SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AC=10, E bisects segment AB, F bisects segment BC, G is the midpoint of segment DC, H bisects segment AD, quadrilateral ABCD is a isosceles trapezoid. Find the perimeter of quadrilateral EFGH.
问题：如图所示，AC=10，E平分线段AB，F是线段BC的中点，G是线段DC的中点，H是线段AD的中点，四边形ABCD是等腰梯形。求EFGH的周长。
Construction (CDL):
Shape(AE,EI,IA)
Shape(AI,IH,HA)
Shape(HI,IO,OJ,JH)
Shape(HJ,JD,DH)
Shape(IE,EK,KO,OI)
Shape(EB,BK,KE)
Shape(KB,BF,FK)
Shape(OK,KF,FL,LO)
Shape(DJ,JG,GD)
Shape(JO,OL,LG,GJ)
Shape(GL,LC,CG)
Shape(LF,FC,CL)
Collinear(AHD)
Collinear(AEB)
Collinear(BFC)
Collinear(DGC)
Collinear(HIE)
Collinear(DJOKB)
Collinear(GLF)
Collinear(HJG)
Collinear(AIOLC)
Collinear(EKF)
Conditions (CDL):
Equal(LengthOfLine(AC),10)
IsMidpointOfLine(E,AB)
IsMidpointOfLine(F,BC)
IsMidpointOfLine(G,DC)
IsMidpointOfLine(H,AD)
IsoscelesTrapezoid(ABCD)
Goal: Value(PerimeterOfQuadrilateral(EFGH))
Answer: 20
Theorem sequence:
quadrilateral_perimeter_formula(1,EFGH)
isosceles_trapezoid_property_diagonal_equal(1,ABCD)
midsegment_of_triangle_judgment_midpoint(1,EH,ABD)
midsegment_of_triangle_judgment_midpoint(1,HG,DAC)
midsegment_of_triangle_judgment_midpoint(1,FE,BCA)
midsegment_of_triangle_judgment_midpoint(1,GF,CDB)
midsegment_of_triangle_property_length(1,EH,ABD)
midsegment_of_triangle_property_length(1,HG,DAC)
midsegment_of_triangle_property_length(1,FE,BCA)
midsegment_of_triangle_property_length(1,GF,CDB)
Problem: As shown in the diagram, AB=3, AC=9, BE=2, AB⊥EB, BC⊥DC. Find the length of line DC.
问题：如图所示，AB=3，AC=9，BE=2，AB⊥EB，BC垂直于DC。求直线DC的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(BC,CD,DE,EB)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),9)
Equal(LengthOfLine(BE),2)
PerpendicularBetweenLine(AB,EB)
PerpendicularBetweenLine(BC,DC)
Goal: Value(LengthOfLine(DC))
Answer: 6
Theorem sequence:
similar_triangle_judgment_aa(1,EAB,DAC)
similar_triangle_property_line_ratio(1,EAB,DAC)
similar_triangle_property_line_ratio(1,ABE,ACD)
Problem: As shown in the diagram, AE=4, the perimeter of △ABC is 30, △DEA is mirror congruent to △DCE. Find the perimeter of △ABD.
问题：如图所示，AE=4，△ABC的周长为30，三角形DEA与三角形DCE是镜像全等三角形。求三角形ABD的周长。
Construction (CDL):
Shape(BD,DA,AB)
Shape(DE,EA,AD)
Shape(ED,DC,CE)
Collinear(BDC)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AE),4)
Equal(PerimeterOfTriangle(ABC),30)
MirrorCongruentBetweenTriangle(DEA,DCE)
Goal: Value(PerimeterOfTriangle(ABD))
Answer: 22
Theorem sequence:
line_addition(1,BD,DC)
line_addition(1,AE,EC)
triangle_perimeter_formula(1,ABD)
triangle_perimeter_formula(1,ABC)
mirror_congruent_triangle_property_line_equal(1,DEA,DCE)
mirror_congruent_triangle_property_line_equal(1,EAD,EDC)
Problem: As shown in the diagram, ∠AOC=36°, CO⊥DO. Find the measure of ∠DOB.
