SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=8, OC=3, the center of ⊙O is O, OC is perpendicular to AC. Find the diameter of ⊙O.
问题：如图所示，AB=8，OC=3，圆O的圆心为O，OC⊥AC。求⊙O的直径。
Construction (CDL):
Shape(OAB,BC,CO,OA)
Shape(OBA,AC,CB)
Shape(OC,CA,AO)
Collinear(ACB)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(OC),3)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(OC,AC)
Goal: Value(DiameterOfCircle(O))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,BCO,OCA)
circle_property_chord_perpendicular_bisect_chord(1,O,OC,BA)
line_addition(1,AC,CB)
right_triangle_judgment_angle(1,OCA)
right_triangle_property_pythagorean(1,OCA)
radius_of_circle_property_length_equal(1,OA,O)
circle_property_length_of_radius_and_diameter(1,O)
Problem: As shown in the diagram, CD=2, the radius of ⊙O is 5, O is the center of circle O, AD⊥OD. Find the length of line AB.
问题：如图所示，CD=2，圆O的半径为5，⊙O的圆心为O，AD垂直于OD。求直线AB的长度。
Construction (CDL):
Shape(OAC,CD,DA)
Shape(OCB,BD,DC)
Shape(OBA,AO,OD,DB)
Shape(OA,AD,DO)
Collinear(ADB)
Collinear(ODC)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(CD),2)
Equal(RadiusOfCircle(O),5)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AD,OD)
Goal: Value(LengthOfLine(AB))
Answer: 8
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
line_addition(1,OD,DC)
right_triangle_judgment_angle(1,ADO)
right_triangle_property_pythagorean(1,ADO)
circle_property_chord_perpendicular_bisect_chord(1,O,OD,AB)
line_addition(1,AD,DB)
Problem: As shown in the diagram, AB=4, ∠COP=60°, O is the center of ⊙O, D is the midpoint of ⌒OBC, PF⊥DF. Find the sum of the length of line CP and the length of line PD.
问题：如图所示，AB=4，∠COP=60°，O是圆O的圆心，D平分⌒OBC，PF垂直于DF。求直线CP的长度与直线PD的长度之和。
Construction (CDL):
Shape(OAE,EO,OA)
Shape(OE,EP,PO)
Shape(PE,EF,FP)
Shape(OEB,BF,FE)
Shape(OBD,DF,FB)
Shape(DP,PF,FD)
Shape(ODC,CP,PD)
Shape(CO,OP,PC)
Shape(OCA,AO,OC)
Collinear(AOPFB)
Collinear(DFE)
Collinear(CPE)
Cocircular(O,AEBDC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(MeasureOfAngle(COP),60)
IsCentreOfCircle(O,O)
IsMidpointOfArc(D,OBC)
PerpendicularBetweenLine(PF,DF)
Goal: Value(Add(LengthOfLine(CP),LengthOfLine(PD)))
Answer: 2*sqrt(2)
Theorem sequence:
adjacent_complementary_angle(1,EFP,PFD)
circle_property_chord_perpendicular_bisect_chord(1,O,OF,ED)
perpendicular_bisector_judgment_per_and_mid(1,PF,ED)
perpendicular_bisector_property_distance_equal(1,PF,ED)
arc_property_center_angle(1,OBC,O)
congruent_arc_judgment_length_equal(1,OBD,ODC)
congruent_arc_property_measure_equal(1,OBD,ODC)
arc_addition_measure(1,OBD,ODC)
circle_property_chord_perpendicular_bisect_arc(1,OED,OFB)
congruent_arc_judgment_length_equal(1,OEB,ODC)
congruent_arc_property_measure_equal(1,OEB,ODC)
arc_property_center_angle(1,OEB,O)
angle_addition(1,COP,POE)
diameter_of_circle_judgment_pass_centre(1,AOB,O)
diameter_of_circle_property_length_equal(1,AB,O)
circle_property_length_of_radius_and_diameter(1,O)
radius_of_circle_property_length_equal(1,OE,O)
radius_of_circle_property_length_equal(1,OC,O)
right_triangle_judgment_angle(1,COE)
right_triangle_property_pythagorean(1,COE)
line_addition(1,CP,PE)
Problem: As shown in the diagram, the radius of ⊙O is 10, O is the center of ⊙O, AD perpendicular bisects OC. Find the length of line AB.
