SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, DC=10, E is the midpoint of segment AD, F is the midpoint of segment BC, G bisects segment EP, H is the midpoint of segment PF, BA and CD are opposite sides of the ▱ ABCD. Find the length of line GH.
问题：如图所示，DC=10，E是线段AD的中点，F是线段BC的中点，G是线段EP的中点，H是线段PF的中点，ABCD是平行四边形。求直线GH的长度。
Construction (CDL):
Shape(DE,EG,GP,PD)
Shape(CP,PH,HF,FC)
Shape(PG,GH,HP)
Shape(GE,EF,FH,HG)
Shape(EA,AB,BF,FE)
Collinear(AED)
Collinear(DPC)
Collinear(CFB)
Collinear(EGP)
Collinear(PHF)
Conditions (CDL):
Equal(LengthOfLine(DC),10)
IsMidpointOfLine(E,AD)
IsMidpointOfLine(F,BC)
IsMidpointOfLine(G,EP)
IsMidpointOfLine(H,PF)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(GH))
Answer: 5
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
midsegment_of_triangle_judgment_midpoint(1,GH,PEF)
midsegment_of_triangle_property_length(1,GH,PEF)
midsegment_of_quadrilateral_judgment_midpoint(1,EF,DABC)
midsegment_of_quadrilateral_property_length(1,EF,DABC)
Problem: As shown in the diagram, the perimeter of △CDE is 5, EO perpendicular bisects AC, ABCD is a parallelogram. Find the perimeter of quadrilateral ABCD.
问题：如图所示，三角形CDE的周长为5，EO是AC的垂直平分线，四边形ABCD是▱。求四边形ABCD的周长。
Construction (CDL):
Shape(AO,OE,EA)
Shape(EO,OC,CE)
Shape(DE,EC,CD)
Shape(AB,BF,FO,OA)
Shape(OF,FC,CO)
Collinear(AED)
Collinear(BFC)
Collinear(AOC)
Collinear(EOF)
Conditions (CDL):
Equal(PerimeterOfTriangle(CDE),5)
IsPerpendicularBisectorOfLine(EO,AC)
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 10
Theorem sequence:
triangle_perimeter_formula(1,CDE)
perpendicular_bisector_property_distance_equal(1,EO,AC)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
line_addition(1,AE,ED)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, EF=3, ∠ABC=60°, AE is parallel to BD, BA and CD are opposite sides of the ▱ ABCD, CF⊥EF. Find the length of line AB.
问题：如图所示，EF=3，∠ABC=60°，AE平行于BD，AD和BC是平行四边形ABCD的一组对边，CF垂直于EF。求直线AB的长度。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DE,EA)
Shape(DB,BC,CD)
Shape(ED,DC,CF,FE)
Collinear(BCF)
Collinear(EDC)
Conditions (CDL):
Equal(LengthOfLine(EF),3)
Equal(MeasureOfAngle(ABC),60)
ParallelBetweenLine(AE,BD)
Parallelogram(ABCD)
PerpendicularBetweenLine(CF,EF)
Goal: Value(LengthOfLine(AB))
Answer: sqrt(3)
Theorem sequence:
line_addition(1,CD,DE)
parallel_property_collinear_extend(1,DC,AB,E)
parallelogram_judgment_parallel_and_parallel(1,ABDE)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,ABDE)
parallel_property_corresponding_angle(2,BA,CD,F)
sine_theorem(1,ECF)
Problem: As shown in the diagram, AD=8, BE=2, DE is the angle bisector of ∠CDA, BA and CD are opposite sides of the parallelogram ABCD. Find the perimeter of quadrilateral ABCD.
问题：如图所示，AD=8，BE=2，DE是∠CDA的角平分线，ABCD是平行四边形。求ABCD的周长。
Construction (CDL):
Shape(AB,BE,ED,DA)
Shape(DE,EC,CD)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BE),2)
IsBisectorOfAngle(DE,CDA)
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 28
Theorem sequence:
line_addition(1,BE,EC)
parallel_property_collinear_extend(3,CB,DA,E)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
parallel_property_alternate_interior_angle(2,CE,DA)
isosceles_triangle_judgment_angle_equal(1,CDE)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, DE=CE, ∠ABC=50°, ∠CAB=80°, quadrilateral ABCD is a ▱. Find the measure of ∠EOC.
