SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BC=2, ∠CAD=30°, the center of ⊙O is O. Find the length of line OB.
问题：如图所示，BC=2，∠CAD=30°，圆O的圆心为O。求直线OB的长度。
Construction (CDL):
Shape(BO,OD,DA,OAB)
Shape(OB,BD,DO)
Shape(AD,DC,CA)
Shape(DB,BC,CD)
Shape(AC,OCA)
Shape(CB,OBC)
Collinear(ODC)
Collinear(BDA)
Cocircular(O,BCA)
Conditions (CDL):
Equal(LengthOfLine(BC),2)
Equal(MeasureOfAngle(CAD),30)
IsCentreOfCircle(O,O)
Goal: Value(LengthOfLine(OB))
Answer: 2
Theorem sequence:
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)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
isosceles_triangle_judgment_angle_equal(1,BCO)
Problem: As shown in the diagram, ∠EDA=55°, the center of ⊙O is O, AB is the diameter of circle O. Find the measure of ∠EAC.
问题：如图所示，∠EDA=55°，O是⊙O的圆心，圆O的直径为AB。求∠EAC的大小。
Construction (CDL):
Shape(AD,ODA)
Shape(DA,AE,ED)
Shape(DE,EO,OB,OBD)
Shape(BO,OE,EC,CB)
Shape(EA,AC,CE)
Shape(CA,OAC)
Shape(BC,OCB)
Collinear(AEOB)
Collinear(DEC)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(MeasureOfAngle(EDA),55)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
Goal: Value(MeasureOfAngle(EAC))
Answer: 35
Theorem sequence:
arc_property_circumference_angle_external(1,OAC,B)
arc_property_circumference_angle_external(1,OAC,D)
diameter_of_circle_property_right_angle(1,ACB,O)
triangle_property_angle_sum(1,CBA)
Problem: As shown in the diagram, DO=AD, ∠DAF=22°, O is the center of ⊙O. Find the measure of ∠GFO.
问题：如图所示，DO=AD，∠DAF=22°，O是⊙O的圆心。求∠GFO的大小。
Construction (CDL):
Shape(DA,AF,ODF)
Shape(FG,GD,ODF)
Shape(GF,FO,OG)
Shape(ED,DG,GE)
Shape(DE,OED)
Shape(EG,GO,OB,OBE)
Shape(BO,OF,OFB)
Collinear(DGO)
Collinear(EGF)
Collinear(ADE)
Collinear(AFOB)
Cocircular(O,FGED)
Conditions (CDL):
Equal(LengthOfLine(DO),LengthOfLine(AD))
Equal(MeasureOfAngle(DAF),22)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(GFO))
Answer: 33
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DAO)
isosceles_triangle_property_angle_equal(1,DAO)
arc_property_center_angle(1,ODF,O)
arc_property_circumference_angle_external(1,ODF,E)
triangle_property_angle_sum(1,EAF)
adjacent_complementary_angle(1,AFG,GFO)
Problem: As shown in the diagram, ∠OCA=45°, O is the center of circle O. Find the measure of ∠ABC.
问题：如图所示，∠OCA=45°，圆O的圆心为O。求∠ABC的大小。
Construction (CDL):
Shape(NC,CO,OA,AB)
Shape(OC,CA,AO)
Shape(CB,OBC)
Shape(BA,OAB)
Shape(AC,OCA)
Cocircular(O,CAB)
Conditions (CDL):
Equal(MeasureOfAngle(OCA),45)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(ABC))
Answer: 45
Theorem sequence:
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OA,O)
isosceles_triangle_judgment_line_equal(1,OCA)
isosceles_triangle_property_angle_equal(1,OCA)
triangle_property_angle_sum(1,OCA)
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_external(1,OCA,B)
Problem: As shown in the diagram, ∠OAE=50°, O is the center of circle O, the diameter of circle O is AB, DC is the tangent to ⊙O, ED is perpendicular to CD. Find the measure of ∠DCB.
