SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, Div(LengthOfLine(AC)=LengthOfLine(BC)), Div(LengthOfLine(EF)=LengthOfLine(FD)), AC=6, AE=7, FD=4, ABDE is a isosceles trapezoid, AE is parallel to CF, BC is perpendicular to AC, EF⊥DF. Find the length of line BD.
问题：如图所示，Div(LengthOfLine(AC)=LengthOfLine(BC))，Div(LengthOfLine(EF)=LengthOfLine(FD))，AC=6，AE=7，FD=4，四边形ABDE是等腰梯形，AE平行于CF，BC垂直于AC，EF垂直于DF。求直线BD的长度。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CF,FE,EA)
Shape(EF,FD,DE)
Collinear(BCFD)
Conditions (CDL):
Equal(Div(LengthOfLine(AC),LengthOfLine(BC)),3/2)
Equal(Div(LengthOfLine(EF),LengthOfLine(FD)),3/2)
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(AE),7)
Equal(LengthOfLine(FD),4)
IsoscelesTrapezoid(ABDE)
ParallelBetweenLine(AE,CF)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(EF,DF)
Goal: Value(LengthOfLine(BD))
Answer: 15
Theorem sequence:
flat_angle(1,BFD)
angle_addition(1,CFE,EFD)
parallel_judgment_corresponding_angle(1,CA,FE,B)
parallelogram_judgment_parallel_and_parallel(1,ACFE)
parallelogram_property_opposite_line_equal(1,CFEA)
line_addition(1,BC,CF)
line_addition(1,BF,FD)
Problem: As shown in the diagram, AB=130, CA=50, BC and CA are the legs of the right triangle BCA. Find the ratio of the length of line AC to the length of line BC.
问题：如图所示，AB=130，CA=50，三角形BCA为直角三角形。求直线AC的长度与直线BC的长度之比。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),130)
Equal(LengthOfLine(CA),50)
RightTriangle(BCA)
Goal: Value(Div(LengthOfLine(AC),LengthOfLine(BC)))
Answer: 5/12
Theorem sequence:
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, AC=3, BD=6, CD=11, Triangle ACE is mirror similar to triangle BED.. Find the ratio of the length of line CE to the length of line AC.
问题：如图所示，AC=3，BD=6，CD=11，△ACE镜像相似于△BED。求直线CE的长度与直线AC的长度之比。
Construction (CDL):
Shape(AC,CE,EA)
Shape(BE,ED,DB)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(BD),6)
Equal(LengthOfLine(CD),11)
MirrorSimilarBetweenTriangle(ACE,BED)
Goal: Value(Div(LengthOfLine(CE),LengthOfLine(AC)))
Answer: 11/9
Theorem sequence:
line_addition(1,CE,ED)
mirror_similar_triangle_property_line_ratio(1,ACE,BED)
mirror_similar_triangle_property_line_ratio(1,EAC,EDB)
Problem: As shown in the diagram, Div(LengthOfLine(AB)=LengthOfLine(CA)), AB=3, △CAB is a right △. Find the length of line BC.
问题：如图所示，Div(LengthOfLine(AB)=LengthOfLine(CA))，AB=3，△CAB为直角△。求直线BC的长度。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(Div(LengthOfLine(AB),LengthOfLine(CA)),1/2)
Equal(LengthOfLine(AB),3)
RightTriangle(CAB)
Goal: Value(LengthOfLine(BC))
Answer: 3*sqrt(5)
Theorem sequence:
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, Div(LengthOfLine(BC)=LengthOfLine(AC)), AC=6, triangle ACB is a right triangle. Find the length of line BC.
问题：如图所示，Div(LengthOfLine(BC)=LengthOfLine(AC))，AC=6，∠ACB是直角三角形ACB的直角。求直线BC的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(Div(LengthOfLine(BC),LengthOfLine(AC)),2/3)
Equal(LengthOfLine(AC),6)
RightTriangle(ACB)
Goal: Value(LengthOfLine(BC))
Answer: 4
Theorem sequence:
Problem: As shown in the diagram, AB=13, CA=5, BC and CA are the legs of the right triangle BCA. Find the ratio of the length of line BC to the length of line AB.
