SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠DOE=110°, the center of circle O is O, O is the incenter of triangle ABC, the tangent to circle O is AD, the tangent to ⊙O is AE. Find the measure of ∠COB.
问题：如图所示，∠DOE=110°，⊙O的圆心为O，三角形ABC的内切圆圆心为O，AD是⊙O的切线，AE是⊙O的切线。求∠COB的大小。
Construction (CDL):
Shape(AD,OED,EA)
Shape(DO,OE,OED)
Shape(OD,ODM,MO)
Shape(EO,ON,ONE)
Shape(OM,OMF,OFN,NO)
Shape(ODM,DB,BM)
Shape(OMF,MB,BF)
Shape(OFN,FC,CN)
Shape(ONE,NC,CE)
Collinear(ADB)
Collinear(BFC)
Collinear(AEC)
Collinear(BMO)
Collinear(ONC)
Cocircular(O,DMFNE)
Conditions (CDL):
Equal(MeasureOfAngle(DOE),110)
IsCentreOfCircle(O,O)
IsIncenterOfTriangle(O,ABC)
IsTangentOfCircle(AD,O)
IsTangentOfCircle(AE,O)
Goal: Value(MeasureOfAngle(COB))
Answer: 125
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AD,O,O)
tangent_of_circle_property_perpendicular(1,AE,O,O)
quadrilateral_property_angle_sum(1,ADOE)
triangle_property_angle_sum(1,ABC)
angle_addition(1,ABO,OBC)
angle_addition(1,BCO,OCA)
triangle_property_angle_sum(1,OBC)
Problem: As shown in the diagram, ∠CIB=130°, I is the center of ⊙I, I is the incenter of △ABC. Find the measure of ∠CAB.
问题：如图所示，∠CIB=130°，I是⊙I的圆心，I是△ABC内切圆的圆心。求∠CAB的大小。
Construction (CDL):
Shape(AD,IED,EA)
Shape(IN,INE,IED,IDM,MI)
Shape(IM,IMF,IFN,NI)
Shape(IDM,DB,BM)
Shape(IMF,MB,BF)
Shape(IFN,FC,CN)
Shape(INE,NC,CE)
Collinear(ADB)
Collinear(BFC)
Collinear(AEC)
Collinear(BMI)
Collinear(INC)
Cocircular(I,DMFNE)
Conditions (CDL):
Equal(MeasureOfAngle(CIB),130)
IsCentreOfCircle(I,I)
IsIncenterOfTriangle(I,ABC)
Goal: Value(MeasureOfAngle(CAB))
Answer: 80
Theorem sequence:
triangle_property_angle_sum(1,ABC)
angle_addition(1,ABI,IBC)
angle_addition(1,BCI,ICA)
triangle_property_angle_sum(1,IBC)
Problem: As shown in the diagram, ∠CAB=70°, O is the center of circle O, O is the incenter of △ABC. Find the measure of ∠COB.
问题：如图所示，∠CAB=70°，圆O的圆心为O，△ABC的内切圆圆心为O。求∠COB的大小。
Construction (CDL):
Shape(AD,OED,EA)
Shape(ON,ONE,OED,ODM,MO)
Shape(OM,OMF,OFN,NO)
Shape(ODM,DB,BM)
Shape(OMF,MB,BF)
Shape(OFN,FC,CN)
Shape(ONE,NC,CE)
Collinear(ADB)
Collinear(BFC)
Collinear(AEC)
Collinear(BMO)
Collinear(ONC)
Cocircular(O,DMFNE)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),70)
IsCentreOfCircle(O,O)
IsIncenterOfTriangle(O,ABC)
Goal: Value(MeasureOfAngle(COB))
Answer: 125
Theorem sequence:
triangle_property_angle_sum(1,ABC)
angle_addition(1,ABO,OBC)
angle_addition(1,BCO,OCA)
triangle_property_angle_sum(1,OBC)
Problem: As shown in the diagram, AB=4, AC=2, BC=5, I is the center of the inscribed circle of △ABC, AC∥IN, BA∥MI. Find the perimeter of △IMN.