问题：如图所示，DE=CE，∠ABC=50°，∠CAB=80°，四边形ABCD是▱。求∠EOC的大小。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(DO,OE,ED)
Shape(EO,OC,CE)
Shape(BC,CO,OB)
Collinear(AOC)
Collinear(BOD)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(DE),LengthOfLine(CE))
Equal(MeasureOfAngle(ABC),50)
Equal(MeasureOfAngle(CAB),80)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(EOC))
Answer: 50
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ABCD,O)
midsegment_of_triangle_judgment_midpoint(1,EO,CDA)
midsegment_of_triangle_property_parallel(1,EO,CDA)
parallel_property_corresponding_angle(2,AD,OE,C)
parallel_property_ipsilateral_internal_angle(1,AD,BC)
angle_addition(1,DAO,OAB)
Problem: As shown in the diagram, AB=18, AD=12, EF=4, BE bisects ∠ABC, ABCD is a parallelogram, FG is perpendicular to CG. Find the length of line CG.
问题：如图所示，AB=18，AD=12，EF=4，BE平分∠ABC，BA和CD是▱ABCD的一组对边，FG垂直于CG。求直线CG的长度。
Construction (CDL):
Shape(ED,DF,FE)
Shape(EF,FC)
Shape(DA,AB,BG,GF,FD)
Shape(FG,GC,CF)
Shape(CG,GB,BC)
Collinear(DFC)
Collinear(ADE)
Collinear(EFGB)
Conditions (CDL):
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(EF),4)
IsBisectorOfAngle(BE,ABC)
Parallelogram(ABCD)
PerpendicularBetweenLine(FG,CG)
Goal: Value(LengthOfLine(CG))
Answer: 8*sqrt(2)
Theorem sequence:
line_addition(1,DF,FC)
line_addition(1,FG,GB)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
parallel_property_collinear_extend(2,AD,BC,E)
parallel_property_collinear_extend(3,DC,AB,F)
parallel_property_alternate_interior_angle(1,DE,BC)
parallel_property_alternate_interior_angle(2,DE,BC)
parallel_property_alternate_interior_angle(1,FC,AB)
isosceles_triangle_judgment_angle_equal(1,CFB)
perpendicular_bisector_judgment_distance_equal(1,CG,FB)
similar_triangle_judgment_aa(1,FED,FBC)
similar_triangle_property_line_ratio(1,EDF,BCF)
similar_triangle_property_line_ratio(1,DFE,CFB)
cosine_theorem(1,GCF)
Problem: As shown in the diagram, AB=3, BC=5, EF perpendicular bisects AC, BA and CD are opposite sides of the parallelogram ABCD. Find the perimeter of △CDE.
问题：如图所示，AB=3，BC=5，EF是AC的垂直平分线，BA和CD是平行四边形ABCD的一组对边。求三角形CDE的周长。
Construction (CDL):
Shape(AB,BC,CF,FA)
Shape(AF,FE,EA)
Shape(EF,FC,CE)
Shape(DE,EC,CD)
Collinear(AFC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(BC),5)
IsPerpendicularBisectorOfLine(EF,AC)
Parallelogram(ABCD)
Goal: Value(PerimeterOfTriangle(CDE))
Answer: 8
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
perpendicular_bisector_property_distance_equal(1,EF,AC)
line_addition(1,AE,ED)
triangle_perimeter_formula(1,CDE)
Problem: As shown in the diagram, AB=10, AD=6, ∠BCA=90°, AD and BC are opposite sides of the ▱ ABCD. Find the length of line OB.
问题：如图所示，AB=10，AD=6，∠BCA=90°，BA和CD是▱ABCD的一组对边。求直线OB的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(OB,BC,CO)
Shape(DO,OC,CD)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AD),6)
Equal(MeasureOfAngle(BCA),90)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(OB))
Answer: 2*sqrt(13)
Theorem sequence:
line_addition(1,AO,OC)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_opposite_line_equal(1,BCDA)
cosine_theorem(1,CAB)
cosine_theorem(1,COB)
Problem: As shown in the diagram, BD=8, the perimeter of ABCD is 24, E bisects segment DC, O is the midpoint of segment BD, ABCD is a ▱. Find the perimeter of △DOE.