问题：如图所示，⊙O的半径为10，O是圆O的圆心，AD垂直平分AB。求直线AB的长度。
Construction (CDL):
Shape(OAB,BD,DO,OA)
Shape(OBC,CD,DB)
Shape(OCA,AD,DC)
Shape(OD,DA,AO)
Collinear(ODC)
Collinear(ADB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(RadiusOfCircle(O),10)
IsCentreOfCircle(O,O)
IsPerpendicularBisectorOfLine(AD,OC)
Goal: Value(LengthOfLine(AB))
Answer: 10*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,BDO,ODA)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
line_addition(1,OD,DC)
right_triangle_judgment_angle(1,ODA)
right_triangle_property_pythagorean(1,ODA)
circle_property_chord_perpendicular_bisect_chord(1,O,OD,BA)
line_addition(1,AD,DB)
Problem: As shown in the diagram, CD=10, OB=6, the center of circle O is O, OP⊥DP. Find the length of line OP.
问题：如图所示，CD=10，OB=6，O是⊙O的圆心，OP⊥DP。求直线OP的长度。
Construction (CDL):
Shape(OAC,CO,OA)
Shape(OC,CP,PO)
Shape(OCB,BP,PC)
Shape(OBD,DP,PB)
Shape(ODA,AO,OP,PD)
Collinear(CPD)
Collinear(AOPB)
Cocircular(O,ACBD)
Conditions (CDL):
Equal(LengthOfLine(CD),10)
Equal(LengthOfLine(OB),6)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(OP,DP)
Goal: Value(LengthOfLine(OP))
Answer: sqrt(11)
Theorem sequence:
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OB,O)
adjacent_complementary_angle(1,CPO,OPD)
circle_property_chord_perpendicular_bisect_chord(1,O,OP,CD)
line_addition(1,CP,PD)
right_triangle_judgment_angle(1,CPO)
right_triangle_property_pythagorean(1,CPO)
Problem: As shown in the diagram, AB=8, the radius of circle O is 5, the center of ⊙O is O, AC⊥OC. Find the length of line OC.
问题：如图所示，AB=8，圆O的半径为5，⊙O的圆心为O，AC垂直于OC。求直线OC的长度。
Construction (CDL):
Shape(OAB,BC,CA)
Shape(OA,AC,CO)
Shape(OBA,AO,OC,CB)
Collinear(ACB)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(RadiusOfCircle(O),5)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AC,OC)
Goal: Value(LengthOfLine(OC))
Answer: 3
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OC,AB)
line_addition(1,AC,CB)
radius_of_circle_property_length_equal(1,OA,O)
right_triangle_judgment_angle(1,ACO)
right_triangle_property_pythagorean(1,ACO)
Problem: As shown in the diagram, AB=2*sqrt(3), the radius of circle O is 2, the center of ⊙O is O, AE is perpendicular to OE. Find the length of line OE.
问题：如图所示，AB=2*sqrt(3)，⊙O的半径为2，⊙O的圆心为O，AE垂直于OE。求直线OE的长度。
Construction (CDL):
Shape(OAB,BE,EA)
Shape(OA,AE,EO)
Shape(OBA,AO,OE,EB)
Collinear(AEB)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),2*sqrt(3))
Equal(RadiusOfCircle(O),2)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AE,OE)
Goal: Value(LengthOfLine(OE))
Answer: 1
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OE,AB)
line_addition(1,AE,EB)
radius_of_circle_property_length_equal(1,OA,O)
right_triangle_judgment_angle(1,AEO)
right_triangle_property_pythagorean(1,AEO)
Problem: As shown in the diagram, ∠HFA=76°, quadrilateral AMNF is mirror similar to quadrilateral AFDE. Find the measure of ∠NFH.
问题：如图所示，∠HFA=76°，AMNF与AFDE镜像相似。求∠NFH的大小。
Construction (CDL):
Shape(AB,BC,CH,HM,MA)
Shape(AM,MH,HF,FA)
Shape(AF,FD,DE,EA)
Shape(HN,NF,FH)
Collinear(CHFD)
Collinear(MHN)
Conditions (CDL):
Equal(MeasureOfAngle(HFA),76)
MirrorSimilarBetweenQuadrilateral(AMNF,AFDE)
Goal: Value(MeasureOfAngle(NFH))
Answer: 28
Theorem sequence:
adjacent_complementary_angle(1,HFA,AFD)
mirror_similar_quadrilateral_property_angle_equal(1,FAMN,FDEA)
angle_addition(1,NFH,HFA)
Problem: As shown in the diagram, AB=8, AD=5, BC=5. Find the length of line CD.
