SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BM=2, MC=2, OE=3, ∠BCD=90°, the center of circle O is O, OG is the perpendicular bisector of BM, CE is the tangent to circle O, the tangent to circle O is DE, quadrilateral OGCE is a rectangle. Find the length of line OC.
问题：如图所示，BM=2，MC=2，OE=3，∠BCD=90°，圆O的圆心为O，OG垂直平分AB，CE是圆O的切线，⊙O的切线为DE，OGCE是长方形。求直线OC的长度。
Construction (CDL):
Shape(BO,OA,OAB)
Shape(AO,OE,OEA)
Shape(ED,DA,OEA)
Shape(EO,OF,OFE)
Shape(FO,OM,OMF)
Shape(OB,BG,GO)
Shape(OG,GM,MO)
Shape(MG,GB,OBM)
Shape(FC,CE,OFE)
Shape(MC,CF,OMF)
Collinear(DAOB)
Collinear(DEC)
Collinear(BGMC)
Collinear(OFC)
Cocircular(O,BMFEA)
Conditions (CDL):
Equal(LengthOfLine(BM),2)
Equal(LengthOfLine(MC),2)
Equal(LengthOfLine(OE),3)
Equal(MeasureOfAngle(BCD),90)
IsCentreOfCircle(O,O)
IsPerpendicularBisectorOfLine(OG,BM)
IsTangentOfCircle(CE,O)
IsTangentOfCircle(DE,O)
Rectangle(OGCE)
Goal: Value(LengthOfLine(OC))
Answer: sqrt(17)
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OE,O)
line_addition(1,BG,GM)
line_addition(1,BG,GC)
line_addition(1,BM,MC)
right_triangle_judgment_angle(1,BGO)
right_triangle_judgment_angle(1,OGC)
right_triangle_property_pythagorean(1,BGO)
right_triangle_property_pythagorean(1,OGC)
Problem: As shown in the diagram, ∠CAB=35°, the center of circle O is O, PA is the tangent to circle O, the tangent to circle O is PB. Find the measure of ∠APB.
问题：如图所示，∠CAB=35°，⊙O的圆心为O，圆O的切线为PA，圆O的切线为PB。求∠APB的大小。
Construction (CDL):
Shape(AP,PB,OAB)
Shape(OA,AB,BO)
Shape(CO,OB,OBC)
Shape(AO,OC,OCA)
Shape(BA,OAB)
Collinear(AOC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),35)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(APB))
Answer: 70
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
tangent_of_circle_property_perpendicular(2,PB,O,O)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
adjacent_complementary_angle(1,COB,BOA)
quadrilateral_property_angle_sum(1,APBO)
Problem: As shown in the diagram, AC=6, OA=8, O is the center of ⊙O, the tangent to circle O is AC. Find the length of line CO.
问题：如图所示，AC=6，OA=8，O是⊙O的圆心，AC是⊙O的切线。求直线CO的长度。
Construction (CDL):
Shape(DO,OC,OCD)
Shape(CO,OD,ODC)
Shape(DA,AC,ODC)
Collinear(ADO)
Cocircular(O,CD)
Conditions (CDL):
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(OA),8)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AC,O)
Goal: Value(LengthOfLine(CO))
Answer: 2*sqrt(7)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AC,O,O)
right_triangle_judgment_angle(1,ACO)
right_triangle_property_pythagorean(1,ACO)
Problem: As shown in the diagram, ∠APC=36°, the center of circle O is O, the tangent to circle O is PA. Find the measure of ∠OCB.
问题：如图所示，∠APC=36°，⊙O的圆心为O，PA是⊙O的切线。求∠OCB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(OCB))
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)
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)
Problem: As shown in the diagram, PA=5, ∠PAQ=120°, O is the center of circle O, the tangent to ⊙O is AP, AQ is the tangent to ⊙O. Find the radius of ⊙O.
