SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, CB=1/2, CG=12, FE=3/2, GE=2, ∠CGA=∠FGE, CBDE is a rectangle, AC and CG are the legs of the right △ ACG, ∠GEF is the right angle of the right triangle GEF. Find the length of line AB.
问题：如图所示，CB=1/2，CG=12，FE=3/2，GE=2，∠CGA=∠FGE，CBDE是长方形，△ACG为直角△，∠GEF是直角三角形GEF的直角。求直线AB的长度。
Construction (CDL):
Shape(AC,CG,GA)
Shape(FG,GE,EF)
Shape(CB,BD,DE,EG,GC)
Collinear(CGE)
Collinear(ACB)
Collinear(FED)
Conditions (CDL):
Equal(LengthOfLine(CB),1/2)
Equal(LengthOfLine(CG),12)
Equal(LengthOfLine(FE),3/2)
Equal(LengthOfLine(GE),2)
Equal(MeasureOfAngle(CGA),MeasureOfAngle(FGE))
Rectangle(CBDE)
RightTriangle(ACG)
RightTriangle(GEF)
Goal: Value(LengthOfLine(AB))
Answer: 19/2
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,ACG,FGE)
mirror_similar_triangle_property_line_ratio(1,ACG,FGE)
mirror_similar_triangle_property_line_ratio(1,GAC,GEF)
line_addition(1,AC,CB)
Problem: As shown in the diagram, AD=8, DB=22, DE=3, AD is perpendicular to ED, DB⊥CB. Find the length of line BC.
问题：如图所示，AD=8，DB=22，DE=3，AD⊥ED，DB⊥CB。求直线BC的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(CE,ED,DB,BC)
Collinear(AEC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(DB),22)
Equal(LengthOfLine(DE),3)
PerpendicularBetweenLine(AD,ED)
PerpendicularBetweenLine(DB,CB)
Goal: Value(LengthOfLine(BC))
Answer: 45/4
Theorem sequence:
line_addition(1,AD,DB)
similar_triangle_judgment_aa(1,EAD,CAB)
similar_triangle_property_line_ratio(1,EAD,CAB)
similar_triangle_property_line_ratio(1,ADE,ABC)
Problem: As shown in the diagram, QR=60, QS=60, ST=120, PQ⊥RQ, QS is perpendicular to TS. Find the length of line PQ.
问题：如图所示，QR=60，QS=60，ST=120，PQ⊥RQ，QS垂直于TS。求直线PQ的长度。
Construction (CDL):
Shape(PQ,QR,RP)
Shape(RQ,QS,ST,TR)
Collinear(PQS)
Collinear(PRT)
Conditions (CDL):
Equal(LengthOfLine(QR),60)
Equal(LengthOfLine(QS),60)
Equal(LengthOfLine(ST),120)
PerpendicularBetweenLine(PQ,RQ)
PerpendicularBetweenLine(QS,TS)
Goal: Value(LengthOfLine(PQ))
Answer: 60
Theorem sequence:
line_addition(1,PQ,QS)
similar_triangle_judgment_aa(1,RPQ,TPS)
similar_triangle_property_line_ratio(1,RPQ,TPS)
similar_triangle_property_line_ratio(1,PQR,PST)
Problem: As shown in the diagram, ∠AOB=96°, O is the center of ⊙O. Find the measure of ∠BCA.
问题：如图所示，∠AOB=96°，⊙O的圆心为O。求∠BCA的大小。
Construction (CDL):
Shape(OCA,AC)
Shape(OBC,CB)
Shape(OAB,BO,OA)
Shape(BC,CA,AO,OB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(AOB),96)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BCA))
Answer: 132
Theorem sequence:
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_internal(1,OBA,C)
Problem: As shown in the diagram, AD=80, BC=120, AD is the altitude of △ ABC, EFGH is a square. Find the length of line EH.