问题：如图所示，AB=13，CA=5，∠BCA是直角△BCA的直角。求直线BC的长度与直线AB的长度之比。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(CA),5)
RightTriangle(BCA)
Goal: Value(Div(LengthOfLine(BC),LengthOfLine(AB)))
Answer: 12/13
Theorem sequence:
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, AB=3, AC=5, AD=1, DE=0.6, △ABC is a right △, ∠AED is the right angle of the right triangle AED. Find the ratio of the length of line CB to the length of line AB.
问题：如图所示，AB=3，AC=5，AD=1，DE=0.6，△ABC为直角△，AE和ED是直角三角形AED的两个直角边。求直线CB的长度与直线AB的长度之比。
Construction (CDL):
Shape(DA,AE,ED)
Shape(DE,EB,BC,CD)
Collinear(AEB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(AD),1)
Equal(LengthOfLine(DE),0.6)
RightTriangle(ABC)
RightTriangle(AED)
Goal: Value(Div(LengthOfLine(CB),LengthOfLine(AB)))
Answer: 4/3
Theorem sequence:
right_triangle_property_pythagorean(1,ABC)
Problem: As shown in the diagram, Div(LengthOfLine(AB)=LengthOfLine(BC)), AB=13, ∠BCA is the right angle of the right △ BCA. Find the length of line CA.
问题：如图所示，Div(LengthOfLine(AB)=LengthOfLine(BC))，AB=13，BC和CA是直角△BCA的两个直角边。求直线CA的长度。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(Div(LengthOfLine(AB),LengthOfLine(BC)),13/12)
Equal(LengthOfLine(AB),13)
RightTriangle(BCA)
Goal: Value(LengthOfLine(CA))
Answer: 5
Theorem sequence:
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, Div(LengthOfLine(DG)=LengthOfLine(CG)), Div(LengthOfLine(EH)=LengthOfLine(HF)), DE=2, DG=15, EH=15, GH=2, AD is parallel to BC, DE∥HC, DG is perpendicular to FG, EH⊥FH. Find the length of line CF.
问题：如图所示，Div(LengthOfLine(DG)=LengthOfLine(CG))，Div(LengthOfLine(EH)=LengthOfLine(HF))，DE=2，DG=15，EH=15，GH=2，AD平行于BC，DE平行于HC，DG⊥FG，EH⊥FH。求直线CF的长度。
Construction (CDL):
Shape(AB,BG,GD,DA)
Shape(DG,GH,HI,ID)
Shape(ED,DI,IE)
Shape(IH,HC,CI)
Shape(EI,IC,CF,FE)
Collinear(ADE)
Collinear(BGHCF)
Collinear(EIH)
Collinear(DIC)
Conditions (CDL):
Equal(Div(LengthOfLine(DG),LengthOfLine(CG)),5/3)
Equal(Div(LengthOfLine(EH),LengthOfLine(HF)),3/4)
Equal(LengthOfLine(DE),2)
Equal(LengthOfLine(DG),15)
Equal(LengthOfLine(EH),15)
Equal(LengthOfLine(GH),2)
ParallelBetweenLine(AD,BC)
ParallelBetweenLine(DE,HC)
PerpendicularBetweenLine(DG,FG)
PerpendicularBetweenLine(EH,FH)
Goal: Value(LengthOfLine(CF))
Answer: 13
Theorem sequence:
line_addition(1,GH,HC)
line_addition(1,HC,CF)
Problem: As shown in the diagram, PA=60, ∠APB=105°, ∠BAP=30°, ∠PBA=45°. Find the length of line BP.