问题：如图所示，BD=8，四边形ABCD的周长为24，E是线段DC的中点，O是线段BD的中点，BA和CD是平行四边形ABCD的一组对边。求三角形DOE的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BC,CO)
Shape(OC,CE,EO)
Shape(DO,OE,ED)
Shape(AO,OD,DA)
Collinear(BOD)
Collinear(AOC)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(BD),8)
Equal(PerimeterOfQuadrilateral(ABCD),24)
IsMidpointOfLine(E,DC)
IsMidpointOfLine(O,BD)
Parallelogram(ABCD)
Goal: Value(PerimeterOfTriangle(DOE))
Answer: 10
Theorem sequence:
quadrilateral_perimeter_formula(1,ABCD)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
line_addition(1,BO,OD)
line_addition(1,CE,ED)
midsegment_of_triangle_judgment_midpoint(1,OE,DBC)
midsegment_of_triangle_property_parallel(1,OE,DBC)
parallel_property_corresponding_angle(1,OE,BC,D)
similar_triangle_judgment_aa(1,EDO,CDB)
similar_triangle_property_line_ratio(1,EDO,CDB)
similar_triangle_property_line_ratio(1,DOE,DBC)
triangle_perimeter_formula(1,DOE)
Problem: As shown in the diagram, AB=4, CE bisects ∠DCB, DE is the angle bisector of ∠ADC, AE is parallel to DC, EB is parallel to DC, DA and CB are opposite sides of the ▱ ADCB. Find the perimeter of quadrilateral ADCB.
问题：如图所示，AB=4，CE平分∠DCB，DE是∠ADC的角平分线，AE∥DC，EB平行于DC，四边形ADCB是平行四边形。求ADCB的周长。
Construction (CDL):
Shape(BE,EC,CB)
Shape(ED,DC,CE)
Shape(EA,AD,DE)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
IsBisectorOfAngle(CE,DCB)
IsBisectorOfAngle(DE,ADC)
ParallelBetweenLine(AE,DC)
ParallelBetweenLine(EB,DC)
Parallelogram(ADCB)
Goal: Value(PerimeterOfQuadrilateral(ADCB))
Answer: 12
Theorem sequence:
line_addition(1,AE,EB)
parallelogram_property_opposite_line_equal(1,DCBA)
parallel_property_alternate_interior_angle(1,EB,DC)
parallel_property_alternate_interior_angle(2,AE,DC)
isosceles_triangle_judgment_angle_equal(1,BEC)
isosceles_triangle_judgment_angle_equal(1,ADE)
quadrilateral_perimeter_formula(1,ADCB)
Problem: As shown in the diagram, the perimeter of quadrilateral ABCD is 60, AD and BC are opposite sides of the parallelogram ABCD, EO⊥DO. Find the perimeter of triangle ABE.
问题：如图所示，四边形ABCD的周长为60，ABCD是平行四边形，EO⊥DO。求三角形ABE的周长。
Construction (CDL):
Shape(DO,OC,CD)
Shape(OB,BC,CO)
Shape(EA,AF,FE)
Shape(EF,FO,OE)
Shape(DE,EO,OD)
Shape(OF,FB,BO)
Shape(AB,BF,FA)
Collinear(AFOC)
Collinear(BOD)
Collinear(AED)
Collinear(BFE)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),60)
Parallelogram(ABCD)
PerpendicularBetweenLine(EO,DO)
Goal: Value(PerimeterOfTriangle(ABE))
Answer: 30
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
line_addition(1,AE,ED)
adjacent_complementary_angle(1,BEO,EOD)
parallelogram_property_diagonal_bisection(1,BCDA,O)
triangle_perimeter_formula(1,ABE)
quadrilateral_perimeter_formula(1,ABCD)
mirror_congruent_triangle_judgment_sas(1,OEB,ODE)
mirror_congruent_triangle_property_line_equal(1,OEB,ODE)
Problem: As shown in the diagram, AB=2, AE=1, BC=4, ABCD is a ▱, BE⊥AE. Find the area of ABCD.