问题：如图所示，AD=80，BC=120，AD是三角形ABC的高，EFGH是正方形。求直线EH的长度。
Construction (CDL):
Shape(AE,EK,KA)
Shape(AK,KH,HA)
Shape(EB,BF,FE)
Shape(EF,FD,DK,KE)
Shape(HK,KD,DG,GH)
Shape(HG,GC,CH)
Collinear(EKH)
Collinear(BFDGC)
Collinear(AKD)
Collinear(AEB)
Collinear(AHC)
Conditions (CDL):
Equal(LengthOfLine(AD),80)
Equal(LengthOfLine(BC),120)
IsAltitudeOfTriangle(AD,ABC)
Square(EFGH)
Goal: Value(LengthOfLine(EH))
Answer: 48
Theorem sequence:
parallel_judgment_corresponding_angle(2,FE,DK,G)
parallel_property_collinear_extend(1,GF,HE,C)
parallel_property_collinear_extend(2,GF,HE,B)
parallel_property_collinear_extend(3,EH,BF,K)
parallel_property_collinear_extend(1,FB,KE,D)
parallelogram_judgment_parallel_and_parallel(1,KEFD)
parallelogram_property_opposite_line_equal(1,EFDK)
line_addition(1,AK,KD)
parallel_property_corresponding_angle(1,EH,BG,A)
parallel_property_corresponding_angle(2,CG,HE,A)
parallel_property_corresponding_angle(2,DF,KE,A)
similar_triangle_judgment_aa(1,AEH,ABC)
altitude_of_triangle_judgment(1,AK,AEH)
triangle_area_formula_common(1,AEH)
triangle_area_formula_common(1,ABC)
similar_triangle_property_line_ratio(1,AEH,ABC)
similar_triangle_property_area_square_ratio(1,AEH,ABC)
Problem: As shown in the diagram, BD=120, CE=30, DC=40, AB is perpendicular to DB, EC⊥DC. Find the length of line AB.
问题：如图所示，BD=120，CE=30，DC=40，AB垂直于DB，EC垂直于DC。求直线AB的长度。
Construction (CDL):
Shape(AB,BD,DA)
Shape(CD,DE,EC)
Collinear(BDC)
Collinear(ADE)
Conditions (CDL):
Equal(LengthOfLine(BD),120)
Equal(LengthOfLine(CE),30)
Equal(LengthOfLine(DC),40)
PerpendicularBetweenLine(AB,DB)
PerpendicularBetweenLine(EC,DC)
Goal: Value(LengthOfLine(AB))
Answer: 90
Theorem sequence:
vertical_angle(1,BDA,CDE)
similar_triangle_judgment_aa(1,ABD,ECD)
similar_triangle_property_line_ratio(1,ABD,ECD)
similar_triangle_property_line_ratio(1,DAB,DEC)
Problem: As shown in the diagram, AC=3, BD=4, CD=11, ∠CEA=∠BED, AC⊥EC, CE is perpendicular to FE, ED is perpendicular to BD. Find the value of tan(AE).
问题：如图所示，AC=3，BD=4，CD=11，∠CEA=∠BED，AC垂直于EC，CE垂直于FE，ED⊥BD。求tan(AE)的值。
Construction (CDL):
Shape(AC,CE,EA)
Shape(BE,ED,DB)
Shape(AE,EF)
Shape(FE,EB)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(BD),4)
Equal(LengthOfLine(CD),11)
Equal(MeasureOfAngle(CEA),MeasureOfAngle(BED))
PerpendicularBetweenLine(AC,EC)
PerpendicularBetweenLine(CE,FE)
PerpendicularBetweenLine(ED,BD)
Goal: Value(Tan(MeasureOfAngle(AEF)))
Answer: tan(I*(log(85)-log(-36+77*I))/2)
Theorem sequence:
line_addition(1,CE,ED)
mirror_similar_triangle_judgment_aa(1,ACE,BED)
mirror_similar_triangle_property_line_ratio(1,ACE,BED)
mirror_similar_triangle_property_line_ratio(1,EAC,EDB)
triangle_property_angle_sum(1,ACE)
sine_theorem(1,CEA)
angle_addition(1,CEA,AEF)
Problem: As shown in the diagram, ∠BOA=140°, the center of ⊙O is O, M bisects arc OAB. Find the measure of ∠BNO.