问题：如图所示，AB=8，AD=5，BC=5。求直线CD的长度。
Construction (CDL):
Collinear(ACDB)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(AD),5)
Equal(LengthOfLine(BC),5)
Goal: Value(LengthOfLine(CD))
Answer: 2
Theorem sequence:
line_addition(1,AD,DB)
line_addition(1,CD,DB)
Problem: As shown in the diagram, AB=13, CD=4, D bisects segment AC. Find the length of line BC.
问题：如图所示，AB=13，CD=4，D平分线段AC。求直线BC的长度。
Construction (CDL):
Collinear(ADCB)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(CD),4)
IsMidpointOfLine(D,AC)
Goal: Value(LengthOfLine(BC))
Answer: 5
Theorem sequence:
line_addition(1,AD,DB)
line_addition(1,DC,CB)
Problem: As shown in the diagram, AB=20, C bisects segment AB, D bisects segment BC. Find the length of line AD.
问题：如图所示，AB=20，C平分线段AB，D是线段BC的中点。求直线AD的长度。
Construction (CDL):
Collinear(ACDB)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
IsMidpointOfLine(C,AB)
IsMidpointOfLine(D,BC)
Goal: Value(LengthOfLine(AD))
Answer: 15
Theorem sequence:
line_addition(1,AC,CD)
line_addition(1,AC,CB)
line_addition(1,AD,DB)
Problem: As shown in the diagram, EC=CF, GE=GF, ∠ABC=70°, ∠EAB=60°. Find the measure of ∠ACG.
问题：如图所示，EC=CF，GE=GF，∠ABC=70°，∠EAB=60°。求∠ACG的大小。
Construction (CDL):
Shape(AB,BC,CE,EA)
Shape(EC,CG,GE)
Shape(GC,CF,FG)
Collinear(BCFD)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(EC),LengthOfLine(CF))
Equal(LengthOfLine(GE),LengthOfLine(GF))
Equal(MeasureOfAngle(ABC),70)
Equal(MeasureOfAngle(EAB),60)
Goal: Value(MeasureOfAngle(ACG))
Answer: 65
Theorem sequence:
triangle_property_angle_sum(1,ABC)
adjacent_complementary_angle(1,BCE,ECD)
mirror_congruent_triangle_judgment_sss(1,ECG,FGC)
mirror_congruent_triangle_property_angle_equal(1,CGE,CFG)
angle_addition(1,ECG,GCF)
Problem: As shown in the diagram, ∠AOC=42°. Find the measure of ∠COB.
问题：如图所示，∠AOC=42°。求∠COB的大小。
Construction (CDL):
Shape(AO,OC)
Shape(CO,OB)
Collinear(AOB)
Conditions (CDL):
Equal(MeasureOfAngle(AOC),42)
Goal: Value(MeasureOfAngle(COB))
Answer: 138
Theorem sequence:
adjacent_complementary_angle(1,AOC,COB)
Problem: As shown in the diagram, AB=5, AC=3, AD is the median of triangle ABC. Find the perimeter of triangle ABD minus the perimeter of △ADC.
问题：如图所示，AB=5，AC=3，AD是△ABC的中线。求三角形ABD的周长减去△ADC的周长。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),3)
IsMedianOfTriangle(AD,ABC)
Goal: Value(Sub(PerimeterOfTriangle(ABD),PerimeterOfTriangle(ADC)))
Answer: 2
Theorem sequence:
triangle_perimeter_formula(1,ABD)
triangle_perimeter_formula(1,ADC)
Problem: As shown in the diagram, AB=9, BC=5, D bisects segment AC. Find the length of line AD.
问题：如图所示，AB=9，BC=5，D是线段AC的中点。求直线AD的长度。
Construction (CDL):
Collinear(ADBC)
Conditions (CDL):
Equal(LengthOfLine(AB),9)
Equal(LengthOfLine(BC),5)
IsMidpointOfLine(D,AC)
Goal: Value(LengthOfLine(AD))
Answer: 7
Theorem sequence:
line_addition(1,AD,DC)
line_addition(1,AB,BC)
Problem: As shown in the diagram, CB=4, DB=7, D bisects segment AC. Find the length of line AB.