问题：如图所示，AB=2，AE=1，BC=4，ABCD是平行四边形，BE⊥AE。求四边形ABCD的面积。
Construction (CDL):
Shape(AB,BE,EA)
Shape(DA,AE,EC,CD)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(AE),1)
Equal(LengthOfLine(BC),4)
Parallelogram(ABCD)
PerpendicularBetweenLine(BE,AE)
Goal: Value(AreaOfQuadrilateral(ABCD))
Answer: 4
Theorem sequence:
altitude_of_quadrilateral_judgment_left_vertex(1,AE,ABCD)
parallelogram_area_formula_common(1,ABCD)
Problem: As shown in the diagram, the perimeter of ABCD is 8, E is the midpoint of segment BC, F is the midpoint of segment DC, O bisects segment AC, AD and BC are opposite sides of the ▱ ABCD. Find the perimeter of quadrilateral OECF.
问题：如图所示，ABCD的周长为8，E是线段BC的中点，F平分线段DC，O平分线段AC，ABCD是▱。求四边形OECF的周长。
Construction (CDL):
Shape(CF,FO,OC)
Shape(CO,OE,EC)
Shape(OF,FD,DA,AO)
Shape(EO,OA,AB,BE)
Collinear(AOC)
Collinear(DFC)
Collinear(BEC)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),8)
IsMidpointOfLine(E,BC)
IsMidpointOfLine(F,DC)
IsMidpointOfLine(O,AC)
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(OECF))
Answer: 4
Theorem sequence:
quadrilateral_perimeter_formula(1,ABCD)
midsegment_of_triangle_judgment_midpoint(1,OE,CAB)
midsegment_of_triangle_judgment_midpoint(1,FO,CDA)
midsegment_of_triangle_property_length(1,OE,CAB)
midsegment_of_triangle_property_length(1,FO,CDA)
line_addition(1,DF,FC)
line_addition(1,CE,EB)
quadrilateral_perimeter_formula(1,CFOE)
Problem: As shown in the diagram, AB=2, AD=1, EF=1, ∠DAE=60°, E bisects segment AB, F bisects segment AC. Find the length of line BD.
问题：如图所示，AB=2，AD=1，EF=1，∠DAE=60°，E是线段AB的中点，F平分线段AC。求直线BD的长度。
Construction (CDL):
Shape(DA,AE,EG,GD)
Shape(FD,DG,GF)
Shape(FG,GB,BC,CF)
Shape(GE,EB,BG)
Collinear(DFC)
Collinear(DGB)
Collinear(AEB)
Collinear(FGE)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(AD),1)
Equal(LengthOfLine(EF),1)
Equal(MeasureOfAngle(DAE),60)
IsMidpointOfLine(E,AB)
IsMidpointOfLine(F,AC)
Goal: Value(LengthOfLine(BD))
Answer: sqrt(3)
Theorem sequence:
cosine_theorem(1,ABD)
Problem: As shown in the diagram, ED=6, FG=2, BG is the angle bisector of ∠ABC, CF is the angle bisector of ∠BCA, EG is parallel to BC, FD is parallel to BC. Find the sum of the length of line EB and the length of line DC.
问题：如图所示，ED=6，FG=2，BG是∠ABC的角平分线，CF是∠BCA的角平分线，EG平行于BC，FD平行于BC。求直线EB的长度与直线DC的长度之和。
Construction (CDL):
Shape(AE,EF,FG,GD,DA)
Shape(EB,BH,HF,FE)
Shape(GF,FH,HG)
Shape(DG,GH,HC,CD)
Shape(HB,BC,CH)
Collinear(EFGD)
Collinear(FHC)
Collinear(GHB)
Collinear(AEB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(ED),6)
Equal(LengthOfLine(FG),2)
IsBisectorOfAngle(BG,ABC)
IsBisectorOfAngle(CF,BCA)
ParallelBetweenLine(EG,BC)
ParallelBetweenLine(FD,BC)
Goal: Value(Add(LengthOfLine(EB),LengthOfLine(DC)))
Answer: 8
Theorem sequence:
line_addition(1,EG,GD)
line_addition(1,DG,GF)
parallel_property_alternate_interior_angle(1,FD,BC)
parallel_property_alternate_interior_angle(2,EG,BC)
isosceles_triangle_judgment_angle_equal(1,EBG)
isosceles_triangle_judgment_angle_equal(1,DFC)
Problem: As shown in the diagram, ∠EAB=36°, AE bisects ∠CAB, AC∥ED, BE⊥AE. Find the measure of ∠DEB.