问题：如图所示，∠AOC=36°，CO⊥DO。求∠DOB的大小。
Construction (CDL):
Shape(AO,OC)
Shape(CO,OD)
Shape(DO,OB)
Collinear(AOB)
Conditions (CDL):
Equal(MeasureOfAngle(AOC),36)
PerpendicularBetweenLine(CO,DO)
Goal: Value(MeasureOfAngle(DOB))
Answer: 54
Theorem sequence:
angle_addition(1,AOC,COD)
adjacent_complementary_angle(1,AOD,DOB)
Problem: As shown in the diagram, the perimeter of ABCD is 20, AD and BC are opposite sides of the parallelogram ABCD, BO is perpendicular to EO. Find the perimeter of triangle ABE.
问题：如图所示，四边形ABCD的周长为20，四边形ABCD是平行四边形，BO⊥EO。求三角形ABE的周长。
Construction (CDL):
Shape(BF,FA,AB)
Shape(AF,FE,EA)
Shape(EF,FO,OE)
Shape(FB,BO,OF)
Shape(OB,BC,CO)
Shape(OC,CD,DO)
Shape(EO,OD,DE)
Collinear(AED)
Collinear(AFOC)
Collinear(BFE)
Collinear(BOD)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),20)
Parallelogram(ABCD)
PerpendicularBetweenLine(BO,EO)
Goal: Value(PerimeterOfTriangle(ABE))
Answer: 10
Theorem sequence:
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)
triangle_perimeter_formula(1,ABE)
quadrilateral_perimeter_formula(1,ABCD)
line_addition(1,AE,ED)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
Problem: As shown in the diagram, AB=AC, BC=5, the perimeter of triangle ABC is 21, ED perpendicular bisects AB, ED is perpendicular to BD. Find the perimeter of △BCE.
问题：如图所示，AB=AC，BC=5，△ABC的周长为21，ED垂直平分AB，ED⊥BD。求三角形BCE的周长。
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(BC),5)
Equal(PerimeterOfTriangle(ABC),21)
IsPerpendicularBisectorOfLine(ED,AB)
PerpendicularBetweenLine(ED,BD)
Goal: Value(PerimeterOfTriangle(BCE))
Answer: 13
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,ED,AB)
line_addition(1,AE,EC)
triangle_perimeter_formula(1,ABC)
triangle_perimeter_formula(1,BCE)
Problem: As shown in the diagram, AC=20, BD=20, E bisects segment AB, F is the midpoint of segment BC, G bisects segment DC, H bisects segment AD. Find the perimeter of EFGH.
问题：如图所示，AC=20，BD=20，E是线段AB的中点，F是线段BC的中点，G是线段DC的中点，H平分线段AD。求四边形EFGH的周长。
Construction (CDL):
Shape(AE,EI,IA)
Shape(AI,IH,HA)
Shape(HI,IO,OJ,JH)
Shape(HJ,JD,DH)
Shape(IE,EK,KO,OI)
Shape(EB,BK,KE)
Shape(KB,BF,FK)
Shape(OK,KF,FL,LO)
Shape(DJ,JG,GD)
Shape(JO,OL,LG,GJ)
Shape(GL,LC,CG)
Shape(LF,FC,CL)
Collinear(AHD)
Collinear(AEB)
Collinear(BFC)
Collinear(DGC)
Collinear(HIE)
Collinear(DJOKB)
Collinear(GLF)
Collinear(HJG)
Collinear(AIOLC)
Collinear(EKF)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BD),20)
IsMidpointOfLine(E,AB)
IsMidpointOfLine(F,BC)
IsMidpointOfLine(G,DC)
IsMidpointOfLine(H,AD)
Goal: Value(PerimeterOfQuadrilateral(EFGH))
Answer: 40
Theorem sequence:
quadrilateral_perimeter_formula(1,EFGH)
midsegment_of_triangle_judgment_midpoint(1,EH,ABD)
midsegment_of_triangle_judgment_midpoint(1,HG,DAC)
midsegment_of_triangle_judgment_midpoint(1,FE,BCA)
midsegment_of_triangle_judgment_midpoint(1,GF,CDB)
midsegment_of_triangle_property_length(1,EH,ABD)
midsegment_of_triangle_property_length(1,HG,DAC)
midsegment_of_triangle_property_length(1,FE,BCA)
midsegment_of_triangle_property_length(1,GF,CDB)
Problem: As shown in the diagram, ∠EFB=55°, ∠EFB=∠HFE, ∠JCD=∠HFE, AD∥EG. Find the measure of ∠IBC.