问题：如图所示，∠OAE=50°，O是圆O的圆心，AB是圆O的直径，⊙O的切线为DC，ED⊥CD。求∠DCB的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(BO,OC,CB)
Shape(BC,OCB)
Shape(CO,OA,AE,OEC)
Shape(ED,DC,OEC)
Shape(EA,OAE)
Collinear(AED)
Collinear(AOB)
Cocircular(O,AECB)
Conditions (CDL):
Equal(MeasureOfAngle(OAE),50)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(DC,O)
PerpendicularBetweenLine(ED,CD)
Goal: Value(MeasureOfAngle(DCB))
Answer: 155
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DC,O,O)
parallel_judgment_ipsilateral_internal_angle(1,DA,CO)
parallel_property_corresponding_angle(1,OC,AD,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)
triangle_property_angle_sum(1,OCB)
angle_addition(1,DCO,OCB)
Problem: As shown in the diagram, AB=2, AC=BC, OC=1, O is the center of circle O, PA is the tangent to circle O, PB is the tangent to ⊙O, OC is perpendicular to BC. Find the measure of ∠APB.
问题：如图所示，AB=2，AC=BC，OC=1，⊙O的圆心为O，⊙O的切线为PA，⊙O的切线为PB，OC⊥BC。求∠APB的大小。
Construction (CDL):
Shape(OA,AC,CO)
Shape(OC,CB,BO)
Shape(AO,OB,OBA)
Shape(BC,CA,OAB)
Shape(AP,PB,OAB)
Collinear(ACB)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(AC),LengthOfLine(BC))
Equal(LengthOfLine(OC),1)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
PerpendicularBetweenLine(OC,BC)
Goal: Value(MeasureOfAngle(APB))
Answer: 90
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
tangent_of_circle_property_perpendicular(2,PB,O,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
line_addition(1,AC,CB)
adjacent_complementary_angle(1,ACO,OCB)
isosceles_triangle_judgment_line_equal(1,COA)
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,COA)
isosceles_triangle_property_angle_equal(1,OAB)
triangle_property_angle_sum(1,OAC)
triangle_property_angle_sum(1,OAB)
quadrilateral_property_angle_sum(1,OAPB)
Problem: As shown in the diagram, ∠ABO=55°, O is the center of ⊙O, TA is the tangent to circle O. Find the measure of ∠BAT.
问题：如图所示，∠ABO=55°，O是圆O的圆心，TA是⊙O的切线。求∠BAT的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OA,AB,BO)
Shape(BA,OAB)
Shape(TA,OAB)
Cocircular(O,AB)
Conditions (CDL):
Equal(MeasureOfAngle(ABO),55)
IsCentreOfCircle(O,O)
IsTangentOfCircle(TA,O)
Goal: Value(MeasureOfAngle(BAT))
Answer: 35
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OAB)
tangent_of_circle_property_perpendicular(1,TA,O,O)
angle_addition(1,OAB,BAT)
Problem: As shown in the diagram, OA=3, OP=6, O is the center of circle O, PA is the tangent to circle O. Find the measure of ∠APB.
问题：如图所示，OA=3，OP=6，⊙O的圆心为O，圆O的切线为PA。求∠APB的大小。
Construction (CDL):
Shape(OA,AB,BO)
Shape(BA,OAB)
Shape(BA,AP)
Shape(AO,OB,OBA)
Shape(AP,PB,OAB)
Collinear(OBP)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(OA),3)
Equal(LengthOfLine(OP),6)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(APB))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
right_triangle_judgment_angle(1,OAP)
right_triangle_property_pythagorean(1,OAP)
cosine_theorem(1,POA)
Problem: As shown in the diagram, OA=2, OC=2, ∠CAO=30°, O is the center of circle O, DC is the tangent to circle O. Find the length of line CD.
问题：如图所示，OA=2，OC=2，∠CAO=30°，O是圆O的圆心，⊙O的切线为DC。求直线CD的长度。
Construction (CDL):
Shape(AC,OCA)
Shape(CA,AO,OC)
Shape(CO,OB,OBC)
Shape(BD,DC,OBC)
Shape(BO,OA,OAB)
Collinear(AOBD)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(OA),2)
Equal(LengthOfLine(OC),2)
Equal(MeasureOfAngle(CAO),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DC,O)
Goal: Value(LengthOfLine(CD))
Answer: 2*sqrt(3)
Theorem sequence:
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
triangle_property_angle_sum(1,COD)
tangent_of_circle_property_perpendicular(2,DC,O,O)
sine_theorem(1,COD)
Problem: As shown in the diagram, AB=4, OA=5, the center of ⊙O is O, AB is the tangent to circle O. Find the length of line OB.
