SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AC=3, DE=5, AD is the angle bisector of ∠BAH, E is the midpoint of segment BC, AD⊥BD. Find the length of line AB.
问题：如图所示，AC=3，DE=5，AD是∠BAH的角平分线，E是线段BC的中点，AD⊥BD。求直线AB的长度。
Construction (CDL):
Shape(HD,DA,AH)
Shape(DF,FA,AD)
Shape(DB,BF,FD)
Shape(FB,BE,EF)
Shape(AF,FE,EC,CA)
Collinear(HDB)
Collinear(HAC)
Collinear(BFA)
Collinear(DFE)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(DE),5)
IsBisectorOfAngle(AD,BAH)
IsMidpointOfLine(E,BC)
PerpendicularBetweenLine(AD,BD))
Goal: Value(LengthOfLine(AB))
Answer: 7
Theorem sequence:
adjacent_complementary_angle(1,HDA,ADB)
triangle_property_angle_sum(1,HDA)
triangle_property_angle_sum(1,ADB)
mirror_congruent_triangle_judgment_aas(1,ADB,AHD)
mirror_congruent_triangle_property_line_equal(1,ADB,AHD)
mirror_congruent_triangle_property_line_equal(1,DBA,DAH)
line_addition(1,HA,AC)
midsegment_of_triangle_judgment_midpoint(1,ED,BCH)
midsegment_of_triangle_property_length(1,ED,BCH)
Problem: As shown in the diagram, CB=4, ∠CAB=30°, D bisects segment AB, E is the midpoint of segment AD, F is the midpoint of segment AC, BC and CA are the legs of the right triangle BCA. Find the length of line EF.
问题：如图所示，CB=4，∠CAB=30°，D是线段AB的中点，E平分线段AD，F平分线段AC，∠BCA是直角△BCA的直角。求直线EF的长度。
Construction (CDL):
Shape(AE,EF,FA)
Shape(ED,DC,CF,FE)
Shape(DB,BC,CD)
Collinear(AEDB)
Collinear(AFC)
Conditions (CDL):
Equal(LengthOfLine(CB),4)
Equal(MeasureOfAngle(CAB),30)
IsMidpointOfLine(D,AB)
IsMidpointOfLine(E,AD)
IsMidpointOfLine(F,AC)
RightTriangle(BCA)
Goal: Value(LengthOfLine(EF))
Answer: 2
Theorem sequence:
sine_theorem(1,BCA)
median_of_triangle_judgment(1,CD,CAB)
right_triangle_property_length_of_median(1,BCA,D)
midsegment_of_triangle_judgment_midpoint(1,EF,ADC)
midsegment_of_triangle_property_length(1,EF,ADC)
Problem: As shown in the diagram, DE=3/2, D is the midpoint of segment AB, E bisects segment AC. Find the length of line BC.
问题：如图所示，DE=3/2，D平分线段AB，E是线段AC的中点。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(DE),3/2)
IsMidpointOfLine(D,AB)
IsMidpointOfLine(E,AC)
Goal: Value(LengthOfLine(BC))
Answer: 3
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,DE,ABC)
midsegment_of_triangle_property_length(1,DE,ABC)
Problem: As shown in the diagram, AB=3, AD=CD, BC=5, ∠CDA=100°, ∠DAB=130°, E is the midpoint of segment AD, F bisects segment CD. Find the length of line EF.
问题：如图所示，AB=3，AD=CD，BC=5，∠CDA=100°，∠DAB=130°，E平分线段AD，F平分线段CD。求直线EF的长度。
Construction (CDL):
Shape(DE,EF,FD)
Shape(EA,AC,CF,FE)
Shape(AB,BC,CA)
Collinear(AED)
Collinear(CFD)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AD),LengthOfLine(CD))
Equal(LengthOfLine(BC),5)
Equal(MeasureOfAngle(CDA),100)
Equal(MeasureOfAngle(DAB),130)
IsMidpointOfLine(E,AD)
IsMidpointOfLine(F,CD)
Goal: Value(LengthOfLine(EF))
Answer: 2
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DAC)
isosceles_triangle_property_angle_equal(1,DAC)
triangle_property_angle_sum(1,DAC)
angle_addition(1,EAC,CAB)
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
midsegment_of_triangle_judgment_midpoint(1,EF,DAC)
midsegment_of_triangle_property_length(1,EF,DAC)
Problem: As shown in the diagram, AD=7, BD=4, CD=3, E is the midpoint of segment AB, F is the midpoint of segment BD, G bisects segment CD, H is the midpoint of segment AC, CD is perpendicular to BD. Find the perimeter of quadrilateral EFGH.