问题：如图所示，∠EFB=55°，∠EFB=∠HFE，∠JCD=∠HFE，AD∥EG。求∠IBC的大小。
Construction (CDL):
Shape(AB,BI)
Shape(IB,BC)
Shape(BC,CJ)
Shape(JC,CD)
Shape(FB,BA)
Shape(EF,FB)
Shape(BF,FC,CB)
Shape(DC,CF)
Shape(CF,FG)
Shape(HF,FE)
Shape(GF,FH)
Collinear(ABCD)
Collinear(EFG)
Collinear(IBF)
Collinear(JCF)
Conditions (CDL):
Equal(MeasureOfAngle(EFB),55)
Equal(MeasureOfAngle(EFB),MeasureOfAngle(HFE))
Equal(MeasureOfAngle(JCD),MeasureOfAngle(HFE))
ParallelBetweenLine(AD,EG)
Goal: Value(MeasureOfAngle(IBC))
Answer: 125
Theorem sequence:
parallel_property_collinear_extend(3,AD,EG,B)
parallel_property_collinear_extend(3,GE,DB,F)
parallel_property_alternate_interior_angle(1,BD,EF)
adjacent_complementary_angle(1,IBC,CBF)
Problem: As shown in the diagram, the length of ⌒OBC is equal to the length of ⌒ODB, ∠CAB=25°, O is the center of ⊙O. Find the measure of ∠BOD.
问题：如图所示，弧OBC的长度与弧ODB的长度相等，∠CAB=25°，圆O的圆心为O。求∠BOD的大小。
Construction (CDL):
Shape(AC,OCA)
Shape(OB,OBC,CA,AO)
Shape(BO,OD,ODB)
Shape(DO,OA,OAD)
Collinear(AOB)
Cocircular(O,DBCA)
Conditions (CDL):
Equal(LengthOfArc(OBC),LengthOfArc(ODB))
Equal(MeasureOfAngle(CAB),25)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BOD))
Answer: 50
Theorem sequence:
arc_property_circumference_angle_external(1,OBC,A)
congruent_arc_judgment_length_equal(1,OBC,ODB)
congruent_arc_property_measure_equal(1,OBC,ODB)
arc_property_center_angle(1,ODB,O)
Problem: As shown in the diagram, ∠DAB=50°, O is the center of ⊙O. Find the measure of ∠COB.
问题：如图所示，∠DAB=50°，O是圆O的圆心。求∠COB的大小。
Construction (CDL):
Shape(BA,OAB)
Shape(AB,BD,DA)
Shape(OD,DC,CO)
Shape(AD,DO,OC,OCA)
Shape(CD,DB,OBC)
Collinear(ADC)
Collinear(ODB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(DAB),50)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(COB))
Answer: 100
Theorem sequence:
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, ∠BOA=100°, O is the center of ⊙O. Find the sum of the measure of ∠CAO and the measure of ∠OBC.
问题：如图所示，∠BOA=100°，O是⊙O的圆心。求∠CAO的大小与∠OBC的大小之和。
Construction (CDL):
Shape(AO,OB,BC,CA)
Shape(AC,OCA)
Shape(CB,OBA)
Shape(BO,OA,OAB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BOA),100)
IsCentreOfCircle(O,O)
Goal: Value(Add(MeasureOfAngle(CAO),MeasureOfAngle(OBC)))
Answer: 50
Theorem sequence:
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
round_angle(1,BOA,AOB)
quadrilateral_property_angle_sum(1,AOBC)
Problem: As shown in the diagram, ∠CBA=25°, O is the center of ⊙O, DA is the diameter of circle O. Find the measure of ∠DAC.