问题：如图所示，∠EAB=36°，AE平分∠CAB，AC∥ED，BE⊥AE。求∠DEB的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,ED,DC,CA)
Shape(EB,BD,DE)
Collinear(BDC)
Conditions (CDL):
Equal(MeasureOfAngle(EAB),36)
IsBisectorOfAngle(AE,CAB)
ParallelBetweenLine(AC,ED)
PerpendicularBetweenLine(BE,AE)
Goal: Value(MeasureOfAngle(DEB))
Answer: 126
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AC,ED)
round_angle(1,AED,DEA)
angle_addition(1,DEB,BEA)
Problem: As shown in the diagram, AB=4, AC=3, BE bisects ∠ABC, CD is the angle bisector of ∠BCA, DE is parallel to BC. Find the length of line DE.
问题：如图所示，AB=4，AC=3，BE是∠ABC的角平分线，CD平分∠BCA，DE平行于BC。求直线DE的长度。
Construction (CDL):
Shape(DF,FA,AD)
Shape(AF,FH,HG,GA)
Shape(AG,GE,EA)
Shape(GH,HC,CG)
Shape(HB,BC,CH)
Shape(HF,FB,BH)
Collinear(DAE)
Collinear(DFHC)
Collinear(EGHB)
Collinear(AFB)
Collinear(AGC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AC),3)
IsBisectorOfAngle(BE,ABC)
IsBisectorOfAngle(CD,BCA)
ParallelBetweenLine(DE,BC)
Goal: Value(LengthOfLine(DE))
Answer: 7
Theorem sequence:
parallel_property_alternate_interior_angle(1,DE,BC)
parallel_property_alternate_interior_angle(2,DE,BC)
isosceles_triangle_judgment_angle_equal(1,ADC)
isosceles_triangle_judgment_angle_equal(1,ABE)
line_addition(1,DA,AE)
Problem: As shown in the diagram, AE=AD, ∠ABC=∠BCA, ∠DAB=50°. Find the measure of ∠EDC.
问题：如图所示，AE=AD，∠ABC=∠BCA，∠DAB=50°。求∠EDC的大小。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DE,EA)
Shape(ED,DC,CE)
Collinear(BDC)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AE),LengthOfLine(AD))
Equal(MeasureOfAngle(ABC),MeasureOfAngle(BCA))
Equal(MeasureOfAngle(DAB),50)
Goal: Value(MeasureOfAngle(EDC))
Answer: 25
Theorem sequence:
adjacent_complementary_angle(1,CED,DEA)
triangle_property_angle_sum(1,EDC)
adjacent_complementary_angle(1,BDA,ADC)
triangle_property_angle_sum(1,DAB)
isosceles_triangle_judgment_line_equal(1,ADE)
isosceles_triangle_property_angle_equal(1,ADE)
angle_addition(1,ADE,EDC)
Problem: As shown in the diagram, PM=2, ∠MOQ=30°, OP is the angle bisector of ∠MOQ, OM∥QP, PM⊥OM. Find the length of line OQ.
问题：如图所示，PM=2，∠MOQ=30°，OP是∠MOQ的角平分线，OM平行于QP，PM垂直于OM。求直线OQ的长度。
Construction (CDL):
Shape(MO,OP,PM)
Shape(PO,OQ,QP)
Conditions (CDL):
Equal(LengthOfLine(PM),2)
Equal(MeasureOfAngle(MOQ),30)
IsBisectorOfAngle(OP,MOQ)
ParallelBetweenLine(OM,QP)
PerpendicularBetweenLine(PM,OM)
Goal: Value(LengthOfLine(OQ))
Answer: 4
Theorem sequence:
angle_addition(1,MOP,POQ)
triangle_property_angle_sum(1,POQ)
parallel_property_alternate_interior_angle(1,OM,QP)
sine_theorem(1,PMO)
sine_theorem(1,OQP)
Problem: As shown in the diagram, AC=14, BC=10, AO is the angle bisector of ∠CAB, BO bisects ∠ABC, BA is parallel to MO, BA∥ON. Find the perimeter of △CNM.