问题：如图所示，AB=4，AC=2，BC=5，I是△ABC的内心，AC平行于IN，BA平行于MI。求三角形IMN的周长。
Construction (CDL):
Shape(AB,BI,IC,CA)
Shape(IB,BM,MI)
Shape(IM,MN,NI)
Shape(IN,NC,CI)
Collinear(BMNC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AC),2)
Equal(LengthOfLine(BC),5)
IsIncenterOfTriangle(I,ABC)
ParallelBetweenLine(AC,IN)
ParallelBetweenLine(BA,MI)
Goal: Value(PerimeterOfTriangle(IMN))
Answer: 5
Theorem sequence:
parallel_property_alternate_interior_angle(1,BA,MI)
parallel_property_alternate_interior_angle(2,AC,IN)
isosceles_triangle_judgment_angle_equal(1,MIB)
isosceles_triangle_judgment_angle_equal(1,NCI)
line_addition(1,BM,MN)
line_addition(1,BN,NC)
triangle_perimeter_formula(1,IMN)
Problem: As shown in the diagram, AB=AC, CD=1, ∠CAB=36°, BD bisects ∠ABC. Find the length of line AC.
问题：如图所示，AB=AC，CD=1，∠CAB=36°，BD平分∠ABC。求直线AC的长度。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(CD),1)
Equal(MeasureOfAngle(CAB),36)
IsBisectorOfAngle(BD,ABC)
Goal: Value(LengthOfLine(AC))
Answer: 1+sqrt(sqrt(5)+5)/sqrt(5-sqrt(5))
Theorem sequence:
triangle_property_angle_sum(1,ABC)
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
sine_theorem(1,DBC)
sine_theorem(1,DAB)
line_addition(1,AD,DC)
Problem: As shown in the diagram, Div(LengthOfLine(BC)=LengthOfLine(AB)), BC=1, A is the center of ⊙A, the center of ⊙C is C, AB⊥CB. Find the ratio of the length of line AP to the length of line AB.
问题：如图所示，Div(LengthOfLine(BC)=LengthOfLine(AB))，BC=1，A是⊙A的圆心，⊙C的圆心为C，AB垂直于CB。求直线AP的长度与直线AB的长度之比。
Construction (CDL):
Shape(DA,AP,APD)
Shape(PA,AD,ADP)
Shape(CD,CDB,BC)
Shape(DC,CB,CBD)
Shape(APD,PB,CDB)
Collinear(APB)
Collinear(ADC)
Cocircular(A,PD)
Cocircular(C,DB)
Conditions (CDL):
Equal(Div(LengthOfLine(BC),LengthOfLine(AB)),1/2)
Equal(LengthOfLine(BC),1)
IsCentreOfCircle(A,A)
IsCentreOfCircle(C,C)
PerpendicularBetweenLine(AB,CB)
Goal: Value(Div(LengthOfLine(AP),LengthOfLine(AB)))
Answer: -1/2+sqrt(5)/2
Theorem sequence:
radius_of_circle_property_length_equal(1,AP,A)
radius_of_circle_property_length_equal(1,AD,A)
radius_of_circle_property_length_equal(1,CD,C)
radius_of_circle_property_length_equal(1,CB,C)
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,ABC)
line_addition(1,AD,DC)
Problem: As shown in the diagram, AC=4, BC=3, ∠ACB is the right angle of the right triangle ACB. Find the value of sin(BA).
问题：如图所示，AC=4，BC=3，∠ACB是直角△ACB的直角。求sin(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BC),3)
RightTriangle(ACB)
Goal: Value(Sin(MeasureOfAngle(BAC)))
Answer: 3/5
Theorem sequence:
right_triangle_property_pythagorean(1,ACB)
sine_theorem(1,BAC)
Problem: As shown in the diagram, AB=10, AC=6, BC and CA are the legs of the right △ BCA. Find the value of sin(CA).
问题：如图所示，AB=10，AC=6，∠BCA是直角△BCA的直角。求sin(CA)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),6)
RightTriangle(BCA)
Goal: Value(Sin(MeasureOfAngle(CAB)))
Answer: 4/5
Theorem sequence:
right_triangle_property_pythagorean(1,BCA)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=10, BC=8, ∠BCA is the right angle of the right triangle BCA. Find the value of sin(CA).