问题：如图所示，∠CBA=25°，⊙O的圆心为O，圆O的直径为DA。求∠DAC的大小。
Construction (CDL):
Shape(DC,OCD)
Shape(DO,OE,EC,CD)
Shape(EA,AC,CE)
Shape(EB,BA,AE)
Shape(BE,EO,OD,ODB)
Shape(AB,OBA)
Shape(CA,OAC)
Collinear(DOEA)
Collinear(BEC)
Cocircular(O,DBAC)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),25)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(DA,O)
Goal: Value(MeasureOfAngle(DAC))
Answer: 65
Theorem sequence:
diameter_of_circle_property_right_angle(1,ACD,O)
arc_property_circumference_angle_external(1,OAC,D)
arc_property_circumference_angle_external(1,OAC,B)
triangle_property_angle_sum(1,DAC)
Problem: As shown in the diagram, ∠EAB=118°, ∠GAB=62°, AD∥BC, AE∥BF. Find the measure of ∠GBC.
问题：如图所示，∠EAB=118°，∠GAB=62°，AD平行于BC，AE∥BF。求∠GBC的大小。
Construction (CDL):
Shape(EA,AG,GF,FE)
Shape(GA,AB,BG)
Shape(GB,BC,CD,DG)
Collinear(AGD)
Collinear(BGF)
Conditions (CDL):
Equal(MeasureOfAngle(EAB),118)
Equal(MeasureOfAngle(GAB),62)
ParallelBetweenLine(AD,BC)
ParallelBetweenLine(AE,BF)
Goal: Value(MeasureOfAngle(GBC))
Answer: 56
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AE,BF)
parallel_property_ipsilateral_internal_angle(1,AD,BC)
angle_addition(1,ABG,GBC)
Problem: As shown in the diagram, ∠CFG=65°, DG is parallel to HI, AB⊥CB. Find the measure of ∠HBE.
问题：如图所示，∠CFG=65°，DG平行于HI，AB垂直于CB。求∠HBE的大小。
Construction (CDL):
Shape(DE,EA)
Shape(AE,EF)
Shape(EF,FC)
Shape(CF,FG)
Shape(GF,FB)
Shape(EB,BF,FE)
Shape(HB,BE)
Shape(BE,ED)
Shape(FB,BI)
Collinear(DEFG)
Collinear(HBI)
Collinear(AEB)
Collinear(BFC)
Conditions (CDL):
Equal(MeasureOfAngle(CFG),65)
ParallelBetweenLine(DG,HI)
PerpendicularBetweenLine(AB,CB)
Goal: Value(MeasureOfAngle(HBE))
Answer: 25
Theorem sequence:
parallel_property_collinear_extend(3,DG,HI,F)
parallel_property_collinear_extend(3,IH,GF,B)
parallel_property_corresponding_angle(1,FG,BI,C)
angle_addition(1,EBF,FBI)
adjacent_complementary_angle(1,HBE,EBI)
Problem: As shown in the diagram, ∠DBF=50°, D bisects segment AB, triangle ADE is mirror congruent to triangle FED. Find the measure of ∠EDF.
问题：如图所示，∠DBF=50°，D平分线段AB，△ADE与△FED是镜像全等三角形。求∠EDF的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BF,FD)
Shape(DF,FE,ED)
Shape(EF,FC,CE)
Collinear(BFC)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(MeasureOfAngle(DBF),50)
IsMidpointOfLine(D,AB)
MirrorCongruentBetweenTriangle(ADE,FED)
Goal: Value(MeasureOfAngle(EDF))
Answer: 50
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,EAD,EDF)
isosceles_triangle_judgment_line_equal(1,DBF)
isosceles_triangle_property_angle_equal(1,DBF)
triangle_property_angle_sum(1,DBF)
adjacent_complementary_angle(1,ADF,FDB)
angle_addition(1,ADE,EDF)
mirror_congruent_triangle_property_angle_equal(1,DEA,DFE)
Problem: As shown in the diagram, △ABC is congruent to △AFE, ∠CAB=70°, EC∥AB. Find the measure of ∠DAB.
