SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=AC, BC=10, ∠CAB=120°, D is the midpoint of segment BC, CF⊥DF, DE⊥BE. Find the sum of the length of line DE and the length of line DF.
问题：如图所示，AB=AC，BC=10，∠CAB=120°，D平分线段BC，CF垂直于DF，DE垂直于BE。求直线DE的长度与直线DF的长度之和。
Construction (CDL):
Shape(AE,ED,DA)
Shape(AD,DF,FA)
Shape(EB,BD,DE)
Shape(FD,DC,CF)
Collinear(AFC)
Collinear(AEB)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(BC),10)
Equal(MeasureOfAngle(CAB),120)
IsMidpointOfLine(D,BC)
PerpendicularBetweenLine(CF,DF)
PerpendicularBetweenLine(DE,BE)
Goal: Value(Add(LengthOfLine(DE),LengthOfLine(DF)))
Answer: 5
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
line_addition(1,BD,DC)
sine_theorem(1,DEB)
sine_theorem(1,DCF)
Problem: As shown in the diagram, BC=10, ∠BCA=90°, ∠CAB=30°. Find the length of line AC.
问题：如图所示，BC=10，∠BCA=90°，∠CAB=30°。求直线AC的长度。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(BC),10)
Equal(MeasureOfAngle(BCA),90)
Equal(MeasureOfAngle(CAB),30)
Goal: Value(LengthOfLine(AC))
Answer: 10*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,CAB)
Problem: As shown in the diagram, AB=4, ∠BCA=30°, AB⊥CB. Find the sum of the length of line AB and the length of line AC.
问题：如图所示，AB=4，∠BCA=30°，AB⊥CB。求直线AB的长度与直线AC的长度之和。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(MeasureOfAngle(BCA),30)
PerpendicularBetweenLine(AB,CB)
Goal: Value(Add(LengthOfLine(AB),LengthOfLine(AC)))
Answer: 12
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, AB=10, AC=10, ∠BAC=120°, AD is the angle bisector of ∠BAC, AE is the angle bisector of ∠BAD, FD is parallel to BA. Find the length of line DF.
问题：如图所示，AB=10，AC=10，∠BAC=120°，AD是∠BAC的角平分线，AE是∠BAD的角平分线，FD平行于BA。求直线DF的长度。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DE,EA)
Shape(AE,EB,BA)
Shape(ED,DF,FE)
Collinear(AEF)
Collinear(CDEB)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),10)
Equal(MeasureOfAngle(BAC),120)
IsBisectorOfAngle(AD,BAC)
IsBisectorOfAngle(AE,BAD)
ParallelBetweenLine(FD,BA)
Goal: Value(LengthOfLine(DF))
Answer: 5
Theorem sequence:
triangle_property_angle_sum(1,ACB)
isosceles_triangle_judgment_line_equal(1,ACB)
isosceles_triangle_property_angle_equal(1,ACB)
isosceles_triangle_property_line_coincidence(3,ACB,D)
parallel_property_alternate_interior_angle(1,FD,BA)
isosceles_triangle_judgment_angle_equal(1,DFA)
sine_theorem(1,ACD)
Problem: As shown in the diagram, MN=3, PC=12, PM=PN, ∠PCM=60°, MD⊥PD. Find the length of line CM.
问题：如图所示，MN=3，PC=12，PM=PN，∠PCM=60°，MD⊥PD。求直线CM的长度。
Construction (CDL):
Shape(PC,CM,MP)
Shape(PM,MD,DP)
Shape(PD,DN,NP)
Collinear(CMDN)
Conditions (CDL):
Equal(LengthOfLine(MN),3)
Equal(LengthOfLine(PC),12)
Equal(LengthOfLine(PM),LengthOfLine(PN))
Equal(MeasureOfAngle(PCM),60)
PerpendicularBetweenLine(MD,PD)
Goal: Value(LengthOfLine(CM))
Answer: 9/2
Theorem sequence:
adjacent_complementary_angle(1,MDP,PDN)
triangle_property_angle_sum(1,PCD)
line_addition(1,MD,DN)
isosceles_triangle_judgment_line_equal(1,PMN)
isosceles_triangle_property_angle_equal(1,PMN)
mirror_congruent_triangle_judgment_aas(1,PMD,PDN)
mirror_congruent_triangle_property_line_equal(1,PMD,PDN)
sine_theorem(1,CDP)
line_addition(1,CM,MD)
Problem: As shown in the diagram, BC=3, ∠DAE=15°, ∠ECB=90°, ED is the perpendicular bisector of BA. Find the length of line AE.