问题：如图所示，PA=60，∠APB=105°，∠BAP=30°，∠PBA=45°。求直线BP的长度。
Construction (CDL):
Shape(AP,PB,BA)
Conditions (CDL):
Equal(LengthOfLine(PA),60)
Equal(MeasureOfAngle(APB),105)
Equal(MeasureOfAngle(BAP),30)
Equal(MeasureOfAngle(PBA),45)
Goal: Value(LengthOfLine(BP))
Answer: 30*sqrt(2)
Theorem sequence:
sine_theorem(1,PBA)
Problem: As shown in the diagram, AO=500, ∠AOB=30°, ∠OBA=90°. Find the length of line BA.
问题：如图所示，AO=500，∠AOB=30°，∠OBA=90°。求直线BA的长度。
Construction (CDL):
Shape(AO,OB,BA)
Conditions (CDL):
Equal(LengthOfLine(AO),500)
Equal(MeasureOfAngle(AOB),30)
Equal(MeasureOfAngle(OBA),90)
Goal: Value(LengthOfLine(BA))
Answer: 250
Theorem sequence:
sine_theorem(1,AOB)
Problem: As shown in the diagram, PA=50, PB=40, ∠BPA=90°. Find the ratio of the length of line BP to the length of line AP.
问题：如图所示，PA=50，PB=40，∠BPA=90°。求直线BP的长度与直线AP的长度之比。
Construction (CDL):
Shape(AB,BP,PA)
Conditions (CDL):
Equal(LengthOfLine(PA),50)
Equal(LengthOfLine(PB),40)
Equal(MeasureOfAngle(BPA),90)
Goal: Value(Div(LengthOfLine(BP),LengthOfLine(AP)))
Answer: 4/5
Theorem sequence:
Problem: As shown in the diagram, PA=80, ∠APB=105°, ∠BAP=30°, ∠PBA=45°. Find the length of line BP.
问题：如图所示，PA=80，∠APB=105°，∠BAP=30°，∠PBA=45°。求直线BP的长度。
Construction (CDL):
Shape(AP,PB,BA)
Conditions (CDL):
Equal(LengthOfLine(PA),80)
Equal(MeasureOfAngle(APB),105)
Equal(MeasureOfAngle(BAP),30)
Equal(MeasureOfAngle(PBA),45)
Goal: Value(LengthOfLine(BP))
Answer: 40*sqrt(2)
Theorem sequence:
sine_theorem(1,PBA)
Problem: As shown in the diagram, CA=20, ∠BCA=45°, ∠CAB=90°. Find the length of line BC.
问题：如图所示，CA=20，∠BCA=45°，∠CAB=90°。求直线BC的长度。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(CA),20)
Equal(MeasureOfAngle(BCA),45)
Equal(MeasureOfAngle(CAB),90)
Goal: Value(LengthOfLine(BC))
Answer: 20*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,BCA)
sine_theorem(1,CAB)
Problem: As shown in the diagram, ∠BDC=90°, ∠DAB=30°, ∠DCB=60°. Find the ratio of the length of line AC to the length of line DC.
问题：如图所示，∠BDC=90°，∠DAB=30°，∠DCB=60°。求直线AC的长度与直线DC的长度之比。
Construction (CDL):
Shape(BD,DC,CB)
Shape(BC,CA,AB)
Collinear(DCA)
Conditions (CDL):
Equal(MeasureOfAngle(BDC),90)
Equal(MeasureOfAngle(DAB),30)
Equal(MeasureOfAngle(DCB),60)
Goal: Value(Div(LengthOfLine(AC),LengthOfLine(DC)))
Answer: 2
Theorem sequence:
flat_angle(1,DCA)
angle_addition(1,DCB,BCA)
triangle_property_angle_sum(1,BDC)
triangle_property_angle_sum(1,BCA)
isosceles_triangle_judgment_angle_equal(1,CAB)
sine_theorem(1,CBD)
Problem: As shown in the diagram, AB=200, ∠ABC=120°, ∠CAB=30°. Find the length of line BC.
