SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BO=6, CD=2, DO=3, triangle ABO is similar to triangle CDO.. Find the length of line AB.
问题：如图所示，BO=6，CD=2，DO=3，三角形ABO相似于三角形CDO。求直线AB的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OC,CD,DO)
Collinear(BOD)
Collinear(AOC)
Conditions (CDL):
Equal(LengthOfLine(BO),6)
Equal(LengthOfLine(CD),2)
Equal(LengthOfLine(DO),3)
SimilarBetweenTriangle(ABO,CDO)
Goal: Value(LengthOfLine(AB))
Answer: 4
Theorem sequence:
similar_triangle_property_line_ratio(1,ABO,CDO)
similar_triangle_property_line_ratio(1,OAB,OCD)
Problem: As shown in the diagram, AB=12, BD=9, CA=4, ∠CMD=120°, M bisects segment AB, CAMH is a kite, DB and DN are one pair of adjacent sides of the kite MBDN. Find the length of line CD.
问题：如图所示，AB=12，BD=9，CA=4，∠CMD=120°，M是线段AB的中点，CAMH是风筝形，MB和MN是风筝形MBDN的一组临边。求直线CD的长度。
Construction (CDL):
Shape(HC,CP,PH)
Shape(PC,CA,AP)
Shape(HP,PM,MH)
Shape(PA,AM,MP)
Shape(HM,MN,NH)
Shape(NM,MQ,QN)
Shape(QM,MB,BQ)
Shape(NQ,QD,DN)
Shape(QB,BD,DQ)
Collinear(CHND)
Collinear(CPM)
Collinear(HPA)
Collinear(NQB)
Collinear(DQM)
Collinear(AMB)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(BD),9)
Equal(LengthOfLine(CA),4)
Equal(MeasureOfAngle(CMD),120)
IsMidpointOfLine(M,AB)
Kite(CAMH)
Kite(MBDN)
Goal: Value(LengthOfLine(CD))
Answer: 19
Theorem sequence:
angle_addition(1,AMC,CMD)
adjacent_complementary_angle(1,AMD,DMB)
kite_property_diagonal_perpendicular_bisection(1,MHCA,P)
kite_property_diagonal_perpendicular_bisection(1,MBDN,Q)
perpendicular_bisector_property_bisector(1,MP,HA)
perpendicular_bisector_property_bisector(1,MQ,BN)
angle_addition(1,AMC,CMH)
angle_addition(1,NMD,DMB)
angle_addition(1,AMH,HMN)
adjacent_complementary_angle(1,AMN,NMB)
cosine_theorem(1,MNH)
line_addition(1,CH,HN)
line_addition(1,CN,ND)
line_addition(1,AM,MB)
Problem: As shown in the diagram, ∠BCA=35°, ∠CDA=80°, triangle BCA is mirror congruent to triangle DAC. Find the measure of ∠DAB.
问题：如图所示，∠BCA=35°，∠CDA=80°，三角形BCA镜像全等于三角形DAC。求∠DAB的大小。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),35)
Equal(MeasureOfAngle(CDA),80)
MirrorCongruentBetweenTriangle(BCA,DAC)
Goal: Value(MeasureOfAngle(DAB))
Answer: 130
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,BCA,DAC)
mirror_congruent_triangle_property_angle_equal(1,CAB,CDA)
triangle_property_angle_sum(1,ABC)
triangle_property_angle_sum(1,ACD)
angle_addition(1,DAC,CAB)
Problem: As shown in the diagram, AB=4, ME is perpendicular to CE, MF⊥DF, NG is perpendicular to CG, NH is perpendicular to DH, ABCD is a square. Find the area of triangle ACD.
问题：如图所示，AB=4，ME⊥CE，MF⊥DF，NG⊥CG，NH垂直于DH，四边形ABCD是正方形。求三角形ACD的面积。
Construction (CDL):
Shape(AB,BE,EM,MA)
Shape(ME,EG,GN,NM)
Shape(NG,GC,CN)
Shape(AM,MF,FD,DA)
Shape(MN,NH,HF,FM)
Shape(NC,CH,HN)
Collinear(AMNC)
Collinear(BEGC)
Collinear(DFHC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
PerpendicularBetweenLine(ME,CE)
PerpendicularBetweenLine(MF,DF)
PerpendicularBetweenLine(NG,CG)
PerpendicularBetweenLine(NH,DH)
Square(ABCD)
Goal: Value(AreaOfTriangle(ACD))
Answer: 8
Theorem sequence:
triangle_area_formula_sine(1,DAC)
Problem: As shown in the diagram, AC=3, BD=2, △ABD is mirror congruent to △ADC, AE⊥DE, DF⊥AF. Find the perimeter of triangle ABC.