问题：如图所示，∠BOA=140°，O是圆O的圆心，M是⌒OAB的中点。求∠BNO的大小。
Construction (CDL):
Shape(ONA,AO,ON)
Shape(OAM,MO,OA)
Shape(OMB,BO,OM)
Shape(OBN,NB)
Shape(NO,OB,BN)
Collinear(NOM)
Cocircular(O,AMBN)
Conditions (CDL):
Equal(MeasureOfAngle(BOA),140)
IsCentreOfCircle(O,O)
IsMidpointOfArc(M,OAB)
Goal: Value(MeasureOfAngle(BNO))
Answer: 35
Theorem sequence:
arc_property_center_angle(1,OAB,O)
arc_property_center_angle(1,OMB,O)
congruent_arc_judgment_length_equal(1,OAM,OMB)
congruent_arc_property_measure_equal(1,OAM,OMB)
arc_addition_measure(1,OAM,OMB)
adjacent_complementary_angle(1,NOB,BOM)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,ON,O)
isosceles_triangle_judgment_line_equal(1,OBN)
isosceles_triangle_property_angle_equal(1,OBN)
triangle_property_angle_sum(1,OBN)
Problem: As shown in the diagram, ∠EAO=20°, the center of circle O is O. Find the measure of ∠DCA.
问题：如图所示，∠EAO=20°，O是圆O的圆心。求∠DCA的大小。
Construction (CDL):
Shape(OCA,AC)
Shape(ODC,CD)
Shape(OBD,DB)
Shape(DC,CE,ED)
Shape(BD,DE,EB)
Shape(CA,AE,EC)
Shape(BE,EA,AO,OB)
Shape(OAB,BO,OA)
Collinear(AOB)
Collinear(AED)
Collinear(CEB)
Cocircular(O,ABDC)
Conditions (CDL):
Equal(MeasureOfAngle(EAO),20)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(DCA))
Answer: 110
Theorem sequence:
diameter_of_circle_judgment_pass_centre(1,AOB,O)
diameter_of_circle_property_right_angle(1,BCA,O)
arc_property_circumference_angle_external(1,OBD,A)
arc_property_circumference_angle_external(1,OBD,C)
angle_addition(1,DCE,ECA)
Problem: As shown in the diagram, ∠BCD=44°, O is the center of ⊙O. Find the measure of ∠DOA.
问题：如图所示，∠BCD=44°，圆O的圆心为O。求∠DOA的大小。
Construction (CDL):
Shape(OCA,AO,OD,DC)
Shape(OAB,BD,DA)
Shape(OA,AD,DO)
Shape(CD,DB,BC)
Shape(OBC,CB)
Collinear(ADC)
Collinear(ODB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),44)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(DOA))
Answer: 88
Theorem sequence:
arc_property_circumference_angle_external(1,OAB,C)
arc_property_center_angle(1,OAB,O)
Problem: As shown in the diagram, ∠COA=120°, O is the center of circle O, BD is the diameter of circle O, AE is perpendicular to OE. Find the measure of ∠CDB.
问题：如图所示，∠COA=120°，O是⊙O的圆心，BD是圆O的直径，AE垂直于OE。求∠CDB的大小。
Construction (CDL):
Shape(ODA,AO,OD)
Shape(OCD,DC)
Shape(OAB,BE,EA)
Shape(OBC,CE,EB)
Shape(DO,OC,CD)
Shape(OA,AE,EO)
Shape(OE,EC,CO)
Collinear(AEC)
Collinear(DOEB)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(COA),120)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(BD,O)
PerpendicularBetweenLine(AE,OE)
Goal: Value(MeasureOfAngle(CDB))
Answer: 30
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OE,AC)
perpendicular_bisector_property_distance_equal(1,OE,AC)
isosceles_triangle_judgment_line_equal(1,OAC)
isosceles_triangle_property_angle_equal(1,OAC)
triangle_property_angle_sum(1,OAC)
triangle_property_angle_sum(1,OEC)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,D)
Problem: As shown in the diagram, BC=7/5, BD=1/2, DE=6/5, DE is parallel to BC. Find the length of line AB.
问题：如图所示，BC=7/5，BD=1/2，DE=6/5，DE平行于BC。求直线AB的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CE,ED)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(BC),7/5)
Equal(LengthOfLine(BD),1/2)
Equal(LengthOfLine(DE),6/5)
ParallelBetweenLine(DE,BC)
Goal: Value(LengthOfLine(AB))
Answer: 7/2
Theorem sequence:
parallel_property_corresponding_angle(1,DE,BC,A)
similar_triangle_judgment_aa(1,EAD,CAB)
similar_triangle_property_line_ratio(1,EAD,CAB)
similar_triangle_property_line_ratio(1,ADE,ABC)
line_addition(1,AD,DB)
Problem: As shown in the diagram, CD=1, DE=3, EB=8, ∠DEC=∠AEB, CD⊥ED, EB is perpendicular to AB. Find the length of line AB.
