SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠BAC=55°, the center of circle O is O, the tangent to circle O is CA. Find the measure of ∠BOA.
问题：如图所示，∠BAC=55°，⊙O的圆心为O，⊙O的切线为CA。求∠BOA的大小。
Construction (CDL):
Shape(OBA,AO,OB)
Shape(OA,AB,BO)
Shape(BA,AC)
Shape(OAB,BA)
Cocircular(O,AB)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),55)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(BOA))
Answer: 110
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CA,O,O)
angle_addition(1,OAB,BAC)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OAB)
triangle_property_angle_sum(1,OAB)
Problem: As shown in the diagram, OP=8, PA=6, O is the center of ⊙O, PA is the tangent to ⊙O. Find the radius of circle O.
问题：如图所示，OP=8，PA=6，O是⊙O的圆心，PA是圆O的切线。求圆O的半径。
Construction (CDL):
Shape(OA,OAB,BO)
Shape(AO,OB,OBA)
Shape(BP,PA,OBA)
Collinear(OBP)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(OP),8)
Equal(LengthOfLine(PA),6)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(RadiusOfCircle(O))
Answer: 2*sqrt(7)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
right_triangle_judgment_angle(1,PAO)
right_triangle_property_pythagorean(1,PAO)
radius_of_circle_property_length_equal(1,OA,O)
Problem: As shown in the diagram, the radius of circle M is 3, the radius of ⊙N is 5, O is the center of circle M, the center of ⊙N is O, AC is the tangent to circle M, BC is the tangent to ⊙M. Find the length of line AB.
问题：如图所示，⊙M的半径为3，圆N的半径为5，O是⊙M的圆心，圆N的圆心为O，AC是⊙O的切线，圆O的切线为BC。求直线AB的长度。
Construction (CDL):
Shape(NBA,AC,CB)
Shape(CA,AD,MCD)
Shape(OC,MCD,DO)
Shape(CO,OE,MEC)
Shape(EB,BC,MEC)
Shape(EO,OD,MDE)
Shape(DA,NAB,BE,MDE)
Collinear(ACB)
Collinear(ADO)
Collinear(OEB)
Cocircular(M,ECD)
Cocircular(N,AB)
Conditions (CDL):
Equal(RadiusOfCircle(M),3)
Equal(RadiusOfCircle(N),5)
IsCentreOfCircle(O,M)
IsCentreOfCircle(O,N)
IsTangentOfCircle(AC,M)
IsTangentOfCircle(BC,M)
Goal: Value(LengthOfLine(AB))
Answer: 8
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BC,M,O)
tangent_of_circle_property_perpendicular(1,AC,M,O)
radius_of_circle_property_length_equal(1,OC,M)
radius_of_circle_property_length_equal(1,OB,N)
radius_of_circle_property_length_equal(1,OA,N)
right_triangle_judgment_angle(1,BCO)
right_triangle_property_pythagorean(1,BCO)
right_triangle_judgment_angle(1,OCA)
right_triangle_property_pythagorean(1,OCA)
line_addition(1,AC,CB)
Problem: As shown in the diagram, ∠BDC=25°, O is the center of ⊙O, DC is the diameter of ⊙O, C bisects ⌒OAB. Find the measure of ∠COA.
问题：如图所示，∠BDC=25°，圆O的圆心为O，DC是⊙O的直径，C是⌒OAB的中点。求∠COA的大小。
Construction (CDL):
Shape(OD,ODA,AO)
Shape(OA,AC,CO)
Shape(OAC,CA)
Shape(BD,DO,OC,OCB)
Shape(OBD,DB)
Collinear(DOC)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(MeasureOfAngle(BDC),25)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(DC,O)
IsMidpointOfArc(C,OAB)
Goal: Value(MeasureOfAngle(COA))
Answer: 50
Theorem sequence:
arc_property_circumference_angle_external(1,OCB,D)
congruent_arc_judgment_length_equal(1,OAC,OCB)
congruent_arc_property_measure_equal(1,OAC,OCB)
arc_property_center_angle(1,OAC,O)
Problem: As shown in the diagram, ∠DBC=33°, BD is the diameter of circle O. Find the measure of ∠CAB.