问题：如图所示，AB=10，BC=8，∠BCA是直角△BCA的直角。求sin(CA)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(BC),8)
RightTriangle(BCA)
Goal: Value(Sin(MeasureOfAngle(CAB)))
Answer: 4/5
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=10, AC=8, triangle ACB is a right triangle. Find the value of sin(CB).
问题：如图所示，AB=10，AC=8，∠ACB是直角三角形ACB的直角。求sin(CB)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),8)
RightTriangle(ACB)
Goal: Value(Sin(MeasureOfAngle(CBA)))
Answer: 4/5
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, AB=3, the ratio of similar triangles ACB and DFE is 2/3, triangle ACB is similar to triangle DFE.. Find the length of line DE.
问题：如图所示，AB=3，△ACB和△DFE的相似比为2/3，△ACB相似于△DFE。求直线DE的长度。
Construction (CDL):
Shape(AO,OC,CA)
Shape(AC,CM,MA)
Shape(DA,AM,MF,FD)
Shape(DF,FE,ED)
Shape(CO,OB,BC)
Shape(CB,BM,MC)
Shape(FM,MB,BE,EF)
Collinear(OAD)
Collinear(OCMF)
Collinear(OBE)
Collinear(AMB)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(RatioOfSimilarTriangle(ACB,DFE),2/3)
SimilarBetweenTriangle(ACB,DFE)
Goal: Value(LengthOfLine(DE))
Answer: 9/2
Theorem sequence:
similar_triangle_property_line_ratio(1,CBA,FED)
Problem: As shown in the diagram, the area of quadrilateral EFGH is 9, The ratio of similar quadrilaterals ABCD and EFGH is 5/3, ABCD is similar to EFGH. Find the area of ABCD.
问题：如图所示，EFGH的面积为9，ABCD和EFGH的相似比为5/3，四边形ABCD相似于四边形EFGH。求ABCD的面积。
Construction (CDL):
Shape(OH,HE,EO)
Shape(OE,EF,FO)
Shape(OF,FG,GO)
Shape(OG,GH,HO)
Shape(HD,DA,AE,EH)
Shape(EA,AB,BF,FE)
Shape(FB,BC,CG,GF)
Shape(GC,CD,DH,HG)
Collinear(DHOFB)
Collinear(AEOGC)
Conditions (CDL):
Equal(AreaOfQuadrilateral(EFGH),9)
Equal(RatioOfSimilarQuadrilateral(ABCD,EFGH),5/3)
SimilarBetweenQuadrilateral(ABCD,EFGH)
Goal: Value(AreaOfQuadrilateral(ABCD))
Answer: 25
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,ABCD,EFGH)
Problem: As shown in the diagram, the area of ABCD is 4, The ratio of similar quadrilaterals ABCD and EFGH is 2/3, quadrilateral ABCD is similar to quadrilateral EFGH. Find the area of EFGH.
问题：如图所示，ABCD的面积为4，四边形ABCD和四边形EFGH的相似比为2/3，ABCD与EFGH相似。求四边形EFGH的面积。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(EF,FG,GH,HE)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ABCD),4)
Equal(RatioOfSimilarQuadrilateral(ABCD,EFGH),2/3)
SimilarBetweenQuadrilateral(ABCD,EFGH)
Goal: Value(AreaOfQuadrilateral(EFGH))
Answer: 9
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,ABCD,EFGH)
Problem: As shown in the diagram, the area of △ABC is 4, the ratio of similar △ABC and △DEF is 1/2, △ABC is similar to △DEF.. Find the area of triangle DEF.
问题：如图所示，三角形ABC的面积为4，三角形ABC和三角形DEF的相似比为1/2，三角形ABC与三角形DEF是相似三角形。求△DEF的面积。
Construction (CDL):
Shape(AB,BC,CA)
Shape(DE,EF,FD)
Conditions (CDL):
Equal(AreaOfTriangle(ABC),4)
Equal(RatioOfSimilarTriangle(ABC,DEF),1/2)
SimilarBetweenTriangle(ABC,DEF)
Goal: Value(AreaOfTriangle(DEF))
Answer: 16
Theorem sequence:
similar_triangle_property_area_square_ratio(1,ABC,DEF)
Problem: As shown in the diagram, AB=2, OB=4, the ratio of similar △OBA and △ODC is 2, OB⊥AB, △OBA is similar to △ODC.. Find the length of line OC.
