear(DOEA)
Collinear(BOC)
Cocircular(O,BDCE)
Conditions (CDL):
Equal(MeasureOfAngle(OAB),20)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(MeasureOfAngle(ODC))
Answer: 55
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
triangle_property_angle_sum(1,ODC)
triangle_property_angle_sum(1,BOA)
vertical_angle(1,BOE,COD)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,ODC)
isosceles_triangle_property_angle_equal(1,ODC)
Problem: As shown in the diagram, ∠CAB=25°, the center of ⊙O is O, the tangent to ⊙O is DC. Find the measure of ∠BDC.
问题：如图所示，∠CAB=25°，圆O的圆心为O，⊙O的切线为DC。求∠BDC的大小。
Construction (CDL):
Shape(EO,OC,OCA)
Shape(CO,OE,EC)
Shape(EA,AB,BE)
Shape(CE,EB,OBC)
Shape(BA,OAB)
Shape(BD,DC,OBC)
Collinear(OEBD)
Collinear(CEA)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(BDC))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DC,O,O)
triangle_property_angle_sum(1,COD)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, ∠APC=36°, the center of ⊙O is O, PA is the tangent to circle O. Find the measure of ∠CBA.
问题：如图所示，∠APC=36°，圆O的圆心为O，PA是⊙O的切线。求∠CBA的大小。
Construction (CDL):
Shape(AP,PC,OAC)
Shape(CO,OA,OAC)
Shape(OC,CB,BO)
Shape(BC,OCB)
Shape(AO,OB,OBA)
Collinear(PCO)
Collinear(AOB)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(APC),36)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(CBA))
Answer: 27
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
triangle_property_angle_sum(1,APO)
arc_property_center_angle(1,OAC,O)
arc_property_circumference_angle_external(1,OAC,B)
Problem: As shown in the diagram, ∠EAD=50°, the center of ⊙O is O, PD is the tangent to ⊙O, the tangent to circle O is PE, △ABC is an isosceles △. Find the measure of ∠EPD.
问题：如图所示，∠EAD=50°，⊙O的圆心为O，⊙O的切线为PD，PE是⊙O的切线，三角形ABC为等腰三角形。求∠EPD的大小。
Construction (CDL):
Shape(AD,DP,PE,EA)
Shape(EP,PD,OED)
Shape(BD,ODB)
Shape(DB,BO,OD)
Shape(DO,OE,OED)
Shape(EO,OC,CE)
Shape(EC,OCE)
Shape(CO,OB,OBC)
Collinear(ADB)
Collinear(CEA)
Collinear(BOC)
Cocircular(O,BCED)
Conditions (CDL):
Equal(MeasureOfAngle(EAD),50)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PD,O)
IsTangentOfCircle(PE,O)
IsoscelesTriangle(ABC)
Goal: Value(MeasureOfAngle(EPD))
Answer: 100
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PD,O,O)
tangent_of_circle_property_perpendicular(1,PE,O,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,OE,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,ODB)
isosceles_triangle_judgment_line_equal(1,OCE)
isosceles_triangle_property_angle_equal(1,ODB)
isosceles_triangle_property_angle_equal(1,OCE)
triangle_property_angle_sum(1,DBO)
triangle_property_angle_sum(1,EOC)
triangle_property_angle_sum(1,ABC)
angle_addition(1,DOE,EOC)
adjacent_complementary_angle(1,BOD,DOC)
quadrilateral_property_angle_sum(1,PDOE)
Problem: As shown in the diagram, ∠BAO=30°, O is the center of circle O, the tangent to circle O is PA, the tangent to circle O is PB. Find the measure of ∠BPA.
问题：如图所示，∠BAO=30°，O是圆O的圆心，圆O的切线为PA，⊙O的切线为PB。求∠BPA的大小。
Construction (CDL):
Shape(BO,OA,OAB)
Shape(AO,OC,CA)
Shape(CO,OB,BC)
Shape(AC,CD,ODA)
Shape(DC,CB,OBD)
Shape(DP,PA,ODA)
Shape(BP,PD,OBD)
Collinear(ACB)
Collinear(OCDP)
Cocircular(O,ABD)
Conditions (CDL):
Equal(MeasureOfAngle(BAO),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(BPA))
Answer: 60
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OBA)
isosceles_triangle_property_angle_equal(1,OBA)
triangle_property_angle_sum(1,AOB)
quadrilateral_property_angle_sum(1,PAOB)
Problem: As shown in the diagram, ∠BSC=30°, O is the center of ⊙O, SC is the tangent to circle O. Find the measure of ∠CAB.
问题：如图所示，∠BSC=30°，圆O的圆心为O，SC是圆O的切线。求∠CAB的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(CA,AO,OC)
Shape(CO,OB,OBC)
Shape(BS,SC,OBC)
Shape(BO,OA,OAB)
Collinear(AOBS)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BSC),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(SC,O)
Goal: Value(MeasureOfAngle(CAB))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(2,SC,O,O)
triangle_property_angle_sum(1,COS)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, ∠BAC=55°, the center of circle O is O, the tangent to ⊙O is PB, the tangent to ⊙O is PC. Find the measure of ∠CPB.