问题：如图所示，CD=1，DE=3，EB=8，∠DEC=∠AEB，CD⊥ED，EB⊥AB。求直线AB的长度。
Construction (CDL):
Shape(CD,DE,EC)
Shape(AE,EB,BA)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(CD),1)
Equal(LengthOfLine(DE),3)
Equal(LengthOfLine(EB),8)
Equal(MeasureOfAngle(DEC),MeasureOfAngle(AEB))
PerpendicularBetweenLine(CD,ED)
PerpendicularBetweenLine(EB,AB)
Goal: Value(LengthOfLine(AB))
Answer: 8/3
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,CDE,AEB)
mirror_similar_triangle_property_line_ratio(1,CDE,AEB)
mirror_similar_triangle_property_line_ratio(1,ECD,EBA)
Problem: As shown in the diagram, AB=6, CD=4, CD is perpendicular to ED, EB is perpendicular to AB, PE⊥BE. Find the length of line PE.
问题：如图所示，AB=6，CD=4，CD⊥ED，EB⊥AB，PE⊥BE。求直线PE的长度。
Construction (CDL):
Shape(CD,DP,PC)
Shape(PD,DE,EP)
Shape(PE,EB,BP)
Shape(AP,PB,BA)
Collinear(CPB)
Collinear(APD)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(CD,ED)
PerpendicularBetweenLine(EB,AB)
PerpendicularBetweenLine(PE,BE)
Goal: Value(LengthOfLine(PE))
Answer: 12/5
Theorem sequence:
similar_triangle_judgment_aa(1,PEB,CDB)
parallel_judgment_ipsilateral_internal_angle(1,DC,BA)
parallel_property_alternate_interior_angle(1,DC,BA)
parallel_property_alternate_interior_angle(2,DC,BA)
similar_triangle_judgment_aa(1,PCD,PBA)
similar_triangle_property_line_ratio(1,PCD,PBA)
similar_triangle_property_line_ratio(1,DPC,APB)
similar_triangle_property_line_ratio(1,EBP,DBC)
similar_triangle_property_line_ratio(1,BPE,BCD)
line_addition(1,CP,PB)
Problem: As shown in the diagram, BD=22, CD=16/5, DE=8, EB is perpendicular to AB, ED is perpendicular to CD. Find the length of line AB.
问题：如图所示，BD=22，CD=16/5，DE=8，EB⊥AB，ED垂直于CD。求直线AB的长度。
Construction (CDL):
Shape(CE,ED,DC)
Shape(AC,CD,DB,BA)
Collinear(ACE)
Collinear(BDE)
Conditions (CDL):
Equal(LengthOfLine(BD),22)
Equal(LengthOfLine(CD),16/5)
Equal(LengthOfLine(DE),8)
PerpendicularBetweenLine(EB,AB)
PerpendicularBetweenLine(ED,CD)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
similar_triangle_judgment_aa(1,CED,AEB)
similar_triangle_property_line_ratio(1,CED,AEB)
similar_triangle_property_line_ratio(1,EDC,EBA)
line_addition(1,ED,DB)
Problem: As shown in the diagram, AB=1/2, BC=1, CE=16, CA⊥BA, CD⊥ED. Find the length of line ED.
问题：如图所示，AB=1/2，BC=1，CE=16，CA垂直于BA，CD垂直于ED。求直线ED的长度。
Construction (CDL):
Shape(CA,AB,BC)
Shape(EC,CD,DE)
Collinear(ACD)
Collinear(BCE)
Conditions (CDL):
Equal(LengthOfLine(AB),1/2)
Equal(LengthOfLine(BC),1)
Equal(LengthOfLine(CE),16)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(CD,ED)
Goal: Value(LengthOfLine(ED))
Answer: 8
Theorem sequence:
vertical_angle(1,BCA,ECD)
similar_triangle_judgment_aa(1,BCA,ECD)
similar_triangle_property_line_ratio(1,CAB,CDE)
similar_triangle_property_line_ratio(1,ABC,DEC)
Problem: As shown in the diagram, CD=2, DE=4, EB=8, ∠DEC=∠AEB, CD is perpendicular to ED, EB is perpendicular to AB. Find the length of line AB.