问题：如图所示，AB=200，∠ABC=120°，∠CAB=30°。求直线BC的长度。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),200)
Equal(MeasureOfAngle(ABC),120)
Equal(MeasureOfAngle(CAB),30)
Goal: Value(LengthOfLine(BC))
Answer: 200
Theorem sequence:
triangle_property_angle_sum(1,CAB)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AC=1200, ∠BCA=90°, ∠CAB=60°. Find the length of line AB.
问题：如图所示，AC=1200，∠BCA=90°，∠CAB=60°。求直线AB的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),1200)
Equal(MeasureOfAngle(BCA),90)
Equal(MeasureOfAngle(CAB),60)
Goal: Value(LengthOfLine(AB))
Answer: 2400
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,ABC)
Problem: As shown in the diagram, BD=12, ∠ABD=15°, ∠BDA=150°, ∠DCA=90°. Find the length of line AC.
问题：如图所示，BD=12，∠ABD=15°，∠BDA=150°，∠DCA=90°。求直线AC的长度。
Construction (CDL):
Shape(AB,BD,DA)
Collinear(BDC)
Shape(AD,DC,CA)
Conditions (CDL):
Equal(LengthOfLine(BD),12)
Equal(MeasureOfAngle(ABD),15)
Equal(MeasureOfAngle(BDA),150)
Equal(MeasureOfAngle(DCA),90)
Goal: Value(LengthOfLine(AC))
Answer: 6
Theorem sequence:
flat_angle(1,BDC)
angle_addition(1,BDA,ADC)
triangle_property_angle_sum(1,ABD)
sine_theorem(1,DAB)
sine_theorem(1,ADC)
Problem: As shown in the diagram, AB=1, MC=MB, ∠CAM=30°, ∠CBM=45°, MD is the perpendicular bisector of CB. Find the length of line CB.
问题：如图所示，AB=1，MC=MB，∠CAM=30°，∠CBM=45°，MD垂直平分AB。求直线CB的长度。
Construction (CDL):
Shape(MC,CD,DM)
Shape(MD,DB,BM)
Shape(MB,BA,AM)
Collinear(CDBA)
Conditions (CDL):
Equal(LengthOfLine(AB),1)
Equal(LengthOfLine(MC),LengthOfLine(MB))
Equal(MeasureOfAngle(CAM),30)
Equal(MeasureOfAngle(CBM),45)
IsPerpendicularBisectorOfLine(MD,CB)
Goal: Value(LengthOfLine(CB))
Answer: -1+sqrt(3)
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,MCB)
isosceles_triangle_property_angle_equal(1,MCB)
triangle_property_angle_sum(1,MBA)
triangle_property_angle_sum(1,MCB)
sine_theorem(1,MCD)
sine_theorem(1,MBA)
sine_theorem(1,BAM)
sine_theorem(1,CBM)
sine_theorem(1,MDA)
Problem: As shown in the diagram, AB=25, ∠CAB=26°, ∠CBD=52°. Find the length of line BC.
问题：如图所示，AB=25，∠CAB=26°，∠CBD=52°。求直线BC的长度。
Construction (CDL):
Shape(CB,BD)
Shape(BC,CA,AB)
Collinear(ABD)
Conditions (CDL):
Equal(LengthOfLine(AB),25)
Equal(MeasureOfAngle(CAB),26)
Equal(MeasureOfAngle(CBD),52)
Goal: Value(LengthOfLine(BC))
Answer: 25
Theorem sequence:
flat_angle(1,ABD)
angle_addition(1,ABC,CBD)
triangle_property_angle_sum(1,BCA)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=20, ∠ABM=105°, ∠MAB=45°. Find the length of line BM.
问题：如图所示，AB=20，∠ABM=105°，∠MAB=45°。求直线BM的长度。
Construction (CDL):
Shape(MA,AB,BM)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(MeasureOfAngle(ABM),105)
Equal(MeasureOfAngle(MAB),45)
Goal: Value(LengthOfLine(BM))
Answer: 20*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,MAB)
sine_theorem(1,BMA)
Problem: As shown in the diagram, AB=40, BC=30, ∠CBA=90°. Find the length of line AC.