问题：如图所示，BC=3，∠DAE=15°，∠ECB=90°，ED是BA的垂直平分线。求直线AE的长度。
Construction (CDL):
Shape(DA,AE,ED)
Shape(BD,DE,EB)
Shape(BE,EC,CB)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(BC),3)
Equal(MeasureOfAngle(DAE),15)
Equal(MeasureOfAngle(ECB),90)
IsPerpendicularBisectorOfLine(ED,BA)
Goal: Value(LengthOfLine(AE))
Answer: 6
Theorem sequence:
triangle_property_angle_sum(1,ACB)
perpendicular_bisector_property_distance_equal(1,ED,BA)
isosceles_triangle_judgment_line_equal(1,EBA)
isosceles_triangle_property_angle_equal(1,EBA)
angle_addition(1,CBE,EBD)
triangle_property_angle_sum(1,BEC)
sine_theorem(1,BEC)
Problem: As shown in the diagram, BC=5, BD=2, triangle ABC is an equilateral triangle, BA is parallel to DE, ED∥AB, FE is perpendicular to DE. Find the length of line DF.
问题：如图所示，BC=5，BD=2，三角形ABC为等边三角形，BA平行于DE，ED平行于AB，FE垂直于DE。求直线DF的长度。
Construction (CDL):
Shape(AB,BD,DE,EA)
Shape(ED,DC,CE)
Shape(EC,CF,FE)
Collinear(AEC)
Collinear(BDCF)
Conditions (CDL):
Equal(LengthOfLine(BC),5)
Equal(LengthOfLine(BD),2)
EquilateralTriangle(ABC)
ParallelBetweenLine(BA,DE)
ParallelBetweenLine(ED,AB)
PerpendicularBetweenLine(FE,DE)
Goal: Value(LengthOfLine(DF))
Answer: 6
Theorem sequence:
line_addition(1,BD,DC)
equilateral_triangle_property_angle(1,ABC)
equilateral_triangle_property_angle(1,BCA)
equilateral_triangle_property_angle(1,CAB)
parallel_property_corresponding_angle(1,ED,AB,C)
parallel_property_corresponding_angle(2,BA,DE,C)
isosceles_triangle_judgment_angle_equal(1,EDC)
isosceles_triangle_judgment_angle_equal(1,CED)
triangle_property_angle_sum(1,EDF)
sine_theorem(1,DFE)
Problem: As shown in the diagram, AB=6, ∠BAC=30°, D is the midpoint of segment AB, AC is perpendicular to BC, AE is perpendicular to DE. Find the length of line DE.
问题：如图所示，AB=6，∠BAC=30°，D是线段AB的中点，AC垂直于BC，AE⊥DE。求直线DE的长度。
Construction (CDL):
Shape(DA,AE,ED)
Shape(BD,DE,EC,CB)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(MeasureOfAngle(BAC),30)
IsMidpointOfLine(D,AB)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(AE,DE)
Goal: Value(LengthOfLine(DE))
Answer: 3/2
Theorem sequence:
line_addition(1,AD,DB)
sine_theorem(1,DAE)
Problem: As shown in the diagram, BC=8, ∠ABC=150°, △BDC is a right △. Find the length of line CD.
问题：如图所示，BC=8，∠ABC=150°，△BDC为直角△。求直线CD的长度。
Construction (CDL):
Shape(CB,BD,DC)
Shape(AB,BC)
Collinear(ABD)
Conditions (CDL):
Equal(LengthOfLine(BC),8)
Equal(MeasureOfAngle(ABC),150)
RightTriangle(BDC)
Goal: Value(LengthOfLine(CD))
Answer: 4
Theorem sequence:
flat_angle(1,ABD)
angle_addition(1,ABC,CBD)
sine_theorem(1,CBD)
Problem: As shown in the diagram, CD=20, ∠ACB=15°, ∠ADB=30°, CB⊥AB. Find the length of line AB.
