SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=10, DG=3, AE is the angle bisector of ∠DAB, F bisects segment CD, ABCD is a parallelogram, DG is perpendicular to FG. Find the length of line AE.
问题：如图所示，AB=10，DG=3，AE平分∠DAB，F平分线段CD，BA和CD是平行四边形ABCD的一组对边，DG垂直于FG。求直线AE的长度。
Construction (CDL):
Shape(AG,GD,DA)
Shape(DG,GF,FD)
Shape(FC,CE,EF)
Shape(AB,BC,CF,FG,GA)
Collinear(AGFE)
Collinear(DFC)
Collinear(BCE)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(DG),3)
IsBisectorOfAngle(AE,DAB)
IsMidpointOfLine(F,CD)
Parallelogram(ABCD)
PerpendicularBetweenLine(DG,FG)
Goal: Value(LengthOfLine(AE))
Answer: 16
Theorem sequence:
line_addition(1,DF,FC)
line_addition(1,AF,FE)
line_addition(1,AG,GF)
adjacent_complementary_angle(1,AGD,DGF)
parallelogram_property_opposite_line_equal(1,ABCD)
parallel_property_collinear_extend(1,CB,DA,E)
parallel_property_collinear_extend(3,DC,AB,F)
parallel_property_alternate_interior_angle(1,EC,DA)
parallel_property_alternate_interior_angle(2,EC,DA)
parallel_property_alternate_interior_angle(2,DF,AB)
isosceles_triangle_judgment_angle_equal(1,DAF)
right_triangle_judgment_angle(1,AGD)
right_triangle_judgment_angle(1,DGF)
right_triangle_property_pythagorean(1,AGD)
right_triangle_property_pythagorean(1,DGF)
congruent_triangle_judgment_aas(1,FCE,FDA)
congruent_triangle_property_line_equal(1,CEF,DAF)
Problem: As shown in the diagram, ∠OAE=60°, quadrilateral ABCD is a parallelogram, AF is perpendicular to DF, BE is perpendicular to AE. Find the measure of ∠ABE.
问题：如图所示，∠OAE=60°，AD和BC是▱ABCD的一组对边，AF⊥DF，BE垂直于AE。求∠ABE的大小。
Construction (CDL):
Shape(BE,EA,AB)
Shape(AE,EO,OA)
Shape(OF,FC,CO)
Shape(AO,OC,CD,DA)
Collinear(BEOC)
Collinear(AOF)
Collinear(FCD)
Conditions (CDL):
Equal(MeasureOfAngle(OAE),60)
Parallelogram(ABCD)
PerpendicularBetweenLine(AF,DF)
PerpendicularBetweenLine(BE,AE)
Goal: Value(MeasureOfAngle(ABE))
Answer: 60
Theorem sequence:
parallel_property_collinear_extend(2,DC,AB,F)
parallel_property_alternate_interior_angle(2,DF,AB)
angle_addition(1,OAE,EAB)
triangle_property_angle_sum(1,BEA)
Problem: As shown in the diagram, AB=15, AD=15, BD=15. Find the measure of ∠ADB.
问题：如图所示，AB=15，AD=15，BD=15。求∠ADB的大小。
Construction (CDL):
Shape(AD,DB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AD),15)
Equal(LengthOfLine(BD),15)
Goal: Value(MeasureOfAngle(ADB))
Answer: 60
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ADB)
isosceles_triangle_judgment_line_equal(1,DBA)
isosceles_triangle_property_angle_equal(1,ADB)
isosceles_triangle_property_angle_equal(1,DBA)
triangle_property_angle_sum(1,ADB)
Problem: As shown in the diagram, AB=4, CE=3, AE bisects ∠DAB, AD and BC are opposite sides of the ▱ ABCD. Find the perimeter of quadrilateral ABCD.
问题：如图所示，AB=4，CE=3，AE是∠DAB的角平分线，BA和CD是平行四边形ABCD的一组对边。求ABCD的周长。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,EC,CD,DA)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(CE),3)
IsBisectorOfAngle(AE,DAB)
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 22
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(1,EB,DA)
isosceles_triangle_judgment_angle_equal(1,BEA)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, ∠DAO=42°, ∠OBC=23°, quadrilateral ABCD is a parallelogram. Find the measure of ∠DOC.