问题：如图所示，AD=7，BD=4，CD=3，E是线段AB的中点，F是线段BD的中点，G是线段CD的中点，H平分线段AC，CD⊥BD。求EFGH的周长。
Construction (CDL):
Shape(AE,EB,BF,FD,DA)
Shape(AD,DG,GC,CH,HA)
Shape(DF,FB,BC,CG,GD)
Shape(EF,FG,GH,HE)
Shape(AE,EH,HA)
Shape(EB,BF,FE)
Shape(FB,BC,CG,GF)
Shape(HG,GC,CH)
Collinear(AEB)
Collinear(AHC)
Collinear(BFD)
Collinear(CGD)
Conditions (CDL):
Equal(LengthOfLine(AD),7)
Equal(LengthOfLine(BD),4)
Equal(LengthOfLine(CD),3)
IsMidpointOfLine(E,AB)
IsMidpointOfLine(F,BD)
IsMidpointOfLine(G,CD)
IsMidpointOfLine(H,AC)
PerpendicularBetweenLine(CD,BD)
Goal: Value(PerimeterOfQuadrilateral(EFGH))
Answer: 12
Theorem sequence:
right_triangle_judgment_angle(1,CDB)
right_triangle_property_pythagorean(1,CDB)
midsegment_of_triangle_judgment_midpoint(1,FE,BDA)
midsegment_of_triangle_judgment_midpoint(1,HG,CAD)
midsegment_of_triangle_judgment_midpoint(1,FG,DBC)
midsegment_of_triangle_judgment_midpoint(1,EH,ABC)
midsegment_of_triangle_property_length(1,FE,BDA)
midsegment_of_triangle_property_length(1,HG,CAD)
midsegment_of_triangle_property_length(1,FG,DBC)
midsegment_of_triangle_property_length(1,EH,ABC)
quadrilateral_perimeter_formula(1,EFGH)
Problem: As shown in the diagram, AE=12, BF=8, D is the midpoint of segment AB, P is the midpoint of segment AF, Q is the midpoint of segment EB, BC and CA are the legs of the right △ BCA. Find the length of line PQ.
问题：如图所示，AE=12，BF=8，D是线段AB的中点，P是线段AF的中点，Q平分线段EB，∠BCA是直角△BCA的直角。求直线PQ的长度。
Construction (CDL):
Shape(CA,AP,PF,FC)
Shape(AD,DP,PA)
Shape(PD,DB,BF,FP)
Shape(DQ,QB,BD)
Shape(AE,EQ,QD,DA)
Shape(PQ,QD,DP)
Collinear(EQB)
Collinear(ADB)
Collinear(APF)
Collinear(EAC)
Collinear(BFC)
Conditions (CDL):
Equal(LengthOfLine(AE),12)
Equal(LengthOfLine(BF),8)
IsMidpointOfLine(D,AB)
IsMidpointOfLine(P,AF)
IsMidpointOfLine(Q,EB)
RightTriangle(BCA)
Goal: Value(LengthOfLine(PQ))
Answer: 2*sqrt(13)
Theorem sequence:
triangle_property_angle_sum(1,BCA)
angle_addition(1,QDA,ADP)
midsegment_of_triangle_judgment_midpoint(1,DP,ABF)
midsegment_of_triangle_judgment_midpoint(1,DQ,BAE)
midsegment_of_triangle_property_parallel(1,DP,ABF)
midsegment_of_triangle_property_parallel(1,DQ,BAE)
midsegment_of_triangle_property_length(1,DP,ABF)
midsegment_of_triangle_property_length(1,DQ,BAE)
parallel_property_collinear_extend(2,EA,QD,C)
parallel_property_corresponding_angle(1,DP,BF,A)
parallel_property_alternate_interior_angle(1,AC,QD)
right_triangle_judgment_angle(1,QDP)
right_triangle_property_pythagorean(1,QDP)
Problem: As shown in the diagram, AB=10, AC=6, D bisects segment AC, E bisects segment AB, F bisects segment CB, ∠BCA is the right angle of the right △ BCA. Find the perimeter of △DEF.