问题：如图所示，AC=3，BD=2，三角形ABD镜像全等于三角形ADC，AE⊥DE，DF垂直于AF。求三角形ABC的周长。
Construction (CDL):
Shape(AE,ED,DA)
Shape(EB,BD,DE)
Shape(AD,DF,FA)
Shape(DC,CF,FD)
Collinear(AEB)
Collinear(AFC)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(BD),2)
MirrorCongruentBetweenTriangle(ABD,ADC)
PerpendicularBetweenLine(AE,DE)
PerpendicularBetweenLine(DF,AF)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 10
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,ABD,ADC)
mirror_congruent_triangle_property_line_equal(1,DAB,DCA)
line_addition(1,BD,DC)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, AN=9/2, PA=3, PN=4, AE is the perpendicular bisector of PQ, NF is the perpendicular bisector of PR. Find the length of line QR.
问题：如图所示，AN=9/2，PA=3，PN=4，AE垂直平分AB，NF垂直平分AB。求直线QR的长度。
Construction (CDL):
Shape(AP,PE,EA)
Shape(AE,EQ,QA)
Shape(EP,PM,ME)
Shape(EM,MN,NQ,QE)
Shape(PD,DM,MP)
Shape(MD,DF,FN,NM)
Shape(FR,RN,NF)
Shape(DO,OF,FD)
Collinear(PDFR)
Collinear(PMN)
Collinear(PEQ)
Collinear(AEMDO)
Collinear(AQNR)
Collinear(OFN)
Conditions (CDL):
Equal(LengthOfLine(AN),9/2)
Equal(LengthOfLine(PA),3)
Equal(LengthOfLine(PN),4)
IsPerpendicularBisectorOfLine(AE,PQ)
IsPerpendicularBisectorOfLine(NF,PR)
Goal: Value(LengthOfLine(QR))
Answer: 11/2
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,NF,PR)
perpendicular_bisector_property_distance_equal(1,AE,PQ)
line_addition(1,AQ,QN)
line_addition(1,QN,NR)
Problem: As shown in the diagram, ∠ABC=40°, ∠DAB=150°, △ABC is mirror congruent to △ACD. Find the measure of ∠BCD.
问题：如图所示，∠ABC=40°，∠DAB=150°，三角形ABC镜像全等于三角形ACD。求∠BCD的大小。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),40)
Equal(MeasureOfAngle(DAB),150)
MirrorCongruentBetweenTriangle(ABC,ACD)
Goal: Value(MeasureOfAngle(BCD))
Answer: 130
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,BCA,DAC)
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, PA=PB, ∠PAB=30°, △ABP is mirror congruent to △APC. Find the measure of ∠CPB.
问题：如图所示，PA=PB，∠PAB=30°，△ABP与△APC是镜像全等三角形。求∠CPB的大小。
Construction (CDL):
Shape(AB,BP,PA)
Shape(AP,PC,CA)
Shape(PB,BC,CP)
Conditions (CDL):
Equal(LengthOfLine(PA),LengthOfLine(PB))
Equal(MeasureOfAngle(PAB),30)
MirrorCongruentBetweenTriangle(ABP,APC)
Goal: Value(MeasureOfAngle(CPB))
Answer: 120
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,ABP,APC)
mirror_congruent_triangle_property_angle_equal(1,ABP,APC)
isosceles_triangle_judgment_line_equal(1,PAB)
isosceles_triangle_judgment_line_equal(1,PCA)
isosceles_triangle_property_angle_equal(1,PAB)
isosceles_triangle_property_angle_equal(1,PCA)
angle_addition(1,BPA,APC)
triangle_property_angle_sum(1,ABP)
triangle_property_angle_sum(1,APC)
round_angle(1,CPB,BPC)
Problem: As shown in the diagram, ∠ABE=86°, quadrilateral BCDA is mirror congruent to quadrilateral BEFC, AD and BC are opposite sides of the parallelogram ABCD, BEFC is a ▱. Find the measure of ∠BEF.