问题：如图所示，∠DAO=42°，∠OBC=23°，四边形ABCD是▱。求∠DOC的大小。
Construction (CDL):
Shape(AO,OD,DA)
Shape(BO,OA,AB)
Shape(OB,BC,CO)
Shape(OC,CD,DO)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(MeasureOfAngle(DAO),42)
Equal(MeasureOfAngle(OBC),23)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(DOC))
Answer: 65
Theorem sequence:
parallel_property_alternate_interior_angle(1,AD,BC)
triangle_property_angle_sum(1,OBC)
adjacent_complementary_angle(1,DOC,COB)
Problem: As shown in the diagram, AC=10, AD=4, BD=6, AD and BC are opposite sides of the parallelogram ABCD. Find the sum of the area of triangle DAC and the area of △ABC.
问题：如图所示，AC=10，AD=4，BD=6，AD和BC是▱ABCD的一组对边。求△DAC的面积与△ABC的面积之和。
Construction (CDL):
Shape(AO,OD,DA)
Shape(DO,OC,CD)
Shape(AB,BO,OA)
Shape(OB,BC,CO)
Collinear(DOB)
Collinear(AOC)
Conditions (CDL):
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(BD),6)
Parallelogram(ABCD)
Goal: Value(Add(AreaOfTriangle(DAC),AreaOfTriangle(ABC)))
Answer: 24
Theorem sequence:
line_addition(1,DO,OB)
line_addition(1,AO,OC)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
parallelogram_property_opposite_line_equal(1,BCDA)
right_triangle_judgment_pythagorean_inverse(1,ODA)
right_triangle_judgment_pythagorean_inverse(1,OBC)
triangle_area_formula_sine(1,AOD)
triangle_area_formula_sine(1,DAO)
triangle_area_formula_sine(1,BCO)
triangle_area_formula_sine(1,COB)
triangle_area_formula_sine(1,ACD)
triangle_area_formula_sine(1,CAB)
Problem: As shown in the diagram, the height of quadrilateral ABCD is 2, BC=3, AD and BC are opposite sides of the ▱ ABCD. Find the area of quadrilateral ABCD.
问题：如图所示，四边形ABCD的高为2，BC=3，BA和CD是▱ABCD的一组对边。求四边形ABCD的面积。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Equal(HeightOfQuadrilateral(ABCD),2)
Equal(LengthOfLine(BC),3)
Parallelogram(ABCD)
Goal: Value(AreaOfQuadrilateral(ABCD))
Answer: 6
Theorem sequence:
parallelogram_area_formula_common(1,ABCD)
Problem: As shown in the diagram, ∠DEB=150°, BE is the angle bisector of ∠ABC, quadrilateral ABCD is a ▱. Find the measure of ∠EAB.
问题：如图所示，∠DEB=150°，BE是∠ABC的角平分线，AD和BC是平行四边形ABCD的一组对边。求∠EAB的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BC,CD,DE)
Collinear(AED)
Conditions (CDL):
Equal(MeasureOfAngle(DEB),150)
IsBisectorOfAngle(BE,ABC)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(EAB))
Answer: 120
Theorem sequence:
adjacent_complementary_angle(1,DEB,BEA)
parallel_property_collinear_extend(3,AD,BC,E)
parallel_property_alternate_interior_angle(2,AE,BC)
triangle_property_angle_sum(1,ABE)
Problem: As shown in the diagram, BD=CD, ∠BCD=70°, BA and CD are opposite sides of the parallelogram ABCD, BE⊥AE. Find the measure of ∠DAE.
问题：如图所示，BD=CD，∠BCD=70°，四边形ABCD是平行四边形，BE⊥AE。求∠DAE的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,ED,DA)
Shape(DE,EB,BC,CD)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(BD),LengthOfLine(CD))
Equal(MeasureOfAngle(BCD),70)
Parallelogram(ABCD)
PerpendicularBetweenLine(BE,AE)
Goal: Value(MeasureOfAngle(DAE))
Answer: 20
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DBC)
isosceles_triangle_property_angle_equal(1,DBC)
parallel_property_alternate_interior_angle(2,AD,BC)
adjacent_complementary_angle(1,BEA,AED)
triangle_property_angle_sum(1,AED)
Problem: As shown in the diagram, Div(LengthOfLine(BC)=LengthOfLine(AB)), ∠AEF=54°, ∠CEH=18°, F is the midpoint of segment AD, H bisects segment BC, AD is parallel to EG, EG is parallel to BC, BA and CD are opposite sides of the parallelogram ABCD, CE⊥BE. Find the measure of ∠EBC.