问题：如图所示，AB=2，OB=4，△OBA和△ODC的相似比为2，OB垂直于AB，三角形OBA与三角形ODC是相似三角形。求直线OC的长度。
Construction (CDL):
Shape(CO,OD,DC)
Shape(AC,CD,DB,BA)
Collinear(ACO)
Collinear(ODB)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(OB),4)
Equal(RatioOfSimilarTriangle(OBA,ODC),2)
PerpendicularBetweenLine(OB,AB)
SimilarBetweenTriangle(OBA,ODC)
Goal: Value(LengthOfLine(OC))
Answer: sqrt(5)
Theorem sequence:
right_triangle_judgment_angle(1,OBA)
right_triangle_property_pythagorean(1,OBA)
similar_triangle_property_line_ratio(1,BAO,DCO)
Problem: As shown in the diagram, AB=3, ∠CAD=60°, the center of circle O is O, the tangent to circle O is AB, AE is the tangent to ⊙O, ∠DCA is the right angle of the right △ DCA. Find the diameter of circle O.
问题：如图所示，AB=3，∠CAD=60°，圆O的圆心为O，圆O的切线为AB，⊙O的切线为AE，△DCA为直角△。求圆O的直径。
Construction (CDL):
Shape(DC,CA,AE,ED)
Shape(EA,AB,OEB)
Shape(OE,EB,BO)
Shape(EO,OB,OBE)
Shape(BE,OEB)
Collinear(CAB)
Collinear(DEA)
Cocircular(O,EB)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(MeasureOfAngle(CAD),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
IsTangentOfCircle(AE,O)
RightTriangle(DCA)
Goal: Value(DiameterOfCircle(O))
Answer: 6*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,CAD,DAB)
tangent_of_circle_property_perpendicular(2,AB,O,O)
tangent_of_circle_property_perpendicular(1,AE,O,O)
tangent_of_circle_property_length_equal(1,AB,AE,O)
quadrilateral_property_angle_sum(1,OEAB)
cosine_theorem(1,ABE)
radius_of_circle_property_length_equal(1,OE,O)
radius_of_circle_property_length_equal(1,OB,O)
cosine_theorem(1,OEB)
circle_property_length_of_radius_and_diameter(1,O)
Problem: As shown in the diagram, AB=8, AC=10, BC=9, the center of circle O is O, the tangent to circle O is AG, AM is the tangent to circle O, the tangent to circle O is BM, BN is the tangent to circle O, the tangent to ⊙O is DF, the tangent to ⊙O is DN, EF is the tangent to ⊙O, the tangent to ⊙O is EG. Find the perimeter of △CED.
问题：如图所示，AB=8，AC=10，BC=9，⊙O的圆心为O，AG是⊙O的切线，AM是⊙O的切线，BM是圆O的切线，BN是⊙O的切线，圆O的切线为DF，DN是圆O的切线，⊙O的切线为EF，⊙O的切线为EG。求三角形CED的周长。
Construction (CDL):
Shape(AM,OGM,GA)
Shape(MB,BN,OMN)
Shape(ND,DF,ONF)
Shape(FE,EG,OFG)
Shape(CE,EF,FD,DC)
Shape(O)
Shape(OMN,ONF,OFG,OGM)
Collinear(AGEC)
Collinear(AMB)
Collinear(BNDC)
Collinear(EFD)
Cocircular(O,MNFG)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(BC),9)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AG,O)
IsTangentOfCircle(AM,O)
IsTangentOfCircle(BM,O)
IsTangentOfCircle(BN,O)
IsTangentOfCircle(DF,O)
IsTangentOfCircle(DN,O)
IsTangentOfCircle(EF,O)
IsTangentOfCircle(EG,O)
Goal: Value(PerimeterOfTriangle(CED))
Answer: 11
Theorem sequence:
tangent_of_circle_property_length_equal(1,EF,EG,O)
tangent_of_circle_property_length_equal(1,DF,DN,O)
tangent_of_circle_property_length_equal(1,BN,BM,O)
tangent_of_circle_property_length_equal(1,AM,AG,O)
line_addition(1,AG,GE)
line_addition(1,AE,EC)
line_addition(1,AM,MB)
line_addition(1,BN,ND)
line_addition(1,BD,DC)
line_addition(1,EF,FD)
triangle_perimeter_formula(1,CED)
Problem: As shown in the diagram, PA=10, the tangent to circle O is CA, CE is the tangent to circle O, the tangent to ⊙O is DB, the tangent to circle O is DE, the tangent to circle O is PA, PB is the tangent to circle O. Find the perimeter of △PCD.