问题：如图所示，CD=2，DE=4，EB=8，∠DEC=∠AEB，CD⊥ED，EB⊥AB。求直线AB的长度。
Construction (CDL):
Shape(CD,DE,EC)
Shape(AE,EB,BA)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(CD),2)
Equal(LengthOfLine(DE),4)
Equal(LengthOfLine(EB),8)
Equal(MeasureOfAngle(DEC),MeasureOfAngle(AEB))
PerpendicularBetweenLine(CD,ED)
PerpendicularBetweenLine(EB,AB)
Goal: Value(LengthOfLine(AB))
Answer: 4
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,CDE,AEB)
mirror_similar_triangle_property_line_ratio(1,CDE,AEB)
mirror_similar_triangle_property_line_ratio(1,ECD,EBA)
Problem: As shown in the diagram, BC=3, CD=1, ED=1/2, ED∥BC. Find the length of line CA.
问题：如图所示，BC=3，CD=1，ED=1/2，ED∥BC。求直线CA的长度。
Construction (CDL):
Shape(BC,CD,DE,EB)
Shape(AE,ED,DA)
Collinear(BEA)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(BC),3)
Equal(LengthOfLine(CD),1)
Equal(LengthOfLine(ED),1/2)
ParallelBetweenLine(ED,BC)
Goal: Value(LengthOfLine(CA))
Answer: 6/5
Theorem sequence:
parallel_property_corresponding_angle(1,ED,BC,A)
similar_triangle_judgment_aa(1,DAE,CAB)
similar_triangle_property_line_ratio(1,AED,ABC)
similar_triangle_property_line_ratio(1,EDA,BCA)
line_addition(1,CD,DA)
Problem: As shown in the diagram, AB=2, CD=5, PC=3, AB∥CD. Find the length of line AC.
问题：如图所示，AB=2，CD=5，PC=3，AB平行于CD。求直线AC的长度。
Construction (CDL):
Shape(PA,AB,BP)
Shape(AC,CD,DB,BA)
Collinear(PAC)
Collinear(PBD)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(CD),5)
Equal(LengthOfLine(PC),3)
ParallelBetweenLine(AB,CD)
Goal: Value(LengthOfLine(AC))
Answer: 9/5
Theorem sequence:
parallel_property_corresponding_angle(1,AB,CD,P)
similar_triangle_judgment_aa(1,BPA,DPC)
similar_triangle_property_line_ratio(1,BPA,DPC)
similar_triangle_property_line_ratio(1,PAB,PCD)
line_addition(1,PA,AC)
Problem: As shown in the diagram, AB=20, CE=15, ED=4, AB⊥CB, DE is perpendicular to CE. Find the length of line BC.
问题：如图所示，AB=20，CE=15，ED=4，AB⊥CB，DE⊥CE。求直线BC的长度。
Construction (CDL):
Shape(AB,BC,CA)
Shape(EC,CD,DE)
Collinear(ACD)
Collinear(BCE)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(CE),15)
Equal(LengthOfLine(ED),4)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(DE,CE)
Goal: Value(LengthOfLine(BC))
Answer: 75
Theorem sequence:
vertical_angle(1,BCA,ECD)
similar_triangle_judgment_aa(1,ABC,DEC)
similar_triangle_property_line_ratio(1,ABC,DEC)
similar_triangle_property_line_ratio(1,CAB,CDE)
Problem: As shown in the diagram, AB=3, AD=6, DE=16, BA∥CE. Find the length of line EC.