问题：如图所示，三角形ABC全等于三角形AFE，∠CAB=70°，EC平行于AB。求∠DAB的大小。
Construction (CDL):
Shape(FE,EC,CD,DF)
Shape(CE,EA,AC)
Shape(DC,CA,AD)
Shape(DA,AB,BD)
Collinear(ADF)
Collinear(CDB)
Conditions (CDL):
CongruentBetweenTriangle(ABC,AFE)
Equal(MeasureOfAngle(CAB),70)
ParallelBetweenLine(EC,AB)
Goal: Value(MeasureOfAngle(DAB))
Answer: 40
Theorem sequence:
parallel_property_alternate_interior_angle(2,EC,AB)
congruent_triangle_property_line_equal(1,BCA,FEA)
isosceles_triangle_judgment_line_equal(1,ACE)
isosceles_triangle_property_angle_equal(1,ACE)
triangle_property_angle_sum(1,ACE)
congruent_triangle_property_angle_equal(1,ABC,AFE)
angle_addition(1,EAC,CAF)
angle_addition(1,CAF,DAB)
Problem: As shown in the diagram, AB=AC, ∠DAB=36°, BD is the altitude of △ BCA, BD is perpendicular to AD. Find the measure of ∠DBC.
问题：如图所示，AB=AC，∠DAB=36°，BD是△BCA的高，BD垂直于AD。求∠DBC的大小。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(DAB),36)
IsAltitudeOfTriangle(BD,BCA)
PerpendicularBetweenLine(BD,AD)
Goal: Value(MeasureOfAngle(DBC))
Answer: 18
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
triangle_property_angle_sum(1,DBC)
Problem: As shown in the diagram, CD=8, the radius of circle O is 5, the center of circle O is O, OD perpendicular bisects AB. Find the length of line AB.
问题：如图所示，CD=8，圆O的半径为5，圆O的圆心为O，OD垂直平分AB。求直线AB的长度。
Construction (CDL):
Shape(AD,DO,OA)
Shape(AO,OC,OCA)
Shape(OD,DB,BO)
Shape(CO,OB,OBC)
Collinear(ADB)
Collinear(COD)
Cocircular(O,BCA)
Conditions (CDL):
Equal(LengthOfLine(CD),8)
Equal(RadiusOfCircle(O),5)
IsCentreOfCircle(O,O)
IsPerpendicularBisectorOfLine(OD,AB)
Goal: Value(LengthOfLine(AB))
Answer: 8
Theorem sequence:
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OA,O)
line_addition(1,CO,OD)
right_triangle_judgment_angle(1,ADO)
right_triangle_property_pythagorean(1,ADO)
line_addition(1,AD,DB)
Problem: As shown in the diagram, ∠COB=78°, O is the center of circle O. Find the measure of ∠CAB.
问题：如图所示，∠COB=78°，O是圆O的圆心。求∠CAB的大小。
Construction (CDL):
Shape(BO,OD,DA,OAB)
Shape(BD,DO,OB)
Shape(AD,DC,CA)
Shape(CD,DB,OBC)
Shape(AC,OCA)
Collinear(BDA)
Collinear(ODC)
Cocircular(O,BCA)
Conditions (CDL):
Equal(MeasureOfAngle(COB),78)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(CAB))
Answer: 39
Theorem sequence:
arc_property_circumference_angle_external(1,OBC,A)
arc_property_center_angle(1,OBC,O)
Problem: As shown in the diagram, ∠ABD=58°, the center of circle O is O, BA is the diameter of circle O. Find the measure of ∠DCB.
问题：如图所示，∠ABD=58°，O是圆O的圆心，BA是⊙O的直径。求∠DCB的大小。
Construction (CDL):
Shape(AD,ODA)
Shape(AO,OE,ED,DA)
Shape(DE,EB,BD)
Shape(DB,OBD)
Shape(CE,EO,OA,OAC)
Shape(EC,CB,BE)
Shape(BC,OCB)
Collinear(AOEB)
Collinear(DEC)
Cocircular(O,CBDA)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),58)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(BA,O)
Goal: Value(MeasureOfAngle(DCB))
Answer: 32
Theorem sequence:
diameter_of_circle_property_right_angle(1,BDA,O)
triangle_property_angle_sum(1,ABD)
arc_property_circumference_angle_external(1,OBD,C)
arc_property_circumference_angle_external(1,OBD,A)
Problem: As shown in the diagram, ∠ABE=80°, ∠DAE=∠ABE, ABCD is a rhombus. Find the measure of ∠CDE.