问题：如图所示，CD=20，∠ACB=15°，∠ADB=30°，CB垂直于AB。求直线AB的长度。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(CD),20)
Equal(MeasureOfAngle(ACB),15)
Equal(MeasureOfAngle(ADB),30)
PerpendicularBetweenLine(CB,AB)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(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, BD=6, ∠ACB=90°, ∠CBA=60°, BD is the angle bisector of ∠CBA, CP is the median of triangle CBD, P is the midpoint of segment BD. Find the length of line CP.
问题：如图所示，BD=6，∠ACB=90°，∠CBA=60°，BD平分∠CBA，CP是△CBD的中线，P是线段BD的中点。求直线CP的长度。
Construction (CDL):
Shape(CB,BP,PC)
Shape(CP,PD,DC)
Shape(DP,PB,BA,AD)
Collinear(CDA)
Collinear(BPD)
Conditions (CDL):
Equal(LengthOfLine(BD),6)
Equal(MeasureOfAngle(ACB),90)
Equal(MeasureOfAngle(CBA),60)
IsBisectorOfAngle(BD,CBA)
IsMedianOfTriangle(CP,CBD)
IsMidpointOfLine(P,BD)
Goal: Value(LengthOfLine(CP))
Answer: 3
Theorem sequence:
right_triangle_judgment_angle(1,DCB)
right_triangle_property_length_of_median(1,DCB,P)
Problem: As shown in the diagram, CD=2, ∠ABC=60°, ∠BCA=90°, △AEB is an equilateral △, CD is the median of △ CAB. Find the length of line EF.
问题：如图所示，CD=2，∠ABC=60°，∠BCA=90°，三角形AEB为等边三角形，CD是△CAB的中线。求直线EF的长度。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Shape(DA,AF,FD)
Shape(DF,FE,EB,BD)
Collinear(ADB)
Collinear(CDF)
Collinear(AFE)
Conditions (CDL):
Equal(LengthOfLine(CD),2)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(BCA),90)
EquilateralTriangle(AEB)
IsMedianOfTriangle(CD,CAB)
Goal: Value(LengthOfLine(EF))
Answer: 2
Theorem sequence:
line_addition(1,AD,DB)
line_addition(1,AF,FE)
equilateral_triangle_property_angle(1,AEB)
equilateral_triangle_property_angle(1,EBA)
triangle_property_angle_sum(1,CDB)
triangle_property_angle_sum(1,DAF)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_length_of_median(1,BCA,D)
isosceles_triangle_judgment_line_equal(1,DBC)
isosceles_triangle_property_angle_equal(1,DBC)
vertical_angle(1,CDB,FDA)
parallel_judgment_corresponding_angle(1,FD,EB,A)
midsegment_of_triangle_judgment_parallel(2,FD,AEB)
Problem: As shown in the diagram, CD=1, ∠DCA=90°, ∠EBD=30°, △AED is mirror congruent to △ADC. Find the length of line BD.
问题：如图所示，CD=1，∠DCA=90°，∠EBD=30°，三角形AED镜像全等于三角形ADC。求直线BD的长度。
Construction (CDL):
Shape(EB,BD,DE)
Shape(AE,ED,DA)
Shape(AD,DC,CA)
Collinear(BEA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(CD),1)
Equal(MeasureOfAngle(DCA),90)
Equal(MeasureOfAngle(EBD),30)
MirrorCongruentBetweenTriangle(AED,ADC)
Goal: Value(LengthOfLine(BD))
Answer: 2
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,AED,ADC)
adjacent_complementary_angle(1,AED,DEB)
mirror_congruent_triangle_property_angle_equal(1,EDA,CAD)
sine_theorem(1,DEB)
Problem: As shown in the diagram, CE=3, △ABC is an equilateral △, D is the midpoint of segment AC, DE⊥CE. Find the length of line AB.
