SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠CAB=30°, the radius of circle O is 2, the center of ⊙O is O, the tangent to ⊙O is PC. Find the length of line PC.
问题：如图所示，∠CAB=30°，圆O的半径为2，圆O的圆心为O，PC是圆O的切线。求直线PC的长度。
Construction (CDL):
Shape(OCA,AC)
Shape(CA,AO,OC)
Shape(CO,OB,OBC)
Shape(OBC,BP,PC)
Shape(BO,OA,OAB)
Collinear(AOBP)
Cocircular(O,BCA)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),30)
Equal(RadiusOfCircle(O),2)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PC,O)
Goal: Value(LengthOfLine(PC))
Answer: 2*sqrt(3)
Theorem sequence:
arc_property_circumference_angle_external(1,OBC,A)
arc_property_center_angle(1,OBC,O)
tangent_of_circle_property_perpendicular(2,PC,O,O)
triangle_property_angle_sum(1,OPC)
radius_of_circle_property_length_equal(1,OC,O)
sine_theorem(1,COP)
Problem: As shown in the diagram, ∠BCO=40°, the center of circle O is O, the tangent to circle O is CB, OC is parallel to AB. Find the measure of ∠BOA.
问题：如图所示，∠BCO=40°，O是⊙O的圆心，圆O的切线为CB，OC∥AB。求∠BOA的大小。
Construction (CDL):
Shape(OM,OMA,AO)
Shape(OA,AB,BO)
Shape(OB,OBM,MO)
Shape(BA,OAB)
Shape(OBM,BC,CM)
Collinear(OMC)
Cocircular(O,ABM)
Conditions (CDL):
Equal(MeasureOfAngle(BCO),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CB,O)
ParallelBetweenLine(OC,AB)
Goal: Value(MeasureOfAngle(BOA))
Answer: 80
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)
parallel_property_alternate_interior_angle(1,OC,AB)
triangle_property_angle_sum(1,OBC)
sine_theorem(1,OAB)
triangle_property_angle_sum(1,OAB)
Problem: As shown in the diagram, ∠BAO=40°, the center of circle O is O, AB is the tangent to ⊙O. Find the measure of ∠CDB.
问题：如图所示，∠BAO=40°，O是圆O的圆心，⊙O的切线为AB。求∠CDB的大小。
Construction (CDL):
Shape(OM,MD,ODB,BO)
Shape(OB,BM,MO)
Shape(DM,MC,CD)
Shape(DC,OCD)
Shape(MB,OBC,CM)
Shape(OBC,BA,AC)
Collinear(OMCA)
Collinear(BMD)
Cocircular(O,BCD)
Conditions (CDL):
Equal(MeasureOfAngle(BAO),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(MeasureOfAngle(CDB))
Answer: 25
Theorem sequence:
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,D)
tangent_of_circle_property_perpendicular(1,AB,O,O)
triangle_property_angle_sum(1,OBA)
Problem: As shown in the diagram, ∠BCA=50°, the center of ⊙O is O, CB is the tangent to ⊙O. Find the measure of ∠DOB.
问题：如图所示，∠BCA=50°，圆O的圆心为O，CB是圆O的切线。求∠DOB的大小。
Construction (CDL):
Shape(ODA,AD)
Shape(DA,AO,OD)
Shape(DO,OB,OBD)
Shape(OBD,BC,CD)
Shape(BO,OA,OAB)
Collinear(AOB)
Collinear(ADC)
Cocircular(O,ABD)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),50)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(DOB))
Answer: -50+180*acos(-sin(2*pi/9))/pi
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CB,O,O)
triangle_property_angle_sum(1,DAO)
triangle_property_angle_sum(1,ABC)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,ODA)
isosceles_triangle_property_angle_equal(1,ODA)
adjacent_complementary_angle(1,AOD,DOB)
Problem: As shown in the diagram, ∠APB=34°, O is the center of circle O, the tangent to ⊙O is PB. Find the measure of ∠BCA.