问题：如图所示，Div(LengthOfLine(BC)=LengthOfLine(AB))，∠AEF=54°，∠CEH=18°，F平分线段AD，H平分线段BC，AD平行于EG，EG平行于BC，AD和BC是平行四边形ABCD的一组对边，CE⊥BE。求∠EBC的大小。
Construction (CDL):
Shape(AE,EF,FA)
Shape(FE,EG,GD,DF)
Shape(EB,BH,HE)
Shape(EH,HC,CE)
Shape(EC,CG,GE)
Collinear(AEB)
Collinear(DGC)
Collinear(BHC)
Collinear(AFD)
Conditions (CDL):
Equal(Div(LengthOfLine(BC),LengthOfLine(AB)),2)
Equal(MeasureOfAngle(AEF),54)
Equal(MeasureOfAngle(CEH),18)
IsMidpointOfLine(F,AD)
IsMidpointOfLine(H,BC)
ParallelBetweenLine(AD,EG)
ParallelBetweenLine(EG,BC)
Parallelogram(ABCD)
PerpendicularBetweenLine(CE,BE)
Goal: Value(MeasureOfAngle(EBC))
Answer: 72
Theorem sequence:
line_addition(1,BH,HC)
median_of_triangle_judgment(1,EH,EBC)
right_triangle_judgment_angle(1,CEB)
right_triangle_property_length_of_median(1,CEB,H)
isosceles_triangle_judgment_line_equal(1,HCE)
isosceles_triangle_property_angle_equal(1,HCE)
triangle_property_angle_sum(1,EBC)
Problem: As shown in the diagram, AB=5, AD=6, OE=2, BA and CD are opposite sides of the parallelogram ABCD. Find the perimeter of ABFE.
问题：如图所示，AB=5，AD=6，OE=2，BA和CD是平行四边形ABCD的一组对边。求ABFE的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OE,EA)
Shape(OB,BF,FO)
Shape(EO,OD,DE)
Shape(DO,OC,CD)
Shape(OF,FC,CO)
Collinear(AED)
Collinear(AOC)
Collinear(BFC)
Collinear(BOD)
Collinear(EOF)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(OE),2)
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(ABFE))
Answer: 15
Theorem sequence:
line_addition(1,AE,ED)
line_addition(1,EO,OF)
parallel_property_alternate_interior_angle(2,AD,BC)
vertical_angle(1,EOD,FOB)
parallelogram_property_diagonal_bisection(1,BCDA,O)
congruent_triangle_judgment_aas(2,EOD,FOB)
congruent_triangle_property_line_equal(1,ODE,OBF)
congruent_triangle_property_line_equal(1,DEO,BFO)
quadrilateral_perimeter_formula(1,ABFE)
Problem: As shown in the diagram, AB=5, the perimeter of triangle OCD is 23, quadrilateral ABCD is a ▱. Find the sum of the length of line AC and the length of line BD.
问题：如图所示，AB=5，△OCD的周长为23，四边形ABCD是平行四边形。求直线AC的长度与直线BD的长度之和。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(DO,OC,CD)
Shape(BC,CO,OB)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(PerimeterOfTriangle(OCD),23)
Parallelogram(ABCD)
Goal: Value(Add(LengthOfLine(AC),LengthOfLine(BD)))
Answer: 36
Theorem sequence:
line_addition(1,AO,OC)
line_addition(1,BO,OD)
parallelogram_property_opposite_line_equal(1,ABCD)
triangle_perimeter_formula(1,DOC)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
Problem: As shown in the diagram, Sub(MeasureOfAngle(ABC)=MeasureOfAngle(DAB)), BA and CD are opposite sides of the ▱ ABCD. Find the measure of ∠CDA.