问题：如图所示，∠ABE=80°，∠DAE=∠ABE，四边形ABCD是菱形。求∠CDE的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,ED,DA)
Shape(DE,EC,CD)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(ABE),80)
Equal(MeasureOfAngle(DAE),MeasureOfAngle(ABE))
Rhombus(ABCD)
Goal: Value(MeasureOfAngle(CDE))
Answer: 30
Theorem sequence:
parallel_property_collinear_extend(3,CB,DA,E)
parallel_property_alternate_interior_angle(1,AD,BE)
isosceles_triangle_judgment_angle_equal(1,ABE)
isosceles_triangle_judgment_line_equal(1,AED)
isosceles_triangle_property_angle_equal(1,AED)
parallelogram_property_opposite_angle_equal(1,DABC)
triangle_property_angle_sum(1,AED)
angle_addition(1,CDE,EDA)
Problem: As shown in the diagram, ∠BCE=60°, ∠FEC=35°, BD∥FE. Find the measure of ∠CBA.
问题：如图所示，∠BCE=60°，∠FEC=35°，BD平行于FE。求∠CBA的大小。
Construction (CDL):
Shape(CB,BA)
Shape(BC,CE,ED,DB)
Shape(FE,EC)
Collinear(ABD)
Conditions (CDL):
Equal(MeasureOfAngle(BCE),60)
Equal(MeasureOfAngle(FEC),35)
ParallelBetweenLine(BD,FE)
Goal: Value(MeasureOfAngle(CBA))
Answer: 25
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,EF,DB)
angle_addition(1,FEC,CED)
quadrilateral_property_angle_sum(1,BCED)
adjacent_complementary_angle(1,DBC,CBA)
Problem: As shown in the diagram, ∠OAB=30°, O is the center of ⊙O. Find the measure of ∠ACB.
问题：如图所示，∠OAB=30°，⊙O的圆心为O。求∠ACB的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(BO,OA,AB)
Shape(BA,AC,CB)
Shape(CA,OAC)
Shape(BC,OCB)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(OAB),30)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(ACB))
Answer: 120
Theorem sequence:
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)
round_angle(1,AOB,BOA)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,C)
Problem: As shown in the diagram, AC=8, ∠ABC=45°, AD is the altitude of △ ABC, BE is the altitude of △ BCA. Find the length of line BF.
问题：如图所示，AC=8，∠ABC=45°，AD是三角形ABC的高，BE是三角形BCA的高。求直线BF的长度。
Construction (CDL):
Shape(AB,BF,FA)
Shape(AF,FE,EA)
Shape(FB,BD,DF)
Shape(FD,DC,CE,EF)
Collinear(AFD)
Collinear(BFE)
Collinear(BDC)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(MeasureOfAngle(ABC),45)
IsAltitudeOfTriangle(AD,ABC)
IsAltitudeOfTriangle(BE,BCA)
Goal: Value(LengthOfLine(BF))
Answer: 8
Theorem sequence:
vertical_angle(1,AFE,DFB)
triangle_property_angle_sum(1,ABD)
triangle_property_angle_sum(1,FBD)
triangle_property_angle_sum(1,AFE)
isosceles_triangle_judgment_angle_equal(1,DAB)
congruent_triangle_judgment_aas(2,FBD,CAD)
congruent_triangle_property_line_equal(1,DFB,DCA)
Problem: As shown in the diagram, AB=AC, ∠CBA=70°. Find the measure of ∠BAC.
问题：如图所示，AB=AC，∠CBA=70°。求∠BAC的大小。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(CBA),70)
Goal: Value(MeasureOfAngle(BAC))
Answer: 40
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ACB)
isosceles_triangle_property_angle_equal(1,ACB)
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, ∠JEB=55°, AB is parallel to CD. Find the measure of ∠KFC.
问题：如图所示，∠JEB=55°，AB平行于CD。求∠KFC的大小。
Construction (CDL):
Shape(AE,EJ)
Shape(JE,EB)
Shape(FE,EA)
Shape(BE,EF)
Shape(CF,FE)
Shape(EF,FD)
Shape(KF,FC)
Shape(DF,FK)
Collinear(CFD)
Collinear(AEB)
Collinear(JEFK)
Conditions (CDL):
Equal(MeasureOfAngle(JEB),55)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(KFC))
Answer: 55
Theorem sequence:
parallel_property_collinear_extend(3,AB,CD,E)
parallel_property_collinear_extend(3,DC,EA,F)
parallel_property_corresponding_angle(1,FC,EA,K)
vertical_angle(1,JEB,FEA)
Problem: As shown in the diagram, AB=4, AB=AC, O is the center of ⊙O, O bisects segment BC, AD is the tangent to ⊙O, the tangent to ⊙O is AE, triangle ABC is an isosceles right triangle. Find the radius of ⊙O.