问题：如图所示，∠DBC=33°，圆O的直径为BD。求∠CAB的大小。
Construction (CDL):
Shape(OAB,BA)
Shape(OBC,CB)
Shape(OCD,DC)
Shape(ODA,AE,ED)
Shape(AB,BE,EA)
Shape(EB,BC,CE)
Shape(DE,EC,CD)
Collinear(AEC)
Collinear(BED)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(DBC),33)
IsDiameterOfCircle(BD,O)
Goal: Value(MeasureOfAngle(CAB))
Answer: 57
Theorem sequence:
diameter_of_circle_property_right_angle(1,BCD,O)
triangle_property_angle_sum(1,DBC)
arc_property_circumference_angle_external(1,OBC,D)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, ∠OBC=50°, O is the center of ⊙O. Find the measure of ∠CAB.
问题：如图所示，∠OBC=50°，O是圆O的圆心。求∠CAB的大小。
Construction (CDL):
Shape(OAB,BA)
Shape(OBC,CB)
Shape(OCA,AC)
Shape(AB,BO,OC,CA)
Shape(OB,BC,CO)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(OBC),50)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(CAB))
Answer: 40
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OBC)
triangle_property_angle_sum(1,OBC)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, the perimeter of quadrilateral DABC is 28, the perimeter of △ABC is 22, quadrilateral DABC is a ▱. Find the length of line AC.
问题：如图所示，DABC的周长为28，三角形ABC的周长为22，DC和AB是平行四边形DABC的一组对边。求直线AC的长度。
Construction (CDL):
Shape(DA,AC,CD)
Shape(CA,AB,BC)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(DABC),28)
Equal(PerimeterOfTriangle(ABC),22)
Parallelogram(DABC)
Goal: Value(LengthOfLine(AC))
Answer: 8
Theorem sequence:
quadrilateral_perimeter_formula(1,DABC)
triangle_perimeter_formula(1,ABC)
parallelogram_property_opposite_line_equal(1,DABC)
parallelogram_property_opposite_line_equal(1,ABCD)
Problem: As shown in the diagram, ∠CAE=45°, AC and EF are opposite sides of the parallelogram AEFC. Find the measure of ∠EFC.
问题：如图所示，∠CAE=45°，四边形AEFC是▱。求∠EFC的大小。
Construction (CDL):
Shape(AE,EF,FC,CA)
Conditions (CDL):
Equal(MeasureOfAngle(CAE),45)
Parallelogram(AEFC)
Goal: Value(MeasureOfAngle(EFC))
Answer: 45
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,AEFC)
Problem: As shown in the diagram, DB=DC, ∠DAB=65°, DC and AB are opposite sides of the parallelogram DABC, DE⊥CE. Find the measure of ∠BCE.
问题：如图所示，DB=DC，∠DAB=65°，AD和BC是平行四边形DABC的一组对边，DE垂直于CE。求∠BCE的大小。
Construction (CDL):
Shape(DA,AB,BE,ED)
Shape(CD,DE,EC)
Shape(CE,EB,BC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(DB),LengthOfLine(DC))
Equal(MeasureOfAngle(DAB),65)
Parallelogram(DABC)
PerpendicularBetweenLine(DE,CE)
Goal: Value(MeasureOfAngle(BCE))
Answer: 25
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,ABCD)
isosceles_triangle_judgment_line_equal(1,DBC)
isosceles_triangle_property_angle_equal(1,DBC)
adjacent_complementary_angle(1,DEC,CEB)
triangle_property_angle_sum(1,CEB)
Problem: As shown in the diagram, AB=4, AD=10, AE bisects ∠DAB, AD and BC are opposite sides of the ▱ ABCD. Find the length of line EC.
问题：如图所示，AB=4，AD=10，AE平分∠DAB，四边形ABCD是平行四边形。求直线EC的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,EC,CD,DA)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AD),10)
IsBisectorOfAngle(AE,DAB)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(EC))
Answer: 6
Theorem sequence:
parallel_property_collinear_extend(3,CB,DA,E)
parallel_property_alternate_interior_angle(1,AD,BE)
isosceles_triangle_judgment_angle_equal(1,BEA)
parallelogram_property_opposite_line_equal(1,BCDA)
line_addition(1,BE,EC)
Problem: As shown in the diagram, ∠FDA=50°, AE bisects ∠DAB, AD and BC are opposite sides of the parallelogram DABC, DF is perpendicular to EF. Find the measure of ∠BCD.