问题：如图所示，AC=14，BC=10，AO平分∠CAB，BO是∠ABC的角平分线，BA∥MO，BA平行于ON。求△CNM的周长。
Construction (CDL):
Shape(OA,AB,BO)
Shape(NA,AO,ON)
Shape(OB,BM,MO)
Shape(NO,OM,MC,CN)
Collinear(MON)
Collinear(ANC)
Collinear(BMC)
Conditions (CDL):
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(BC),10)
IsBisectorOfAngle(AO,CAB)
IsBisectorOfAngle(BO,ABC)
ParallelBetweenLine(BA,MO)
ParallelBetweenLine(BA,ON)
Goal: Value(PerimeterOfTriangle(CNM))
Answer: 24
Theorem sequence:
triangle_perimeter_formula(1,CNM)
parallel_property_alternate_interior_angle(1,BA,MO)
parallel_property_alternate_interior_angle(2,BA,ON)
isosceles_triangle_judgment_angle_equal(1,NAO)
isosceles_triangle_judgment_angle_equal(1,MOB)
line_addition(1,AN,NC)
line_addition(1,MO,ON)
line_addition(1,BM,MC)
Problem: As shown in the diagram, BC=16, BO is the angle bisector of ∠ABC, CO bisects ∠BCA, AC is parallel to OE, BA is parallel to DO. Find the perimeter of triangle ODE.
问题：如图所示，BC=16，BO平分∠ABC，CO平分∠BCA，AC∥OE，BA平行于DO。求△ODE的周长。
Construction (CDL):
Shape(AB,BO,OC,CA)
Shape(OB,BD,DO)
Shape(OD,DE,EO)
Shape(OE,EC,CO)
Collinear(BDEC)
Conditions (CDL):
Equal(LengthOfLine(BC),16)
IsBisectorOfAngle(BO,ABC)
IsBisectorOfAngle(CO,BCA)
ParallelBetweenLine(AC,OE)
ParallelBetweenLine(BA,DO)
Goal: Value(PerimeterOfTriangle(ODE))
Answer: 16
Theorem sequence:
line_addition(1,BD,DE)
line_addition(1,BE,EC)
parallel_property_alternate_interior_angle(1,BA,DO)
parallel_property_alternate_interior_angle(2,AC,OE)
isosceles_triangle_judgment_angle_equal(1,DOB)
isosceles_triangle_judgment_angle_equal(1,ECO)
triangle_perimeter_formula(1,ODE)
Problem: As shown in the diagram, ∠FDG=30°, BD bisects ∠ABC, ED∥BC. Find the measure of ∠AEG.
问题：如图所示，∠FDG=30°，BD是∠ABC的角平分线，ED∥BC。求∠AEG的大小。
Construction (CDL):
Shape(AG,GD)
Shape(DF,FC)
Shape(AE,EG,GA)
Shape(DG,GF,FD)
Shape(FB,BC,CF)
Shape(GE,EB,BF,FG)
Collinear(AGFC)
Collinear(EGD)
Collinear(BFD)
Collinear(AEB)
Conditions (CDL):
Equal(MeasureOfAngle(FDG),30)
IsBisectorOfAngle(BD,ABC)
ParallelBetweenLine(ED,BC)
Goal: Value(MeasureOfAngle(AEG))
Answer: 60
Theorem sequence:
parallel_property_alternate_interior_angle(2,ED,BC)
triangle_property_angle_sum(1,EBD)
adjacent_complementary_angle(1,AEG,GEB)
Problem: As shown in the diagram, AB=5, EF=2, AE is the angle bisector of ∠DAB, DF bisects ∠CDA, AD∥BC, AD∥BE, AD is parallel to FC, BA∥CD. Find the length of line AD.