问题：如图所示，CB=4，DB=7，D平分线段AC。求直线AB的长度。
Construction (CDL):
Collinear(ADCB)
Conditions (CDL):
Equal(LengthOfLine(CB),4)
Equal(LengthOfLine(DB),7)
IsMidpointOfLine(D,AC)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
line_addition(1,AD,DB)
line_addition(1,DC,CB)
Problem: As shown in the diagram, ∠ABF=80°, ∠EAB=110°, quadrilateral EABF is mirror congruent to quadrilateral EFQP. Find the sum of the measure of ∠DEP and the measure of ∠QFC.
问题：如图所示，∠ABF=80°，∠EAB=110°，EABF镜像全等于EFQP。求∠DEP的大小与∠QFC的大小之和。
Construction (CDL):
Shape(AB,BF,FE,EA)
Shape(EF,FQ,QP,PE)
Shape(DE,EP,PQ,QF,FC,CD)
Collinear(AED)
Collinear(BFC)
Conditions (CDL):
Equal(MeasureOfAngle(ABF),80)
Equal(MeasureOfAngle(EAB),110)
MirrorCongruentBetweenQuadrilateral(EABF,EFQP)
Goal: Value(Add(MeasureOfAngle(DEP),MeasureOfAngle(QFC)))
Answer: 20
Theorem sequence:
quadrilateral_property_angle_sum(1,ABFE)
quadrilateral_property_angle_sum(1,ABCD)
mirror_congruent_quadrilateral_property_angle_equal(1,EABF,EFQP)
mirror_congruent_quadrilateral_property_angle_equal(1,FEAB,FQPE)
quadrilateral_property_angle_sum(1,DEFC)
angle_addition(1,DEP,PEF)
angle_addition(1,EFQ,QFC)
Problem: As shown in the diagram, ∠AOC=120°. Find the measure of ∠COB.
问题：如图所示，∠AOC=120°。求∠COB的大小。
Construction (CDL):
Shape(AO,OC)
Shape(CO,OB)
Collinear(AOB)
Conditions (CDL):
Equal(MeasureOfAngle(AOC),120)
Goal: Value(MeasureOfAngle(COB))
Answer: 60
Theorem sequence:
adjacent_complementary_angle(1,AOC,COB)
Problem: As shown in the diagram, Sub(LengthOfLine(MC)=LengthOfLine(NC)), M is the midpoint of segment AC, N is the midpoint of segment BC. Find the length of line AC minus the length of line BC.
问题：如图所示，Sub(LengthOfLine(MC)=LengthOfLine(NC))，M平分线段AC，N是线段BC的中点。求直线AC的长度减去直线BC的长度。
Construction (CDL):
Collinear(AMCNB)
Conditions (CDL):
Equal(Sub(LengthOfLine(MC),LengthOfLine(NC)),2)
IsMidpointOfLine(M,AC)
IsMidpointOfLine(N,BC)
Goal: Value(Sub(LengthOfLine(AC),LengthOfLine(BC)))
Answer: 4
Theorem sequence:
line_addition(1,AM,MC)
line_addition(1,CN,NB)
Problem: As shown in the diagram, . Find the sum of the measure of ∠JAB, the measure of ∠ABC, the measure of ∠BCI, the measure of ∠HDJ, the measure of ∠JEI and the measure of ∠IFH.
问题：如图所示，。求∠JAB的大小、∠ABC的大小、∠BCI的大小、∠HDJ的大小、∠JEI的大小与∠IFH的大小之和。
Construction (CDL):
Shape(DA,AJ,JD)
Shape(HD,DJ,JH)
Shape(FH,HI,IF)
Shape(HJ,JE,EI,IH)
Shape(CI,IE,EJ,JA,AB,BC)
Collinear(DHIC)
Collinear(DJE)
Collinear(FIE)
Collinear(FHJA)
Conditions (CDL):
Goal: Value(Add(MeasureOfAngle(JAB),MeasureOfAngle(ABC),MeasureOfAngle(BCI),MeasureOfAngle(HDJ),MeasureOfAngle(JEI),MeasureOfAngle(IFH)))
Answer: 360
Theorem sequence:
vertical_angle(1,AJD,FJE)
triangle_property_angle_sum(1,FJE)
triangle_property_angle_sum(1,DAJ)
quadrilateral_property_angle_sum(1,ABCD)
angle_addition(1,HDJ,JDA)
angle_addition(1,DAJ,JAB)
Problem: As shown in the diagram, CD=4, EF=8, E bisects segment AC, F is the midpoint of segment BD. Find the length of line AB.