问题：如图所示，AB=4，AB=AC，O是圆O的圆心，O平分线段BC，⊙O的切线为AD，AE是圆O的切线，AB和AC是等腰直角△ABC的腰。求圆O的半径。
Construction (CDL):
Shape(DB,BM,ODM)
Shape(MO,OD,ODM)
Shape(DO,ON,ND)
Shape(DN,ONE,OED)
Shape(NC,CE,ONE)
Shape(EA,AD,OED)
Collinear(BMONC)
Collinear(ADB)
Collinear(AEC)
Cocircular(O,NEDM)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AB),LengthOfLine(AC))
IsCentreOfCircle(O,O)
IsMidpointOfLine(O,BC)
IsTangentOfCircle(AD,O)
IsTangentOfCircle(AE,O)
IsoscelesRightTriangle(ABC)
Goal: Value(RadiusOfCircle(O))
Answer: 2
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AD,O,O)
parallel_judgment_ipsilateral_internal_angle(1,AC,DO)
midsegment_of_triangle_judgment_parallel(1,OD,BCA)
midsegment_of_triangle_property_length(1,OD,BCA)
radius_of_circle_property_length_equal(1,OD,O)
Problem: As shown in the diagram, ∠FCD=20°, the center of ⊙O is O, G bisects segment EF. Find the measure of ∠DOE.
问题：如图所示，∠FCD=20°，O是圆O的圆心，G平分线段EF。求∠DOE的大小。
Construction (CDL):
Shape(EO,OC,OCE)
Shape(EG,GO,OE)
Shape(DG,GE,OED)
Shape(CO,OF,FC)
Shape(OG,GF,FO)
Shape(FG,GD,ODF)
Shape(CF,OFC)
Collinear(COGD)
Collinear(EGF)
Cocircular(O,EDFC)
Conditions (CDL):
Equal(MeasureOfAngle(FCD),20)
IsCentreOfCircle(O,O)
IsMidpointOfLine(G,EF)
Goal: Value(MeasureOfAngle(DOE))
Answer: 40
Theorem sequence:
radius_of_circle_property_length_equal(1,OE,O)
radius_of_circle_property_length_equal(1,OF,O)
mirror_congruent_triangle_judgment_sss(1,OEG,OGF)
mirror_congruent_triangle_property_angle_equal(1,OEG,OGF)
arc_property_circumference_angle_external(1,ODF,C)
arc_property_center_angle(1,ODF,O)
Problem: As shown in the diagram, ∠OBA=20°, AB is parallel to CD. Find the measure of ∠OCD.
问题：如图所示，∠OBA=20°，AB∥CD。求∠OCD的大小。
Construction (CDL):
Shape(AO,OB,BA)
Shape(OC,CD,DO)
Collinear(AOD)
Collinear(BOC)
Conditions (CDL):
Equal(MeasureOfAngle(OBA),20)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(OCD))
Answer: 20
Theorem sequence:
parallel_property_alternate_interior_angle(2,AB,CD)
Problem: As shown in the diagram, ∠BCA=70°, ∠DBA=100°. Find the measure of ∠CAB.
问题：如图所示，∠BCA=70°，∠DBA=100°。求∠CAB的大小。
Construction (CDL):
Shape(CA,AB,BC)
Shape(DB,BA)
Collinear(CBD)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),70)
Equal(MeasureOfAngle(DBA),100)
Goal: Value(MeasureOfAngle(CAB))
Answer: 30
Theorem sequence:
adjacent_complementary_angle(1,DBA,ABC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠ABC=50°, the center of circle O is O. Find the measure of ∠AOC.