问题：如图所示，AB=4，OA=5，圆O的圆心为O，⊙O的切线为AB。求直线OB的长度。
Construction (CDL):
Shape(BO,OC,OCB)
Shape(CO,OB,OBC)
Shape(CA,AB,OCB)
Collinear(ACO)
Cocircular(O,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(OA),5)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(LengthOfLine(OB))
Answer: 3
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
right_triangle_judgment_angle(1,ABO)
right_triangle_property_pythagorean(1,ABO)
Problem: As shown in the diagram, AB=2, AC=1, ∠BDA=90°, ∠CAB=90°. Find the length of line AD.
问题：如图所示，AB=2，AC=1，∠BDA=90°，∠CAB=90°。求直线AD的长度。
Construction (CDL):
Shape(CA,AD,DC)
Shape(DA,AB,BD)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(AC),1)
Equal(MeasureOfAngle(BDA),90)
Equal(MeasureOfAngle(CAB),90)
Goal: Value(LengthOfLine(AD))
Answer: 2*sqrt(5)/5
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
cosine_theorem(1,BCA)
sine_theorem(1,ABD)
Problem: As shown in the diagram, OB=4, OC=4, ∠BOC=60°, the center of ⊙O is O, AC is the tangent to ⊙O. Find the length of line AO.
问题：如图所示，OB=4，OC=4，∠BOC=60°，⊙O的圆心为O，AC是⊙O的切线。求直线AO的长度。
Construction (CDL):
Shape(BO,OC,OCB)
Shape(CO,OB,OBC)
Shape(CA,AB,OCB)
Collinear(ABO)
Cocircular(O,BC)
Conditions (CDL):
Equal(LengthOfLine(OB),4)
Equal(LengthOfLine(OC),4)
Equal(MeasureOfAngle(BOC),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AC,O)
Goal: Value(LengthOfLine(AO))
Answer: 8
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AC,O,O)
triangle_property_angle_sum(1,OCA)
sine_theorem(1,OCA)
Problem: As shown in the diagram, ∠DAO=35°, O is the center of circle O, the diameter of circle O is AB, CD is the tangent to ⊙O. Find the measure of ∠BCD.
问题：如图所示，∠DAO=35°，O是圆O的圆心，⊙O的直径为AB，CD是圆O的切线。求∠BCD的大小。
Construction (CDL):
Shape(AD,ODA)
Shape(DA,AO,OD)
Shape(DO,OB,OBD)
Shape(BO,OA,OAB)
Shape(BC,CD,OBD)
Collinear(AOBC)
Cocircular(O,ABD)
Conditions (CDL):
Equal(MeasureOfAngle(DAO),35)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(CD,O)
Goal: Value(MeasureOfAngle(BCD))
Answer: 20
Theorem sequence:
arc_property_center_angle(1,OBD,O)
arc_property_circumference_angle_external(1,OBD,A)
tangent_of_circle_property_perpendicular(2,CD,O,O)
triangle_property_angle_sum(1,DOC)
Problem: As shown in the diagram, ∠BAP=40°, the center of circle O is O, the tangent to ⊙O is AP. Find the measure of ∠APB.
问题：如图所示，∠BAP=40°，O是圆O的圆心，圆O的切线为AP。求∠APB的大小。
Construction (CDL):
Shape(BA,AP,OBP)
Shape(BP,PO,OB)
Shape(PB,OBP)
Shape(BO,OP,OPB)
Collinear(OBA)
Cocircular(O,BP)
Conditions (CDL):
Equal(MeasureOfAngle(BAP),40)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AP,O)
Goal: Value(MeasureOfAngle(APB))
Answer: 25
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OP,O)
tangent_of_circle_property_perpendicular(2,AP,O,O)
triangle_property_angle_sum(1,POA)
isosceles_triangle_judgment_line_equal(1,OBP)
isosceles_triangle_property_angle_equal(1,OBP)
triangle_property_angle_sum(1,POB)
angle_addition(1,APB,BPO)
Problem: As shown in the diagram, AB=4, AO=6, O is the center of circle O, AB is the tangent to circle O. Find the length of line OB.