问题：如图所示，∠ABE=86°，BCDA镜像全等于BEFC，四边形ABCD是▱，BC和EF是平行四边形BEFC的一组对边。求∠BEF的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(BE,EF,FC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(ABE),86)
MirrorCongruentBetweenQuadrilateral(BCDA,BEFC)
Parallelogram(ABCD)
Parallelogram(BEFC)
Goal: Value(MeasureOfAngle(BEF))
Answer: 137
Theorem sequence:
mirror_congruent_quadrilateral_property_angle_equal(1,BCDA,BEFC)
angle_addition(1,ABC,CBE)
parallel_property_ipsilateral_internal_angle(1,BC,EF)
Problem: As shown in the diagram, ∠ABC=40°, ∠DAC=60°, △ABC is mirror congruent to △ACD. Find the measure of ∠BCD.
问题：如图所示，∠ABC=40°，∠DAC=60°，△ABC与△ACD是镜像全等三角形。求∠BCD的大小。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),40)
Equal(MeasureOfAngle(DAC),60)
MirrorCongruentBetweenTriangle(ABC,ACD)
Goal: Value(MeasureOfAngle(BCD))
Answer: 160
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,ABC,ACD)
mirror_congruent_triangle_property_angle_equal(1,BCA,DAC)
angle_addition(1,DAC,CAB)
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, ME=CE, ∠ADE=70°, triangle DME is mirror congruent to triangle DEA. Find the measure of ∠ABC.
问题：如图所示，ME=CE，∠ADE=70°，△DME镜像全等于△DEA。求∠ABC的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DM,ME,ED)
Shape(DB,BM,MD)
Shape(EM,MC,CE)
Collinear(ADB)
Collinear(BMC)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(ME),LengthOfLine(CE))
Equal(MeasureOfAngle(ADE),70)
MirrorCongruentBetweenTriangle(DME,DEA)
Goal: Value(MeasureOfAngle(ABC))
Answer: 70
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,DME,DEA)
mirror_congruent_triangle_property_angle_equal(1,MED,ADE)
mirror_congruent_triangle_property_angle_equal(1,EDM,EAD)
angle_addition(1,MED,DEA)
adjacent_complementary_angle(1,CEM,MEA)
isosceles_triangle_judgment_line_equal(1,EMC)
isosceles_triangle_property_angle_equal(1,EMC)
triangle_property_angle_sum(1,DME)
triangle_property_angle_sum(1,EMC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, AO=2, ∠AOG=15°, ∠GON=45°, OG perpendicular bisects BA, ON is the perpendicular bisector of CB. Find the length of line AC.
问题：如图所示，AO=2，∠AOG=15°，∠GON=45°，OG垂直平分AB，ON垂直平分AB。求直线AC的长度。
Construction (CDL):
Shape(AO,OH,HA)
Shape(AH,HG,GA)
Shape(HO,OP,PH)
Shape(HP,PB,BG,GH)
Shape(PO,OQ,QP)
Shape(BP,PQ,QN,NB)
Shape(QO,OC,CQ)
Shape(NQ,QC,CN)
Collinear(AHPQC)
Collinear(AGB)
Collinear(BPO)
Collinear(OHG)
Collinear(BNC)
Collinear(OQN)
Conditions (CDL):
Equal(LengthOfLine(AO),2)
Equal(MeasureOfAngle(AOG),15)
Equal(MeasureOfAngle(GON),45)
IsPerpendicularBisectorOfLine(OG,BA)
IsPerpendicularBisectorOfLine(ON,CB)
Goal: Value(LengthOfLine(AC))
Answer: 2*sqrt(2)
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,OG,BA)
perpendicular_bisector_property_distance_equal(1,ON,CB)
perpendicular_bisector_property_bisector(1,OG,BA)
perpendicular_bisector_property_bisector(1,ON,CB)
angle_addition(1,GOB,BON)
angle_addition(1,AOG,GON)
angle_addition(1,AON,NOC)
cosine_theorem(1,OCA)
Problem: As shown in the diagram, PM=PN, the perimeter of triangle BMN is 36, the perimeter of △PMN is 60, BA perpendicular bisects PM. Find the length of line AM.