问题：如图所示，AB=10，AC=6，D是线段AC的中点，E平分线段AB，F是线段CB的中点，△BCA为直角△。求三角形DEF的周长。
Construction (CDL):
Shape(CD,DF,FC)
Shape(DA,AE,ED)
Shape(FD,DE,EF)
Shape(FE,EB,BF)
Collinear(CDA)
Collinear(CFB)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),6)
IsMidpointOfLine(D,AC)
IsMidpointOfLine(E,AB)
IsMidpointOfLine(F,CB))
RightTriangle(BCA)
Goal: Value(PerimeterOfTriangle(DEF))
Answer: 12
Theorem sequence:
right_triangle_property_pythagorean(1,BCA)
midsegment_of_triangle_judgment_midpoint(1,DF,CAB)
midsegment_of_triangle_judgment_midpoint(1,ED,ABC)
midsegment_of_triangle_judgment_midpoint(1,FE,BCA)
midsegment_of_triangle_property_length(1,DF,CAB)
midsegment_of_triangle_property_length(1,ED,ABC)
midsegment_of_triangle_property_length(1,FE,BCA)
triangle_perimeter_formula(1,FDE)
Problem: As shown in the diagram, AC=14, ED=3, AE is the angle bisector of ∠CAB, D bisects segment BC, BE is perpendicular to AE. Find the length of line AB.
问题：如图所示，AC=14，ED=3，AE平分∠CAB，D平分线段BC，BE垂直于AE。求直线AB的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,EF,FA)
Shape(EB,BD,DE)
Shape(ED,DC,CF,FE)
Collinear(AFC)
Collinear(BEF)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(ED),3)
IsBisectorOfAngle(AE,CAB)
IsMidpointOfLine(D,BC)
PerpendicularBetweenLine(BE,AE))
Goal: Value(LengthOfLine(AB))
Answer: 8
Theorem sequence:
adjacent_complementary_angle(1,BEA,AEF)
triangle_property_angle_sum(1,ABE)
triangle_property_angle_sum(1,AEF)
mirror_congruent_triangle_judgment_aas(1,AEF,ABE)
mirror_congruent_triangle_property_line_equal(1,AEF,ABE)
mirror_congruent_triangle_property_line_equal(1,EFA,EAB)
midsegment_of_triangle_judgment_midpoint(1,DE,BCF)
midsegment_of_triangle_property_length(1,DE,BCF)
line_addition(1,AF,FC)
Problem: As shown in the diagram, DE=40, D is the midpoint of segment AC, E is the midpoint of segment CB. Find the length of line AB.
问题：如图所示，DE=40，D平分线段AC，E是线段CB的中点。求直线AB的长度。
Construction (CDL):
Shape(DC,CE,ED)
Shape(AD,DE,EB,BA)
Collinear(CDA)
Collinear(CEB)
Conditions (CDL):
Equal(LengthOfLine(DE),40)
IsMidpointOfLine(D,AC)
IsMidpointOfLine(E,CB)
Goal: Value(LengthOfLine(AB))
Answer: 80
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,ED,CBA)
midsegment_of_triangle_property_length(1,ED,CBA)
Problem: As shown in the diagram, AD=DC, DE=10, ∠DAB=∠CDE. Find the length of line AB.
问题：如图所示，AD=DC，DE=10，∠DAB=∠CDE。求直线AB的长度。
Construction (CDL):
Shape(AB,BE,ED,DA)
Shape(CD,DE,EC)
Collinear(ADC)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(DC))
Equal(LengthOfLine(DE),10)
Equal(MeasureOfAngle(DAB),MeasureOfAngle(CDE))
Goal: Value(LengthOfLine(AB))
Answer: 20
Theorem sequence:
parallel_judgment_corresponding_angle(1,DE,AB,C)
midsegment_of_triangle_judgment_parallel(1,DE,CAB)
midsegment_of_triangle_property_length(1,DE,CAB)
Problem: As shown in the diagram, AB=4, DM=1, AD is the angle bisector of ∠CAB, M bisects segment BC, BD is perpendicular to AD. Find the length of line AC.