问题：如图所示，CE=3，△ABC为等边△，D平分线段AC，DE⊥CE。求直线AB的长度。
Construction (CDL):
Shape(AB,BE,ED,DA)
Shape(DE,EC,CD)
Collinear(ADC)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(CE),3)
EquilateralTriangle(ABC)
IsMidpointOfLine(D,AC)
PerpendicularBetweenLine(DE,CE)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
triangle_property_angle_sum(1,DEC)
equilateral_triangle_property_angle(1,CAB)
sine_theorem(1,CDE)
line_addition(1,AD,DC)
Problem: As shown in the diagram, AE=2, ED=EC, triangle ABC is an equilateral triangle, D bisects segment BC, CA is parallel to IE, DF is perpendicular to BF. Find the length of line AF.
问题：如图所示，AE=2，ED=EC，△ABC为等边△，D平分线段BC，CA平行于IE，DF垂直于BF。求直线AF的长度。
Construction (CDL):
Shape(FB,BD,DF)
Shape(AF,FD,DH,HA)
Shape(EA,AH,HE)
Shape(EH,HC,CE)
Shape(CH,HD,DC)
Shape(EC,CI,IE)
Collinear(BFAE)
Collinear(AHC)
Collinear(EHD)
Collinear(BDCI)
Conditions (CDL):
Equal(LengthOfLine(AE),2)
Equal(LengthOfLine(ED),LengthOfLine(EC))
EquilateralTriangle(ABC)
IsMidpointOfLine(D,BC)
ParallelBetweenLine(CA,IE)
PerpendicularBetweenLine(DF,BF)
Goal: Value(LengthOfLine(AF))
Answer: 3
Theorem sequence:
equilateral_triangle_property_angle(1,ABC)
equilateral_triangle_property_angle(1,BCA)
equilateral_triangle_property_angle(1,CAB)
parallel_property_corresponding_angle(1,CA,IE,B)
parallel_property_corresponding_angle(2,EI,AC,B)
triangle_property_angle_sum(1,FBD)
triangle_property_angle_sum(1,ECI)
triangle_property_angle_sum(1,BDE)
isosceles_triangle_judgment_angle_equal(1,BIE)
isosceles_triangle_judgment_angle_equal(1,IEB)
isosceles_triangle_judgment_line_equal(1,EDC)
isosceles_triangle_property_angle_equal(1,EDC)
adjacent_complementary_angle(1,BDH,HDC)
adjacent_complementary_angle(1,BCE,ECI)
mirror_congruent_triangle_judgment_sas(1,ECI,EBD)
mirror_congruent_triangle_property_line_equal(1,ECI,EBD)
line_addition(1,BF,FA)
line_addition(1,BA,AE)
line_addition(1,BD,DC)
line_addition(1,BC,CI)
sine_theorem(1,BDF)
Problem: As shown in the diagram, BD=500, ∠BDE=60°, ∠DBA=150°, DE is perpendicular to BE. Find the length of line DE.
问题：如图所示，BD=500，∠BDE=60°，∠DBA=150°，DE垂直于BE。求直线DE的长度。
Construction (CDL):
Shape(EC,CB,BD,DE)
Shape(DB,BA)
Collinear(ABCE)
Conditions (CDL):
Equal(LengthOfLine(BD),500)
Equal(MeasureOfAngle(BDE),60)
Equal(MeasureOfAngle(DBA),150)
PerpendicularBetweenLine(DE,BE)
Goal: Value(LengthOfLine(DE))
Answer: 250
Theorem sequence:
triangle_property_angle_sum(1,EBD)
sine_theorem(1,DEB)
Problem: As shown in the diagram, CD=36, ∠ACB=30°, △BAC is mirror congruent to △DCA, BA is perpendicular to CA. Find the length of line BA.
问题：如图所示，CD=36，∠ACB=30°，三角形BAC镜像全等于三角形DCA，BA⊥CA。求直线BA的长度。
Construction (CDL):
Shape(BA,AC,CB)
Shape(AD,DC,CA)
Conditions (CDL):
Equal(LengthOfLine(CD),36)
Equal(MeasureOfAngle(ACB),30)
MirrorCongruentBetweenTriangle(BAC,DCA)
PerpendicularBetweenLine(BA,CA)
Goal: Value(LengthOfLine(BA))
Answer: 18
Theorem sequence:
sine_theorem(1,BAC)
mirror_congruent_triangle_property_line_equal(1,ACB,ADC)
Problem: As shown in the diagram, AB=5, ∠CBA=60°, AC⊥BC. Find the length of line BC.