问题：如图所示，AB=4，AO=6，O是⊙O的圆心，⊙O的切线为AB。求直线OB的长度。
Construction (CDL):
Shape(BO,OC,OCB)
Shape(CO,OB,OBC)
Shape(CA,AB,OCB)
Collinear(ACO)
Cocircular(O,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AO),6)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
Goal: Value(LengthOfLine(OB))
Answer: 2*sqrt(5)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,O,O)
right_triangle_judgment_angle(1,ABO)
right_triangle_property_pythagorean(1,ABO)
Problem: As shown in the diagram, AB=8, AH=HB, OA=5, O is the center of ⊙O, AH is perpendicular to OH. Find the length of line HC.
问题：如图所示，AB=8，AH=HB，OA=5，O是圆O的圆心，AH⊥OH。求直线HC的长度。
Construction (CDL):
Shape(OA,AH,HO)
Shape(CH,HA,OAH)
Shape(BH,HC,OCB)
Shape(AO,OH,HB,OBA)
Collinear(OHC)
Collinear(AHB)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(AH),LengthOfLine(HB))
Equal(LengthOfLine(OA),5)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AH,OH)
Goal: Value(LengthOfLine(HC))
Answer: 2
Theorem sequence:
line_addition(1,AH,HB)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
right_triangle_judgment_angle(1,AHO)
right_triangle_property_pythagorean(1,AHO)
line_addition(1,OH,HC)
Problem: As shown in the diagram, ∠CAO=35°, the center of ⊙O is O, the tangent to ⊙O is PC. Find the measure of ∠BPC.
问题：如图所示，∠CAO=35°，O是圆O的圆心，⊙O的切线为PC。求∠BPC的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(CA,AO,OC)
Shape(CO,OB,OBC)
Shape(BP,PC,OBC)
Collinear(AOBP)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAO),35)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PC,O)
Goal: Value(MeasureOfAngle(BPC))
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,PC,O,O)
triangle_property_angle_sum(1,COP)
Problem: As shown in the diagram, PA=2, PO=5, the center of circle O is O, the tangent to ⊙O is PB. Find the length of line PB.
问题：如图所示，PA=2，PO=5，O是圆O的圆心，圆O的切线为PB。求直线PB的长度。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(BP,PA,OBA)
Shape(BO,OA,OAB)
Collinear(PAO)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(PA),2)
Equal(LengthOfLine(PO),5)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PB,O)
Goal: Value(LengthOfLine(PB))
Answer: 4
Theorem sequence:
line_addition(1,PA,AO)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
right_triangle_judgment_angle(1,OBP)
right_triangle_property_pythagorean(1,OBP)
Problem: As shown in the diagram, ∠BOC=76°, the center of ⊙O is O, EC is the tangent to ⊙O. Find the measure of ∠BCE.
问题：如图所示，∠BOC=76°，O是⊙O的圆心，EC是圆O的切线。求∠BCE的大小。
Construction (CDL):
Shape(AC,CO,OB,BA)
Shape(OC,CB,BO)
Shape(BC,OCB)
Shape(BC,CE)
Shape(AB,OBA)
Shape(CA,OAC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BOC),76)
IsCentreOfCircle(O,O)
IsTangentOfCircle(EC,O)
Goal: Value(MeasureOfAngle(BCE))
Answer: 38
Theorem sequence:
tangent_of_circle_property_perpendicular(1,EC,O,O)
angle_addition(1,OCB,BCE)
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)
Problem: As shown in the diagram, AC=10, CO=10, OB=10, O is the center of circle O, AE is the tangent to ⊙O. Find the length of line AE.
问题：如图所示，AC=10，CO=10，OB=10，O是⊙O的圆心，圆O的切线为AE。求直线AE的长度。
Construction (CDL):
Shape(EA,AC,OEC)
Shape(CO,OE,OEC)
Shape(EO,OB,OBE)
Shape(BO,OC,OCB)
Collinear(ACOB)
Cocircular(O,ECB)
Conditions (CDL):
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(CO),10)
Equal(LengthOfLine(OB),10)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AE,O)
Goal: Value(LengthOfLine(AE))
Answer: 10*sqrt(3)
Theorem sequence:
radius_of_circle_property_length_equal(1,OE,O)
radius_of_circle_property_length_equal(1,OB,O)
line_addition(1,AC,CO)
tangent_of_circle_property_perpendicular(1,AE,O,O)
right_triangle_judgment_angle(1,OEA)
right_triangle_property_pythagorean(1,OEA)
Problem: As shown in the diagram, ∠ABC=50°, O is the center of ⊙O, the tangent to circle O is EA. Find the measure of ∠EAC.