问题：如图所示，∠FDA=50°，AE是∠DAB的角平分线，DABC是▱，DF垂直于EF。求∠BCD的大小。
Construction (CDL):
Shape(DA,AF,FD)
Shape(DF,FE,ED)
Shape(EF,FA,AB,BC,CE)
Collinear(AFE)
Collinear(DEC)
Conditions (CDL):
Equal(MeasureOfAngle(FDA),50)
IsBisectorOfAngle(AE,DAB)
Parallelogram(DABC)
PerpendicularBetweenLine(DF,EF)
Goal: Value(MeasureOfAngle(BCD))
Answer: 80
Theorem sequence:
adjacent_complementary_angle(1,AFD,DFE)
triangle_property_angle_sum(1,DAF)
angle_addition(1,DAE,EAB)
parallelogram_property_opposite_angle_equal(1,ABCD)
Problem: As shown in the diagram, ∠FCD=130°, ∠FCD=∠EAB, DC∥EF, EF is parallel to AB, DE is perpendicular to AE. Find the measure of ∠CDE.
问题：如图所示，∠FCD=130°，∠FCD=∠EAB，DC∥EF，EF∥AB，DE垂直于AE。求∠CDE的大小。
Construction (CDL):
Shape(DE,EF,FC,CD)
Shape(EA,AB,BF,FE)
Conditions (CDL):
Equal(MeasureOfAngle(FCD),130)
Equal(MeasureOfAngle(FCD),MeasureOfAngle(EAB))
ParallelBetweenLine(DC,EF)
ParallelBetweenLine(EF,AB)
PerpendicularBetweenLine(DE,AE)
Goal: Value(MeasureOfAngle(CDE))
Answer: 140
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,FE,CD)
parallel_property_ipsilateral_internal_angle(1,BA,FE)
quadrilateral_property_angle_sum(1,EABF)
angle_addition(1,DEF,FEA)
quadrilateral_property_angle_sum(1,DEFC)
Problem: As shown in the diagram, the perimeter of quadrilateral ABCD is 18, BA and CD are opposite sides of the ▱ ABCD, EO⊥DO. Find the perimeter of triangle ABE.
问题：如图所示，四边形ABCD的周长为18，ABCD是平行四边形，EO⊥DO。求三角形ABE的周长。
Construction (CDL):
Shape(AB,BF,FA)
Shape(AF,FE,EA)
Shape(EF,FO,OE)
Shape(FB,BO,OF)
Shape(OB,BC,CO)
Shape(EO,OD,DE)
Shape(DO,OC,CD)
Collinear(AFOC)
Collinear(BFE)
Collinear(AED)
Collinear(BOD)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),18)
Parallelogram(ABCD)
PerpendicularBetweenLine(EO,DO)
Goal: Value(PerimeterOfTriangle(ABE))
Answer: 9
Theorem sequence:
parallelogram_property_diagonal_bisection(1,BCDA,O)
adjacent_complementary_angle(1,BOE,EOD)
perpendicular_bisector_judgment_per_and_mid(1,EO,BD)
perpendicular_bisector_property_distance_equal(1,EO,BD)
quadrilateral_perimeter_formula(1,ABCD)
triangle_perimeter_formula(1,ABE)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
line_addition(1,AE,ED)
Problem: As shown in the diagram, the perimeter of triangle ABE is 12, ABCD is a ▱, EO is perpendicular to DO. Find the perimeter of ABCD.