问题：如图所示，PM=PN，△BMN的周长为36，△PMN的周长为60，BA垂直平分AB。求直线AM的长度。
Construction (CDL):
Shape(PA,AB,BP)
Shape(AM,MB,BA)
Shape(BM,MN,NB)
Collinear(PAM)
Collinear(PBN)
Conditions (CDL):
Equal(LengthOfLine(PM),LengthOfLine(PN))
Equal(PerimeterOfTriangle(BMN),36)
Equal(PerimeterOfTriangle(PMN),60)
IsPerpendicularBisectorOfLine(BA,PM)
Goal: Value(LengthOfLine(AM))
Answer: 12
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,BA,PM)
line_addition(1,PB,BN)
triangle_perimeter_formula(1,BMN)
triangle_perimeter_formula(1,PMN)
line_addition(1,PA,AM)
Problem: As shown in the diagram, AB=4, AC=5, DM perpendicular bisects BC. Find the perimeter of triangle ADB.
问题：如图所示，AB=4，AC=5，DM垂直平分AB。求三角形ADB的周长。
Construction (CDL):
Shape(AD,DB,BA)
Shape(BD,DM,MB)
Shape(DC,CM,MD)
Collinear(ADC)
Collinear(BMC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AC),5)
IsPerpendicularBisectorOfLine(DM,BC)
Goal: Value(PerimeterOfTriangle(ADB))
Answer: 9
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,DM,BC)
line_addition(1,AD,DC)
triangle_perimeter_formula(1,ADB)
Problem: As shown in the diagram, ∠ABE=110°, BCDA is mirror congruent to BEFC, ABCD is a parallelogram, BC and EF are opposite sides of the ▱ BEFC. Find the measure of ∠EFC.
问题：如图所示，∠ABE=110°，BCDA镜像全等于BEFC，ABCD是平行四边形，四边形BEFC是▱。求∠EFC的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(BE,EF,FC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(ABE),110)
MirrorCongruentBetweenQuadrilateral(BCDA,BEFC)
Parallelogram(ABCD)
Parallelogram(BEFC)
Goal: Value(MeasureOfAngle(EFC))
Answer: 55
Theorem sequence:
angle_addition(1,ABC,CBE)
mirror_congruent_quadrilateral_property_angle_equal(1,BCDA,BEFC)
parallelogram_property_opposite_angle_equal(1,BEFC)
Problem: As shown in the diagram, AD=4, BC=6, AD perpendicular bisects BC. Find the sum of the area of triangle ABE, the area of △FBD and the area of △CEF.
问题：如图所示，AD=4，BC=6，AD垂直平分AB。求△ABE的面积、三角形FBD的面积与△CEF的面积之和。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BF,FE)
Shape(FB,BD,DF)
Shape(AE,EC,CA)
Shape(EF,FC,CE)
Shape(FD,DC,CF)
Collinear(AEFD)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(BC),6)
IsPerpendicularBisectorOfLine(AD,BC)
Goal: Value(Add(AreaOfTriangle(ABE),AreaOfTriangle(FBD),AreaOfTriangle(CEF)))
Answer: 6
Theorem sequence:
line_addition(1,BD,DC)
altitude_of_triangle_judgment(2,BD,BEA)
altitude_of_triangle_judgment(3,CD,CEF)
line_addition(1,AE,EF)
line_addition(1,AF,FD)
triangle_area_formula_common(1,BEA)
triangle_area_formula_sine(1,DFB)
triangle_area_formula_common(1,CEF)
Problem: As shown in the diagram, ∠ADE=60°, ∠EAD=40°, quadrilateral ABCM is mirror congruent to quadrilateral AMDE. Find the measure of ∠ABC.
问题：如图所示，∠ADE=60°，∠EAD=40°，ABCM与AMDE镜像全等。求∠ABC的大小。
Construction (CDL):
Shape(AB,BC,CM,MA)
Shape(AM,MD,DA)
Shape(AD,DE,EA)
Collinear(CMD)
Conditions (CDL):
Equal(MeasureOfAngle(ADE),60)
Equal(MeasureOfAngle(EAD),40)
MirrorCongruentBetweenQuadrilateral(ABCM,AMDE)
Goal: Value(MeasureOfAngle(ABC))
Answer: 80
Theorem sequence:
triangle_property_angle_sum(1,ADE)
mirror_congruent_quadrilateral_property_angle_equal(1,BCMA,EAMD)
Problem: As shown in the diagram, AB=AD, ∠DAB=110°, AM is the perpendicular bisector of BN, CM perpendicular bisects NB, AE⊥DE. Find the measure of ∠BCA.