问题：如图所示，PA=5，∠PAQ=120°，O是圆O的圆心，⊙O的切线为AP，⊙O的切线为AQ。求⊙O的半径。
Construction (CDL):
Shape(PO,OQ,OQP)
Shape(DO,OP,OPD)
Shape(PA,AD,OPD)
Shape(QO,OD,ODQ)
Shape(DA,AQ,ODQ)
Collinear(ODA)
Cocircular(O,PDQ)
Conditions (CDL):
Equal(LengthOfLine(PA),5)
Equal(MeasureOfAngle(PAQ),120)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AP,O)
IsTangentOfCircle(AQ,O)
Goal: Value(RadiusOfCircle(O))
Answer: 5*sqrt(3)
Theorem sequence:
radius_of_circle_property_length_equal(1,OP,O)
tangent_of_circle_property_perpendicular(1,AP,O,O)
tangent_of_circle_property_perpendicular(2,AQ,O,O)
tangent_of_circle_property_length_equal(1,AP,AQ,O)
quadrilateral_property_angle_sum(1,OPAQ)
angle_addition(1,QOA,AOP)
mirror_congruent_triangle_judgment_hl(1,OPA,OAQ)
mirror_congruent_triangle_property_angle_equal(1,OPA,OAQ)
cosine_theorem(1,OPA)
cosine_theorem(1,PAO)
Problem: As shown in the diagram, ∠ACB=70°, O is the center of ⊙O, the tangent to ⊙O is PA, PB is the tangent to circle O. Find the measure of ∠BPA.
问题：如图所示，∠ACB=70°，O是⊙O的圆心，圆O的切线为PA，PB是⊙O的切线。求∠BPA的大小。
Construction (CDL):
Shape(CA,OAC)
Shape(BC,OCB)
Shape(AC,CB,BO,OA)
Shape(AO,OB,OBA)
Shape(BP,PA,OBA)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),70)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(BPA))
Answer: 40
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,OBA,O)
arc_property_circumference_angle_external(1,OBA,C)
quadrilateral_property_angle_sum(1,AOBP)
Problem: As shown in the diagram, ∠BAO=40°, the center of circle O is O, AB is the tangent to circle O. Find the measure of ∠BCO.
问题：如图所示，∠BAO=40°，O是圆O的圆心，AB是圆O的切线。求∠BCO的大小。
Construction (CDL):
Shape(BO,OC,OCB)
Shape(OB,BC,CO)
Shape(CB,OBC)
Shape(BA,AC,OBC)
Collinear(OCA)
Cocircular(O,BC)
Conditions (CDL):
Equal(MeasureOfAngle(BAO),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(MeasureOfAngle(BCO))
Answer: 65
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AB,O,O)
triangle_property_angle_sum(1,OBC)
triangle_property_angle_sum(1,OBA)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OBC)
Problem: As shown in the diagram, AB=11, the radius of circle O is 5, O is the center of ⊙O, AM is the tangent to ⊙O, the tangent to circle O is AN, BN is the tangent to circle O, the tangent to ⊙O is DE, DM is the tangent to ⊙O, quadrilateral ABCD is a square. Find the length of line DE.
问题：如图所示，AB=11，圆O的半径为5，圆O的圆心为O，AM是⊙O的切线，AN是⊙O的切线，BN是圆O的切线，圆O的切线为DE，⊙O的切线为DM，ABCD是正方形。求直线DE的长度。
Construction (CDL):
Shape(MA,AN,OMN)
Shape(NO,OM,OMN)
Shape(ED,DM,OEM)
Shape(NB,BC,CD,DE,ONE)
Shape(MO,ON,ONE,OEM)
Collinear(DMA)
Collinear(ANB)
Cocircular(O,EMN)
Conditions (CDL):
Equal(LengthOfLine(AB),11)
Equal(RadiusOfCircle(O),5)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AM,O)
IsTangentOfCircle(AN,O)
IsTangentOfCircle(BN,O)
IsTangentOfCircle(DE,O)
IsTangentOfCircle(DM,O)
Square(ABCD)
Goal: Value(LengthOfLine(DE))
Answer: 6
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AM,O,O)
tangent_of_circle_property_perpendicular(2,AN,O,O)
radius_of_circle_property_length_equal(1,OM,O)
radius_of_circle_property_length_equal(1,ON,O)
tangent_of_circle_property_length_equal(1,AM,AN,O)
tangent_of_circle_property_length_equal(1,DM,DE,O)
quadrilateral_property_angle_sum(1,ANOM)
parallelogram_judgment_angle_and_angle(1,ANOM)
kite_judgment_equal_and_equal(1,ANOM)
rhombus_judgment_parallelogram_and_kite(1,ANOM)
line_addition(1,DM,MA)
Problem: As shown in the diagram, ∠ABC=65°, O is the center of circle O, CB is the tangent to circle O. Find the measure of ∠ADB.