问题：如图所示，△ABE的周长为12，AD和BC是▱ABCD的一组对边，EO⊥DO。求ABCD的周长。
Construction (CDL):
Shape(AB,BF,FA)
Shape(AF,FE,EA)
Shape(EF,FO,OE)
Shape(FB,BO,OF)
Shape(OB,BC,CO)
Shape(EO,OD,DE)
Shape(DO,OC,CD)
Collinear(AFOC)
Collinear(BFE)
Collinear(AED)
Collinear(BOD)
Conditions (CDL):
Equal(PerimeterOfTriangle(ABE),12)
Parallelogram(ABCD)
PerpendicularBetweenLine(EO,DO)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 24
Theorem sequence:
parallelogram_property_diagonal_bisection(1,BCDA,O)
adjacent_complementary_angle(1,BOE,EOD)
perpendicular_bisector_judgment_per_and_mid(1,EO,BD)
perpendicular_bisector_property_distance_equal(1,EO,BD)
quadrilateral_perimeter_formula(1,ABCD)
triangle_perimeter_formula(1,ABE)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
line_addition(1,AE,ED)
Problem: As shown in the diagram, AB=10, AD=6, CF=3, quadrilateral ABCD is a ▱, AE⊥CE, AF⊥CF. Find the length of line CE.
问题：如图所示，AB=10，AD=6，CF=3，四边形ABCD是▱，AE⊥CE，AF垂直于CF。求直线CE的长度。
Construction (CDL):
Shape(FB,BC,CF)
Shape(GF,FC,CG)
Shape(AE,EG,GA)
Shape(GC,CD,DA,AG)
Collinear(BFGA)
Collinear(EAD)
Collinear(EGC)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(CF),3)
Parallelogram(ABCD)
PerpendicularBetweenLine(AE,CE)
PerpendicularBetweenLine(AF,CF)
Goal: Value(LengthOfLine(CE))
Answer: 5
Theorem sequence:
altitude_of_quadrilateral_judgment_right_vertex(1,CF,DABC)
altitude_of_quadrilateral_judgment_left_vertex(5,CE,CDAB)
parallelogram_area_formula_common(1,CDAB)
parallelogram_area_formula_common(1,DABC)
Problem: As shown in the diagram, CF=CD, ∠ABC=72°, ABCD is a ▱. Find the measure of ∠CFA.
问题：如图所示，CF=CD，∠ABC=72°，ABCD是平行四边形。求∠CFA的大小。
Construction (CDL):
Shape(AB,BC,CF,FA)
Shape(FC,CD,DF)
Collinear(AFD)
Conditions (CDL):
Equal(LengthOfLine(CF),LengthOfLine(CD))
Equal(MeasureOfAngle(ABC),72)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(CFA))
Answer: 108
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BCDA)
isosceles_triangle_judgment_line_equal(1,CDF)
isosceles_triangle_property_angle_equal(1,CDF)
adjacent_complementary_angle(1,DFC,CFA)
Problem: As shown in the diagram, AB=5, AC=5, BC=6, △ABC is an isosceles △, EN is parallel to CA, ME is parallel to AB, ANEM is a ▱. Find the perimeter of quadrilateral ANEM.
问题：如图所示，AB=5，AC=5，BC=6，AB和AC是等腰△ABC的腰，EN∥CA，ME平行于AB，AM和NE是▱ANEM的一组对边。求四边形ANEM的周长。
Construction (CDL):
Shape(AN,NE,EM,MA)
Shape(NB,BE,EN)
Shape(ME,EC,CM)
Collinear(ANB)
Collinear(BEC)
Collinear(CMA)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),6)
IsoscelesTriangle(ABC)
ParallelBetweenLine(EN,CA)
ParallelBetweenLine(ME,AB)
Parallelogram(ANEM)
Goal: Value(PerimeterOfQuadrilateral(ANEM))
Answer: 10
Theorem sequence:
isosceles_triangle_property_angle_equal(1,ABC)
parallel_property_corresponding_angle(1,EN,CA,B)
isosceles_triangle_judgment_angle_equal(1,NBE)
quadrilateral_perimeter_formula(1,ANEM)
parallelogram_property_opposite_line_equal(1,ANEM)
parallelogram_property_opposite_line_equal(1,NEMA)
line_addition(1,AN,NB)
Problem: As shown in the diagram, Add(LengthOfLine(AC)=LengthOfLine(BD)), BC=4, AD and BC are opposite sides of the parallelogram ABCD. Find the perimeter of triangle OBC.