问题：如图所示，AB=3，AD=6，DE=16，BA∥CE。求直线EC的长度。
Construction (CDL):
Shape(AD,DB,BA)
Shape(EA,AB,BC,CE)
Collinear(DAE)
Collinear(DBC)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(DE),16)
ParallelBetweenLine(BA,CE)
Goal: Value(LengthOfLine(EC))
Answer: 8
Theorem sequence:
parallel_property_corresponding_angle(1,BA,CE,D)
similar_triangle_judgment_aa(1,ADB,EDC)
similar_triangle_property_line_ratio(1,DBA,DCE)
similar_triangle_property_line_ratio(1,BAD,CED)
Problem: As shown in the diagram, AB=12, CD=40, DA=20, CA is perpendicular to BA, CD is perpendicular to ED. Find the length of line ED.
问题：如图所示，AB=12，CD=40，DA=20，CA垂直于BA，CD⊥ED。求直线ED的长度。
Construction (CDL):
Shape(EC,CD,DE)
Shape(BE,ED,DA,AB)
Collinear(CEB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(CD),40)
Equal(LengthOfLine(DA),20)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(CD,ED)
Goal: Value(LengthOfLine(ED))
Answer: 8
Theorem sequence:
similar_triangle_judgment_aa(1,ECD,BCA)
similar_triangle_property_line_ratio(1,ECD,BCA)
similar_triangle_property_line_ratio(1,CDE,CAB)
line_addition(1,CD,DA)
Problem: As shown in the diagram, AB=1/2, AC=1, CD=8, CA⊥BA, CD⊥ED. Find the length of line ED.
问题：如图所示，AB=1/2，AC=1，CD=8，CA垂直于BA，CD⊥ED。求直线ED的长度。
Construction (CDL):
Shape(AB,BC,CA)
Shape(EC,CD,DE)
Collinear(ACD)
Collinear(BCE)
Conditions (CDL):
Equal(LengthOfLine(AB),1/2)
Equal(LengthOfLine(AC),1)
Equal(LengthOfLine(CD),8)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(CD,ED)
Goal: Value(LengthOfLine(ED))
Answer: 4
Theorem sequence:
vertical_angle(1,BCA,ECD)
similar_triangle_judgment_aa(1,BCA,ECD)
similar_triangle_property_line_ratio(1,BCA,ECD)
similar_triangle_property_line_ratio(1,CAB,CDE)
Problem: As shown in the diagram, BC=11, CA=4, EC=8, AB⊥DB, AC is perpendicular to EC. Find the length of line DB.
问题：如图所示，BC=11，CA=4，EC=8，AB⊥DB，AC⊥EC。求直线DB的长度。
Construction (CDL):
Shape(AC,CE,EA)
Shape(EC,CB,BD,DE)
Collinear(ACB)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(BC),11)
Equal(LengthOfLine(CA),4)
Equal(LengthOfLine(EC),8)
PerpendicularBetweenLine(AB,DB)
PerpendicularBetweenLine(AC,EC)
Goal: Value(LengthOfLine(DB))
Answer: 30
Theorem sequence:
similar_triangle_judgment_aa(1,EAC,DAB)
similar_triangle_property_line_ratio(1,EAC,DAB)
similar_triangle_property_line_ratio(1,ACE,ABD)
line_addition(1,AC,CB)
Problem: As shown in the diagram, ∠BCO=35°, ∠CDA=85°, the center of ⊙O is O. Find the measure of ∠BAO.
问题：如图所示，∠BCO=35°，∠CDA=85°，O是圆O的圆心。求∠BAO的大小。
Construction (CDL):
Shape(OCB,BD,DO,OC)
Shape(CO,OD,DC)
Shape(DB,BA,AD)
Shape(OBA,AB)
Shape(OAC,CD,DA)
Collinear(BDC)
Collinear(ODA)
Cocircular(O,BAC)
Conditions (CDL):
Equal(MeasureOfAngle(BCO),35)
Equal(MeasureOfAngle(CDA),85)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BAO))
Answer: 60
Theorem sequence:
adjacent_complementary_angle(1,ODC,CDA)
triangle_property_angle_sum(1,ODC)
arc_property_center_angle(1,OAC,O)
arc_property_circumference_angle_external(1,OAC,B)
adjacent_complementary_angle(1,CDA,ADB)
triangle_property_angle_sum(1,DBA)
Problem: As shown in the diagram, BA=5, CA=1, DC=2, AB is perpendicular to EB, AC is perpendicular to DC. Find the length of line EB.