问题：如图所示，∠ABC=65°，O是圆O的圆心，CB是圆O的切线。求∠ADB的大小。
Construction (CDL):
Shape(AB,OBE,OEA)
Shape(AO,OB,BA)
Shape(BO,OD,DB)
Shape(BD,ODB)
Shape(DO,OA,OAD)
Shape(BC,CE,OBE)
Collinear(AOD)
Cocircular(O,ADBE)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),65)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(ADB))
Answer: 65
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CB,O,O)
angle_addition(1,OBA,ABC)
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)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,D)
triangle_property_angle_sum(1,AOB)
Problem: As shown in the diagram, CO=sqrt(3), △ABC is an equilateral △, O is the center of circle O, OF perpendicular bisects EC, BC is the tangent to ⊙O. Find the length of line CE.
问题：如图所示，CO=sqrt(3)，△ABC为等边△，O是圆O的圆心，OF是EC的垂直平分线，⊙O的切线为BC。求直线CE的长度。
Construction (CDL):
Shape(EA,AB,BC,OEC)
Shape(CF,FE,OEC)
Shape(OE,EF,FO)
Shape(OF,FC,CO)
Shape(EO,OC,OCE)
Collinear(AEFC)
Cocircular(O,EC)
Conditions (CDL):
Equal(LengthOfLine(CO),sqrt(3))
EquilateralTriangle(ABC)
IsCentreOfCircle(O,O)
IsPerpendicularBisectorOfLine(OF,EC)
IsTangentOfCircle(BC,O)
Goal: Value(LengthOfLine(CE))
Answer: 3
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BC,O,O)
angle_addition(1,BCF,FCO)
equilateral_triangle_property_angle(1,CAB)
radius_of_circle_property_length_equal(1,OE,O)
radius_of_circle_property_length_equal(1,OC,O)
cosine_theorem(1,COE)
Problem: As shown in the diagram, ∠APB=30°, the center of ⊙O is O, the tangent to circle O is PA. Find the measure of ∠BCA.
问题：如图所示，∠APB=30°，圆O的圆心为O，PA是⊙O的切线。求∠BCA的大小。
Construction (CDL):
Shape(AP,PB,OAB)
Shape(BO,OA,OAB)
Shape(AO,OC,CA)
Shape(AC,OCA)
Shape(CO,OB,OBC)
Collinear(PBOC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(APB),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
triangle_property_angle_sum(1,AOC)
triangle_property_angle_sum(1,APO)
adjacent_complementary_angle(1,POA,AOC)
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)
Problem: As shown in the diagram, DC=6, ED=4, ∠ECB=60°, the center of ⊙O is O, BC is the tangent to ⊙O, CF⊥DF, EF⊥CF. Find the area of triangle DCE.
问题：如图所示，DC=6，ED=4，∠ECB=60°，O是圆O的圆心，BC是圆O的切线，CF垂直于DF，EF⊥CF。求三角形DCE的面积。
Construction (CDL):
Shape(O)
Shape(CD,ODC)
Shape(DC,CF,FD)
Shape(EF,FC,CE)
Shape(DF,FE,OED)
Shape(EC,OCG,OGE)
Shape(CB,BG,OCG)
Collinear(DFE)
Cocircular(O,DCGE)
Conditions (CDL):
Equal(LengthOfLine(DC),6)
Equal(LengthOfLine(ED),4)
Equal(MeasureOfAngle(ECB),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BC,O)
PerpendicularBetweenLine(CF,DF)
PerpendicularBetweenLine(EF,CF)
Goal: Value(AreaOfTriangle(DCE))
Answer: 6*sqrt(3)
Theorem sequence:
circle_property_angle_of_osculation(1,OCE,B)
arc_property_circumference_angle_external(1,OCE,D)
sine_theorem(1,CFD)
altitude_of_triangle_judgment(1,CF,CED)
triangle_area_formula_common(1,CED)
Problem: As shown in the diagram, ∠ECF=70°, I is the center of circle I, CE is the tangent to circle I, the tangent to ⊙I is CF. Find the measure of ∠FDE.