问题：如图所示，Add(LengthOfLine(AC)=LengthOfLine(BD))，BC=4，AD和BC是平行四边形ABCD的一组对边。求三角形OBC的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BC,CO)
Shape(AO,OD,DA)
Shape(DO,OC,CD)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(Add(LengthOfLine(AC),LengthOfLine(BD)),10)
Equal(LengthOfLine(BC),4)
Parallelogram(ABCD)
Goal: Value(PerimeterOfTriangle(OBC))
Answer: 9
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,OBC)
Problem: As shown in the diagram, AB=5, BC=13, ABCD is a parallelogram, CA is perpendicular to BA. Find the area of ABCD.
问题：如图所示，AB=5，BC=13，AD和BC是▱ABCD的一组对边，CA⊥BA。求ABCD的面积。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(BC),13)
Parallelogram(ABCD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(AreaOfQuadrilateral(ABCD))
Answer: 60
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
cosine_theorem(1,BCA)
parallelogram_area_formula_sine(1,ABCD)
Problem: As shown in the diagram, BN=6, ∠NDC=∠ADM, DABC is a parallelogram. Find the perimeter of DABC.
问题：如图所示，BN=6，∠NDC=∠ADM，四边形DABC是▱。求四边形DABC的周长。
Construction (CDL):
Shape(ND,DC,CN)
Shape(DM,MA,AD)
Shape(DA,AB,BC,CD)
Collinear(NDM)
Collinear(MAB)
Collinear(NCB)
Conditions (CDL):
Equal(LengthOfLine(BN),6)
Equal(MeasureOfAngle(NDC),MeasureOfAngle(ADM))
Parallelogram(DABC)
Goal: Value(PerimeterOfQuadrilateral(DABC))
Answer: 12
Theorem sequence:
parallel_property_collinear_extend(1,CB,DA,N)
parallel_property_collinear_extend(2,BA,CD,M)
parallel_property_corresponding_angle(1,DC,MB,N)
parallel_property_corresponding_angle(2,NB,DA,M)
isosceles_triangle_judgment_angle_equal(1,CND)
isosceles_triangle_judgment_angle_equal(1,ADM)
isosceles_triangle_judgment_angle_equal(1,BNM)
line_addition(1,MA,AB)
line_addition(1,NC,CB)
quadrilateral_perimeter_formula(1,DABC)
Problem: As shown in the diagram, ∠ABC=110°, DABC is a ▱. Find the sum of the measure of ∠FED and the measure of ∠DFE.
问题：如图所示，∠ABC=110°，四边形DABC是平行四边形。求∠FED的大小与∠DFE的大小之和。
Construction (CDL):
Shape(ED,DF,FE)
Shape(DA,AB,BC,CD)
Collinear(EDC)
Collinear(FDA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),110)
Parallelogram(DABC)
Goal: Value(Add(MeasureOfAngle(FED),MeasureOfAngle(DFE)))
Answer: 70
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,DABC)
vertical_angle(1,EDF,CDA)
triangle_property_angle_sum(1,EDF)
Problem: As shown in the diagram, AB=3, BC=5, EF=AB, BA and CD are opposite sides of the parallelogram ABCD. Find the perimeter of quadrilateral EFCD.
问题：如图所示，AB=3，BC=5，EF=AB，ABCD是▱。求四边形EFCD的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(EA,AO,OE)
Shape(OB,BF,FO)
Shape(EO,OD,DE)
Shape(OF,FC,CO)
Shape(DO,OC,CD)
Collinear(AED)
Collinear(BFC)
Collinear(AOC)
Collinear(BOD)
Collinear(EOF)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(BC),5)
Equal(LengthOfLine(EF),LengthOfLine(AB))
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(EFCD))
Answer: 11
Theorem sequence:
parallel_property_collinear_extend(3,AD,BC,E)
parallel_property_collinear_extend(3,CB,EA,F)
parallel_property_alternate_interior_angle(1,AE,FC)
parallel_property_alternate_interior_angle(2,AE,FC)
parallelogram_property_diagonal_bisection(1,ABCD,O)
congruent_triangle_judgment_aas(3,OFC,OEA)
congruent_triangle_property_line_equal(1,OFC,OEA)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
line_addition(1,AE,ED)
quadrilateral_perimeter_formula(1,EFCD)
Problem: As shown in the diagram, Add(LengthOfLine(AC)=LengthOfLine(BD)), Add(PerimeterOfTriangle(ABO)=PerimeterOfTriangle(AOD)), BA and CD are opposite sides of the parallelogram ABCD. Find the perimeter of quadrilateral ABCD.