问题：如图所示，PA=10，圆O的切线为CA，圆O的切线为CE，DB是圆O的切线，⊙O的切线为DE，⊙O的切线为PA，PB是⊙O的切线。求三角形PCD的周长。
Construction (CDL):
Shape(PC,CE,ED,DP)
Shape(EC,CA,OEA)
Shape(BD,DE,OBE)
Shape(OAB,OBE,OEA)
Shape(O)
Collinear(ACP)
Collinear(BDP)
Collinear(CED)
Cocircular(O,ABE)
Conditions (CDL):
Equal(LengthOfLine(PA),10)
IsTangentOfCircle(CA,O)
IsTangentOfCircle(CE,O)
IsTangentOfCircle(DB,O)
IsTangentOfCircle(DE,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(PerimeterOfTriangle(PCD))
Answer: 20
Theorem sequence:
tangent_of_circle_property_length_equal(1,PA,PB,O)
tangent_of_circle_property_length_equal(1,CE,CA,O)
tangent_of_circle_property_length_equal(1,DE,DB,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, AB=10, CD=12, O is the center of ⊙O, the tangent to ⊙O is AE, AF is the tangent to circle O, BF is the tangent to ⊙O, the tangent to ⊙O is BG, CG is the tangent to ⊙O, the tangent to ⊙O is CH, the tangent to circle O is DE, the tangent to circle O is DH. Find the perimeter of ABCD.
问题：如图所示，AB=10，CD=12，圆O的圆心为O，AE是⊙O的切线，⊙O的切线为AF，BF是⊙O的切线，BG是⊙O的切线，圆O的切线为CG，⊙O的切线为CH，DE是⊙O的切线，DH是⊙O的切线。求四边形ABCD的周长。
Construction (CDL):
Shape(EA,AF,OEF)
Shape(HD,DE,OHE)
Shape(FB,BG,OFG)
Shape(GC,CH,OGH)
Shape(O)
Shape(OEF,OFG,OGH,OHE)
Collinear(AED)
Collinear(AFB)
Collinear(BGC)
Collinear(CHD)
Cocircular(O,EFGH)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(CD),12)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AE,O)
IsTangentOfCircle(AF,O)
IsTangentOfCircle(BF,O)
IsTangentOfCircle(BG,O)
IsTangentOfCircle(CG,O)
IsTangentOfCircle(CH,O)
IsTangentOfCircle(DE,O)
IsTangentOfCircle(DH,O)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 44
Theorem sequence:
tangent_of_circle_property_length_equal(1,AE,AF,O)
tangent_of_circle_property_length_equal(1,DE,DH,O)
tangent_of_circle_property_length_equal(1,CH,CG,O)
tangent_of_circle_property_length_equal(1,BG,BF,O)
line_addition(1,AE,ED)
line_addition(1,AF,FB)
line_addition(1,BG,GC)
line_addition(1,CH,HD)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, AB=5, AC=3, the center of ⊙O is O, the tangent to ⊙O is AC, the tangent to circle O is AP, BD is the tangent to ⊙O, the tangent to ⊙O is BP. Find the length of line BD.