问题：如图所示，∠ECF=70°，I是圆I的圆心，⊙O的切线为CE，CF是⊙O的切线。求∠FDE的大小。
Construction (CDL):
Shape(FA,AD,IFD)
Shape(IF,FD,DE,EI)
Shape(FI,IE,IEF)
Shape(ED,IDE)
Shape(EC,CF,IEF)
Shape(DB,BE,IDE)
Collinear(CFA)
Collinear(BEC)
Collinear(ADB)
Cocircular(I,FDE)
Conditions (CDL):
Equal(MeasureOfAngle(ECF),70)
IsCentreOfCircle(I,I)
IsTangentOfCircle(CE,I)
IsTangentOfCircle(CF,I)
Goal: Value(MeasureOfAngle(FDE))
Answer: 55
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CF,I,I)
tangent_of_circle_property_perpendicular(1,CE,I,I)
quadrilateral_property_angle_sum(1,FIEC)
arc_property_center_angle(1,IEF,I)
arc_property_circumference_angle_external(1,IEF,D)
Problem: As shown in the diagram, ∠ABH=50°, O is the center of circle O, HD is the tangent to ⊙O, BH is perpendicular to DH. Find the measure of ∠ABD.
问题：如图所示，∠ABH=50°，O是⊙O的圆心，HD是圆O的切线，BH⊥DH。求∠ABD的大小。
Construction (CDL):
Shape(AO,OD,ODA)
Shape(DO,OB,BD)
Shape(DB,BC,OCD)
Shape(CB,OBC)
Shape(CH,HD,OCD)
Shape(BO,OA,OAB)
Collinear(AOB)
Collinear(BCH)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(ABH),50)
IsCentreOfCircle(O,O)
IsTangentOfCircle(HD,O)
PerpendicularBetweenLine(BH,DH)
Goal: Value(MeasureOfAngle(ABD))
Answer: 25
Theorem sequence:
tangent_of_circle_property_perpendicular(2,HD,O,O)
quadrilateral_property_angle_sum(1,DOBH)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,OBD)
isosceles_triangle_property_angle_equal(1,OBD)
triangle_property_angle_sum(1,DOB)
Problem: As shown in the diagram, ∠OCB=70°, O is the center of circle O, AB is the tangent to circle O. Find the measure of ∠CAB.
问题：如图所示，∠OCB=70°，⊙O的圆心为O，AB是⊙O的切线。求∠CAB的大小。
Construction (CDL):
Shape(CO,OB,OBC)
Shape(OC,CB,BO)
Shape(BC,OCB)
Shape(CA,AB,OCB)
Collinear(ACO)
Cocircular(O,CB)
Conditions (CDL):
Equal(MeasureOfAngle(OCB),70)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(MeasureOfAngle(CAB))
Answer: 50
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
triangle_property_angle_sum(1,OCB)
triangle_property_angle_sum(1,BOA)
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)
Problem: As shown in the diagram, ∠CAP=40°, ∠PCA=100°, the center of circle O is O, PA is the tangent to ⊙O. Find the measure of arc OCB.