问题：如图所示，∠APB=34°，⊙O的圆心为O，⊙O的切线为PB。求∠BCA的大小。
Construction (CDL):
Shape(OCA,AC)
Shape(CA,AO,OB,BC)
Shape(OA,AB,BO)
Shape(CB,OBC)
Shape(BA,OAB)
Shape(AP,PB,OAB)
Collinear(PAO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(APB),34)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 28
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PB,O,O)
triangle_property_angle_sum(1,OPB)
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
Problem: As shown in the diagram, OB=4, ∠OPA=30°, the center of circle O is O, the tangent to circle O is PA. Find the length of line BP.
问题：如图所示，OB=4，∠OPA=30°，O是⊙O的圆心，⊙O的切线为PA。求直线BP的长度。
Construction (CDL):
Shape(OA,OAB,BO)
Shape(AO,OB,OBA)
Shape(OBA,BP,PA)
Collinear(OBP)
Cocircular(O,BA)
Conditions (CDL):
Equal(LengthOfLine(OB),4)
Equal(MeasureOfAngle(OPA),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(LengthOfLine(BP))
Answer: 4
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OA,O)
sine_theorem(1,OPA)
line_addition(1,OB,BP)
Problem: As shown in the diagram, AD=18, the center of circle O is O, AD is the tangent to ⊙O, AE is the tangent to circle O, the tangent to ⊙O is BF, CF is the tangent to circle O. Find the perimeter of △ABC.
问题：如图所示，AD=18，O是⊙O的圆心，AD是圆O的切线，AE是圆O的切线，圆O的切线为BF，CF是圆O的切线。求△ABC的周长。
Construction (CDL):
Shape(AB,BF,FC,CA)
Shape(FB,BE,OFE)
Shape(DC,CF,ODF)
Shape(OFE,EO,OF)
Shape(ODF,FO,OD)
Shape(DO,OE,OED)
Collinear(ABE)
Collinear(ACD)
Collinear(BFC)
Cocircular(O,DFE)
Conditions (CDL):
Equal(LengthOfLine(AD),18)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AD,O)
IsTangentOfCircle(AE,O)
IsTangentOfCircle(BF,O)
IsTangentOfCircle(CF,O)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 36
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AE,O,O)
tangent_of_circle_property_perpendicular(1,AD,O,O)
tangent_of_circle_judgment_perpendicular(1,CD,O,O)
tangent_of_circle_judgment_perpendicular(1,BE,O,O)
tangent_of_circle_property_length_equal(1,AE,AD,O)
tangent_of_circle_property_length_equal(1,BF,BE,O)
tangent_of_circle_property_length_equal(1,CF,CD,O)
line_addition(1,AB,BE)
line_addition(1,AC,CD)
line_addition(1,BF,FC)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, the perimeter of triangle ACB is 24, O is the center of circle O, AD is the tangent to ⊙O, AE is the tangent to circle O, BF is the tangent to ⊙O, the tangent to ⊙O is CF. Find the length of line AD.
问题：如图所示，△ACB的周长为24，O是⊙O的圆心，⊙O的切线为AD，圆O的切线为AE，圆O的切线为BF，圆O的切线为CF。求直线AD的长度。
Construction (CDL):
Shape(AC,CF,FB,BA)
Shape(FC,CE,OFE)
Shape(BF,ODF,DB)
Shape(OFE,EO,OF)
Shape(ODF,FO,OD)
Shape(DO,OE,OED)
Collinear(ACE)
Collinear(ABD)
Collinear(CFB)
Cocircular(O,DFE)
Conditions (CDL):
Equal(PerimeterOfTriangle(ACB),24)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AD,O)
IsTangentOfCircle(AE,O)
IsTangentOfCircle(BF,O)
IsTangentOfCircle(CF,O)
Goal: Value(LengthOfLine(AD))
Answer: 12
Theorem sequence:
triangle_perimeter_formula(1,ACB)
tangent_of_circle_property_perpendicular(1,AE,O,O)
tangent_of_circle_property_perpendicular(1,AD,O,O)
tangent_of_circle_judgment_perpendicular(1,CE,O,O)
tangent_of_circle_judgment_perpendicular(1,BD,O,O)
tangent_of_circle_property_length_equal(1,AD,AE,O)
tangent_of_circle_property_length_equal(1,CF,CE,O)
tangent_of_circle_property_length_equal(1,BF,BD,O)
line_addition(1,AC,CE)
line_addition(1,AB,BD)
line_addition(1,CF,FB)
Problem: As shown in the diagram, ∠ADC=122°, O is the center of ⊙O, the tangent to ⊙O is CD. Find the measure of ∠DCO.