问题：如图所示，∠BAC=55°，O是圆O的圆心，PB是⊙O的切线，⊙O的切线为PC。求∠CPB的大小。
Construction (CDL):
Shape(AB,OBA)
Shape(CA,OAC)
Shape(OB,BA,AC,CO)
Shape(BO,OC,OCB)
Shape(CP,PB,OCB)
Cocircular(O,BAC)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),55)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PB,O)
IsTangentOfCircle(PC,O)
Goal: Value(MeasureOfAngle(CPB))
Answer: 70
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PB,O,O)
tangent_of_circle_property_perpendicular(1,PC,O,O)
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,A)
quadrilateral_property_angle_sum(1,BOCP)
Problem: As shown in the diagram, ∠BAC=25°, O is the center of ⊙O, the tangent to circle O is PA, PB is the tangent to ⊙O. Find the measure of ∠BPA.
问题：如图所示，∠BAC=25°，圆O的圆心为O，PA是⊙O的切线，PB是圆O的切线。求∠BPA的大小。
Construction (CDL):
Shape(CO,OA,OAC)
Shape(AO,OB,BA)
Shape(BO,OC,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:
tangent_of_circle_property_perpendicular(2,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,A)
adjacent_complementary_angle(1,AOB,BOC)
quadrilateral_property_angle_sum(1,AOBP)
Problem: As shown in the diagram, ∠CAD=30°, O is the center of circle O, AC is the tangent to circle O. Find the measure of ∠OBC.
问题：如图所示，∠CAD=30°，O是圆O的圆心，AC是圆O的切线。求∠OBC的大小。
Construction (CDL):
Shape(BO,OD,ODB)
Shape(CB,OBC)
Shape(OB,BC,CO)
Shape(DO,OC,OCD)
Shape(CA,AD,OCD)
Collinear(BODA)
Cocircular(O,BCD)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AC,O)
Goal: Value(MeasureOfAngle(OBC))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AC,O,O)
triangle_property_angle_sum(1,OCA)
arc_property_center_angle(1,OCD,O)
arc_property_circumference_angle_external(1,OCD,B)
Problem: As shown in the diagram, ∠DAC=50°, the center of circle O is O, the tangent to ⊙O is CA. Find the measure of ∠ADO.
问题：如图所示，∠DAC=50°，圆O的圆心为O，圆O的切线为CA。求∠ADO的大小。
Construction (CDL):
Shape(AO,OD,ODA)
Shape(DO,OA,AD)
Shape(DA,OAD)
Shape(AC,CD,OAD)
Cocircular(O,AD)
Conditions (CDL):
Equal(MeasureOfAngle(DAC),50)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(ADO))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CA,O,O)
angle_addition(1,OAD,DAC)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,OAD)
isosceles_triangle_property_angle_equal(1,OAD)
Problem: As shown in the diagram, the radius of ⊙C is 5, the radius of ⊙D is 3, O is the center of circle C, O is the center of ⊙D. Find the area of the circle C minus the area of the ⊙D.
问题：如图所示，⊙C的半径为5，圆D的半径为3，O是⊙C的圆心，⊙D的圆心为O。求⊙C的面积减去圆D的面积。
Construction (CDL):
Shape(C)
Shape(D)
Collinear(AEOFB)
Cocircular(C,AB)
Cocircular(D,EF)
Conditions (CDL):
Equal(RadiusOfCircle(C),5)
Equal(RadiusOfCircle(D),3)
IsCentreOfCircle(O,C)
IsCentreOfCircle(O,D)
Goal: Value(Sub(AreaOfCircle(C),AreaOfCircle(D)))
Answer: 16*pi
Theorem sequence:
circle_area_formula(1,C)
circle_area_formula(1,D)
Problem: As shown in the diagram, ∠ACD=40°, O is the center of ⊙O, the tangent to circle O is CD. Find the measure of ∠DAC.
问题：如图所示，∠ACD=40°，O是圆O的圆心，CD是圆O的切线。求∠DAC的大小。
Construction (CDL):
Shape(BO,OA,OAB)
Shape(AO,OD,DA)
Shape(DO,OB,OBD)
Shape(AD,ODA)
Shape(BC,CD,OBD)
Collinear(AOBC)
Cocircular(O,ABD)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CD,O)
Goal: Value(MeasureOfAngle(DAC))
Answer: 25
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CD,O,O)
triangle_property_angle_sum(1,DOC)
arc_property_center_angle(1,OBD,O)
arc_property_circumference_angle_external(1,OBD,A)
Problem: As shown in the diagram, ∠CAB=26°, the center of ⊙O is O, DC is the tangent to circle O. Find the measure of ∠BDC.
问题：如图所示，∠CAB=26°，O是⊙O的圆心，DC是圆O的切线。求∠BDC的大小。
Construction (CDL):
Shape(BO,OA,OAB)
Shape(BD,DC,OBC)
Shape(CO,OB,OBC)
Shape(AO,OC,CA)
Shape(AC,OCA)
Collinear(DBOA)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),26)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(BDC))
Answer: 38
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DC,O,O)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
triangle_property_angle_sum(1,DCO)