问题：如图所示，CD=4，EF=8，E是线段AC的中点，F平分线段BD。求直线AB的长度。
Construction (CDL):
Collinear(AECDFB)
Conditions (CDL):
Equal(LengthOfLine(CD),4)
Equal(LengthOfLine(EF),8)
IsMidpointOfLine(E,AC)
IsMidpointOfLine(F,BD)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
line_addition(1,AE,EC)
line_addition(1,AE,EF)
line_addition(1,AC,CD)
line_addition(1,AD,DF)
line_addition(1,AF,FB)
Problem: As shown in the diagram, ∠GDH=30°, ∠HAP=45°, ∠MFN=90°, ∠NCG=90°. Find the sum of the measure of ∠GHP and the measure of ∠QMN.
问题：如图所示，∠GDH=30°，∠HAP=45°，∠MFN=90°，∠NCG=90°。求∠GHP的大小与∠QMN的大小之和。
Construction (CDL):
Shape(AP,PH,HA)
Shape(GH,HP,PQ,QM,MN,NG)
Shape(MQ,QB,BM)
Shape(CG,GN,NC)
Shape(GD,DH,HG)
Shape(FN,NM,MF)
Shape(PE,EQ,QP)
Shape(AH,HD)
Collinear(AHGC)
Collinear(APQB)
Collinear(BMNC)
Collinear(DGNF)
Collinear(DHPE)
Collinear(FMQE)
Conditions (CDL):
Equal(MeasureOfAngle(GDH),30)
Equal(MeasureOfAngle(HAP),45)
Equal(MeasureOfAngle(MFN),90)
Equal(MeasureOfAngle(NCG),90)
Goal: Value(Add(MeasureOfAngle(GHP),MeasureOfAngle(QMN)))
Answer: 210
Theorem sequence:
triangle_property_angle_sum(1,GDH)
adjacent_complementary_angle(1,DHG,GHP)
triangle_property_angle_sum(1,FNM)
adjacent_complementary_angle(1,QMN,NMF)
vertical_angle(1,HGD,CGN)
vertical_angle(1,GNC,FNM)
triangle_property_angle_sum(1,CGN)
Problem: As shown in the diagram, NP=PM, ∠DBM=20°, ∠NCD=90°, A is the center of circle A. Find the measure of ∠PDB.
问题：如图所示，NP=PM，∠DBM=20°，∠NCD=90°，A是圆A的圆心。求∠PDB的大小。
Construction (CDL):
Shape(ANE,EA,AN)
Shape(AEM,MA,AE)
Shape(ANE,NP,PE)
Shape(AEM,EP,PM)
Shape(NC,CD,DP,PN)
Shape(MP,PD,DB,BM)
Collinear(ANC)
Collinear(AEPD)
Collinear(AMB)
Collinear(CDB)
Cocircular(A,NEM)
Conditions (CDL):
Equal(LengthOfLine(NP),LengthOfLine(PM))
Equal(MeasureOfAngle(DBM),20)
Equal(MeasureOfAngle(NCD),90)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfAngle(PDB))
Answer: 125
Theorem sequence:
radius_of_circle_property_length_equal(1,AN,A)
radius_of_circle_property_length_equal(1,AM,A)
mirror_congruent_triangle_judgment_sss(1,ANP,APM)
mirror_congruent_triangle_property_angle_equal(1,ANP,APM)
angle_addition(1,MAE,EAN)
triangle_property_angle_sum(1,ADB)
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, ∠DBC=70°, ∠DCE=25°, ∠EDC=25°, CD bisects ∠BCE. Find the measure of ∠CDB.
问题：如图所示，∠DBC=70°，∠DCE=25°，∠EDC=25°，CD是∠BCE的角平分线。求∠CDB的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CD)
Shape(ED,DC,CE)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(MeasureOfAngle(DBC),70)
Equal(MeasureOfAngle(DCE),25)
Equal(MeasureOfAngle(EDC),25)
IsBisectorOfAngle(CD,BCE)
Goal: Value(MeasureOfAngle(CDB))
Answer: 85
Theorem sequence:
triangle_property_angle_sum(1,DBC)
Problem: As shown in the diagram, triangle AED is congruent to triangle ACB, AB=5, AC=6, BC=7. Find the length of line DE.