问题：如图所示，Add(LengthOfLine(AC)=LengthOfLine(BD))，Add(PerimeterOfTriangle(ABO)=PerimeterOfTriangle(AOD))，BA和CD是平行四边形ABCD的一组对边。求四边形ABCD的周长。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BC,CO)
Shape(DO,OC,CD)
Shape(AO,OD,DA)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(Add(LengthOfLine(AC),LengthOfLine(BD)),11)
Equal(Add(PerimeterOfTriangle(ABO),PerimeterOfTriangle(AOD)),19)
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 16
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ABCD,O)
line_addition(1,AO,OC)
line_addition(1,BO,OD)
triangle_perimeter_formula(1,ABO)
triangle_perimeter_formula(1,AOD)
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 area of quadrilateral ABCD is 28, BC=4, DM=6, BM is the angle bisector of ∠ABC, quadrilateral DABC is a ▱. Find the area of △ABE.
问题：如图所示，ABCD的面积为28，BC=4，DM=6，BM平分∠ABC，DC和AB是▱DABC的一组对边。求三角形ABE的面积。
Construction (CDL):
Shape(DA,AE,EM,MD)
Shape(EA,AB,BE)
Shape(ME,EC,CM)
Shape(CE,EB,BC)
Collinear(AEC)
Collinear(MEB)
Collinear(DMC)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ABCD),28)
Equal(LengthOfLine(BC),4)
Equal(LengthOfLine(DM),6)
IsBisectorOfAngle(BM,ABC)
Parallelogram(DABC)
Goal: Value(AreaOfTriangle(ABE))
Answer: 10
Theorem sequence:
parallel_property_collinear_extend(3,DC,AB,M)
parallel_property_alternate_interior_angle(1,MC,AB)
parallel_property_alternate_interior_angle(2,MC,AB)
isosceles_triangle_judgment_angle_equal(1,CMB)
line_addition(1,DM,MC)
parallelogram_property_opposite_line_equal(1,DABC)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_area_formula_sine(1,DABC)
parallel_property_ipsilateral_internal_angle(1,AD,BC)
cosine_theorem(1,BCA)
similar_triangle_judgment_aa(1,EAB,ECM)
similar_triangle_property_line_ratio(1,EAB,ECM)
similar_triangle_property_line_ratio(1,BEA,MEC)
line_addition(1,AE,EC)
cosine_theorem(1,ABC)
triangle_area_formula_sine(1,ABE)
Problem: As shown in the diagram, the perimeter of quadrilateral ABCD is 18, FE perpendicular bisects AC, BA and CD are opposite sides of the parallelogram ABCD. Find the perimeter of △CDF.
问题：如图所示，ABCD的周长为18，FE垂直平分AB，ABCD是平行四边形。求三角形CDF的周长。
Construction (CDL):
Shape(AB,BC,CE,EA)
Shape(AE,EF,FA)
Shape(FE,EC,CF)
Shape(FC,CD,DF)
Collinear(AEC)
Collinear(AFD)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),18)
IsPerpendicularBisectorOfLine(FE,AC)
Parallelogram(ABCD)
Goal: Value(PerimeterOfTriangle(CDF))
Answer: 9
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,FE,AC)
line_addition(1,AF,FD)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
quadrilateral_perimeter_formula(1,ABCD)
triangle_perimeter_formula(1,CDF)
Problem: As shown in the diagram, AB=AC, AD=5, AD=DB, the perimeter of △EBC is 24, AD is perpendicular to ED. Find the length of line BC.