问题：如图所示，AB=AD，∠DAB=110°，AM垂直平分AB，CM垂直平分AB，AE⊥DE。求∠BCA的大小。
Construction (CDL):
Shape(DA,AE,ED)
Shape(EA,AN,NE)
Shape(NA,AM,MN)
Shape(MA,AB,BM)
Shape(NM,MC,CN)
Shape(MB,BC,CM)
Collinear(DENC)
Collinear(NMB)
Collinear(AMC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AD))
Equal(MeasureOfAngle(DAB),110)
IsPerpendicularBisectorOfLine(AM,BN)
IsPerpendicularBisectorOfLine(CM,NB)
PerpendicularBetweenLine(AE,DE)
Goal: Value(MeasureOfAngle(BCA))
Answer: 35
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,AM,BN)
perpendicular_bisector_property_bisector(1,AM,BN)
perpendicular_bisector_property_bisector(1,CM,NB)
isosceles_triangle_judgment_line_equal(1,AND)
adjacent_complementary_angle(1,CEA,AED)
altitude_of_triangle_judgment(1,AE,AND)
isosceles_triangle_property_line_coincidence(1,AND,E)
angle_addition(1,EAN,NAC)
angle_addition(1,DAE,EAC)
angle_addition(1,DAC,CAB)
triangle_property_angle_sum(1,ACE)
Problem: As shown in the diagram, AD=10, BE=6, △DFE is mirror congruent to △DEC, ABCD is a rectangle. Find the length of line CD.
问题：如图所示，AD=10，BE=6，三角形DFE与三角形DEC是镜像全等三角形，四边形ABCD是矩形。求直线CD的长度。
Construction (CDL):
Shape(AB,BE,EF,FA)
Shape(FE,ED,DF)
Shape(DE,EC,CD)
Collinear(AFD)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BE),6)
MirrorCongruentBetweenTriangle(DFE,DEC)
Rectangle(ABCD)
Goal: Value(LengthOfLine(CD))
Answer: 4
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,FED,CDE)
mirror_congruent_triangle_property_angle_equal(1,EDF,ECD)
angle_addition(1,CDE,EDF)
triangle_property_angle_sum(1,FED)
isosceles_triangle_judgment_angle_equal(1,CDE)
parallelogram_property_opposite_line_equal(1,BCDA)
line_addition(1,BE,EC)
Problem: As shown in the diagram, ∠ABC=60°, ∠BCA=50°, ∠CAB=70°, AM perpendicular bisects DP, AQ perpendicular bisects PF, BM perpendicular bisects PD, BN is the perpendicular bisector of EP, CN perpendicular bisects PE, CQ perpendicular bisects FP. Find the sum of the measure of ∠ADB, the measure of ∠BEC and the measure of ∠CFA.
问题：如图所示，∠ABC=60°，∠BCA=50°，∠CAB=70°，AM垂直平分AB，AQ垂直平分AB，BM是PD的垂直平分线，BN是EP的垂直平分线，CN是PE的垂直平分线，CQ垂直平分AB。求∠ADB的大小、∠BEC的大小与∠CFA的大小之和。
Construction (CDL):
Shape(AD,DM,MA)
Shape(DB,BM,MD)
Shape(AM,MP,PA)
Shape(MB,BP,PM)
Shape(AP,PQ,QA)
Shape(AQ,QF,FA)
Shape(CF,FQ,QC)
Shape(CQ,QP,PC)
Shape(CP,PN,NC)
Shape(CN,NE,EC)
Shape(PB,BN,NP)
Shape(NB,BE,EN)
Collinear(DMP)
Collinear(BMA)
Collinear(BNC)
Collinear(PNE)
Collinear(AQC)
Collinear(PQF)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(BCA),50)
Equal(MeasureOfAngle(CAB),70)
IsPerpendicularBisectorOfLine(AM,DP)
IsPerpendicularBisectorOfLine(AQ,PF)
IsPerpendicularBisectorOfLine(BM,PD)
IsPerpendicularBisectorOfLine(BN,EP)
IsPerpendicularBisectorOfLine(CN,PE)
IsPerpendicularBisectorOfLine(CQ,FP)
Goal: Value(Add(MeasureOfAngle(ADB),MeasureOfAngle(BEC),MeasureOfAngle(CFA)))
Answer: 360
Theorem sequence:
perpendicular_bisector_property_bisector(1,AM,DP)
perpendicular_bisector_property_bisector(1,BM,PD)
perpendicular_bisector_property_bisector(1,BN,EP)
perpendicular_bisector_property_bisector(1,CN,PE)
perpendicular_bisector_property_bisector(1,CQ,FP)
perpendicular_bisector_property_bisector(1,AQ,PF)
angle_addition(1,CAP,PAB)
angle_addition(1,ABP,PBC)
angle_addition(1,BCP,PCA)
triangle_property_angle_sum(1,ADB)
triangle_property_angle_sum(1,BEC)
triangle_property_angle_sum(1,CFA)
Problem: As shown in the diagram, AC=4, ∠BCA=30°, ∠CAB=30°, AN perpendicular bisects OF, AP is the perpendicular bisector of OE, AO is perpendicular to CO. Find the length of line EF.