问题：如图所示，AB=4，DM=1，AD是∠CAB的角平分线，M平分线段BC，BD垂直于AD。求直线AC的长度。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DN,NA)
Shape(DB,BM,MD)
Shape(DM,MC,CN,ND)
Collinear(ANC)
Collinear(BDN)
Collinear(BMC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(DM),1)
IsBisectorOfAngle(AD,CAB)
IsMidpointOfLine(M,BC)
PerpendicularBetweenLine(BD,AD)
Goal: Value(LengthOfLine(AC))
Answer: 6
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADN)
triangle_property_angle_sum(1,ABD)
triangle_property_angle_sum(1,ADN)
mirror_congruent_triangle_judgment_aas(1,ADN,ABD)
mirror_congruent_triangle_property_line_equal(1,ADN,ABD)
mirror_congruent_triangle_property_line_equal(1,DNA,DAB)
line_addition(1,AN,NC)
midsegment_of_triangle_judgment_midpoint(1,MD,BCN)
midsegment_of_triangle_property_length(1,MD,BCN)
Problem: As shown in the diagram, AC=4, AD=6, BC=3, M bisects segment BD, BA and ED are opposite sides of the ▱ ABED, BC⊥AC. Find the length of line CM.
问题：如图所示，AC=4，AD=6，BC=3，M平分线段BD，BA和ED是平行四边形ABED的一组对边，BC垂直于AC。求直线CM的长度。
Construction (CDL):
Shape(AB,BN,NA)
Shape(NB,BC,CN)
Shape(AN,NM,MD,DA)
Shape(MN,NC,CM)
Shape(MC,CD,DM)
Shape(CE,ED,DC)
Collinear(BNMD)
Collinear(ANC)
Collinear(BCE)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(BC),3)
IsMidpointOfLine(M,BD)
Parallelogram(ABED)
PerpendicularBetweenLine(BC,AC)
Goal: Value(LengthOfLine(CM))
Answer: 5/2
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
parallelogram_property_opposite_line_equal(1,ABED)
parallelogram_property_opposite_line_equal(1,BEDA)
line_addition(1,BC,CE)
midsegment_of_triangle_judgment_midpoint(1,CM,BED)
midsegment_of_triangle_property_length(1,CM,BED)
Problem: As shown in the diagram, the perimeter of △ABC is 1, D is the midpoint of segment AB, E bisects segment AC, F bisects segment BC, X is the midpoint of segment DE, Y bisects segment DF, Z bisects segment EF. Find the perimeter of △XYZ.
问题：如图所示，三角形ABC的周长为1，D平分线段AB，E平分线段AC，F平分线段BC，X是线段DE的中点，Y是线段DF的中点，Z平分线段EF。求△XYZ的周长。
Construction (CDL):
Shape(AD,DX,XE,EA)
Shape(DB,BF,FY,YD)
Shape(EZ,ZF,FC,CE)
Shape(DY,YX,XD)
Shape(YF,FZ,ZY)
Shape(XZ,ZE,EX)
Shape(XY,YZ,ZX)
Collinear(ADB)
Collinear(AEC)
Collinear(BFC)
Collinear(DYF)
Collinear(DXE)
Collinear(FZE)
Conditions (CDL):
Equal(PerimeterOfTriangle(ABC),1)
IsMidpointOfLine(D,AB)
IsMidpointOfLine(E,AC)
IsMidpointOfLine(F,BC)
IsMidpointOfLine(X,DE)
IsMidpointOfLine(Y,DF)
IsMidpointOfLine(Z,EF)
Goal: Value(PerimeterOfTriangle(XYZ))
Answer: 1/4
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,YX,DFE)
midsegment_of_triangle_judgment_midpoint(1,ZY,FED)
midsegment_of_triangle_judgment_midpoint(1,XZ,EDF)
midsegment_of_triangle_judgment_midpoint(1,DE,ABC)
midsegment_of_triangle_judgment_midpoint(1,FD,BCA)
midsegment_of_triangle_judgment_midpoint(1,EF,CAB)
midsegment_of_triangle_property_length(1,YX,DFE)
midsegment_of_triangle_property_length(1,ZY,FED)
midsegment_of_triangle_property_length(1,XZ,EDF)
midsegment_of_triangle_property_length(1,DE,ABC)
midsegment_of_triangle_property_length(1,FD,BCA)
midsegment_of_triangle_property_length(1,EF,CAB)
triangle_perimeter_formula(1,XYZ)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, AB=AC, CE=2, D is the midpoint of segment AB, E is the midpoint of segment AC, F bisects segment BC. Find the perimeter of ADFE.