问题：如图所示，AB=AC，AD=5，AD=DB，三角形EBC的周长为24，AD垂直于ED。求直线BC的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BE,ED)
Shape(EB,BC,CE)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(AD),5)
Equal(LengthOfLine(AD),LengthOfLine(DB))
Equal(PerimeterOfTriangle(EBC),24)
PerpendicularBetweenLine(AD,ED)
Goal: Value(LengthOfLine(BC))
Answer: 14
Theorem sequence:
line_addition(1,AD,DB)
perpendicular_bisector_judgment_per_and_mid(1,ED,AB)
perpendicular_bisector_property_distance_equal(1,ED,AB)
line_addition(1,AE,EC)
triangle_perimeter_formula(1,EBC)
Problem: As shown in the diagram, AB=AC, ∠DAE=50°, DE perpendicular bisects AB. Find the measure of ∠DBC.
问题：如图所示，AB=AC，∠DAE=50°，DE垂直平分AB。求∠DBC的大小。
Construction (CDL):
Shape(AE,ED,DA)
Shape(EB,BD,DE)
Shape(DB,BC,CD)
Collinear(AEB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(DAE),50)
IsPerpendicularBisectorOfLine(DE,AB)
Goal: Value(MeasureOfAngle(DBC))
Answer: 15
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,DE,AB)
isosceles_triangle_judgment_line_equal(1,DAB)
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,DAB)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
angle_addition(1,EBD,DBC)
Problem: As shown in the diagram, AB=AC, ∠ABD=∠DBC, ∠CAB=36°. Find the measure of ∠CDB.
问题：如图所示，AB=AC，∠ABD=∠DBC，∠CAB=36°。求∠CDB的大小。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(ABD),MeasureOfAngle(DBC))
Equal(MeasureOfAngle(CAB),36)
Goal: Value(MeasureOfAngle(CDB))
Answer: 72
Theorem sequence:
angle_addition(1,ABD,DBC)
triangle_property_angle_sum(1,ABC)
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,DBC)
Problem: As shown in the diagram, AB=AC, BD=DC, ∠ABC=70°. Find the measure of ∠CAB.
问题：如图所示，AB=AC，BD=DC，∠ABC=70°。求∠CAB的大小。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(BD),LengthOfLine(DC))
Equal(MeasureOfAngle(ABC),70)
Goal: Value(MeasureOfAngle(CAB))
Answer: 40
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, the height of quadrilateral ABCD is 4, BC=6, 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4，BC=6，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),4)
Equal(LengthOfLine(BC),6)
Parallelogram(ABCD)
Goal: Value(AreaOfQuadrilateral(ABCD))
Answer: 24
Theorem sequence:
parallelogram_area_formula_common(1,ABCD)
Problem: As shown in the diagram, AB=AC, CD=CB, ∠CAD=40°. Find the measure of ∠DCA.
问题：如图所示，AB=AC，CD=CB，∠CAD=40°。求∠DCA的大小。
Construction (CDL):
Shape(AD,DC,CA)
Shape(DB,BC,CD)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(CD),LengthOfLine(CB))
Equal(MeasureOfAngle(CAD),40)
Goal: Value(MeasureOfAngle(DCA))
Answer: 30
Theorem sequence:
triangle_property_angle_sum(1,ABC)
isosceles_triangle_judgment_line_equal(1,CDB)
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,CDB)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,DBC)
angle_addition(1,BCD,DCA)
Problem: As shown in the diagram, Div(LengthOfLine(AE)=LengthOfLine(ED)), AE=2, BE is the angle bisector of ∠ABC, ABCD is a ▱. Find the perimeter of ABCD.
问题：如图所示，Div(LengthOfLine(AE)=LengthOfLine(ED))，AE=2，BE是∠ABC的角平分线，AD和BC是▱ABCD的一组对边。求四边形ABCD的周长。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BC,CD,DE)
Collinear(AED)
Conditions (CDL):
Equal(Div(LengthOfLine(AE),LengthOfLine(ED)),2)
Equal(LengthOfLine(AE),2)
IsBisectorOfAngle(BE,ABC)
Parallelogram(ABCD)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 10
Theorem sequence:
parallel_property_collinear_extend(3,AD,BC,E)
parallel_property_alternate_interior_angle(2,AE,BC)
isosceles_triangle_judgment_angle_equal(1,ABE)
line_addition(1,AE,ED)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
quadrilateral_perimeter_formula(1,ABCD)