问题：如图所示，AC=4，∠BCA=30°，∠CAB=30°，AN是OF的垂直平分线，AP垂直平分AB，AO⊥CO。求直线EF的长度。
Construction (CDL):
Shape(AO,OM,MA)
Shape(AM,MN,NA)
Shape(MO,OP,PM)
Shape(MP,PE,EN,NM)
Shape(AN,NF,FA)
Shape(FN,NE,EF)
Shape(PO,OB,BP)
Shape(EP,PB,BC,CE)
Collinear(ANEC)
Collinear(AMPB)
Collinear(OMNF)
Collinear(OPE)
Collinear(OBC)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(BCA),30)
Equal(MeasureOfAngle(CAB),30)
IsPerpendicularBisectorOfLine(AN,OF)
IsPerpendicularBisectorOfLine(AP,OE)
PerpendicularBetweenLine(AO,CO)
Goal: Value(LengthOfLine(EF))
Answer: 2
Theorem sequence:
sine_theorem(1,AOC)
perpendicular_bisector_property_distance_equal(1,AP,OE)
perpendicular_bisector_property_distance_equal(1,AN,OF)
triangle_property_angle_sum(1,AOC)
perpendicular_bisector_property_bisector(1,AN,OF)
cosine_theorem(1,AEF)
Problem: As shown in the diagram, the radius of ⊙O is 2, O is the center of circle O, the center of circle O is X, Y is the center of ⊙O, AO⊥NO, FO⊥GO, GO is perpendicular to AO, NO is perpendicular to FO. Find the sum of the area of the sector OAG and the area of the sector OFN.
问题：如图所示，⊙O的半径为2，O是⊙O的圆心，⊙O的圆心为X，Y是圆O的圆心，AO⊥NO，FO垂直于GO，GO⊥AO，NO⊥FO。求扇形OAG的面积与扇形OFN的面积之和。
Construction (CDL):
Shape(OAG,GH,XBH,BA)
Shape(XBH,HP,YCP,CB)
Shape(YCP,PO,OC)
Shape(ONA,AB,XMB,MN)
Shape(XMB,BC,YQC,QM)
Shape(YQC,CO,OQ)
Shape(OGF,FE,XHE,HG)
Shape(XHE,ED,YPD,PH)
Shape(YPD,DO,OP)
Shape(OFN,NM,XEM,EF)
Shape(XEM,MQ,YDQ,DE)
Shape(OD,YDQ,QO)
Shape(X)
Shape(Y)
Collinear(ABCODEF)
Collinear(GHPOQMN)
Cocircular(O,AGFN)
Cocircular(X,BHEM)
Cocircular(Y,CPDQ)
Conditions (CDL):
Equal(RadiusOfCircle(O),2)
IsCentreOfCircle(O,O)
IsCentreOfCircle(X,O)
IsCentreOfCircle(Y,O)
PerpendicularBetweenLine(AO,NO)
PerpendicularBetweenLine(FO,GO)
PerpendicularBetweenLine(GO,AO)
PerpendicularBetweenLine(NO,FO)
Goal: Value(Add(AreaOfSector(OAG),AreaOfSector(OFN)))
Answer: 2*pi
Theorem sequence:
arc_property_center_angle(1,OAG,O)
arc_property_center_angle(1,OFN,O)
sector_area_formula(1,OAG)
sector_area_formula(1,OFN)
Problem: As shown in the diagram, EF=6, MA perpendicular bisects PE, NB perpendicular bisects FP. Find the perimeter of triangle PMN.