问题：如图所示，AB=5，EF=2，AE平分∠DAB，DF是∠CDA的角平分线，AD∥BC，AD平行于BE，AD平行于FC，BA平行于CD。求直线AD的长度。
Construction (CDL):
Shape(AG,GD,DA)
Shape(GF,FE,EG)
Shape(AB,BF,FG,GA)
Shape(DG,GE,EC,CD)
Collinear(AGE)
Collinear(FGD)
Collinear(BFEC)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(EF),2)
IsBisectorOfAngle(AE,DAB)
IsBisectorOfAngle(DF,CDA)
ParallelBetweenLine(AD,BC)
ParallelBetweenLine(AD,BE)
ParallelBetweenLine(AD,FC)
ParallelBetweenLine(BA,CD)
Goal: Value(LengthOfLine(AD))
Answer: 8
Theorem sequence:
line_addition(1,BF,FE)
line_addition(1,CF,FB)
parallelogram_judgment_parallel_and_parallel(1,DABC)
parallel_property_alternate_interior_angle(1,AD,BE)
parallel_property_alternate_interior_angle(2,AD,FC)
isosceles_triangle_judgment_angle_equal(1,CDF)
isosceles_triangle_judgment_angle_equal(1,BEA)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
Problem: As shown in the diagram, BC=8, BF bisects ∠ABC, D is the midpoint of segment BC, E bisects segment AC, BA∥DF. Find the length of line DF.
问题：如图所示，BC=8，BF是∠ABC的角平分线，D平分线段BC，E平分线段AC，BA∥DF。求直线DF的长度。
Construction (CDL):
Shape(AB,BF,FE,EA)
Shape(FB,BD,DF)
Shape(EF,FD,DC,CE)
Collinear(AEC)
Collinear(EFD)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(BC),8)
IsBisectorOfAngle(BF,ABC)
IsMidpointOfLine(D,BC)
IsMidpointOfLine(E,AC)
ParallelBetweenLine(BA,DF)
Goal: Value(LengthOfLine(DF))
Answer: 4
Theorem sequence:
line_addition(1,BD,DC)
parallel_property_alternate_interior_angle(1,BA,DF)
isosceles_triangle_judgment_angle_equal(1,DFB)
Problem: As shown in the diagram, AD=6, the perimeter of triangle AED is 16, BD bisects ∠ABC, ED∥BC. Find the length of line AB.
问题：如图所示，AD=6，△AED的周长为16，BD平分∠ABC，ED∥BC。求直线AB的长度。
Construction (CDL):
Shape(AE,ED,DA)
Shape(DE,EB,BD)
Shape(DB,BC,CD)
Collinear(AEB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AD),6)
Equal(PerimeterOfTriangle(AED),16)
IsBisectorOfAngle(BD,ABC)
ParallelBetweenLine(ED,BC)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
triangle_perimeter_formula(1,AED)
parallel_property_alternate_interior_angle(2,ED,BC)
isosceles_triangle_judgment_angle_equal(1,EBD)
line_addition(1,AE,EB)
Problem: As shown in the diagram, CE=3/2, triangle ABC is an equilateral triangle, BD is the angle bisector of ∠ABC, DE is perpendicular to CE. Find the length of line AB.
问题：如图所示，CE=3/2，△ABC为等边△，BD平分∠ABC，DE⊥CE。求直线AB的长度。
Construction (CDL):
Shape(DE,EC,CD)
Shape(DB,BE,ED)
Shape(AB,BD,DA)
Collinear(ADC)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(CE),3/2)
EquilateralTriangle(ABC)
IsBisectorOfAngle(BD,ABC)
PerpendicularBetweenLine(DE,CE)
Goal: Value(LengthOfLine(AB))
Answer: 6
Theorem sequence:
isosceles_triangle_property_line_coincidence(3,BCA,D)
equilateral_triangle_property_angle(1,CAB)
triangle_property_angle_sum(1,DEC)
sine_theorem(1,CDE)
line_addition(1,AD,DC)
Problem: As shown in the diagram, AB=BC, BC=CA, BD=CE, CA=AB, ∠ABC=60°, ∠ABC=∠BCA, ∠BCA=∠CAB, ∠CAB=∠ABC, ∠FAB=15°. Find the measure of ∠AFE.