问题：如图所示，∠ADC=122°，O是圆O的圆心，圆O的切线为CD。求∠DCO的大小。
Construction (CDL):
Shape(DC,CB,ODB)
Shape(ODB,BO,OD)
Shape(DO,OA,AD)
Shape(DA,OAD)
Shape(AO,OB,OBA)
Collinear(CBOA)
Cocircular(O,ADB)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),122)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CD,O)
Goal: Value(MeasureOfAngle(DCO))
Answer: 26
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
angle_addition(1,ADO,ODC)
tangent_of_circle_property_perpendicular(1,CD,O,O)
sine_theorem(1,OAD)
adjacent_complementary_angle(1,BOD,DOA)
triangle_property_angle_sum(1,DOA)
triangle_property_angle_sum(1,DCO)
Problem: As shown in the diagram, ∠ACD=55°, O is the center of circle O, DC is the tangent to circle O, CD is perpendicular to AD. Find the measure of ∠CAO.
问题：如图所示，∠ACD=55°，O是圆O的圆心，⊙O的切线为DC，CD垂直于AD。求∠CAO的大小。
Construction (CDL):
Shape(CD,DM,OCM)
Shape(OCM,MA,AC)
Shape(OMA,AM)
Shape(OC,CA,AO)
Shape(CO,OB,OBC)
Shape(BO,OA,OAB)
Collinear(AOB)
Collinear(AMD)
Cocircular(O,BCMA)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),55)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
PerpendicularBetweenLine(CD,AD)
Goal: Value(MeasureOfAngle(CAO))
Answer: 35
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DC,O,O)
angle_addition(1,OCA,ACD)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OA,O)
sine_theorem(1,OCA)
Problem: As shown in the diagram, AC=5, BD=3, the center of circle O is O, AC is the tangent to ⊙O, AP is the tangent to ⊙O, the tangent to ⊙O is BD, BP is the tangent to ⊙O. Find the length of line AB.
问题：如图所示，AC=5，BD=3，O是⊙O的圆心，AC是⊙O的切线，圆O的切线为AP，BD是⊙O的切线，圆O的切线为BP。求直线AB的长度。
Construction (CDL):
Shape(CA,AP,OCP)
Shape(PB,BD,OPD)
Shape(OCP,OPD,ODC)
Shape(O)
Collinear(APB)
Cocircular(O,CPD)
Conditions (CDL):
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BD),3)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AC,O)
IsTangentOfCircle(AP,O)
IsTangentOfCircle(BD,O)
IsTangentOfCircle(BP,O)
Goal: Value(LengthOfLine(AB))
Answer: 8
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, PA=4, O is the center of ⊙O, EC is the tangent to circle O, FC is the tangent to circle O, the tangent to circle O is PA, the tangent to circle O is PB. Find the perimeter of triangle PFE.
问题：如图所示，PA=4，O是圆O的圆心，EC是圆O的切线，⊙O的切线为FC，PA是圆O的切线，PB是⊙O的切线。求三角形PFE的周长。
Construction (CDL):
Shape(PF,FC,CE,EP)
Shape(CF,FB,OCB)
Shape(AE,EC,OAC)
Shape(OAC,OCB,BO,OA)
Shape(AO,OB,OBA)
Collinear(PEA)
Collinear(PFB)
Collinear(ECF)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(PA),4)
IsCentreOfCircle(O,O)
IsTangentOfCircle(EC,O)
IsTangentOfCircle(FC,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(PerimeterOfTriangle(PFE))
Answer: 8
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
tangent_of_circle_judgment_perpendicular(1,EA,O,O)
tangent_of_circle_judgment_perpendicular(1,FB,O,O)
tangent_of_circle_property_length_equal(1,PA,PB,O)
tangent_of_circle_property_length_equal(1,EC,EA,O)
tangent_of_circle_property_length_equal(1,FC,FB,O)
line_addition(1,PE,EA)
line_addition(1,PF,FB)
line_addition(1,EC,CF)
triangle_perimeter_formula(1,PFE)
Problem: As shown in the diagram, ∠BCD=120°, O is the center of ⊙O, the tangent to circle O is PD. Find the measure of ∠ADP.