问题：如图所示，∠CAP=40°，∠PCA=100°，O是圆O的圆心，PA是圆O的切线。求⌒OCB的角度。
Construction (CDL):
Shape(O)
Shape(AB,OBA)
Shape(BA,AC,CB)
Shape(BC,OCB)
Shape(CA,OAC)
Shape(AP,PC,OAC)
Collinear(BCP)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(CAP),40)
Equal(MeasureOfAngle(PCA),100)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfArc(OCB))
Answer: 120
Theorem sequence:
circle_property_angle_of_osculation(1,OAC,P)
arc_property_circumference_angle_external(1,OAC,B)
arc_property_circumference_angle_external(1,OCB,A)
adjacent_complementary_angle(1,PCA,ACB)
triangle_property_angle_sum(1,BAC)
Problem: As shown in the diagram, OC=2, ∠ABC=120°, the center of ⊙O is O, the tangent to ⊙O is AB. Find the length of line BC.
问题：如图所示，OC=2，∠ABC=120°，圆O的圆心为O，圆O的切线为AB。求直线BC的长度。
Construction (CDL):
Shape(EA,AB,OEB)
Shape(EO,OC,OCE)
Shape(BO,OE,OEB)
Shape(OB,BD,DO)
Shape(OD,DC,CO)
Shape(CD,DB,OBC)
Collinear(AEOC)
Collinear(CDB)
Cocircular(O,EBC)
Conditions (CDL):
Equal(LengthOfLine(OC),2)
Equal(MeasureOfAngle(ABC),120)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(LengthOfLine(BC))
Answer: 2*sqrt(3)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
angle_addition(1,ABO,OBD)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OBC)
triangle_property_angle_sum(1,OBC)
sine_theorem(1,BCO)
Problem: As shown in the diagram, ∠ABC=25°, the center of circle O is O, the tangent to circle O is DC. Find the measure of ∠CDA.
问题：如图所示，∠ABC=25°，O是圆O的圆心，圆O的切线为DC。求∠CDA的大小。
Construction (CDL):
Shape(CD,DA,OCA)
Shape(AO,OC,OCA)
Shape(CO,OB,BC)
Shape(CB,OBC)
Shape(BO,OA,OAB)
Collinear(DAOB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(CDA))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DC,O,O)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OBC)
triangle_property_angle_sum(1,COB)
triangle_property_angle_sum(1,CDO)
adjacent_complementary_angle(1,DOC,COB)
Problem: As shown in the diagram, AB=15, BC=9, CA=12, the center of ⊙O is O, the diameter of circle O is DC, the diameter of ⊙O is FE, AD is the tangent to circle O. Find the length of line FE.
问题：如图所示，AB=15，BC=9，CA=12，⊙O的圆心为O，DC是圆O的直径，FE是⊙O的直径，⊙O的切线为AD。求直线FE的长度。
Construction (CDL):
Shape(DB,BE,ODE)
Shape(CE,OEC)
Shape(OE,EC,CO)
Shape(EO,OD,ODE)
Shape(DO,OF,OFD)
Shape(OC,CF,FO)
Shape(FC,OCF)
Shape(FA,AD,OFD)
Collinear(BDA)
Collinear(EOF)
Collinear(COD)
Collinear(CFA)
Collinear(BEC)
Cocircular(O,ECFD)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(BC),9)
Equal(LengthOfLine(CA),12)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(DC,O)
IsDiameterOfCircle(FE,O)
IsTangentOfCircle(AD,O)
Goal: Value(LengthOfLine(FE))
Answer: 36/5
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AD,O,O)
right_triangle_judgment_pythagorean_inverse(1,BCA)
mirror_similar_triangle_judgment_aa(1,CAD,BCA)
mirror_similar_triangle_property_line_ratio(1,ADC,ABC)
mirror_similar_triangle_property_line_ratio(1,DCA,CAB)
diameter_of_circle_property_length_equal(1,FE,O)
diameter_of_circle_property_length_equal(1,DC,O)
Problem: As shown in the diagram, ∠OAC=30°, O is the center of circle O, the tangent to ⊙O is PC. Find the measure of ∠CPB.