问题：如图所示，∠ABC=50°，O是圆O的圆心，圆O的切线为EA。求∠EAC的大小。
Construction (CDL):
Shape(EA,AC)
Shape(AB,BC,CO,OA)
Shape(AO,OC,CA)
Shape(AC,OCA)
Shape(BA,OAB)
Shape(CB,OBC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),50)
IsCentreOfCircle(O,O)
IsTangentOfCircle(EA,O)
Goal: Value(MeasureOfAngle(EAC))
Answer: 50
Theorem sequence:
tangent_of_circle_property_perpendicular(2,EA,O,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_external(1,OCA,B)
isosceles_triangle_judgment_line_equal(1,OCA)
isosceles_triangle_property_angle_equal(1,OCA)
triangle_property_angle_sum(1,AOC)
angle_addition(1,EAC,CAO)
Problem: As shown in the diagram, ∠ABO=65°, O is the center of ⊙O, CA is the tangent to ⊙O. Find the measure of ∠BAC.
问题：如图所示，∠ABO=65°，⊙O的圆心为O，CA是⊙O的切线。求∠BAC的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OA,AB,BO)
Shape(BA,OAB)
Shape(CA,OAB)
Cocircular(O,AB)
Conditions (CDL):
Equal(MeasureOfAngle(ABO),65)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(BAC))
Answer: 25
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OAB)
tangent_of_circle_property_perpendicular(1,CA,O,O)
angle_addition(1,OAB,BAC)
Problem: As shown in the diagram, ∠BCA=40°, O is the center of ⊙O, the diameter of ⊙O is BC, the tangent to circle O is DA. Find the measure of ∠DAC.
问题：如图所示，∠BCA=40°，O是⊙O的圆心，圆O的直径为BC，DA是⊙O的切线。求∠DAC的大小。
Construction (CDL):
Shape(DA,AC)
Shape(OCA,AC)
Shape(OAB,BA)
Shape(BO,OC,CA,AB)
Shape(OBC,CO,OB)
Collinear(BOC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),40)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(BC,O)
IsTangentOfCircle(DA,O)
Goal: Value(MeasureOfAngle(DAC))
Answer: 50
Theorem sequence:
diameter_of_circle_property_right_angle(1,CAB,O)
triangle_property_angle_sum(1,BCA)
arc_property_circumference_angle_external(1,OCA,B)
circle_property_angle_of_osculation(2,OCA,D)
Problem: As shown in the diagram, ∠APO=36°, O is the center of circle O, the tangent to ⊙O is PA. Find the measure of ∠POA.
问题：如图所示，∠APO=36°，圆O的圆心为O，圆O的切线为PA。求∠POA的大小。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(OBA,AO,OB)
Shape(AP,PB,OAB)
Collinear(OBP)
Cocircular(O,AB)
Conditions (CDL):
Equal(MeasureOfAngle(APO),36)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(POA))
Answer: 54
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
triangle_property_angle_sum(1,OAP)
Problem: As shown in the diagram, AC=4, BC=3, the center of circle O is O, AC is the diameter of circle O, triangle ACB is a right triangle. Find the length of line AD.
问题：如图所示，AC=4，BC=3，圆O的圆心为O，圆O的直径为AC，三角形ACB为直角三角形。求直线AD的长度。
Construction (CDL):
Shape(ODA,AD)
Shape(OAC,CO,OA)
Shape(AO,OC,CD,DA)
Shape(OCD,DC)
Shape(CB,BD,OCD)
Collinear(AOC)
Collinear(ADB)
Cocircular(O,ACD)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BC),3)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AC,O)
RightTriangle(ACB)
Goal: Value(LengthOfLine(AD))
Answer: 16/5
Theorem sequence:
diameter_of_circle_property_right_angle(1,CDA,O)
right_triangle_judgment_angle(1,CDA)
cosine_theorem(1,ACB)
cosine_theorem(1,ACD)
right_triangle_property_pythagorean(1,ACB)
right_triangle_property_pythagorean(1,CDA)
Problem: As shown in the diagram, AB=2, the radius of ⊙O is 2, the center of circle O is O, EB is the diameter of ⊙O, AD is the tangent to ⊙O, AC is perpendicular to BC. Find the length of line BC.