问题：如图所示，Sub(MeasureOfAngle(ABC)=MeasureOfAngle(DAB))，四边形ABCD是平行四边形。求∠CDA的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Equal(Sub(MeasureOfAngle(ABC),MeasureOfAngle(DAB)),40)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(CDA))
Answer: 110
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,ABCD)
parallelogram_property_opposite_angle_equal(1,BCDA)
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, the perimeter of ABCD is 32, Sub(PerimeterOfTriangle(AOD)=PerimeterOfTriangle(ABO)), ABCD is a ▱. Find the length of line AB.
问题：如图所示，四边形ABCD的周长为32，Sub(PerimeterOfTriangle(AOD)=PerimeterOfTriangle(ABO))，四边形ABCD是▱。求直线AB的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(BC,CO,OB)
Shape(OC,CD,DO)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),32)
Equal(Sub(PerimeterOfTriangle(AOD),PerimeterOfTriangle(ABO)),4)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(AB))
Answer: 6
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
parallelogram_property_diagonal_bisection(1,BCDA,O)
quadrilateral_perimeter_formula(1,ABCD)
triangle_perimeter_formula(1,ABO)
triangle_perimeter_formula(1,AOD)
Problem: As shown in the diagram, AB=6, AD=9, ∠DAB=120°, AE is the angle bisector of ∠DAB, AD and BC are opposite sides of the parallelogram ABCD. Find the perimeter of triangle EFC.
问题：如图所示，AB=6，AD=9，∠DAB=120°，AE是∠DAB的角平分线，四边形ABCD是▱。求△EFC的周长。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,EC,CD,DA)
Shape(EF,FC,CE)
Collinear(BEC)
Collinear(DCF)
Collinear(AEF)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AD),9)
Equal(MeasureOfAngle(DAB),120)
IsBisectorOfAngle(AE,DAB)
Parallelogram(ABCD)
Goal: Value(PerimeterOfTriangle(EFC))
Answer: 9
Theorem sequence:
parallel_property_collinear_extend(2,DC,AB,F)
parallel_property_collinear_extend(3,CB,DA,E)
parallel_property_ipsilateral_internal_angle(1,AD,BC)
parallel_property_alternate_interior_angle(1,CF,AB)
parallel_property_alternate_interior_angle(2,CF,AB)
parallel_property_alternate_interior_angle(1,EB,DA)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
isosceles_triangle_judgment_angle_equal(1,BEA)
isosceles_triangle_judgment_angle_equal(1,DAF)
triangle_property_angle_sum(1,ABE)
line_addition(1,BE,EC)
line_addition(1,DC,CF)
isosceles_triangle_judgment_angle_equal(1,EFC)
triangle_perimeter_formula(1,EFC)
Problem: As shown in the diagram, Add(LengthOfLine(AC)=LengthOfLine(BD)), the perimeter of △OCD is 20, quadrilateral ABCD is a parallelogram. Find the length of line AB.
问题：如图所示，Add(LengthOfLine(AC)=LengthOfLine(BD))，三角形OCD的周长为20，BA和CD是▱ABCD的一组对边。求直线AB的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(BC,CO,OB)
Shape(OC,CD,DO)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(Add(LengthOfLine(AC),LengthOfLine(BD)),26)
Equal(PerimeterOfTriangle(OCD),20)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(AB))
Answer: 7
Theorem sequence:
line_addition(1,AO,OC)
line_addition(1,BO,OD)
triangle_perimeter_formula(1,OCD)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
parallelogram_property_opposite_line_equal(1,ABCD)
Problem: As shown in the diagram, AB=6, E is the midpoint of segment BC, ABCD is a ▱. Find the length of line OE.
问题：如图所示，AB=6，E是线段BC的中点，BA和CD是平行四边形ABCD的一组对边。求直线OE的长度。
Construction (CDL):
Shape(AO,OD,DA)
Shape(DO,OC,CD)
Shape(AB,BO,OA)
Shape(CO,OE,EC)
Shape(OB,BE,EO)
Collinear(AOC)
Collinear(DOB)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
IsMidpointOfLine(E,BC)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(OE))
Answer: 3
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ABCD,O)
midsegment_of_triangle_judgment_midpoint(1,OE,CAB)
midsegment_of_triangle_property_length(1,OE,CAB)
Problem: As shown in the diagram, the perimeter of triangle BCD is 18, E bisects segment AD, quadrilateral ABCD is a ▱. Find the perimeter of triangle DEO.