问题：如图所示，∠BCD=120°，⊙O的圆心为O，PD是⊙O的切线。求∠ADP的大小。
Construction (CDL):
Shape(DP,PA,ODA)
Shape(ODA,AD)
Shape(OD,DA,AO)
Shape(OCD,DC)
Shape(BC,CD,DO,OB)
Shape(OBC,CB)
Shape(BO,OA,OAB)
Collinear(PAOB)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),120)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PD,O)
Goal: Value(MeasureOfAngle(ADP))
Answer: 30
Theorem sequence:
round_arc(1,ODB,OBD)
arc_property_circumference_angle_external(1,ODB,C)
arc_property_circumference_angle_external(1,OBD,A)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
sine_theorem(1,ODA)
tangent_of_circle_property_perpendicular(1,PD,O,O)
angle_addition(1,ODA,ADP)
Problem: As shown in the diagram, ∠ABC=53°, the center of circle O is O, DC is the tangent to ⊙O. Find the measure of ∠ADC.
问题：如图所示，∠ABC=53°，圆O的圆心为O，DC是⊙O的切线。求∠ADC的大小。
Construction (CDL):
Shape(OCA,AC)
Shape(CA,AO,OC)
Shape(CO,OB,BC)
Shape(CB,OBC)
Shape(OBC,BD,DC)
Shape(BO,OA,OAB)
Collinear(AOBD)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),53)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(ADC))
Answer: 16
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
sine_theorem(1,OBC)
triangle_property_angle_sum(1,OBC)
tangent_of_circle_property_perpendicular(2,DC,O,O)
triangle_property_angle_sum(1,ODC)
Problem: As shown in the diagram, ∠CDO=40°, the center of circle O is O, the tangent to ⊙O is DC. Find the measure of ∠OAC.
问题：如图所示，∠CDO=40°，O是⊙O的圆心，DC是⊙O的切线。求∠OAC的大小。
Construction (CDL):
Shape(OAC,CA)
Shape(OA,AC,CO)
Shape(OC,OCB,BO)
Shape(AO,OB,OBA)
Shape(OCB,CD,DB)
Collinear(AOBD)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(CDO),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(OAC))
Answer: 25
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DC,O,O)
triangle_property_angle_sum(1,OCD)
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,A)
Problem: As shown in the diagram, ∠ABM=140°, the center of circle O is O, MB is the tangent to ⊙O. Find the measure of ∠BCA.
问题：如图所示，∠ABM=140°，圆O的圆心为O，MB是⊙O的切线。求∠BCA的大小。
Construction (CDL):
Shape(CB,BM)
Shape(OBC,CB)
Shape(AO,OB,BC,CA)
Shape(AB,BO,OA)
Shape(OAB,BA)
Shape(AC,OCA)
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)
sine_theorem(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, PA=12, the center of circle O is O, CE is the tangent to circle O, the tangent to circle O is DE, PA is the tangent to circle O, PB is the tangent to ⊙O. Find the perimeter of triangle PCD.