问题：如图所示，BA=5，CA=1，DC=2，AB垂直于EB，AC垂直于DC。求直线EB的长度。
Construction (CDL):
Shape(AC,CD,DA)
Shape(DC,CB,BE,ED)
Collinear(ADE)
Collinear(ACB)
Conditions (CDL):
Equal(LengthOfLine(BA),5)
Equal(LengthOfLine(CA),1)
Equal(LengthOfLine(DC),2)
PerpendicularBetweenLine(AB,EB)
PerpendicularBetweenLine(AC,DC)
Goal: Value(LengthOfLine(EB))
Answer: 10
Theorem sequence:
similar_triangle_judgment_aa(1,DAC,EAB)
similar_triangle_property_line_ratio(1,DAC,EAB)
similar_triangle_property_line_ratio(1,ACD,ABE)
Problem: As shown in the diagram, AB=2, BD=15, OB=6, OB⊥AB, OD⊥CD. Find the length of line CD.
问题：如图所示，AB=2，BD=15，OB=6，OB⊥AB，OD垂直于CD。求直线CD的长度。
Construction (CDL):
Shape(AO,OB,BA)
Shape(AB,BD,DC,CA)
Collinear(OAC)
Collinear(OBD)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(BD),15)
Equal(LengthOfLine(OB),6)
PerpendicularBetweenLine(OB,AB)
PerpendicularBetweenLine(OD,CD)
Goal: Value(LengthOfLine(CD))
Answer: 7
Theorem sequence:
similar_triangle_judgment_aa(1,AOB,COD)
similar_triangle_property_line_ratio(1,AOB,COD)
similar_triangle_property_line_ratio(1,OBA,ODC)
line_addition(1,OB,BD)
Problem: As shown in the diagram, AC=1/2, BC=1, EF=5, △ABC is similar to △DEF.. Find the length of line DF.
问题：如图所示，AC=1/2，BC=1，EF=5，三角形ABC相似于三角形DEF。求直线DF的长度。
Construction (CDL):
Shape(AB,BC,CA)
Shape(DE,EF,FD)
Conditions (CDL):
Equal(LengthOfLine(AC),1/2)
Equal(LengthOfLine(BC),1)
Equal(LengthOfLine(EF),5)
SimilarBetweenTriangle(ABC,DEF)
Goal: Value(LengthOfLine(DF))
Answer: 5/2
Theorem sequence:
similar_triangle_property_line_ratio(1,ABC,DEF)
similar_triangle_property_line_ratio(1,BCA,EFD)
Problem: As shown in the diagram, AO=3/2, OB=4, OM=5, AO⊥BO, NM is perpendicular to BM. Find the length of line NM.
问题：如图所示，AO=3/2，OB=4，OM=5，AO垂直于BO，NM垂直于BM。求直线NM的长度。
Construction (CDL):
Shape(BA,AO,OB)
Shape(NM,MO,OA,AN)
Collinear(NAB)
Collinear(MOB)
Conditions (CDL):
Equal(LengthOfLine(AO),3/2)
Equal(LengthOfLine(OB),4)
Equal(LengthOfLine(OM),5)
PerpendicularBetweenLine(AO,BO)
PerpendicularBetweenLine(NM,BM)
Goal: Value(LengthOfLine(NM))
Answer: 27/8
Theorem sequence:
similar_triangle_judgment_aa(1,AOB,NMB)
similar_triangle_property_line_ratio(1,AOB,NMB)
similar_triangle_property_line_ratio(1,BAO,BNM)
line_addition(1,MO,OB)
Problem: As shown in the diagram, ∠CAB=50°, the center of circle O is O. Find the measure of ∠OBC.
问题：如图所示，∠CAB=50°，圆O的圆心为O。求∠OBC的大小。
Construction (CDL):
Shape(OCA,AE,EO,OC)
Shape(OE,EC,CO)
Shape(EA,AB,BE)
Shape(EB,BC,CE)
Shape(OAB,BA)
Shape(OBC,CB)
Collinear(AEC)
Collinear(BEO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),50)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(OBC))
Answer: 40
Theorem sequence:
arc_property_circumference_angle_external(1,OBC,A)
arc_property_center_angle(1,OBC,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,OBC)
Problem: As shown in the diagram, AC=3, BA=15, EA=2, triangle ECA is similar to triangle DAB.. Find the length of line DB.