问题：如图所示，AB=40，BC=30，∠CBA=90°。求直线AC的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),40)
Equal(LengthOfLine(BC),30)
Equal(MeasureOfAngle(CBA),90)
Goal: Value(LengthOfLine(AC))
Answer: 50
Theorem sequence:
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, CD=60, ∠ACB=30°, ∠ADB=60°, ∠CBA=90°. Find the length of line AB.
问题：如图所示，CD=60，∠ACB=30°，∠ADB=60°，∠CBA=90°。求直线AB的长度。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(CD),60)
Equal(MeasureOfAngle(ACB),30)
Equal(MeasureOfAngle(ADB),60)
Equal(MeasureOfAngle(CBA),90)
Goal: Value(LengthOfLine(AB))
Answer: 30*sqrt(3)
Theorem sequence:
flat_angle(1,CDB)
angle_addition(1,CDA,ADB)
triangle_property_angle_sum(1,ACD)
isosceles_triangle_judgment_angle_equal(1,DAC)
sine_theorem(1,ADB)
Problem: As shown in the diagram, AC=50, ∠BAD=30°, ∠BCD=60°, ∠CDB=90°. Find the length of line BD.
问题：如图所示，AC=50，∠BAD=30°，∠BCD=60°，∠CDB=90°。求直线BD的长度。
Construction (CDL):
Shape(BA,AC,CB)
Shape(BC,CD,DB)
Collinear(ACD)
Conditions (CDL):
Equal(LengthOfLine(AC),50)
Equal(MeasureOfAngle(BAD),30)
Equal(MeasureOfAngle(BCD),60)
Equal(MeasureOfAngle(CDB),90)
Goal: Value(LengthOfLine(BD))
Answer: 25*sqrt(3)
Theorem sequence:
flat_angle(1,ACD)
angle_addition(1,ACB,BCD)
triangle_property_angle_sum(1,BAC)
isosceles_triangle_judgment_angle_equal(1,CBA)
sine_theorem(1,BCD)
Problem: As shown in the diagram, AO=500, ∠AOB=30°, ∠OBA=90°. Find the length of line AB.
问题：如图所示，AO=500，∠AOB=30°，∠OBA=90°。求直线AB的长度。
Construction (CDL):
Shape(AO,OB,BA)
Conditions (CDL):
Equal(LengthOfLine(AO),500)
Equal(MeasureOfAngle(AOB),30)
Equal(MeasureOfAngle(OBA),90)
Goal: Value(LengthOfLine(AB))
Answer: 250
Theorem sequence:
sine_theorem(1,AOB)
Problem: As shown in the diagram, MN=200, ∠MNP=60°, ∠PMN=90°. Find the length of line PM.
问题：如图所示，MN=200，∠MNP=60°，∠PMN=90°。求直线PM的长度。
Construction (CDL):
Shape(PM,MN,NP)
Conditions (CDL):
Equal(LengthOfLine(MN),200)
Equal(MeasureOfAngle(MNP),60)
Equal(MeasureOfAngle(PMN),90)
Goal: Value(LengthOfLine(PM))
Answer: 200*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,PMN)
sine_theorem(1,MNP)
Problem: As shown in the diagram, AB=14, ∠ABM=105°, ∠MAB=30°. Find the length of line BM.
问题：如图所示，AB=14，∠ABM=105°，∠MAB=30°。求直线BM的长度。
Construction (CDL):
Shape(MA,AB,BM)
Conditions (CDL):
Equal(LengthOfLine(AB),14)
Equal(MeasureOfAngle(ABM),105)
Equal(MeasureOfAngle(MAB),30)
Goal: Value(LengthOfLine(BM))
Answer: 7*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,MAB)
sine_theorem(1,BMA)
Problem: As shown in the diagram, AC=50, ∠ABC=30°, ∠BCA=90°. Find the length of line BC.