问题：如图所示，AB=2，圆O的半径为2，O是⊙O的圆心，EB是⊙O的直径，⊙O的切线为AD，AC⊥BC。求直线BC的长度。
Construction (CDL):
Shape(ODE,EO,OD)
Shape(OBD,DO,OB)
Shape(BC,CD,OBD)
Shape(AC,CB,BA)
Collinear(EOBA)
Collinear(DCA)
Cocircular(O,BDE)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(RadiusOfCircle(O),2)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(EB,O)
IsTangentOfCircle(AD,O)
PerpendicularBetweenLine(AC,BC)
Goal: Value(LengthOfLine(BC))
Answer: 1
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OD,O)
tangent_of_circle_property_perpendicular(2,AD,O,O)
parallel_judgment_corresponding_angle(1,CB,DO,A)
midsegment_of_triangle_judgment_parallel(2,CB,ADO)
midsegment_of_triangle_property_length(1,CB,ADO)
Problem: As shown in the diagram, ∠EGF=40°, triangle ABC is an equilateral triangle, ED∥GF. Find the measure of ∠LDF.
问题：如图所示，∠EGF=40°，三角形ABC为等边三角形，ED∥GF。求∠LDF的大小。
Construction (CDL):
Shape(AE,ED,DA)
Shape(LD,DF)
Shape(DE,EG,GF,FD)
Shape(GB,BC,CF,FG)
Collinear(AEGB)
Collinear(EDL)
Collinear(ADFC)
Conditions (CDL):
Equal(MeasureOfAngle(EGF),40)
EquilateralTriangle(ABC)
ParallelBetweenLine(ED,GF)
Goal: Value(MeasureOfAngle(LDF))
Answer: 80
Theorem sequence:
equilateral_triangle_property_angle(1,ABC)
parallel_property_corresponding_angle(1,ED,GF,A)
triangle_property_angle_sum(1,AED)
vertical_angle(1,EDA,LDF)
Problem: As shown in the diagram, ∠DAB=20°, BD perpendicular bisects CA. Find the measure of ∠DBC.
问题：如图所示，∠DAB=20°，BD是CA的垂直平分线。求∠DBC的大小。
Construction (CDL):
Shape(DA,AB,BD)
Shape(CD,DB,BC)
Collinear(ADC)
Conditions (CDL):
Equal(MeasureOfAngle(DAB),20)
IsPerpendicularBisectorOfLine(BD,CA)
Goal: Value(MeasureOfAngle(DBC))
Answer: 70
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,BD,CA)
isosceles_triangle_judgment_line_equal(1,BCA)
isosceles_triangle_property_angle_equal(1,BCA)
triangle_property_angle_sum(1,CDB)
Problem: As shown in the diagram, ∠ABC=20°, ∠BCA=30°, PM is the perpendicular bisector of AB, QN perpendicular bisects CA. Find the measure of ∠QAP.
问题：如图所示，∠ABC=20°，∠BCA=30°，PM垂直平分AB，QN是CA的垂直平分线。求∠QAP的大小。
Construction (CDL):
Shape(AM,MP,PA)
Shape(MB,BP,PM)
Shape(AP,PQ,QA)
Shape(NA,AQ,QN)
Shape(CN,NQ,QC)
Collinear(ANC)
Collinear(AMB)
Collinear(BPQC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),20)
Equal(MeasureOfAngle(BCA),30)
IsPerpendicularBisectorOfLine(PM,AB)
IsPerpendicularBisectorOfLine(QN,CA)
Goal: Value(MeasureOfAngle(QAP))
Answer: 80
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,PM,AB)
perpendicular_bisector_property_distance_equal(1,QN,CA)
isosceles_triangle_judgment_line_equal(1,PAB)
isosceles_triangle_judgment_line_equal(1,QCA)
isosceles_triangle_property_angle_equal(1,PAB)
isosceles_triangle_property_angle_equal(1,QCA)
angle_addition(1,NAQ,QAP)
angle_addition(1,NAP,PAM)
triangle_property_angle_sum(1,BCA)
Problem: As shown in the diagram, ∠DAC=52°, B is the center of ⊙B, △ACB is a right △. Find the measure of ∠CDA.