问题：如图所示，PA=12，O是圆O的圆心，圆O的切线为CE，⊙O的切线为DE，PA是圆O的切线，圆O的切线为PB。求△PCD的周长。
Construction (CDL):
Shape(PC,CE,ED,DP)
Shape(EC,CA,OEA)
Shape(BD,DE,OBE)
Shape(OAB,BO,OA)
Shape(OB,OBE,OEA,AO)
Collinear(PCA)
Collinear(PDB)
Collinear(CED)
Cocircular(O,BEA)
Conditions (CDL):
Equal(LengthOfLine(PA),12)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CE,O)
IsTangentOfCircle(DE,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(PerimeterOfTriangle(PCD))
Answer: 24
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
tangent_of_circle_judgment_perpendicular(1,DB,O,O)
tangent_of_circle_judgment_perpendicular(1,CA,O,O)
tangent_of_circle_property_length_equal(1,PA,PB,O)
tangent_of_circle_property_length_equal(1,CA,CE,O)
tangent_of_circle_property_length_equal(1,DE,DB,O)
line_addition(1,PC,CA)
line_addition(1,PD,DB)
line_addition(1,CE,ED)
triangle_perimeter_formula(1,PCD)
Problem: As shown in the diagram, ∠CAD=35°, O is the center of circle O, DC is the tangent to ⊙O. Find the measure of ∠ADC.
问题：如图所示，∠CAD=35°，O是⊙O的圆心，圆O的切线为DC。求∠ADC的大小。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(AO,OC,CA)
Shape(OB,OBC,CO)
Shape(AC,OCA)
Shape(OBC,BD,DC)
Collinear(AOBD)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),35)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(ADC))
Answer: 20
Theorem sequence:
arc_property_circumference_angle_external(1,OBC,A)
arc_property_center_angle(1,OBC,O)
tangent_of_circle_property_perpendicular(2,DC,O,O)
triangle_property_angle_sum(1,ODC)
Problem: As shown in the diagram, ∠CAD=25°, O is the center of circle O, DC is the tangent to circle O. Find the measure of ∠ADC.
问题：如图所示，∠CAD=25°，圆O的圆心为O，DC是圆O的切线。求∠ADC的大小。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(AO,OC,CA)
Shape(OB,OBC,CO)
Shape(AC,OCA)
Shape(OBC,BD,DC)
Collinear(AOBD)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(ADC))
Answer: 40
Theorem sequence:
arc_property_circumference_angle_external(1,OBC,A)
arc_property_center_angle(1,OBC,O)
tangent_of_circle_property_perpendicular(2,DC,O,O)
triangle_property_angle_sum(1,ODC)
Problem: As shown in the diagram, PA=6, ∠APB=60°, ∠APO=∠OPB, ∠PCA=90°, △PBA is an equilateral △, O is the center of ⊙O, the tangent to circle O is PA, PB is the tangent to ⊙O. Find the length of line OC.
问题：如图所示，PA=6，∠APB=60°，∠APO=∠OPB，∠PCA=90°，△PBA为等边△，O是圆O的圆心，PA是圆O的切线，PB是圆O的切线。求直线OC的长度。
Construction (CDL):
Shape(OA,AC,CO)
Shape(OC,CB,BO)
Shape(OAD,DC,CA)
Shape(CD,ODB,BC)
Shape(OAD,AP,PD)
Shape(DP,PB,ODB)
Shape(OB,OBA,AO)
Collinear(PDCO)
Collinear(ACB)
Cocircular(O,ADB)
Conditions (CDL):
Equal(LengthOfLine(PA),6)
Equal(MeasureOfAngle(APB),60)
Equal(MeasureOfAngle(APO),MeasureOfAngle(OPB))
Equal(MeasureOfAngle(PCA),90)
EquilateralTriangle(PBA)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(LengthOfLine(OC))
Answer: sqrt(3)
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
angle_addition(1,APO,OPB)
triangle_property_angle_sum(1,OAC)
triangle_property_angle_sum(1,OAP)
sine_theorem(1,ACO)
sine_theorem(1,COA)
sine_theorem(1,APC)
sine_theorem(1,APO)
Problem: As shown in the diagram, ∠DBA=24°, the center of ⊙O is O, BA is the diameter of circle O, the tangent to circle O is CA. Find the measure of ∠ACO.