问题：如图所示，AB=AC，CE=2，D平分线段AB，E平分线段AC，F是线段BC的中点。求四边形ADFE的周长。
Construction (CDL):
Shape(AD,DF,FE,EA)
Shape(DB,BF,FD)
Shape(EF,FC,CE)
Collinear(ADB)
Collinear(AEC)
Collinear(BFC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(CE),2)
IsMidpointOfLine(D,AB)
IsMidpointOfLine(E,AC)
IsMidpointOfLine(F,BC)
Goal: Value(PerimeterOfQuadrilateral(ADFE))
Answer: 8
Theorem sequence:
line_addition(1,AD,DB)
line_addition(1,AE,EC)
quadrilateral_perimeter_formula(1,ADFE)
midsegment_of_triangle_judgment_midpoint(1,FD,BCA)
midsegment_of_triangle_judgment_midpoint(1,EF,CAB)
midsegment_of_triangle_property_length(1,FD,BCA)
midsegment_of_triangle_property_length(1,EF,CAB)
Problem: As shown in the diagram, FD=16, D is the midpoint of segment BC, E bisects segment AC, F is the midpoint of segment AB, BH is perpendicular to AH. Find the length of line HE.
问题：如图所示，FD=16，D平分线段BC，E是线段AC的中点，F是线段AB的中点，BH⊥AH。求直线HE的长度。
Construction (CDL):
Shape(AF,FG,GA)
Shape(AG,GE,EA)
Shape(FB,BD,DF)
Shape(FD,DH,HG,GF)
Shape(EG,GH,HE)
Shape(EH,HC,CE)
Collinear(AFB)
Collinear(AEC)
Collinear(FGE)
Collinear(AGH)
Collinear(BDHC)
Conditions (CDL):
Equal(LengthOfLine(FD),16)
IsMidpointOfLine(D,BC)
IsMidpointOfLine(E,AC)
IsMidpointOfLine(F,AB)
PerpendicularBetweenLine(BH,AH)
Goal: Value(LengthOfLine(HE))
Answer: 16
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,DF,BCA)
midsegment_of_triangle_property_length(1,DF,BCA)
adjacent_complementary_angle(1,BHG,GHC)
right_triangle_judgment_angle(1,AHC)
median_of_triangle_judgment(1,HE,HCA)
right_triangle_property_length_of_median(1,AHC,E)
Problem: As shown in the diagram, DE=2, D bisects segment AO, E is the midpoint of segment BO, F is the midpoint of segment CO. Find the length of line AB.
问题：如图所示，DE=2，D平分线段AO，E是线段BO的中点，F平分线段CO。求直线AB的长度。
Construction (CDL):
Shape(AB,BE,ED,DA)
Shape(DE,EO,OD)
Shape(AD,DF,FC,CA)
Shape(DO,OF,FD)
Shape(EB,BC,CF,FE)
Shape(OE,EF,FO)
Collinear(OEB)
Collinear(OFC)
Collinear(ODA)
Conditions (CDL):
Equal(LengthOfLine(DE),2)
IsMidpointOfLine(D,AO)
IsMidpointOfLine(E,BO)
IsMidpointOfLine(F,CO)
Goal: Value(LengthOfLine(AB))
Answer: 4
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,DE,OAB)
midsegment_of_triangle_property_length(1,DE,OAB)
Problem: As shown in the diagram, BC=6, BF bisects ∠ABC, D bisects segment BC, E is the midpoint of segment AC. Find the length of line DF.