问题：如图所示，∠DAC=52°，圆B的圆心为B，AC和CB是直角三角形ACB的两个直角边。求∠CDA的大小。
Construction (CDL):
Shape(DA,AC,BDC)
Shape(BDC,CD)
Shape(BD,DC,CB)
Collinear(ADB)
Cocircular(B,DC)
Conditions (CDL):
Equal(MeasureOfAngle(DAC),52)
IsCentreOfCircle(B,B)
RightTriangle(ACB)
Goal: Value(MeasureOfAngle(CDA))
Answer: 109
Theorem sequence:
triangle_property_angle_sum(1,BDC)
triangle_property_angle_sum(1,ACB)
radius_of_circle_property_length_equal(1,BD,B)
radius_of_circle_property_length_equal(1,BC,B)
isosceles_triangle_judgment_line_equal(1,BDC)
isosceles_triangle_property_angle_equal(1,BDC)
adjacent_complementary_angle(1,BDC,CDA)
Problem: As shown in the diagram, AB=AC, ∠CAB=100°, CE bisects ∠BCA, AD is the median of triangle ABC. Find the measure of ∠AFC.
问题：如图所示，AB=AC，∠CAB=100°，CE是∠BCA的角平分线，AD是△ABC的中线。求∠AFC的大小。
Construction (CDL):
Shape(AE,EF,FA)
Shape(EB,BD,DF,FE)
Shape(AF,FC,CA)
Shape(FD,DC,CF)
Collinear(AEB)
Collinear(AFD)
Collinear(EFC)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(CAB),100)
IsBisectorOfAngle(CE,BCA))
IsMedianOfTriangle(AD,ABC)
Goal: Value(MeasureOfAngle(AFC))
Answer: 110
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
triangle_property_angle_sum(1,AFC)
triangle_property_angle_sum(1,FDC)
triangle_property_angle_sum(1,CAD)
triangle_property_angle_sum(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
mirror_congruent_triangle_judgment_sas(1,BDA,CAD)
mirror_congruent_triangle_property_angle_equal(1,DAB,DCA)
adjacent_complementary_angle(1,BDA,FDC)
adjacent_complementary_angle(1,AFC,CFD)
Problem: As shown in the diagram, BA=BC, AD is the angle bisector of ∠CAE, DE perpendicular bisects AB. Find the measure of ∠EBD.
问题：如图所示，BA=BC，AD平分∠CAE，DE是AB的垂直平分线。求∠EBD的大小。
Construction (CDL):
Shape(BD,DE,EB)
Shape(AE,ED,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(BA),LengthOfLine(BC))
IsBisectorOfAngle(AD,CAE)
IsPerpendicularBisectorOfLine(DE,AB)
Goal: Value(MeasureOfAngle(EBD))
Answer: 36
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,DE,AB)
angle_addition(1,CAD,DAE)
triangle_property_angle_sum(1,BCA)
isosceles_triangle_judgment_line_equal(1,DAB)
isosceles_triangle_judgment_line_equal(1,BCA)
isosceles_triangle_property_angle_equal(1,DAB)
isosceles_triangle_property_angle_equal(1,BCA)
Problem: As shown in the diagram, ∠ACD=20°, the center of circle B is B, AC and CB are the legs of the right △ ACB. Find the measure of ∠BAC.
问题：如图所示，∠ACD=20°，B是⊙B的圆心，三角形ACB为直角三角形。求∠BAC的大小。
Construction (CDL):
Shape(DA,AC,BDC)
Shape(BDC,CD)
Shape(BD,DC,CB)
Collinear(ADB)
Cocircular(B,DC)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),20)
IsCentreOfCircle(B,B)
RightTriangle(ACB)
Goal: Value(MeasureOfAngle(BAC))
Answer: 50
Theorem sequence:
angle_addition(1,ACD,DCB)
triangle_property_angle_sum(1,BDC)
triangle_property_angle_sum(1,ACB)
radius_of_circle_property_length_equal(1,BD,B)
radius_of_circle_property_length_equal(1,BC,B)
isosceles_triangle_judgment_line_equal(1,BDC)
isosceles_triangle_property_angle_equal(1,BDC)