问题：如图所示，AB=5，∠CBA=60°，AC⊥BC。求直线BC的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(MeasureOfAngle(CBA),60)
PerpendicularBetweenLine(AC,BC)
Goal: Value(LengthOfLine(BC))
Answer: 5/2
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,BAC)
Problem: As shown in the diagram, AC=3/2, ∠ABC=30°, BC is perpendicular to AC. Find the length of line AB.
问题：如图所示，AC=3/2，∠ABC=30°，BC⊥AC。求直线AB的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),3/2)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(BC,AC)
Goal: Value(LengthOfLine(AB))
Answer: 3
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, AB=4, AB=AD, ∠ADB=15°, ∠BCA=90°. Find the length of line AC.
问题：如图所示，AB=4，AB=AD，∠ADB=15°，∠BCA=90°。求直线AC的长度。
Construction (CDL):
Shape(BC,CA,AB)
Shape(BA,AD,DB)
Collinear(CAD)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AB),LengthOfLine(AD))
Equal(MeasureOfAngle(ADB),15)
Equal(MeasureOfAngle(BCA),90)
Goal: Value(LengthOfLine(AC))
Answer: 2*sqrt(3)
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ADB)
isosceles_triangle_property_angle_equal(1,ADB)
triangle_property_angle_sum(1,BCA)
triangle_property_angle_sum(1,BAD)
adjacent_complementary_angle(1,CAB,BAD)
sine_theorem(1,ABC)
Problem: As shown in the diagram, BC=18, ∠CAB=30°, AB is perpendicular to CB. Find the length of line AC.
问题：如图所示，BC=18，∠CAB=30°，AB垂直于CB。求直线AC的长度。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(BC),18)
Equal(MeasureOfAngle(CAB),30)
PerpendicularBetweenLine(AB,CB)
Goal: Value(LengthOfLine(AC))
Answer: 36
Theorem sequence:
sine_theorem(1,CAB)
Problem: As shown in the diagram, BC=1, ∠BAD=15°, ∠CBD=60°, ∠DCB=90°. Find the length of line AD.
问题：如图所示，BC=1，∠BAD=15°，∠CBD=60°，∠DCB=90°。求直线AD的长度。
Construction (CDL):
Shape(BA,AD,DB)
Shape(BD,DC,CB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(BC),1)
Equal(MeasureOfAngle(BAD),15)
Equal(MeasureOfAngle(CBD),60)
Equal(MeasureOfAngle(DCB),90)
Goal: Value(LengthOfLine(AD))
Answer: 2
Theorem sequence:
triangle_property_angle_sum(1,BAD)
triangle_property_angle_sum(1,BDC)
sine_theorem(1,BDC)
adjacent_complementary_angle(1,ADB,BDC)
isosceles_triangle_judgment_angle_equal(1,DBA)
Problem: As shown in the diagram, AD=6, ∠DCB=90°, ∠EAD=30°, DE perpendicular bisects BA. Find the length of line CD.
问题：如图所示，AD=6，∠DCB=90°，∠EAD=30°，DE是BA的垂直平分线。求直线CD的长度。
Construction (CDL):
Shape(EA,AD,DE)
Shape(BE,ED,DB)
Shape(BD,DC,CB)
Collinear(AEB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AD),6)
Equal(MeasureOfAngle(DCB),90)
Equal(MeasureOfAngle(EAD),30)
IsPerpendicularBisectorOfLine(DE,BA)
Goal: Value(LengthOfLine(CD))
Answer: 3
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,DE,BA)
isosceles_triangle_judgment_line_equal(1,DBA)
isosceles_triangle_property_angle_equal(1,DBA)
triangle_property_angle_sum(1,ADB)
adjacent_complementary_angle(1,ADB,BDC)
triangle_property_angle_sum(1,BDC)
sine_theorem(1,DCB)
Problem: As shown in the diagram, triangle ACB is congruent to triangle DFE, the area of quadrilateral DABE is 32, AB=16, ∠ACB=90°, ∠CBA=30°, quadrilateral DABE is a ▱. Find the length of line BE.