问题：如图所示，AC=50，∠ABC=30°，∠BCA=90°。求直线BC的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),50)
Equal(MeasureOfAngle(ABC),30)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(LengthOfLine(BC))
Answer: 50*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,CAB)
Problem: As shown in the diagram, AB=30, ∠BCA=30°, ∠BCD=90°, ∠CAB=90°, ∠CDA=90°. Find the length of line DA.
问题：如图所示，AB=30，∠BCA=30°，∠BCD=90°，∠CAB=90°，∠CDA=90°。求直线DA的长度。
Construction (CDL):
Shape(BC,CA,AB)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(LengthOfLine(AB),30)
Equal(MeasureOfAngle(BCA),30)
Equal(MeasureOfAngle(BCD),90)
Equal(MeasureOfAngle(CAB),90)
Equal(MeasureOfAngle(CDA),90)
Goal: Value(LengthOfLine(DA))
Answer: 45
Theorem sequence:
angle_addition(1,BCA,ACD)
triangle_property_angle_sum(1,BCA)
sine_theorem(1,ABC)
sine_theorem(1,ACD)
Problem: As shown in the diagram, AB=30, ∠BDM=90°, ∠BNA=90°, ∠CAB=90°, ∠CAM=45°, ∠MBD=60°. Find the length of line BM.
问题：如图所示，AB=30，∠BDM=90°，∠BNA=90°，∠CAB=90°，∠CAM=45°，∠MBD=60°。求直线BM的长度。
Construction (CDL):
Shape(CA,AN,NM,MC)
Shape(AB,BN,NA)
Shape(NB,BM,MN)
Shape(MB,BD,DM)
Collinear(ABD)
Collinear(ANM)
Conditions (CDL):
Equal(LengthOfLine(AB),30)
Equal(MeasureOfAngle(BDM),90)
Equal(MeasureOfAngle(BNA),90)
Equal(MeasureOfAngle(CAB),90)
Equal(MeasureOfAngle(CAM),45)
Equal(MeasureOfAngle(MBD),60)
Goal: Value(LengthOfLine(BM))
Answer: 30+30*sqrt(3)
Theorem sequence:
angle_addition(1,CAN,NAB)
triangle_property_angle_sum(1,NAB)
triangle_property_angle_sum(1,NBM)
angle_addition(1,NBM,MBD)
adjacent_complementary_angle(1,ABN,NBD)
adjacent_complementary_angle(1,MNB,BNA)
sine_theorem(1,BNA)
sine_theorem(1,BMN)
Problem: As shown in the diagram, AB=5, AC=4, ∠BCA=90°. Find the value of sin(AB).
问题：如图所示，AB=5，AC=4，∠BCA=90°。求sin(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Sin(MeasureOfAngle(ABC)))
Answer: 4/5
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, AB=4, BC=1, ∠BCA=90°. Find the value of cos(CA).
问题：如图所示，AB=4，BC=1，∠BCA=90°。求cos(CA)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(BC),1)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Cos(MeasureOfAngle(CAB)))
Answer: sqrt(15)/4
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, OM=5, PM=12, O is the center of circle O, the tangent to ⊙O is PM. Find the value of sin(MP).
问题：如图所示，OM=5，PM=12，⊙O的圆心为O，圆O的切线为PM。求sin(MP)的值。
Construction (CDL):
Shape(PA,OMA,MP)
Shape(AO,OM,OMA)
Shape(MO,OA,OAM)
Collinear(PAO)
Cocircular(O,AM)
Conditions (CDL):
Equal(LengthOfLine(OM),5)
Equal(LengthOfLine(PM),12)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PM,O)
Goal: Value(Sin(MeasureOfAngle(MPO)))
Answer: 5/13
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PM,O,O)
cosine_theorem(1,MPO)
cosine_theorem(1,POM)
Problem: As shown in the diagram, AB=4, AD=3, quadrilateral ABCD is a rectangle. Find the value of cos(AB).