问题：如图所示，AB=BC，BC=CA，BD=CE，CA=AB，∠ABC=60°，∠ABC=∠BCA，∠BCA=∠CAB，∠CAB=∠ABC，∠FAB=15°。求∠AFE的大小。
Construction (CDL):
Shape(AB,BF,FA)
Shape(AF,FE,EA)
Shape(FB,BD,DF)
Shape(EF,FD,DC,CE)
Collinear(AFD)
Collinear(BFE)
Collinear(AEC)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(BC))
Equal(LengthOfLine(BC),LengthOfLine(CA))
Equal(LengthOfLine(BD),LengthOfLine(CE))
Equal(LengthOfLine(CA),LengthOfLine(AB))
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(ABC),MeasureOfAngle(BCA))
Equal(MeasureOfAngle(BCA),MeasureOfAngle(CAB))
Equal(MeasureOfAngle(CAB),MeasureOfAngle(ABC))
Equal(MeasureOfAngle(FAB),15)
Goal: Value(MeasureOfAngle(AFE))
Answer: 60
Theorem sequence:
congruent_triangle_judgment_sas(1,BDA,CEB)
congruent_triangle_property_angle_equal(1,ABD,BCE)
angle_addition(1,ABF,FBD)
triangle_property_angle_sum(1,ABF)
adjacent_complementary_angle(1,BFA,AFE)
Problem: As shown in the diagram, ∠DAE=60°. Find the sum of the measure of ∠CDE and the measure of ∠DEB.
问题：如图所示，∠DAE=60°。求∠CDE的大小与∠DEB的大小之和。
Construction (CDL):
Shape(DA,AE,ED)
Shape(CD,DE,EB,BC)
Collinear(ADC)
Collinear(AEB)
Conditions (CDL):
Equal(MeasureOfAngle(DAE),60)
Goal: Value(Add(MeasureOfAngle(CDE),MeasureOfAngle(DEB)))
Answer: 240
Theorem sequence:
triangle_property_angle_sum(1,DAE)
adjacent_complementary_angle(1,CDE,EDA)
adjacent_complementary_angle(1,AED,DEB)
Problem: As shown in the diagram, ∠ABC=120-x°, ∠BCA=120-y°, ∠CAB=60°. Find the value of x+y.
问题：如图所示，∠ABC=120-x°，∠BCA=120-y°，∠CAB=60°。求x+y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),120-x)
Equal(MeasureOfAngle(BCA),120-y)
Equal(MeasureOfAngle(CAB),60)
Goal: Value(x+y)
Answer: 120
Theorem sequence:
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, AB=40, BC=40, ∠CBA=60°. Find the length of line AC.
问题：如图所示，AB=40，BC=40，∠CBA=60°。求直线AC的长度。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),40)
Equal(LengthOfLine(BC),40)
Equal(MeasureOfAngle(CBA),60)
Goal: Value(LengthOfLine(AC))
Answer: 40
Theorem sequence:
triangle_property_angle_sum(1,CBA)
isosceles_triangle_judgment_line_equal(1,BAC)
isosceles_triangle_property_angle_equal(1,BAC)
isosceles_triangle_judgment_angle_equal(1,CBA)
Problem: As shown in the diagram, triangle ABC is congruent to triangle FDE, CD=2, ∠BCA=90°, ∠CAB=30°, ∠DEF=90°, EC is perpendicular to DC. Find the length of line AB.
问题：如图所示，△ABC全等于△FDE，CD=2，∠BCA=90°，∠CAB=30°，∠DEF=90°，EC垂直于DC。求直线AB的长度。
Construction (CDL):
Shape(DE,EC,CD)
Shape(CE,EM,MC)
Shape(CM,MG,GH,HC)
Shape(MA,AG,GM)
Shape(HG,GF,FH)
Shape(CH,HB,BC)
Shape(FG,GA)
Shape(BH,HF)
Shape(DC,CB)
Shape(AM,ME)
Collinear(EMGF)
Collinear(AMC)
Collinear(AGHB)
Collinear(DCHF)
Conditions (CDL):
CongruentBetweenTriangle(ABC,FDE)
Equal(LengthOfLine(CD),2)
Equal(MeasureOfAngle(BCA),90)
Equal(MeasureOfAngle(CAB),30)
Equal(MeasureOfAngle(DEF),90)
PerpendicularBetweenLine(EC,DC)
Goal: Value(LengthOfLine(AB))
Answer: 8
Theorem sequence:
congruent_triangle_property_angle_equal(1,ABC,FDE)
triangle_property_angle_sum(1,EFD)
triangle_property_angle_sum(1,ECD)
congruent_triangle_property_line_equal(1,CAB,EFD)
sine_theorem(1,DEC)
sine_theorem(1,DEF)