问题：如图所示，BC=6，BF平分∠ABC，D平分线段BC，E平分线段AC。求直线DF的长度。
Construction (CDL):
Shape(AB,BF,FE,EA)
Shape(FB,BD,DF)
Shape(EF,FD,DC,CE)
Collinear(BDC)
Collinear(EFD)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
IsBisectorOfAngle(BF,ABC)
IsMidpointOfLine(D,BC)
IsMidpointOfLine(E,AC)
Goal: Value(LengthOfLine(DF))
Answer: 3
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,ED,CAB)
midsegment_of_triangle_property_parallel(1,ED,CAB)
parallel_property_collinear_extend(3,ED,AB,F)
parallel_property_alternate_interior_angle(1,FD,AB)
isosceles_triangle_judgment_angle_equal(1,DFB)
line_addition(1,BD,DC)
Problem: As shown in the diagram, BC=6, BF bisects ∠ABC, D bisects segment BC, E is the midpoint of segment AC. Find the length of line DF.
问题：如图所示，BC=6，BF平分∠ABC，D平分线段BC，E是线段AC的中点。求直线DF的长度。
Construction (CDL):
Shape(AB,BF,FE,EA)
Shape(FB,BD,DF)
Shape(EF,FD,DC,CE)
Collinear(BDC)
Collinear(AEC)
Collinear(EFD)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
IsBisectorOfAngle(BF,ABC)
IsMidpointOfLine(D,BC)
IsMidpointOfLine(E,AC)
Goal: Value(LengthOfLine(DF))
Answer: 3
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,ED,CAB)
midsegment_of_triangle_property_parallel(1,ED,CAB)
parallel_property_collinear_extend(3,ED,AB,F)
parallel_property_alternate_interior_angle(1,FD,AB)
isosceles_triangle_judgment_angle_equal(1,DFB)
line_addition(1,BD,DC)
Problem: As shown in the diagram, AD=BC, ∠FEP=18°, E is the midpoint of segment AB, F bisects segment CD, P is the midpoint of segment BD. Find the measure of ∠PFE.
问题：如图所示，AD=BC，∠FEP=18°，E是线段AB的中点，F是线段CD的中点，P是线段BD的中点。求∠PFE的大小。
Construction (CDL):
Shape(DA,AE,EG,GD)
Shape(FD,DG,GF)
Shape(FG,GP,PF)
Shape(PG,GE,EP)
Shape(BP,PE,EB)
Shape(CF,FP,PB,BC)
Collinear(DFC)
Collinear(AEB)
Collinear(DGPB)
Collinear(FGE)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(BC))
Equal(MeasureOfAngle(FEP),18)
IsMidpointOfLine(E,AB)
IsMidpointOfLine(F,CD)
IsMidpointOfLine(P,BD)
Goal: Value(MeasureOfAngle(PFE))
Answer: 18
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,PE,BDA)
midsegment_of_triangle_judgment_midpoint(1,PF,DBC)
midsegment_of_triangle_property_length(1,PE,BDA)
midsegment_of_triangle_property_length(1,PF,DBC)
isosceles_triangle_judgment_line_equal(1,PFE)
isosceles_triangle_property_angle_equal(1,PFE)
Problem: As shown in the diagram, AB=6, AC=9, AD bisects ∠BAE, M bisects segment BC, AD⊥BD. Find the length of line MD.
问题：如图所示，AB=6，AC=9，AD平分∠BAE，M是线段BC的中点，AD垂直于BD。求直线MD的长度。
Construction (CDL):
Shape(ED,DA,AE)
Shape(AD,DF,FA)
Shape(FD,DB,BF)
Shape(CA,AF,FM,MC)
Shape(MF,FB,BM)
Collinear(EAC)
Collinear(EDB)
Collinear(BFA)
Collinear(DFM)
Collinear(BMC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AC),9)
IsBisectorOfAngle(AD,BAE)
IsMidpointOfLine(M,BC)
PerpendicularBetweenLine(AD,BD)
Goal: Value(LengthOfLine(MD))
Answer: 15/2
Theorem sequence:
adjacent_complementary_angle(1,EDA,ADB)
mirror_congruent_triangle_judgment_aas(2,EDA,BAD)
mirror_congruent_triangle_property_line_equal(1,DAE,DBA)
mirror_congruent_triangle_property_line_equal(1,AED,ADB)
line_addition(1,EA,AC)
midsegment_of_triangle_judgment_midpoint(1,MD,BCE)
midsegment_of_triangle_property_length(1,MD,BCE)
Problem: As shown in the diagram, AD=BC, ∠BCA=87°, ∠DAC=15°, E is the midpoint of segment AB, F is the midpoint of segment CD, G bisects segment AC. Find the measure of ∠GEF.