问题：如图所示，AB=4，AD=3，四边形ABCD是长方形。求cos(AB)的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AD),3)
Rectangle(ABCD)
Goal: Value(Cos(MeasureOfAngle(ABD)))
Answer: 4/5
Theorem sequence:
cosine_theorem(1,ABD)
cosine_theorem(1,BDA)
Problem: As shown in the diagram, OA=3, PA=4, the center of circle O is O, PA is the tangent to ⊙O. Find the value of cos(AP).
问题：如图所示，OA=3，PA=4，⊙O的圆心为O，圆O的切线为PA。求cos(AP)的值。
Construction (CDL):
Shape(PB,OAB,AP)
Shape(BO,OA,OAB)
Shape(AO,OB,OBA)
Collinear(OBP)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(OA),3)
Equal(LengthOfLine(PA),4)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(Cos(MeasureOfAngle(APO)))
Answer: 4/5
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
cosine_theorem(1,APO)
cosine_theorem(1,POA)
Problem: As shown in the diagram, AB=10, AC=8, CD is perpendicular to BD, BC and CA are the legs of the right △ BCA. Find the value of sin(DC).
问题：如图所示，AB=10，AC=8，CD⊥BD，BC和CA是直角△BCA的两个直角边。求sin(DC)的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),8)
PerpendicularBetweenLine(CD,BD)
RightTriangle(BCA)
Goal: Value(Sin(MeasureOfAngle(DCA)))
Answer: 4/5
Theorem sequence:
adjacent_complementary_angle(1,ADC,CDB)
sine_theorem(1,ABC)
triangle_property_angle_sum(1,ADC)
triangle_property_angle_sum(1,CAB)
Problem: As shown in the diagram, AB=8, BP=2, the radius of circle O is 5, O is the center of circle O, OD⊥AD. Find the value of tan(OP).
问题：如图所示，AB=8，BP=2，圆O的半径为5，O是圆O的圆心，OD垂直于AD。求tan(OP)的值。
Construction (CDL):
Shape(AO,OD,DA)
Shape(AB,OBA)
Shape(DO,OB,BD)
Shape(BO,OC,OCB)
Shape(CP,PB,OCB)
Shape(CO,OA,OAC)
Collinear(ADBP)
Collinear(OCP)
Cocircular(O,CBA)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(BP),2)
Equal(RadiusOfCircle(O),5)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(OD,AD)
Goal: Value(Tan(MeasureOfAngle(OPA)))
Answer: 1/2
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OA,O)
adjacent_complementary_angle(1,PDO,ODA)
mirror_congruent_triangle_judgment_hl(1,BDO,AOD)
mirror_congruent_triangle_property_line_equal(1,OBD,ODA)
line_addition(1,AD,DB)
line_addition(1,DB,BP)
right_triangle_judgment_angle(1,BDO)
right_triangle_judgment_angle(1,PDO)
right_triangle_property_pythagorean(1,BDO)
right_triangle_property_pythagorean(1,PDO)
cosine_theorem(1,PDO)
Problem: As shown in the diagram, AC=8, BC=6, ED=CE, AB is the diameter of ⊙O, AE is perpendicular to CE. Find the value of sin(DB).
问题：如图所示，AC=8，BC=6，ED=CE，圆O的直径为AB，AE垂直于CE。求sin(DB)的值。
Construction (CDL):
Shape(AC,OCA)
Shape(AO,OE,EC,CA)
Shape(CE,EB,BC)
Shape(CB,OBC)
Shape(ED,DB,BE)
Shape(BD,ODB)
Shape(DE,EO,OA,AD)
Shape(DA,OAD)
Collinear(AOEB)
Collinear(CED)
Cocircular(O,ADBC)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(ED),LengthOfLine(CE))
IsDiameterOfCircle(AB,O)
PerpendicularBetweenLine(AE,CE)
Goal: Value(Sin(MeasureOfAngle(DBE)))
Answer: 4/5
Theorem sequence:
adjacent_complementary_angle(1,AEC,CEB)
adjacent_complementary_angle(1,CEB,BED)
diameter_of_circle_property_right_angle(1,BCA,O)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
sine_theorem(1,ABC)
mirror_congruent_triangle_judgment_sas(1,EBC,EDB)
mirror_congruent_triangle_property_angle_equal(1,BCE,BED)
Problem: As shown in the diagram, AC=4, BC=3, CD is the altitude of △ CAB, BC⊥AC. Find the value of cos(BC).