问题：如图所示，AC=3，BA=15，EA=2，三角形ECA相似于三角形DAB。求直线DB的长度。
Construction (CDL):
Shape(EC,CA,AE)
Shape(DA,AB,BD)
Collinear(CAB)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(BA),15)
Equal(LengthOfLine(EA),2)
SimilarBetweenTriangle(ECA,DAB)
Goal: Value(LengthOfLine(DB))
Answer: 10
Theorem sequence:
similar_triangle_property_line_ratio(1,ECA,DAB)
similar_triangle_property_line_ratio(1,CAE,ABD)
Problem: As shown in the diagram, BE=20, CD=20, CE=10, AB is perpendicular to EB, DC⊥EC. Find the length of line AB.
问题：如图所示，BE=20，CD=20，CE=10，AB垂直于EB，DC⊥EC。求直线AB的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(CE,ED,DC)
Collinear(AED)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(BE),20)
Equal(LengthOfLine(CD),20)
Equal(LengthOfLine(CE),10)
PerpendicularBetweenLine(AB,EB)
PerpendicularBetweenLine(DC,EC)
Goal: Value(LengthOfLine(AB))
Answer: 40
Theorem sequence:
vertical_angle(1,BEA,CED)
similar_triangle_judgment_aa(1,ABE,DCE)
similar_triangle_property_line_ratio(1,ABE,DCE)
similar_triangle_property_line_ratio(1,EAB,EDC)
Problem: As shown in the diagram, DE=30, DG=600, EF=20, GB=140, AG is perpendicular to DG, FE⊥DE. Find the length of line AB.
问题：如图所示，DE=30，DG=600，EF=20，GB=140，AG⊥DG，FE⊥DE。求直线AB的长度。
Construction (CDL):
Shape(AG,GE,EF,FA)
Shape(FE,ED,DF)
Shape(GB,BC,CD,DE,EG)
Collinear(AGB)
Collinear(AFD)
Collinear(GED)
Conditions (CDL):
Equal(LengthOfLine(DE),30)
Equal(LengthOfLine(DG),600)
Equal(LengthOfLine(EF),20)
Equal(LengthOfLine(GB),140)
PerpendicularBetweenLine(AG,DG)
PerpendicularBetweenLine(FE,DE)
Goal: Value(LengthOfLine(AB))
Answer: 540
Theorem sequence:
similar_triangle_judgment_aa(1,FED,AGD)
similar_triangle_property_line_ratio(1,FED,AGD)
similar_triangle_property_line_ratio(1,DFE,DAG)
line_addition(1,AG,GB)
Problem: As shown in the diagram, BE=8, DE=4, EC=22, BC is perpendicular to AC, BE⊥DE. Find the length of line AC.
问题：如图所示，BE=8，DE=4，EC=22，BC⊥AC，BE垂直于DE。求直线AC的长度。
Construction (CDL):
Shape(DB,BE,ED)
Shape(AD,DE,EC,CA)
Collinear(BDA)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(BE),8)
Equal(LengthOfLine(DE),4)
Equal(LengthOfLine(EC),22)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(BE,DE)
Goal: Value(LengthOfLine(AC))
Answer: 15
Theorem sequence:
similar_triangle_judgment_aa(1,DBE,ABC)
similar_triangle_property_line_ratio(1,DBE,ABC)
similar_triangle_property_line_ratio(1,BED,BCA)
line_addition(1,BE,EC)
Problem: As shown in the diagram, BE=6, DE=2, EC=15, BC is perpendicular to AC, BE⊥DE. Find the length of line AC.
问题：如图所示，BE=6，DE=2，EC=15，BC垂直于AC，BE⊥DE。求直线AC的长度。
Construction (CDL):
Shape(DB,BE,ED)
Shape(AD,DE,EC,CA)
Collinear(BDA)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(BE),6)
Equal(LengthOfLine(DE),2)
Equal(LengthOfLine(EC),15)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(BE,DE)
Goal: Value(LengthOfLine(AC))
Answer: 7
Theorem sequence:
similar_triangle_judgment_aa(1,DBE,ABC)
similar_triangle_property_line_ratio(1,DBE,ABC)
similar_triangle_property_line_ratio(1,BED,BCA)
line_addition(1,BE,EC)