问题：如图所示，AB=5，AC=3，O是⊙O的圆心，圆O的切线为AC，⊙O的切线为AP，BD是圆O的切线，⊙O的切线为BP。求直线BD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(AB),5)
Equal(LengthOfLine(AC),3)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AC,O)
IsTangentOfCircle(AP,O)
IsTangentOfCircle(BD,O)
IsTangentOfCircle(BP,O)
Goal: Value(LengthOfLine(BD))
Answer: 2
Theorem sequence:
tangent_of_circle_property_length_equal(1,AC,AP,O)
tangent_of_circle_property_length_equal(1,BP,BD,O)
line_addition(1,AP,PB)
Problem: As shown in the diagram, AB=3, ∠CAD=60°, O is the center of ⊙O, the tangent to ⊙O is AB, AE is the tangent to circle O, triangle DCA is a right triangle. Find the diameter of circle O.
问题：如图所示，AB=3，∠CAD=60°，O是圆O的圆心，AB是圆O的切线，⊙O的切线为AE，三角形DCA为直角三角形。求⊙O的直径。
Construction (CDL):
Shape(DC,CA,AE,ED)
Shape(EA,AB,OEB)
Shape(OE,EB,BO)
Shape(EO,OB,OBE)
Shape(BE,OEB)
Collinear(CAB)
Collinear(DEA)
Cocircular(O,EB)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(MeasureOfAngle(CAD),60)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AB,O)
IsTangentOfCircle(AE,O)
RightTriangle(DCA)
Goal: Value(DiameterOfCircle(O))
Answer: 6*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,CAD,DAB)
tangent_of_circle_property_perpendicular(2,AB,O,O)
tangent_of_circle_property_perpendicular(1,AE,O,O)
tangent_of_circle_property_length_equal(1,AB,AE,O)
quadrilateral_property_angle_sum(1,OEAB)
cosine_theorem(1,ABE)
radius_of_circle_property_length_equal(1,OE,O)
radius_of_circle_property_length_equal(1,OB,O)
cosine_theorem(1,OEB)
circle_property_length_of_radius_and_diameter(1,O)
Problem: As shown in the diagram, BD=1/5, the radius of circle O is 1, the center of circle O is O, AB is perpendicular to OB. Find the length of line AC.
问题：如图所示，BD=1/5，圆O的半径为1，圆O的圆心为O，AB⊥OB。求直线AC的长度。
Construction (CDL):
Shape(OA,AB,BO)
Shape(OB,BC,CO)
Shape(BA,OAD,DB)
Shape(CB,BD,ODC)
Shape(OC,OCA,AO)
Collinear(ABC)
Collinear(OBD)
Cocircular(O,ADC)
Conditions (CDL):
Equal(LengthOfLine(BD),1/5)
Equal(RadiusOfCircle(O),1)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AB,OB)
Goal: Value(LengthOfLine(AC))
Answer: 6/5
Theorem sequence:
adjacent_complementary_angle(1,ABO,OBC)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OC,O)
line_addition(1,AB,BC)
line_addition(1,OB,BD)
right_triangle_judgment_angle(1,ABO)
right_triangle_judgment_angle(1,OBC)
right_triangle_property_pythagorean(1,ABO)
right_triangle_property_pythagorean(1,OBC)
Problem: As shown in the diagram, AB=16, the radius of circle O is 10, O is the center of circle O, AC⊥OC. Find the length of line CD.
问题：如图所示，AB=16，圆O的半径为10，O是⊙O的圆心，AC⊥OC。求直线CD的长度。
Construction (CDL):
Shape(OA,AC,CO)
Shape(OC,CB,BO)
Shape(CA,OAD,DC)
Shape(BC,CD,ODB)
Shape(OB,OBA,AO)
Collinear(ACB)
Collinear(OCD)
Cocircular(O,ADB)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
Equal(RadiusOfCircle(O),10)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AC,OC)
Goal: Value(LengthOfLine(CD))
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,ACO,OCB)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OB,O)
line_addition(1,AC,CB)
line_addition(1,OC,CD)
right_triangle_judgment_angle(1,ACO)
right_triangle_judgment_angle(1,OCB)
right_triangle_property_pythagorean(1,ACO)
right_triangle_property_pythagorean(1,OCB)
Problem: As shown in the diagram, the measure of ⌒DCA is 120, the measure of ⌒OBA is 120, the radius of ⊙O is 2, the radius of circle O is equal to the radius of circle D, the center of circle D is D, O is the center of ⊙O. Find the length of line BC.