问题：如图所示，三角形AED与三角形ACB是全等三角形，AB=5，AC=6，BC=7。求直线DE的长度。
Construction (CDL):
Shape(AE,EF,FA)
Shape(AF,FD,DA)
Shape(FC,CD,DF)
Shape(AD,DB,BA)
Collinear(EFD)
Collinear(CDB)
Collinear(AFC)
Conditions (CDL):
CongruentBetweenTriangle(AED,ACB)
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(BC),7)
Goal: Value(LengthOfLine(DE))
Answer: 7
Theorem sequence:
congruent_triangle_property_line_equal(1,AED,ACB)
Problem: As shown in the diagram, ∠CAD=15°, ∠EBC=70°, BE∥DA. Find the measure of ∠BCA.
问题：如图所示，∠CAD=15°，∠EBC=70°，BE平行于DA。求∠BCA的大小。
Construction (CDL):
Shape(EB,BA,AE)
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),15)
Equal(MeasureOfAngle(EBC),70)
ParallelBetweenLine(BE,DA)
Goal: Value(MeasureOfAngle(BCA))
Answer: 95
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,BE,DA)
adjacent_complementary_angle(1,BDA,ADC)
triangle_property_angle_sum(1,ADC)
Problem: As shown in the diagram, AB=10, C is the midpoint of segment AO, D is the midpoint of segment OB. Find the length of line CD.
问题：如图所示，AB=10，C是线段AO的中点，D平分线段OB。求直线CD的长度。
Construction (CDL):
Collinear(ACODB)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
IsMidpointOfLine(C,AO)
IsMidpointOfLine(D,OB)
Goal: Value(LengthOfLine(CD))
Answer: 5
Theorem sequence:
line_addition(1,AC,CO)
line_addition(1,OD,DB)
line_addition(1,AO,OB)
line_addition(1,CO,OD)
Problem: As shown in the diagram, ∠ABC=112°, ∠MDN=32°, △NAM is mirror congruent to △NMD, MN is parallel to BC. Find the measure of ∠CND.
问题：如图所示，∠ABC=112°，∠MDN=32°，三角形NAM镜像全等于三角形NMD，MN∥BC。求∠CND的大小。
Construction (CDL):
Shape(AM,MN,NA)
Shape(ME,EN,NM)
Shape(MD,DE,EM)
Shape(NE,EB,BC,CN)
Collinear(AMEB)
Collinear(ANC)
Collinear(DEN)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),112)
Equal(MeasureOfAngle(MDN),32)
MirrorCongruentBetweenTriangle(NAM,NMD)
ParallelBetweenLine(MN,BC)
Goal: Value(MeasureOfAngle(CND))
Answer: 108
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,AMN,DNM)
parallel_property_corresponding_angle(1,MN,BC,A)
triangle_property_angle_sum(1,AMN)
mirror_congruent_triangle_property_angle_equal(1,NAM,NMD)
angle_addition(1,DNM,MNA)
adjacent_complementary_angle(1,CND,DNA)
Problem: As shown in the diagram, ∠ACE=50°. Find the sum of the measure of ∠ABD and the measure of ∠BDE.
问题：如图所示，∠ACE=50°。求∠ABD的大小与∠BDE的大小之和。
Construction (CDL):
Shape(AB,BD,DE,EF,FA)
Shape(BC,CD,DB)
Collinear(CBA)
Collinear(CDE)
Conditions (CDL):
Equal(MeasureOfAngle(ACE),50)
Goal: Value(Add(MeasureOfAngle(ABD),MeasureOfAngle(BDE)))
Answer: 230
Theorem sequence:
triangle_property_angle_sum(1,BCD)
adjacent_complementary_angle(1,ABD,DBC)
adjacent_complementary_angle(1,CDB,BDE)
Problem: As shown in the diagram, AB=10, NB=2, M bisects segment AB. Find the length of line MN.