问题：如图所示，三角形ACB与三角形DFE是全等三角形，DABE的面积为32，AB=16，∠ACB=90°，∠CBA=30°，DABE是平行四边形。求直线BE的长度。
Construction (CDL):
Shape(AC,CB,BA)
Shape(DF,FE,ED)
Shape(DA,AB,BE,ED)
Collinear(CFBE)
Conditions (CDL):
CongruentBetweenTriangle(ACB,DFE)
Equal(AreaOfQuadrilateral(DABE),32)
Equal(LengthOfLine(AB),16)
Equal(MeasureOfAngle(ACB),90)
Equal(MeasureOfAngle(CBA),30)
Parallelogram(DABE)
Goal: Value(LengthOfLine(BE))
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,CBA,ABE)
parallelogram_area_formula_sine(1,ABED)
Problem: As shown in the diagram, AC=500, BD=300, BF=600, CD=600, CF=300, ZC=500, ∠ZFB=90°. Find the length of line ZB.
问题：如图所示，AC=500，BD=300，BF=600，CD=600，CF=300，ZC=500，∠ZFB=90°。求直线ZB的长度。
Construction (CDL):
Shape(EC,CA)
Shape(EC,CF,FB,BE)
Shape(ZC,CE,EZ)
Shape(BD,DE,EB)
Collinear(ZEB)
Collinear(CED)
Collinear(ZCFA)
Conditions (CDL):
Equal(LengthOfLine(AC),500)
Equal(LengthOfLine(BD),300)
Equal(LengthOfLine(BF),600)
Equal(LengthOfLine(CD),600)
Equal(LengthOfLine(CF),300)
Equal(LengthOfLine(ZC),500)
Equal(MeasureOfAngle(ZFB),90)
Goal: Value(LengthOfLine(ZB))
Answer: 1000
Theorem sequence:
line_addition(1,ZC,CF)
right_triangle_judgment_angle(1,ZFB)
right_triangle_property_pythagorean(1,ZFB)
Problem: As shown in the diagram, BC=6, Mul(LengthOfLine(CQ)=3), BQ is the angle bisector of ∠PBC, EF is the midsegment of △ ABC. Find the sum of the length of line EP and the length of line BP.
问题：如图所示，BC=6，Mul(LengthOfLine(CQ)=3)，BQ平分∠PBC，EF是三角形ABC的中位线。求直线EP的长度与直线BP的长度之和。
Construction (CDL):
Shape(AE,EP,PF,FA)
Shape(ED,DP,PE)
Shape(EB,BD,DE)
Shape(PD,DQ,QN,NF,FP)
Shape(DB,BQ,QD)
Shape(QB,BC,CQ)
Shape(NQ,QC,CN)
Shape(MF,FN,NM)
Collinear(AEB)
Collinear(AFNC)
Collinear(EPFM)
Collinear(EDQC)
Collinear(BDP)
Collinear(BQNM)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
Equal(Mul(LengthOfLine(CQ),3),LengthOfLine(CE))
IsBisectorOfAngle(BQ,PBC)
IsMidsegmentOfTriangle(EF,ABC)
Goal: Value(Add(LengthOfLine(EP),LengthOfLine(BP)))
Answer: 12
Theorem sequence:
midsegment_of_triangle_property_parallel(1,EF,ABC)
parallel_property_collinear_extend(2,EF,BC,M)
parallel_property_alternate_interior_angle(1,EF,BC)
parallel_property_alternate_interior_angle(2,EM,BC)
isosceles_triangle_judgment_angle_equal(1,PBM)
line_addition(1,EP,PM)
line_addition(1,EQ,QC)
similar_triangle_judgment_aa(1,QBC,QME)
similar_triangle_property_line_ratio(1,QBC,QME)
similar_triangle_property_line_ratio(1,BCQ,MEQ)
Problem: As shown in the diagram, BC=12, the perimeter of triangle ABC is 32, BQ is the angle bisector of ∠ABE, CP is the angle bisector of ∠DCA, BQ is the median of triangle BEA, CP is the median of △ CAD, AP⊥CP, EQ⊥BQ. Find the length of line PQ.