问题：如图所示，AD=BC，∠BCA=87°，∠DAC=15°，E是线段AB的中点，F平分线段CD，G平分线段AC。求∠GEF的大小。
Construction (CDL):
Shape(DA,AG,GF,FD)
Shape(FG,GH,HF)
Shape(CF,FH,HC)
Shape(AE,EG,GA)
Shape(HG,GE,EH)
Shape(CH,HE,EB,BC)
Collinear(DFC)
Collinear(AGHC)
Collinear(AEB)
Collinear(FHE)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(BC))
Equal(MeasureOfAngle(BCA),87)
Equal(MeasureOfAngle(DAC),15)
IsMidpointOfLine(E,AB)
IsMidpointOfLine(F,CD)
IsMidpointOfLine(G,AC)
Goal: Value(MeasureOfAngle(GEF))
Answer: 36
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,FG,CDA)
midsegment_of_triangle_judgment_midpoint(1,EG,ABC)
midsegment_of_triangle_property_parallel(1,FG,CDA)
midsegment_of_triangle_property_parallel(1,EG,ABC)
midsegment_of_triangle_property_length(1,FG,CDA)
midsegment_of_triangle_property_length(1,EG,ABC)
parallel_property_corresponding_angle(2,AD,GF,H)
parallel_property_ipsilateral_internal_angle(1,CB,GE)
angle_addition(1,FGC,CGE)
isosceles_triangle_judgment_line_equal(1,GEF)
isosceles_triangle_property_angle_equal(1,GEF)
triangle_property_angle_sum(1,FGE)
Problem: As shown in the diagram, OP=CD, PC=PD, ∠AOB=30°, OE is perpendicular to PE. Find the measure of ∠DPO.
问题：如图所示，OP=CD，PC=PD，∠AOB=30°，OE⊥PE。求∠DPO的大小。
Construction (CDL):
Shape(PO,OC,CP)
Shape(PC,CE,EP)
Shape(PE,ED,DP)
Shape(AP,PD,DB,BA)
Collinear(OPA)
Collinear(OCEDB)
Conditions (CDL):
Equal(LengthOfLine(OP),LengthOfLine(CD))
Equal(LengthOfLine(PC),LengthOfLine(PD))
Equal(MeasureOfAngle(AOB),30)
PerpendicularBetweenLine(OE,PE)
Goal: Value(MeasureOfAngle(DPO))
Answer: 105
Theorem sequence:
sine_theorem(1,POE)
isosceles_triangle_judgment_line_equal(1,PCD)
altitude_of_triangle_judgment(1,PE,PCD)
isosceles_triangle_property_line_coincidence(1,PCD,E)
line_addition(1,CE,ED)
isosceles_triangle_judgment_line_equal(1,EDP)
isosceles_triangle_property_angle_equal(1,EDP)
triangle_property_angle_sum(1,EDP)
triangle_property_angle_sum(1,POE)
angle_addition(1,DPE,EPO)
Problem: As shown in the diagram, OA=OB, OA=OC, ∠CAO=36°, BC is parallel to AO. Find the measure of ∠CBO.
问题：如图所示，OA=OB，OA=OC，∠CAO=36°，BC平行于AO。求∠CBO的大小。
Construction (CDL):
Shape(BD,DC,CB)
Shape(CD,DO,OC)
Shape(DA,AO,OD)
Collinear(BDO)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(OA),LengthOfLine(OB))
Equal(LengthOfLine(OA),LengthOfLine(OC))
Equal(MeasureOfAngle(CAO),36)
ParallelBetweenLine(BC,AO)
Goal: Value(MeasureOfAngle(CBO))
Answer: 72
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,OCB)
isosceles_triangle_judgment_line_equal(1,OCA)
isosceles_triangle_property_angle_equal(1,OCB)
isosceles_triangle_property_angle_equal(1,OCA)
parallel_property_alternate_interior_angle(2,BC,AO)
angle_addition(1,OCD,DCB)
Problem: As shown in the diagram, the area of triangle ABC is 65, the area of triangle ABC is Add(AreaOfTriangle(DCA),AreaOfTriangle(DAB)), AB=13, AC=13, ∠CMD=90°, AN is perpendicular to DN. Find the sum of the length of line DN and the length of line DM.