问题：如图所示，∠DBA=24°，⊙O的圆心为O，圆O的直径为BA，CA是圆O的切线。求∠ACO的大小。
Construction (CDL):
Shape(BO,OD,DB)
Shape(DO,OA,OAD)
Shape(OAD,AC,CD)
Collinear(BOA)
Collinear(ODC)
Cocircular(O,BAD)
Conditions (CDL):
Equal(MeasureOfAngle(DBA),24)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(BA,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(ACO))
Answer: 42
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CA,O,O)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,ODB)
isosceles_triangle_property_angle_equal(1,ODB)
triangle_property_angle_sum(1,BOD)
triangle_property_angle_sum(1,OAC)
flat_angle(1,BOA)
angle_addition(1,BOD,DOA)
Problem: As shown in the diagram, ∠CAB=30°, O is the center of circle O, the diameter of ⊙O is BA, the tangent to ⊙O is DC. Find the measure of ∠ADC.
问题：如图所示，∠CAB=30°，⊙O的圆心为O，圆O的直径为BA，DC是圆O的切线。求∠ADC的大小。
Construction (CDL):
Shape(OCA,AC)
Shape(OC,CA,AO)
Shape(CO,OB,OBC)
Shape(BD,DC,OBC)
Collinear(AOBD)
Cocircular(O,CAB)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(BA,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(ADC))
Answer: 30
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,COD)
Problem: As shown in the diagram, ∠ABO=38°, O is the center of circle O, BA is the tangent to ⊙O. Find the measure of ∠CDA.
问题：如图所示，∠ABO=38°，O是⊙O的圆心，⊙O的切线为BA。求∠CDA的大小。
Construction (CDL):
Shape(OA,OAC,CO)
Shape(OAC,AB,BC)
Shape(DA,AO,OC,CD)
Collinear(OCB)
Cocircular(O,ACD)
Conditions (CDL):
Equal(MeasureOfAngle(ABO),38)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BA,O)
Goal: Value(MeasureOfAngle(CDA))
Answer: 26
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BA,O,O)
triangle_property_angle_sum(1,OAB)
arc_property_center_angle(1,OAC,O)
arc_property_circumference_angle_external(1,OAC,D)
Problem: As shown in the diagram, ∠BPA=50°, the center of circle O is O, PA is the tangent to ⊙O, the tangent to circle O is PB. Find the measure of ∠ACB.
问题：如图所示，∠BPA=50°，⊙O的圆心为O，⊙O的切线为PA，PB是圆O的切线。求∠ACB的大小。
Construction (CDL):
Shape(BP,PA,OBA)
Shape(AO,OB,OBA)
Shape(OA,AC,CB,BO)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),50)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(ACB))
Answer: 65
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, QP=2*sqrt(2), quadrilateral SRQP is a square. Find the area of SRQP.
问题：如图所示，QP=2*sqrt(2)，SRQP是正方形。求SRQP的面积。
Construction (CDL):
Shape(SR,RQ,QP,PS)
Conditions (CDL):
Equal(LengthOfLine(QP),2*sqrt(2))
Square(SRQP)
Goal: Value(AreaOfQuadrilateral(SRQP))
Answer: 8
Theorem sequence:
parallelogram_area_formula_sine(1,PSRQ)
Problem: As shown in the diagram, ∠CAB=25°, O is the center of ⊙O, the diameter of ⊙O is AB, EC is the tangent to ⊙O. Find the measure of ∠AEC.
问题：如图所示，∠CAB=25°，O是⊙O的圆心，AB是圆O的直径，圆O的切线为EC。求∠AEC的大小。
Construction (CDL):
Shape(CA,AO,OC)
Shape(CO,OB,OBC)
Shape(OCA,AC)
Shape(OBC,BE,EC)
Collinear(AOBE)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),25)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(EC,O)
Goal: Value(MeasureOfAngle(AEC))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(2,EC,O,O)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
triangle_property_angle_sum(1,COE)
Problem: As shown in the diagram, CD=3, ∠DAB=30°, the center of circle O is O, AB is the diameter of circle O, the tangent to circle O is CD. Find the length of line AB.