问题：如图所示，∠OAC=30°，⊙O的圆心为O，PC是圆O的切线。求∠CPB的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OA,AC,CO)
Shape(CA,OAC)
Shape(BO,OC,OCB)
Shape(CP,PB,OCB)
Collinear(AOBP)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(OAC),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PC,O)
Goal: Value(MeasureOfAngle(CPB))
Answer: 30
Theorem sequence:
arc_property_center_angle(1,OCB,O)
arc_property_circumference_angle_external(1,OCB,A)
tangent_of_circle_property_perpendicular(1,PC,O,O)
triangle_property_angle_sum(1,OCP)
Problem: As shown in the diagram, ∠APB=30°, O is the center of ⊙O, the tangent to circle O is PA, PB is the tangent to ⊙O. Find the measure of ∠ACB.
问题：如图所示，∠APB=30°，O是圆O的圆心，圆O的切线为PA，圆O的切线为PB。求∠ACB的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OA,AC,CB,BO)
Shape(CA,OAC)
Shape(BC,OCB)
Shape(AP,PB,OCB,OAC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(APB),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(ACB))
Answer: 105
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
tangent_of_circle_property_perpendicular(2,PB,O,O)
quadrilateral_property_angle_sum(1,OAPB)
round_angle(1,AOB,BOA)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,C)
Problem: As shown in the diagram, PA=5, the center of ⊙O is O, the tangent to circle O is CA, CE is the tangent to circle O, the tangent to circle O is DB, DE is the tangent to circle O, the tangent to circle O is PA, the tangent to ⊙O is PB. Find the perimeter of triangle PCD.
问题：如图所示，PA=5，O是圆O的圆心，CA是⊙O的切线，圆O的切线为CE，DB是圆O的切线，DE是圆O的切线，PA是圆O的切线，PB是⊙O的切线。求三角形PCD的周长。
Construction (CDL):
Shape(O)
Shape(EC,CA,OEA)
Shape(BD,DE,OBE)
Shape(CE,ED,DP,PC)
Collinear(ACP)
Collinear(BDP)
Collinear(CED)
Cocircular(O,ABE)
Conditions (CDL):
Equal(LengthOfLine(PA),5)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
IsTangentOfCircle(CE,O)
IsTangentOfCircle(DB,O)
IsTangentOfCircle(DE,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(PerimeterOfTriangle(PCD))
Answer: 10
Theorem sequence:
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,DB,DE,O)
line_addition(1,AC,CP)
line_addition(1,BD,DP)
line_addition(1,CE,ED)
triangle_perimeter_formula(1,PCD)
Problem: As shown in the diagram, AP=3, OA=4, O is the center of ⊙O, PA is the tangent to circle O. Find the ratio of the length of line AP to the length of line PO.
问题：如图所示，AP=3，OA=4，⊙O的圆心为O，⊙O的切线为PA。求直线AP的长度与直线PO的长度之比。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(BO,OA,OAB)
Shape(AP,PB,OAB)
Collinear(OBP)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AP),3)
Equal(LengthOfLine(OA),4)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(Div(LengthOfLine(AP),LengthOfLine(PO)))
Answer: 3/5
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
right_triangle_judgment_angle(1,OAP)
right_triangle_property_pythagorean(1,OAP)
Problem: As shown in the diagram, ∠CAO=25°, O is the center of circle O, the tangent to ⊙O is DC. Find the measure of ∠BDC.
问题：如图所示，∠CAO=25°，⊙O的圆心为O，圆O的切线为DC。求∠BDC的大小。
Construction (CDL):
Shape(BD,DC,OBC)
Shape(CO,OB,OBC)
Shape(BO,OA,OAB)
Shape(OC,CA,AO)
Shape(AC,OCA)
Collinear(DBOA)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAO),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(BDC))
Answer: 40
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)
Problem: As shown in the diagram, ∠NBM=60°, O is the center of ⊙O, AM is the tangent to ⊙O, the tangent to circle O is AN. Find the measure of ∠MAN.