问题：如图所示，BC=12，三角形ABC的周长为32，BQ是∠ABE的角平分线，CP平分∠DCA，BQ是△BEA的中线，CP是△CAD的中线，AP⊥CP，EQ垂直于BQ。求直线PQ的长度。
Construction (CDL):
Shape(AB,BD,DE,EC,CA)
Shape(AB,BQ,QA)
Shape(QB,BD,DE,EQ)
Shape(AP,PC,CA)
Shape(CP,PD,DE,EC)
Shape(PQ)
Shape(AP,PD,DE,EQ,QA)
Collinear(BDEC)
Collinear(APD)
Collinear(AQE)
Conditions (CDL):
Equal(LengthOfLine(BC),12)
Equal(PerimeterOfTriangle(ABC),32)
IsBisectorOfAngle(BQ,ABE)
IsBisectorOfAngle(CP,DCA)
IsMedianOfTriangle(BQ,BEA)
IsMedianOfTriangle(CP,CAD)
PerpendicularBetweenLine(AP,CP)
PerpendicularBetweenLine(EQ,BQ)
Goal: Value(LengthOfLine(PQ))
Answer: 4
Theorem sequence:
triangle_perimeter_formula(1,ABC)
perpendicular_bisector_judgment_per_and_mid(1,CP,AD)
perpendicular_bisector_judgment_per_and_mid(1,BQ,EA)
perpendicular_bisector_property_distance_equal(1,CP,AD)
perpendicular_bisector_property_distance_equal(1,BQ,EA)
midsegment_of_triangle_judgment_midpoint(1,PQ,ADE)
line_addition(1,BD,DE)
line_addition(1,BD,DC)
midsegment_of_triangle_property_length(1,PQ,ADE)
Problem: As shown in the diagram, AB=7, MN=3, AN bisects ∠CAB, M bisects segment BC, BN is perpendicular to AN. Find the length of line AC.
问题：如图所示，AB=7，MN=3，AN是∠CAB的角平分线，M是线段BC的中点，BN垂直于AN。求直线AC的长度。
Construction (CDL):
Shape(AB,BN,NA)
Shape(NB,BM,MN)
Shape(AN,ND,DA)
Shape(NM,MC,CD,DN)
Collinear(BMC)
Collinear(BND)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),7)
Equal(LengthOfLine(MN),3)
IsBisectorOfAngle(AN,CAB)
IsMidpointOfLine(M,BC)
PerpendicularBetweenLine(BN,AN)
Goal: Value(LengthOfLine(AC))
Answer: 13
Theorem sequence:
triangle_property_angle_sum(1,ABN)
triangle_property_angle_sum(1,AND)
adjacent_complementary_angle(1,BNA,AND)
mirror_congruent_triangle_judgment_aas(1,AND,ABN)
mirror_congruent_triangle_property_line_equal(1,AND,ABN)
mirror_congruent_triangle_property_line_equal(1,NDA,NAB)
midsegment_of_triangle_judgment_midpoint(1,MN,BCD)
midsegment_of_triangle_property_length(1,MN,BCD)
line_addition(1,AD,DC)
Problem: As shown in the diagram, AB=13, AC=5, AD is the angle bisector of ∠BAC, AE is the median of △ ACB, CF⊥AF. Find the length of line EF.
问题：如图所示，AB=13，AC=5，AD平分∠BAC，AE是三角形ACB的中线，CF⊥AF。求直线EF的长度。
Construction (CDL):
Shape(AC,CF,FA)
Shape(AF,FH,HA)
Shape(AH,HG,GA)
Shape(CD,DF,FC)
Shape(FD,DE,EF)
Shape(HF,FE,EH)
Shape(GH,HE,EB,BG)
Collinear(AGB)
Collinear(AHE)
Collinear(AFD)
Collinear(CFHG)
Collinear(CDEB)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),5)
IsBisectorOfAngle(AD,BAC)
IsMedianOfTriangle(AE,ACB)
PerpendicularBetweenLine(CF,AF)
Goal: Value(LengthOfLine(EF))
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,CFA,AFH)
triangle_property_angle_sum(1,ACF)
triangle_property_angle_sum(1,AFG)
mirror_congruent_triangle_judgment_aas(1,AFG,ACF)
mirror_congruent_triangle_property_line_equal(1,AFG,ACF)
mirror_congruent_triangle_property_line_equal(1,FGA,FAC)
line_addition(1,AG,GB)
midsegment_of_triangle_judgment_midpoint(1,EF,CBG)
midsegment_of_triangle_property_length(1,EF,CBG)
Problem: As shown in the diagram, AB=7, DE=1, AE bisects ∠CAB, D is the midpoint of segment BC, AE is perpendicular to CE. Find the length of line AC.