问题：如图所示，∠ABC=50°，⊙O的圆心为O。求∠AOC的大小。
Construction (CDL):
Shape(BA,OAB)
Shape(BC,CO,OA,AB)
Shape(AO,OC,OCA)
Shape(CB,OBC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),50)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(AOC))
Answer: 100
Theorem sequence:
arc_property_center_angle(1,OCA,O)
arc_property_circumference_angle_external(1,OCA,B)
Problem: As shown in the diagram, ∠AEF=30°, ∠DCA=90°, ∠FBD=45°. Find the measure of ∠DFB.
问题：如图所示，∠AEF=30°，∠DCA=90°，∠FBD=45°。求∠DFB的大小。
Construction (CDL):
Shape(FB,BD,DF)
Shape(EF,FA,AE)
Shape(AF,FD,DC,CA)
Collinear(AFB)
Collinear(EAC)
Collinear(BDC)
Collinear(EFD)
Conditions (CDL):
Equal(MeasureOfAngle(AEF),30)
Equal(MeasureOfAngle(DCA),90)
Equal(MeasureOfAngle(FBD),45)
Goal: Value(MeasureOfAngle(DFB))
Answer: 15
Theorem sequence:
adjacent_complementary_angle(1,BDF,FDC)
triangle_property_angle_sum(1,FBD)
triangle_property_angle_sum(1,EDC)
Problem: As shown in the diagram, ∠BEC=100°, ED bisects ∠BEC, AB∥CD. Find the measure of ∠CDE.
问题：如图所示，∠BEC=100°，ED是∠BEC的角平分线，AB∥CD。求∠CDE的大小。
Construction (CDL):
Shape(CE,EA)
Shape(EC,CD,DE)
Shape(BE,ED)
Collinear(AEB)
Conditions (CDL):
Equal(MeasureOfAngle(BEC),100)
IsBisectorOfAngle(ED,BEC)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(CDE))
Answer: 50
Theorem sequence:
parallel_property_collinear_extend(3,AB,CD,E)
parallel_property_alternate_interior_angle(1,EB,CD)
angle_addition(1,BED,DEC)
Problem: As shown in the diagram, ∠CEA=35°, ∠DEC=90°, AB is parallel to CD. Find the measure of ∠CDE.
问题：如图所示，∠CEA=35°，∠DEC=90°，AB∥CD。求∠CDE的大小。
Construction (CDL):
Shape(CE,EA)
Shape(EC,CD,DE)
Shape(BE,ED)
Collinear(AEB)
Conditions (CDL):
Equal(MeasureOfAngle(CEA),35)
Equal(MeasureOfAngle(DEC),90)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(CDE))
Answer: 55
Theorem sequence:
parallel_property_collinear_extend(3,AB,CD,E)
parallel_property_ipsilateral_internal_angle(1,DC,EA)
angle_addition(1,DEC,CEA)
Problem: As shown in the diagram, ∠ACD=80°, AD is the angle bisector of ∠BAC, AB is parallel to CD. Find the measure of ∠CDA.
问题：如图所示，∠ACD=80°，AD平分∠BAC，AB∥CD。求∠CDA的大小。
Construction (CDL):
Shape(BA,AD)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),80)
IsBisectorOfAngle(AD,BAC)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(CDA))
Answer: 50
Theorem sequence:
triangle_property_angle_sum(1,ACD)
parallel_property_alternate_interior_angle(1,AB,CD)
Problem: As shown in the diagram, ∠BAD=70°, AD is the angle bisector of ∠BAC, AB is parallel to CD. Find the measure of ∠ACD.
问题：如图所示，∠BAD=70°，AD平分∠BAC，AB平行于CD。求∠ACD的大小。
Construction (CDL):
Shape(EA,AF)
Shape(FA,AB)
Shape(CA,AE)
Shape(AC,CD,DA)
Shape(BA,AD)
Shape(AD,DH)
Shape(GC,CA)
Shape(IC,CG)
Shape(DC,CI)
Shape(JD,DC)
Shape(HD,DJ)
Collinear(EAB)
Collinear(GCDH)
Collinear(FACI)
Collinear(ADJ)
Conditions (CDL):
Equal(MeasureOfAngle(BAD),70)
IsBisectorOfAngle(AD,BAC)
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(ACD))
Answer: 40
Theorem sequence:
triangle_property_angle_sum(1,ACD)
parallel_property_alternate_interior_angle(1,AB,CD)