问题：如图所示，⌒DCA的角度为120，弧OBA的角度为120，圆O的半径为2，⊙O的半径与⊙D的半径相等，圆D的圆心为D，O是圆O的圆心。求直线BC的长度。
Construction (CDL):
Shape(BC,CD,OBD)
Shape(OBD,DB)
Shape(DC,DCA,ODA)
Shape(ODA,AD)
Shape(OD,DA,AO)
Shape(OB,BD,DO)
Cocircular(D,CA)
Cocircular(O,BDA)
Conditions (CDL):
Equal(MeasureOfArc(DCA),120)
Equal(MeasureOfArc(OBA),120)
Equal(RadiusOfCircle(O),2)
Equal(RadiusOfCircle(O),RadiusOfCircle(D))
IsCentreOfCircle(D,D)
IsCentreOfCircle(O,O)
Goal: Value(LengthOfLine(BC))
Answer: 2*sqrt(3)
Theorem sequence:
arc_property_center_angle(1,DCA,D)
arc_property_center_angle(1,OBA,O)
radius_of_circle_property_length_equal(1,DC,D)
radius_of_circle_property_length_equal(1,DA,D)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OA,O)
isosceles_triangle_judgment_line_equal(1,ODA)
isosceles_triangle_judgment_line_equal(1,DAO)
equilateral_triangle_judgment_isosceles_and_isosceles(1,ODA)
equilateral_triangle_property_angle(1,ODA)
equilateral_triangle_property_angle(1,DAO)
angle_addition(1,AOD,DOB)
cosine_theorem(1,OBD)
triangle_property_angle_sum(1,OBD)
sine_theorem(1,OBD)
round_angle(1,CDB,BDC)
angle_addition(1,ODA,ADC)
angle_addition(1,BDO,ODC)
cosine_theorem(1,DBC)
Problem: As shown in the diagram, ∠ABO=20°, ∠OCA=30°, the center of ⊙O is O. Find the measure of ∠COB.
问题：如图所示，∠ABO=20°，∠OCA=30°，圆O的圆心为O。求∠COB的大小。
Construction (CDL):
Shape(OAB,BA)
Shape(AB,BO,OA)
Shape(AO,OC,CA)
Shape(AC,OCA)
Shape(OB,OBC,CO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABO),20)
Equal(MeasureOfAngle(OCA),30)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(COB))
Answer: 100
Theorem sequence:
angle_addition(1,CAO,OAB)
radius_of_circle_property_length_equal(1,OA,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,OAB)
isosceles_triangle_judgment_line_equal(1,OCA)
isosceles_triangle_property_angle_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OCA)
arc_property_circumference_angle_external(1,OBC,A)
arc_property_center_angle(1,OBC,O)
Problem: As shown in the diagram, AB=1, E is the midpoint of segment AD, F is the midpoint of segment BC, quadrilateral ABCD is a rectangle, quadrilateral EABF is a rectangle, ABCD is similar to EABF. Find the area of quadrilateral ABCD.
问题：如图所示，AB=1，E平分线段AD，F是线段BC的中点，ABCD是长方形，EABF是矩形，四边形ABCD与四边形EABF相似。求四边形ABCD的面积。
Construction (CDL):
Shape(AB,BF,FE,EA)
Shape(EF,FC,CD,DE)
Collinear(AED)
Collinear(BFC)
Conditions (CDL):
Equal(LengthOfLine(AB),1)
IsMidpointOfLine(E,AD)
IsMidpointOfLine(F,BC)
Rectangle(ABCD)
Rectangle(EABF)
SimilarBetweenQuadrilateral(ABCD,EABF)
Goal: Value(AreaOfQuadrilateral(ABCD))
Answer: sqrt(2)
Theorem sequence:
line_addition(1,AE,ED)
parallelogram_property_opposite_line_equal(1,BCDA)
similar_quadrilateral_property_line_ratio(1,ABCD,EABF)
similar_quadrilateral_property_line_ratio(1,BCDA,ABFE)
parallelogram_area_formula_sine(1,ABCD)
Problem: As shown in the diagram, ∠ABC=110°, ∠CAB=45°, △CAB is mirror similar to △BPA.. Find the measure of ∠ABP.