问题：如图所示，三角形BCD的周长为18，E是线段AD的中点，ABCD是平行四边形。求三角形DEO的周长。
Construction (CDL):
Shape(AO,OE,EA)
Shape(DO,OC,CD)
Shape(AB,BO,OA)
Shape(EO,OD,DE)
Shape(OB,BC,CO)
Collinear(AOC)
Collinear(DOB)
Collinear(AED)
Conditions (CDL):
Equal(PerimeterOfTriangle(BCD),18)
IsMidpointOfLine(E,AD)
Parallelogram(ABCD)
Goal: Value(PerimeterOfTriangle(DEO))
Answer: 9
Theorem sequence:
line_addition(1,DO,OB)
line_addition(1,AE,ED)
parallelogram_property_opposite_line_equal(1,BCDA)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
midsegment_of_triangle_judgment_midpoint(1,OE,ACD)
midsegment_of_triangle_property_length(1,OE,ACD)
triangle_perimeter_formula(1,EOD)
triangle_perimeter_formula(1,CDB)
Problem: As shown in the diagram, the perimeter of △CDE is 8, BA and CD are opposite sides of the ▱ ABCD, EO is perpendicular to BO. Find the perimeter of ABCD.
问题：如图所示，△CDE的周长为8，BA和CD是▱ABCD的一组对边，EO⊥BO。求ABCD的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BE,EO)
Shape(OE,EF,FO)
Shape(AO,OD,DA)
Shape(DO,OF,FD)
Shape(FC,CD,DF)
Shape(FE,EC,CF)
Collinear(AOFC)
Collinear(BOD)
Collinear(BEC)
Collinear(EFD)
Conditions (CDL):
Equal(PerimeterOfTriangle(CDE),8)
Parallelogram(ABCD)
PerpendicularBetweenLine(EO,BO)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 16
Theorem sequence:
line_addition(1,BE,EC)
triangle_perimeter_formula(1,CDE)
parallelogram_property_diagonal_bisection(1,BCDA,O)
adjacent_complementary_angle(1,DOE,EOB)
mirror_congruent_triangle_judgment_sas(1,OBE,OED)
mirror_congruent_triangle_property_line_equal(1,OBE,OED)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, the perimeter of quadrilateral ABCD is 18, quadrilateral ABCD is a ▱, EO is perpendicular to DO. Find the perimeter of triangle ABE.
问题：如图所示，ABCD的周长为18，BA和CD是平行四边形ABCD的一组对边，EO垂直于DO。求三角形ABE的周长。
Construction (CDL):
Shape(AB,BF,FA)
Shape(AF,FE,EA)
Shape(FB,BO,OF)
Shape(EF,FO,OE)
Shape(EO,OD,DE)
Shape(DO,OC,CD)
Shape(OB,BC,CO)
Collinear(AFOC)
Collinear(AED)
Collinear(BOD)
Collinear(BFE)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),18)
Parallelogram(ABCD)
PerpendicularBetweenLine(EO,DO)
Goal: Value(PerimeterOfTriangle(ABE))
Answer: 9
Theorem sequence:
line_addition(1,AE,ED)
quadrilateral_perimeter_formula(1,ABCD)
parallelogram_property_diagonal_bisection(1,BCDA,O)
adjacent_complementary_angle(1,BOE,EOD)
mirror_congruent_triangle_judgment_sas(1,OEB,ODE)
mirror_congruent_triangle_property_line_equal(1,OEB,ODE)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
triangle_perimeter_formula(1,ABE)
Problem: As shown in the diagram, the perimeter of quadrilateral ABCD is 20, quadrilateral ABCD is a ▱, EO is perpendicular to DO. Find the perimeter of △ABE.