问题：如图所示，AB=7，DE=1，AE是∠CAB的角平分线，D是线段BC的中点，AE垂直于CE。求直线AC的长度。
Construction (CDL):
Shape(AF,FE,EA)
Shape(AE,EC,CA)
Shape(FB,BD,DE,EF)
Shape(ED,DC,CE)
Collinear(BFA)
Collinear(BDC)
Collinear(CEF)
Conditions (CDL):
Equal(LengthOfLine(AB),7)
Equal(LengthOfLine(DE),1)
IsBisectorOfAngle(AE,CAB)
IsMidpointOfLine(D,BC)
PerpendicularBetweenLine(AE,CE)
Goal: Value(LengthOfLine(AC))
Answer: 5
Theorem sequence:
adjacent_complementary_angle(1,FEA,AEC)
triangle_property_angle_sum(1,AFE)
triangle_property_angle_sum(1,AEC)
mirror_congruent_triangle_judgment_aas(1,AEC,AFE)
mirror_congruent_triangle_property_line_equal(1,AEC,AFE)
mirror_congruent_triangle_property_line_equal(1,ECA,EAF)
line_addition(1,BF,FA)
midsegment_of_triangle_judgment_midpoint(1,ED,CFB)
midsegment_of_triangle_property_length(1,ED,CFB)
Problem: As shown in the diagram, EF=15, E is the midpoint of segment CA, F is the midpoint of segment CB. Find the length of line AB.
问题：如图所示，EF=15，E是线段CA的中点，F平分线段CB。求直线AB的长度。
Construction (CDL):
Shape(EC,CF,FE)
Shape(AE,EF,FB,BA)
Collinear(CEA)
Collinear(CFB)
Conditions (CDL):
Equal(LengthOfLine(EF),15)
IsMidpointOfLine(E,CA)
IsMidpointOfLine(F,CB)
Goal: Value(LengthOfLine(AB))
Answer: 30
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,FE,CBA)
midsegment_of_triangle_property_length(1,FE,CBA)
Problem: As shown in the diagram, AB=12, BC=20, BF bisects ∠ABC, D is the midpoint of segment AB, BF⊥AF. Find the length of line EF.
问题：如图所示，AB=12，BC=20，BF是∠ABC的角平分线，D是线段AB的中点，BF垂直于AF。求直线EF的长度。
Construction (CDL):
Shape(AD,DF,FA)
Shape(DB,BF,FD)
Shape(FB,BG,GF)
Shape(AF,FE,EA)
Shape(FG,GC,CE,EF)
Collinear(BGC)
Collinear(BDA)
Collinear(DFE)
Collinear(AEC)
Collinear(AFG)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(BC),20)
IsBisectorOfAngle(BF,ABC)
IsMidpointOfLine(D,AB)
PerpendicularBetweenLine(BF,AF)
Goal: Value(LengthOfLine(EF))
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,GFB,BFA)
triangle_property_angle_sum(1,FBG)
triangle_property_angle_sum(1,FAB)
mirror_congruent_triangle_judgment_aas(1,BFA,BGF)
mirror_congruent_triangle_property_line_equal(1,BFA,BGF)
mirror_congruent_triangle_property_line_equal(1,FAB,FBG)
line_addition(1,BG,GC)
midsegment_of_triangle_judgment_midpoint(1,DF,ABG)
midsegment_of_triangle_property_parallel(1,DF,ABG)
parallel_property_collinear_extend(1,GB,FD,C)
parallel_property_collinear_extend(2,DF,GC,E)
midsegment_of_triangle_judgment_parallel(1,FE,AGC)
midsegment_of_triangle_property_length(1,FE,AGC)