问题：如图所示，∠ABC=110°，∠CAB=45°，△CAB与△BPA是镜像相似三角形。求∠ABP的大小。
Construction (CDL):
Shape(BC,CP,PB)
Shape(BP,PA,AB)
Collinear(APC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),110)
Equal(MeasureOfAngle(CAB),45)
MirrorSimilarBetweenTriangle(CAB,BPA)
Goal: Value(MeasureOfAngle(ABP))
Answer: 25
Theorem sequence:
triangle_property_angle_sum(1,BCA)
mirror_similar_triangle_property_angle_equal(1,CAB,BPA)
Problem: As shown in the diagram, AB=6, AC=4, AD=3, △ADE is similar to △ABC.. Find the length of line AE.
问题：如图所示，AB=6，AC=4，AD=3，三角形ADE与三角形ABC是相似三角形。求直线AE的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CE,ED)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(AD),3)
SimilarBetweenTriangle(ADE,ABC)
Goal: Value(LengthOfLine(AE))
Answer: 2
Theorem sequence:
similar_triangle_property_line_ratio(1,DEA,BCA)
similar_triangle_property_line_ratio(1,EAD,CAB)
Problem: As shown in the diagram, the area of △ABC is 9, the ratio of similar triangles ADE and ABC is 1/3, △ADE is similar to △ABC.. Find the area of triangle ADE.
问题：如图所示，三角形ABC的面积为9，△ADE和△ABC的相似比为1/3，△ADE相似于△ABC。求△ADE的面积。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CE,ED)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(AreaOfTriangle(ABC),9)
Equal(RatioOfSimilarTriangle(ADE,ABC),1/3)
SimilarBetweenTriangle(ADE,ABC)
Goal: Value(AreaOfTriangle(ADE))
Answer: 1
Theorem sequence:
similar_triangle_property_area_square_ratio(1,ADE,ABC)
Problem: As shown in the diagram, AB=12, AO=9, CD=4, triangle ABO is similar to triangle CDO.. Find the length of line OC.
问题：如图所示，AB=12，AO=9，CD=4，三角形ABO相似于三角形CDO。求直线OC的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OC,CD,DO)
Collinear(BOD)
Collinear(AOC)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AO),9)
Equal(LengthOfLine(CD),4)
SimilarBetweenTriangle(ABO,CDO)
Goal: Value(LengthOfLine(OC))
Answer: 3
Theorem sequence:
similar_triangle_property_line_ratio(1,BOA,DOC)
similar_triangle_property_line_ratio(1,OAB,OCD)
Problem: As shown in the diagram, AB=9, AD=3, DE=2, △ADE is similar to △ABC.. Find the length of line BC.
问题：如图所示，AB=9，AD=3，DE=2，三角形ADE相似于三角形ABC。求直线BC的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CE,ED)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AB),9)
Equal(LengthOfLine(AD),3)
Equal(LengthOfLine(DE),2)
SimilarBetweenTriangle(ADE,ABC)
Goal: Value(LengthOfLine(BC))
Answer: 6
Theorem sequence:
similar_triangle_property_line_ratio(1,ADE,ABC)
similar_triangle_property_line_ratio(1,EAD,CAB)
Problem: As shown in the diagram, ∠APC=65°, ∠CAB=75°, △ABC is mirror similar to △APC.. Find the measure of ∠ABC.
问题：如图所示，∠APC=65°，∠CAB=75°，三角形ABC与三角形APC是镜像相似三角形。求∠ABC的大小。
Construction (CDL):
Shape(AP,PC,CA)
Shape(PB,BC,CP)
Collinear(APB)
Conditions (CDL):
Equal(MeasureOfAngle(APC),65)
Equal(MeasureOfAngle(CAB),75)
MirrorSimilarBetweenTriangle(ABC,APC)
Goal: Value(MeasureOfAngle(ABC))
Answer: 40
Theorem sequence:
triangle_property_angle_sum(1,APC)
mirror_similar_triangle_property_angle_equal(1,BCA,CAP)