问题：如图所示，ABCD的周长为20，四边形ABCD是平行四边形，EO⊥DO。求△ABE的周长。
Construction (CDL):
Shape(AB,BF,FA)
Shape(AF,FE,EA)
Shape(FB,BO,OF)
Shape(EF,FO,OE)
Shape(EO,OD,DE)
Shape(DO,OC,CD)
Shape(OB,BC,CO)
Collinear(AFOC)
Collinear(AED)
Collinear(BOD)
Collinear(BFE)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),20)
Parallelogram(ABCD)
PerpendicularBetweenLine(EO,DO)
Goal: Value(PerimeterOfTriangle(ABE))
Answer: 10
Theorem sequence:
line_addition(1,AE,ED)
quadrilateral_perimeter_formula(1,ABCD)
parallelogram_property_diagonal_bisection(1,BCDA,O)
adjacent_complementary_angle(1,BOE,EOD)
mirror_congruent_triangle_judgment_sas(1,OEB,ODE)
mirror_congruent_triangle_property_line_equal(1,OEB,ODE)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
triangle_perimeter_formula(1,ABE)
Problem: As shown in the diagram, BE=12, CE=5, BE bisects ∠ABC, CE is the angle bisector of ∠BCD, AD and BC are opposite sides of the parallelogram ABCD. Find the height of △EBC.
问题：如图所示，BE=12，CE=5，BE平分∠ABC，CE平分∠BCD，AD和BC是平行四边形ABCD的一组对边。求△EBC的高。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BC,CE)
Shape(EC,CD,DE)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(BE),12)
Equal(LengthOfLine(CE),5)
IsBisectorOfAngle(BE,ABC)
IsBisectorOfAngle(CE,BCD)
Parallelogram(ABCD)
Goal: Value(HeightOfTriangle(EBC))
Answer: 60/13
Theorem sequence:
angle_addition(1,ABE,EBC)
angle_addition(1,BCE,ECD)
parallel_property_ipsilateral_internal_angle(1,BA,CD)
triangle_property_angle_sum(1,EBC)
right_triangle_judgment_angle(1,CEB)
right_triangle_property_pythagorean(1,CEB)
triangle_area_formula_sine(1,EBC)
triangle_area_formula_common(1,EBC)
Problem: As shown in the diagram, AB=6, BC=10, BE=BC, CE=CD, AD and BC are opposite sides of the ▱ ABCD. Find the length of line DE.
问题：如图所示，AB=6，BC=10，BE=BC，CE=CD，BA和CD是▱ABCD的一组对边。求直线DE的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BC,CE)
Shape(EC,CD,DE)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(BC),10)
Equal(LengthOfLine(BE),LengthOfLine(BC))
Equal(LengthOfLine(CE),LengthOfLine(CD))
Parallelogram(ABCD)
Goal: Value(LengthOfLine(DE))
Answer: 18/5
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
parallel_property_collinear_extend(3,AD,BC,E)
parallel_property_alternate_interior_angle(1,ED,BC)
isosceles_triangle_judgment_line_equal(1,BCE)
isosceles_triangle_judgment_line_equal(1,CDE)
isosceles_triangle_property_angle_equal(1,BCE)
isosceles_triangle_property_angle_equal(1,CDE)
similar_triangle_judgment_aa(1,BCE,CDE)
similar_triangle_property_line_ratio(1,BCE,CDE)
similar_triangle_property_line_ratio(1,EBC,ECD)
Problem: As shown in the diagram, AC=4, AC=DC, BD=6, quadrilateral ABCD is a ▱. Find the perimeter of △ABO.
问题：如图所示，AC=4，AC=DC，BD=6，ABCD是▱。求三角形ABO的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(BC,CO,OB)
Shape(OC,CD,DO)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(AC),LengthOfLine(DC))
Equal(LengthOfLine(BD),6)
Parallelogram(ABCD)
Goal: Value(PerimeterOfTriangle(ABO))
Answer: 9
Theorem sequence:
line_addition(1,AO,OC)
line_addition(1,BO,OD)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
parallelogram_property_opposite_line_equal(1,ABCD)
triangle_perimeter_formula(1,ABO)
Problem: As shown in the diagram, AB=5, the perimeter of triangle OCD is 23, ABCD is a ▱. Find the sum of the length of line AC and the length of line BD.