问题：如图所示，∠NBM=60°，⊙O的圆心为O，AM是圆O的切线，AN是圆O的切线。求∠MAN的大小。
Construction (CDL):
Shape(MB,OBM)
Shape(BN,ONB)
Shape(NB,BM,MO,ON)
Shape(NO,OM,OMN)
Shape(MA,AN,OMN)
Cocircular(O,BMN)
Conditions (CDL):
Equal(MeasureOfAngle(NBM),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AM,O)
IsTangentOfCircle(AN,O)
Goal: Value(MeasureOfAngle(MAN))
Answer: 60
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AN,O,O)
tangent_of_circle_property_perpendicular(1,AM,O,O)
arc_property_center_angle(1,OMN,O)
arc_property_circumference_angle_external(1,OMN,B)
quadrilateral_property_angle_sum(1,NOMA)
Problem: As shown in the diagram, ∠DAE=28°, ∠PBC=26°, O is the center of circle O, BC is the tangent to circle O. Find the measure of ∠CDP.
问题：如图所示，∠DAE=28°，∠PBC=26°，圆O的圆心为O，⊙O的切线为BC。求∠CDP的大小。
Construction (CDL):
Shape(PB,BC,OPC)
Shape(PE,OEP)
Shape(EP,PO,OD,ODE)
Shape(DA,AE,ODE)
Shape(CP,OPC)
Shape(OP,PC,CO)
Shape(DO,OC,CD)
Shape(DC,OCD)
Collinear(BPEA)
Collinear(POD)
Collinear(CDA)
Cocircular(O,PCDE)
Conditions (CDL):
Equal(MeasureOfAngle(DAE),28)
Equal(MeasureOfAngle(PBC),26)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BC,O)
Goal: Value(MeasureOfAngle(CDP))
Answer: 36
Theorem sequence:
triangle_property_angle_sum(1,BCA)
tangent_of_circle_property_perpendicular(2,BC,O,O)
angle_addition(1,BCO,OCD)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OD,O)
isosceles_triangle_judgment_line_equal(1,OCD)
isosceles_triangle_property_angle_equal(1,OCD)
Problem: As shown in the diagram, ∠ACB=120°, the center of ⊙O is O, BC is the tangent to ⊙O. Find the measure of ∠BAC.
问题：如图所示，∠ACB=120°，圆O的圆心为O，BC是⊙O的切线。求∠BAC的大小。
Construction (CDL):
Shape(AO,OD,ODA)
Shape(OA,AC,CO)
Shape(DO,OC,OCD)
Shape(CA,OAC)
Shape(CB,BD,OCD)
Collinear(AODB)
Cocircular(O,ACD)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),120)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BC,O)
Goal: Value(MeasureOfAngle(BAC))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BC,O,O)
angle_addition(1,ACO,OCB)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OAC)
isosceles_triangle_property_angle_equal(1,OAC)
Problem: As shown in the diagram, ∠CDB=20°, the center of circle O is O, EC is the tangent to ⊙O. Find the measure of ∠BEC.
问题：如图所示，∠CDB=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(CA,AO,OC)
Shape(CO,OF,FC)
Shape(CF,FB,OBC)
Shape(BE,EC,OBC)
Shape(DF,FO,OA,OAD)
Shape(FD,DB,BF)
Shape(BD,ODB)
Collinear(AOFBE)
Collinear(CFD)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(MeasureOfAngle(CDB),20)
IsCentreOfCircle(O,O)
IsTangentOfCircle(EC,O)
Goal: Value(MeasureOfAngle(BEC))
Answer: 50
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,D)
triangle_property_angle_sum(1,COE)
Problem: As shown in the diagram, ∠PCA=55°, O is the center of ⊙O, PC is the tangent to ⊙O. Find the measure of ∠BAC.
问题：如图所示，∠PCA=55°，⊙O的圆心为O，PC是圆O的切线。求∠BAC的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OA,AC,CO)
Shape(OC,CB,BO)
Shape(BC,OCB)
Shape(CA,OAD,ODC)
Shape(DP,PC,ODC)
Collinear(AOB)
Cocircular(O,ADCB)
Conditions (CDL):
Equal(MeasureOfAngle(PCA),55)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PC,O)
Goal: Value(MeasureOfAngle(BAC))
Answer: 35
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PC,O,O)
angle_addition(1,PCA,ACO)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OAC)
isosceles_triangle_property_angle_equal(1,OAC)