问题：如图所示，AB=10，NB=2，M是线段AB的中点。求直线MN的长度。
Construction (CDL):
Collinear(AMNB)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(NB),2)
IsMidpointOfLine(M,AB)
Goal: Value(LengthOfLine(MN))
Answer: 3
Theorem sequence:
line_addition(1,AM,MB)
line_addition(1,MN,NB)
Problem: As shown in the diagram, ED=6, D bisects segment BC, E bisects segment AC. Find the length of line AB.
问题：如图所示，ED=6，D是线段BC的中点，E平分线段AC。求直线AB的长度。
Construction (CDL):
Collinear(AECDB)
Conditions (CDL):
Equal(LengthOfLine(ED),6)
IsMidpointOfLine(D,BC)
IsMidpointOfLine(E,AC)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
line_addition(1,AE,EC)
line_addition(1,CD,DB)
line_addition(1,EC,CD)
line_addition(1,AC,CB)
Problem: As shown in the diagram, AB=12, AM=Mul(LengthOfLine(BN),3), M is the midpoint of segment AB. Find the length of line MN.
问题：如图所示，AB=12，AM=Mul(LengthOfLine(BN),3)，M平分线段AB。求直线MN的长度。
Construction (CDL):
Collinear(AMNB)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AM),Mul(LengthOfLine(BN),3))
IsMidpointOfLine(M,AB)
Goal: Value(LengthOfLine(MN))
Answer: 4
Theorem sequence:
line_addition(1,AM,MB)
line_addition(1,MN,NB)
Problem: As shown in the diagram, ∠DPB=20°, ∠EAB=50°, BP bisects ∠ABC, DP is the angle bisector of ∠ADC. Find the measure of ∠BCD.
问题：如图所示，∠DPB=20°，∠EAB=50°，BP平分∠ABC，DP平分∠ADC。求∠BCD的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,EP,PA)
Shape(PE,ED,DP)
Shape(PD,DC,CP)
Shape(EB,BM,MD,DE)
Shape(DM,MC,CD)
Collinear(AED)
Collinear(BEP)
Collinear(PDM)
Collinear(BMC)
Conditions (CDL):
Equal(MeasureOfAngle(DPB),20)
Equal(MeasureOfAngle(EAB),50)
IsBisectorOfAngle(BP,ABC)
IsBisectorOfAngle(DP,ADC)
Goal: Value(MeasureOfAngle(BCD))
Answer: 10
Theorem sequence:
quadrilateral_property_angle_sum(1,ABCD)
round_angle(1,ADC,CDA)
angle_addition(1,ADP,PDC)
adjacent_complementary_angle(1,PDC,CDM)
triangle_property_angle_sum(1,DMC)
angle_addition(1,ABP,PBC)
adjacent_complementary_angle(1,BMP,PMC)
triangle_property_angle_sum(1,PBM)
Problem: As shown in the diagram, ∠ABC=Mul(3,MeasureOfAngle(DBC))°, ∠BCD=130°, ∠CDA=Mul(3,MeasureOfAngle(CDB))°. Find the measure of ∠DAB.
问题：如图所示，∠ABC=Mul(3,MeasureOfAngle(DBC))°，∠BCD=130°，∠CDA=Mul(3,MeasureOfAngle(CDB))°。求∠DAB的大小。
Construction (CDL):
Shape(AB,BD,DA)
Shape(CD,DB,BC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),Mul(3,MeasureOfAngle(DBC)))
Equal(MeasureOfAngle(BCD),130)
Equal(MeasureOfAngle(CDA),Mul(3,MeasureOfAngle(CDB)))
Goal: Value(MeasureOfAngle(DAB))
Answer: 80
Theorem sequence:
triangle_property_angle_sum(1,ABD)
triangle_property_angle_sum(1,CDB)
angle_addition(1,ABD,DBC)
angle_addition(1,CDB,BDA)
Problem: As shown in the diagram, AB=12, AD=Mul(1/3,LengthOfLine(AB)), C bisects segment AD. Find the length of line AC.
问题：如图所示，AB=12，AD=Mul(1/3,LengthOfLine(AB))，C是线段AD的中点。求直线AC的长度。
Construction (CDL):
Collinear(ACDB)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AD),Mul(1/3,LengthOfLine(AB)))
IsMidpointOfLine(C,AD)
Goal: Value(LengthOfLine(AC))
Answer: 2
Theorem sequence:
line_addition(1,AC,CD)
Problem: As shown in the diagram, ∠CAB=35°, ∠DBA=69°. Find the measure of ∠BCA.