问题：如图所示，△ABC的面积为65，△ABC的面积为Add(AreaOfTriangle(DCA),AreaOfTriangle(DAB))，AB=13，AC=13，∠CMD=90°，AN⊥DN。求直线DN的长度与直线DM的长度之和。
Construction (CDL):
Shape(CM,MD,DC)
Shape(DN,NB,BD)
Shape(MA,AD,DM)
Shape(DA,AN,ND)
Collinear(CDB)
Collinear(CMA)
Collinear(BNA)
Conditions (CDL):
Equal(AreaOfTriangle(ABC),65)
Equal(AreaOfTriangle(ABC),Add(AreaOfTriangle(DCA),AreaOfTriangle(DAB)))
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),13)
Equal(MeasureOfAngle(CMD),90)
PerpendicularBetweenLine(AN,DN)
Goal: Value(Add(LengthOfLine(DN),LengthOfLine(DM)))
Answer: 10
Theorem sequence:
altitude_of_triangle_judgment(1,DN,DAB)
altitude_of_triangle_judgment(1,DM,DCA)
triangle_area_formula_common(1,DCA)
triangle_area_formula_common(1,DAB)
Problem: As shown in the diagram, AB=AC, ∠CBA=70°, CD∥BA. Find the measure of ∠DCA.
问题：如图所示，AB=AC，∠CBA=70°，CD平行于BA。求∠DCA的大小。
Construction (CDL):
Shape(DC,CA,AD)
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(CBA),70)
ParallelBetweenLine(CD,BA)
Goal: Value(MeasureOfAngle(DCA))
Answer: 40
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ACB)
isosceles_triangle_property_angle_equal(1,ACB)
triangle_property_angle_sum(1,CBA)
parallel_property_alternate_interior_angle(1,CD,BA)
Problem: As shown in the diagram, ∠CAB=30°, A is the center of ⊙A. Find the measure of ∠ABC.
问题：如图所示，∠CAB=30°，圆A的圆心为A。求∠ABC的大小。
Construction (CDL):
Shape(CA,AB,BC)
Shape(CB,ABC)
Shape(AC,ACB,BA)
Cocircular(A,BC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),30)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfAngle(ABC))
Answer: 75
Theorem sequence:
radius_of_circle_property_length_equal(1,AB,A)
radius_of_circle_property_length_equal(1,AC,A)
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,CAB)
Problem: As shown in the diagram, AB=AC, ∠EAD=40°, DE perpendicular bisects CA. Find the measure of ∠BCD.
问题：如图所示，AB=AC，∠EAD=40°，DE是CA的垂直平分线。求∠BCD的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DC,CE)
Shape(DB,BC,CD)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(EAD),40)
IsPerpendicularBisectorOfLine(DE,CA)
Goal: Value(MeasureOfAngle(BCD))
Answer: 30
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
perpendicular_bisector_property_distance_equal(1,DE,CA)
isosceles_triangle_judgment_line_equal(1,DCA)
isosceles_triangle_property_angle_equal(1,DCA)
angle_addition(1,BCD,DCE)
Problem: As shown in the diagram, AD=CD, ∠CDA=40°, AB∥CD. Find the measure of ∠CAE.
问题：如图所示，AD=CD，∠CDA=40°，AB平行于CD。求∠CAE的大小。
Construction (CDL):
Shape(EC,CA,AE)
Shape(BA,AD,DB)
Shape(DA,AC,CD)
Collinear(EAB)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(CD))
Equal(MeasureOfAngle(CDA),40)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(CAE))
Answer: 70
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DAC)
isosceles_triangle_property_angle_equal(1,DAC)
triangle_property_angle_sum(1,DAC)
parallel_property_alternate_interior_angle(1,AB,CD)
angle_addition(1,BAD,DAC)
adjacent_complementary_angle(1,BAC,CAE)