问题：如图所示，AB=5，△OCD的周长为23，ABCD是平行四边形。求直线AC的长度与直线BD的长度之和。
Construction (CDL):
Shape(AB,BO,OA)
Shape(BC,CO,OB)
Shape(DO,OC,CD)
Shape(AO,OD,DA)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(PerimeterOfTriangle(OCD),23)
Parallelogram(ABCD)
Goal: Value(Add(LengthOfLine(AC),LengthOfLine(BD)))
Answer: 36
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
triangle_perimeter_formula(1,OCD)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
line_addition(1,AO,OC)
line_addition(1,BO,OD)
Problem: As shown in the diagram, OE=4, E is the midpoint of segment BC, quadrilateral ABCD is a parallelogram. Find the length of line AB.
问题：如图所示，OE=4，E是线段BC的中点，四边形ABCD是平行四边形。求直线AB的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(DO,OC,CD)
Shape(OB,BE,EO)
Shape(OE,EC,CO)
Collinear(AOC)
Collinear(BOD)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(OE),4)
IsMidpointOfLine(E,BC)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(AB))
Answer: 8
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ABCD,O)
line_addition(1,AO,OC)
line_addition(1,BE,EC)
similar_triangle_judgment_sas(1,CAB,COE)
similar_triangle_property_line_ratio(1,BCA,ECO)
similar_triangle_property_line_ratio(1,CAB,COE)
Problem: As shown in the diagram, AB=4, BC=6, OE=2, quadrilateral ABCD is a parallelogram. Find the perimeter of quadrilateral EFCD.
问题：如图所示，AB=4，BC=6，OE=2，四边形ABCD是平行四边形。求四边形EFCD的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BF,FO)
Shape(OF,FC,CO)
Shape(OC,CD,DO)
Shape(OD,DE,EO)
Shape(OE,EA,AO)
Collinear(AOC)
Collinear(BOD)
Collinear(EOF)
Collinear(AED)
Collinear(BFC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(OE),2)
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(EFCD))
Answer: 14
Theorem sequence:
parallel_property_alternate_interior_angle(1,AD,BC)
parallel_property_alternate_interior_angle(2,AD,BC)
vertical_angle(1,FOB,EOD)
parallelogram_property_diagonal_bisection(1,BCDA,O)
parallelogram_property_diagonal_bisection(1,ABCD,O)
vertical_angle(1,COF,AOE)
congruent_triangle_judgment_aas(2,FOB,EOD)
congruent_triangle_judgment_aas(2,FCO,EAO)
congruent_triangle_property_line_equal(1,COF,AOE)
congruent_triangle_property_line_equal(1,OBF,ODE)
line_addition(1,BF,FC)
line_addition(1,EO,OF)
parallelogram_property_opposite_line_equal(1,ABCD)
quadrilateral_perimeter_formula(1,EFCD)
Problem: As shown in the diagram, AC=12, AD=8, AD=BD, AD and BC are opposite sides of the ▱ ABCD. Find the perimeter of triangle AOD.
问题：如图所示，AC=12，AD=8，AD=BD，AD和BC是▱ABCD的一组对边。求△AOD的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BC,CO)
Shape(OC,CD,DO)
Shape(OD,DA,AO)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(AD),LengthOfLine(BD))
Parallelogram(ABCD)
Goal: Value(PerimeterOfTriangle(AOD))
Answer: 18
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
line_addition(1,AO,OC)
line_addition(1,BO,OD)
triangle_perimeter_formula(1,AOD)
Problem: As shown in the diagram, ∠BCD=110°, BE bisects ∠ABC, quadrilateral ABCD is a ▱. Find the measure of ∠BEA.
问题：如图所示，∠BCD=110°，BE是∠ABC的角平分线，四边形ABCD是▱。求∠BEA的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BC,CD,DE)
Collinear(AED)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),110)
IsBisectorOfAngle(BE,ABC)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(BEA))
Answer: 35
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,ABCD)
parallel_property_ipsilateral_internal_angle(1,AD,BC)
angle_addition(1,ABE,EBC)
triangle_property_angle_sum(1,ABE)
Problem: As shown in the diagram, ∠ABC=60°, ABCD is a parallelogram. Find the measure of ∠CDA.
问题：如图所示，∠ABC=60°，四边形ABCD是▱。求∠CDA的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),60)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(CDA))
Answer: 60
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BCDA)