问题：如图所示，∠CAB=35°，∠DBA=69°。求∠BCA的大小。
Construction (CDL):
Shape(DB,BA)
Shape(AB,BC,CA)
Collinear(DBC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),35)
Equal(MeasureOfAngle(DBA),69)
Goal: Value(MeasureOfAngle(BCA))
Answer: 34
Theorem sequence:
adjacent_complementary_angle(1,DBA,ABC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠ABE=45°, ∠ADC=75°, ∠BAD=110°. Find the measure of ∠DCB.
问题：如图所示，∠ABE=45°，∠ADC=75°，∠BAD=110°。求∠DCB的大小。
Construction (CDL):
Shape(AD,DC,CB,BA)
Shape(AB,BE)
Collinear(EBC)
Conditions (CDL):
Equal(MeasureOfAngle(ABE),45)
Equal(MeasureOfAngle(ADC),75)
Equal(MeasureOfAngle(BAD),110)
Goal: Value(MeasureOfAngle(DCB))
Answer: 40
Theorem sequence:
adjacent_complementary_angle(1,CBA,ABE)
quadrilateral_property_angle_sum(1,ADCB)
Problem: As shown in the diagram, Add(MeasureOfAngle(DAB)=MeasureOfAngle(ABC)), ∠DEC=40°. Find the sum of the measure of ∠EDA and the measure of ∠BCE.
问题：如图所示，Add(MeasureOfAngle(DAB)=MeasureOfAngle(ABC))，∠DEC=40°。求∠EDA的大小与∠BCE的大小之和。
Construction (CDL):
Shape(DE,EC,CD)
Shape(DA,AB,BC,CE,ED)
Conditions (CDL):
Equal(Add(MeasureOfAngle(DAB),MeasureOfAngle(ABC)),140)
Equal(MeasureOfAngle(DEC),40)
Goal: Value(Add(MeasureOfAngle(EDA),MeasureOfAngle(BCE)))
Answer: 80
Theorem sequence:
triangle_property_angle_sum(1,ECD)
quadrilateral_property_angle_sum(1,DABC)
angle_addition(1,CDE,EDA)
angle_addition(1,BCE,ECD)
Problem: As shown in the diagram, ∠BOA=20°, ∠COA=90°. Find the measure of ∠DOC.
问题：如图所示，∠BOA=20°，∠COA=90°。求∠DOC的大小。
Construction (CDL):
Shape(DO,OC)
Shape(CO,OB)
Shape(BO,OA)
Shape(AO,OD)
Collinear(DOB)
Conditions (CDL):
Equal(MeasureOfAngle(BOA),20)
Equal(MeasureOfAngle(COA),90)
Goal: Value(MeasureOfAngle(DOC))
Answer: 110
Theorem sequence:
angle_addition(1,COB,BOA)
adjacent_complementary_angle(1,DOC,COB)
Problem: As shown in the diagram, ∠ABC=40°, ∠BCA=60°. Find the sum of the measure of ∠AEF and the measure of ∠EFA.
问题：如图所示，∠ABC=40°，∠BCA=60°。求∠AEF的大小与∠EFA的大小之和。
Construction (CDL):
Shape(AE,EF,FA)
Shape(EB,BC,CF,FE)
Collinear(AEB)
Collinear(AFC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),40)
Equal(MeasureOfAngle(BCA),60)
Goal: Value(Add(MeasureOfAngle(AEF),MeasureOfAngle(EFA)))
Answer: 100
Theorem sequence:
triangle_property_angle_sum(1,AEF)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠EFB=122°, △EFB is mirror congruent to △EBA. Find the sum of the measure of ∠CAB and the measure of ∠EAD.
问题：如图所示，∠EFB=122°，△EFB与△EBA是镜像全等三角形。求∠CAB的大小与∠EAD的大小之和。
Construction (CDL):
Shape(FB,BE,EF)
Shape(BA,AE,EB)
Shape(BC,CA,AB)
Shape(EA,AD,DE)
Collinear(CAD)
Conditions (CDL):
Equal(MeasureOfAngle(EFB),122)
MirrorCongruentBetweenTriangle(EFB,EBA)
Goal: Value(Add(MeasureOfAngle(CAB),MeasureOfAngle(EAD)))
Answer: 58
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,FBE,AEB)
angle_addition(1,CAB,BAE)
adjacent_complementary_angle(1,CAE,EAD)