问题：如图所示，CD=3，∠DAB=30°，圆O的圆心为O，圆O的直径为AB，CD是⊙O的切线。求直线AB的长度。
Construction (CDL):
Shape(OD,DA,AO)
Shape(DO,OB,BD)
Shape(OBD,DB)
Shape(ODA,AD)
Shape(BC,CD,OBD)
Collinear(AOBC)
Cocircular(O,ABD)
Conditions (CDL):
Equal(LengthOfLine(CD),3)
Equal(MeasureOfAngle(DAB),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(CD,O)
Goal: Value(LengthOfLine(AB))
Answer: 2*sqrt(3)
Theorem sequence:
arc_property_center_angle(1,ODA,O)
arc_property_center_angle(1,OBD,O)
arc_property_circumference_angle_external(1,OBD,A)
arc_property_circumference_angle_external(1,ODA,B)
tangent_of_circle_property_perpendicular(2,CD,O,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,ODA)
isosceles_triangle_property_angle_equal(1,ODA)
triangle_property_angle_sum(1,ODA)
triangle_property_angle_sum(1,DAB)
triangle_property_angle_sum(1,DOC)
sine_theorem(1,ABD)
sine_theorem(1,DAC)
Problem: As shown in the diagram, OP=10, PA=8, O is the center of ⊙O, the tangent to circle O is PA. Find the length of line OA.
问题：如图所示，OP=10，PA=8，O是圆O的圆心，PA是⊙O的切线。求直线OA的长度。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OBA,BP,PA)
Collinear(OBP)
Cocircular(O,BA)
Conditions (CDL):
Equal(LengthOfLine(OP),10)
Equal(LengthOfLine(PA),8)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(LengthOfLine(OA))
Answer: 6
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
right_triangle_judgment_angle(1,PAO)
right_triangle_property_pythagorean(1,PAO)
Problem: As shown in the diagram, the radius of circle O is 8, the radius of ⊙P is sqrt(39), O is the center of ⊙O, P is the center of circle P. Find the area of the ⊙O minus the area of the circle P.
问题：如图所示，圆O的半径为8，圆P的半径为sqrt(39)，圆O的圆心为O，⊙P的圆心为P。求⊙O的面积减去⊙P的面积。
Construction (CDL):
Shape(OAB,OBA)
Shape(PCD,PDC)
Cocircular(O,AB)
Cocircular(P,CD)
Conditions (CDL):
Equal(RadiusOfCircle(O),8)
Equal(RadiusOfCircle(P),sqrt(39))
IsCentreOfCircle(O,O)
IsCentreOfCircle(P,P)
Goal: Value(Sub(AreaOfCircle(O),AreaOfCircle(P)))
Answer: 25*pi
Theorem sequence:
circle_area_formula(1,O)
circle_area_formula(1,P)
Problem: As shown in the diagram, ∠DOB=40°, O is the center of ⊙O, CB is the tangent to circle O. Find the measure of ∠BCO.
问题：如图所示，∠DOB=40°，⊙O的圆心为O，CB是⊙O的切线。求∠BCO的大小。
Construction (CDL):
Shape(OB,OBD,DO)
Shape(OBD,BC,CD)
Collinear(ODC)
Cocircular(O,BD)
Conditions (CDL):
Equal(MeasureOfAngle(DOB),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(BCO))
Answer: 50
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CB,O,O)
triangle_property_angle_sum(1,OBC)
Problem: As shown in the diagram, BO=2, ∠ECD=30°, the center of circle O is O, BOCE is a square. Find the length of line CD.
问题：如图所示，BO=2，∠ECD=30°，⊙O的圆心为O，四边形BOCE是正方形。求直线CD的长度。
Construction (CDL):
Shape(OC,CB,BO)
Shape(OCB,BC)
Shape(OCB,CE,EB)
Shape(EC,CD,DE)
Collinear(BED)
Cocircular(O,CB)
Conditions (CDL):
Equal(LengthOfLine(BO),2)
Equal(MeasureOfAngle(ECD),30)
IsCentreOfCircle(O,O)
Square(BOCE)
Goal: Value(LengthOfLine(CD))
Answer: 4*sqrt(3)/3
Theorem sequence:
triangle_property_angle_sum(1,ECD)
flat_angle(1,DEB)
angle_addition(1,DEC,CEB)
sine_theorem(1,CDE)