问题：如图所示，AC=4，BC=3，CD是三角形CAB的高，BC垂直于AC。求cos(BC)的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BC),3)
IsAltitudeOfTriangle(CD,CAB)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Cos(MeasureOfAngle(BCD)))
Answer: 4/5
Theorem sequence:
right_triangle_judgment_angle(1,ADC)
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,ADC)
right_triangle_property_pythagorean(1,CDB)
right_triangle_property_pythagorean(1,BCA)
line_addition(1,AD,DB)
sine_theorem(1,BCD)
Problem: As shown in the diagram, AB=5, AC=3, BC=4. Find the value of tan(AB).
问题：如图所示，AB=5，AC=3，BC=4。求tan(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(BC),4)
Goal: Value(Tan(MeasureOfAngle(ABC)))
Answer: 3/4
Theorem sequence:
right_triangle_judgment_pythagorean_inverse(1,BCA)
sine_theorem(1,ABC)
Problem: As shown in the diagram, AB=13, BC=5, ∠BCA=90°. Find the value of sin(CA).
问题：如图所示，AB=13，BC=5，∠BCA=90°。求sin(CA)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(BC),5)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(Sin(MeasureOfAngle(CAB)))
Answer: 5/13
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=5, AC=3, ∠BCA=90°, CD⊥BD. Find the value of tan(BC).
问题：如图所示，AB=5，AC=3，∠BCA=90°，CD垂直于BD。求tan(BC)的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),3)
Equal(MeasureOfAngle(BCA),90)
PerpendicularBetweenLine(CD,BD)
Goal: Value(Tan(MeasureOfAngle(BCD)))
Answer: 4/3
Theorem sequence:
line_addition(1,AD,DB)
adjacent_complementary_angle(1,ADC,CDB)
right_triangle_judgment_angle(1,ADC)
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,ADC)
right_triangle_property_pythagorean(1,CDB)
right_triangle_property_pythagorean(1,BCA)
sine_theorem(1,BCD)
Problem: As shown in the diagram, AP=3, OA=2, ∠OAP=90°. Find the value of tan(PO).
问题：如图所示，AP=3，OA=2，∠OAP=90°。求tan(PO)的值。
Construction (CDL):
Shape(OA,AP,PO)
Conditions (CDL):
Equal(LengthOfLine(AP),3)
Equal(LengthOfLine(OA),2)
Equal(MeasureOfAngle(OAP),90)
Goal: Value(Tan(MeasureOfAngle(POA)))
Answer: 3/2
Theorem sequence:
right_triangle_judgment_angle(1,OAP)
right_triangle_property_pythagorean(1,OAP)
sine_theorem(1,POA)
Problem: As shown in the diagram, AC=CD, AD=5000, OA=3000, ∠COA=90°. Find the length of line CD.
问题：如图所示，AC=CD，AD=5000，OA=3000，∠COA=90°。求直线CD的长度。
Construction (CDL):
Shape(OA,AC,CO)
Shape(CA,AD,DC)
Collinear(OCD)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(CD))
Equal(LengthOfLine(AD),5000)
Equal(LengthOfLine(OA),3000)
Equal(MeasureOfAngle(COA),90)
Goal: Value(LengthOfLine(CD))
Answer: 3125
Theorem sequence:
line_addition(1,OC,CD)
right_triangle_judgment_angle(1,COA)
right_triangle_judgment_angle(1,DOA)
right_triangle_property_pythagorean(1,COA)
right_triangle_property_pythagorean(1,DOA)