问题：如图所示，EF=6，MA是PE的垂直平分线，NB垂直平分AB。求△PMN的周长。
Construction (CDL):
Shape(MO,ON,NM)
Shape(AE,EM,MA)
Shape(AM,MP,PA)
Shape(PM,MN,NP)
Shape(PN,NB,BP)
Shape(BN,NF,FB)
Collinear(OMA)
Collinear(ONB)
Collinear(EMNF)
Collinear(EAP)
Collinear(PBF)
Conditions (CDL):
Equal(LengthOfLine(EF),6)
IsPerpendicularBisectorOfLine(MA,PE)
IsPerpendicularBisectorOfLine(NB,FP)
Goal: Value(PerimeterOfTriangle(PMN))
Answer: 6
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,MA,PE)
perpendicular_bisector_property_distance_equal(1,NB,FP)
line_addition(1,EM,MN)
line_addition(1,EN,NF)
triangle_perimeter_formula(1,PMN)
Problem: As shown in the diagram, Add(LengthOfLine(AB)=LengthOfLine(BC)), DE perpendicular bisects CA. Find the perimeter of triangle DBC.
问题：如图所示，Add(LengthOfLine(AB)=LengthOfLine(BC))，DE垂直平分AB。求三角形DBC的周长。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CD)
Shape(ED,DC,CE)
Collinear(AEC)
Collinear(ADB)
Conditions (CDL):
Equal(Add(LengthOfLine(AB),LengthOfLine(BC)),10)
IsPerpendicularBisectorOfLine(DE,CA)
Goal: Value(PerimeterOfTriangle(DBC))
Answer: 10
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,DE,CA)
line_addition(1,AD,DB)
triangle_perimeter_formula(1,DBC)
Problem: As shown in the diagram, ∠CAB=110°, PM perpendicular bisects AB, QN is the perpendicular bisector of CA. Find the measure of ∠QAP.
问题：如图所示，∠CAB=110°，PM垂直平分AB，QN是CA的垂直平分线。求∠QAP的大小。
Construction (CDL):
Shape(AM,MP,PA)
Shape(MB,BP,PM)
Shape(AP,PQ,QA)
Shape(AQ,QN,NA)
Shape(NQ,QC,CN)
Collinear(AMB)
Collinear(ANC)
Collinear(BPQC)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),110)
IsPerpendicularBisectorOfLine(PM,AB)
IsPerpendicularBisectorOfLine(QN,CA)
Goal: Value(MeasureOfAngle(QAP))
Answer: 40
Theorem sequence:
triangle_property_angle_sum(1,ABC)
angle_addition(1,CAQ,QAP)
angle_addition(1,CAP,PAB)
perpendicular_bisector_property_distance_equal(1,PM,AB)
perpendicular_bisector_property_distance_equal(1,QN,CA)
isosceles_triangle_judgment_line_equal(1,PAB)
isosceles_triangle_judgment_line_equal(1,QCA)
isosceles_triangle_property_angle_equal(1,PAB)
isosceles_triangle_property_angle_equal(1,QCA)
Problem: As shown in the diagram, DC=BC, ∠DAB=70°, BE perpendicular bisects DA, DC is parallel to AB. Find the measure of ∠BCD.
问题：如图所示，DC=BC，∠DAB=70°，BE垂直平分AB，DC平行于AB。求∠BCD的大小。
Construction (CDL):
Shape(CD,DB,BC)
Shape(DE,EB,BD)
Shape(EA,AB,BE)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(DC),LengthOfLine(BC))
Equal(MeasureOfAngle(DAB),70)
IsPerpendicularBisectorOfLine(BE,DA)
ParallelBetweenLine(DC,AB)
Goal: Value(MeasureOfAngle(BCD))
Answer: 100
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,BE,DA)
isosceles_triangle_judgment_line_equal(1,CDB)
isosceles_triangle_judgment_line_equal(1,BDA)
isosceles_triangle_property_angle_equal(1,CDB)
isosceles_triangle_property_angle_equal(1,BDA)
triangle_property_angle_sum(1,CDB)
triangle_property_angle_sum(1,BDA)
parallel_property_alternate_interior_angle(1,DC,AB)
Problem: As shown in the diagram, OC=OE, ∠BAO=70°, AB is parallel to OD. Find the measure of ∠OCE.
问题：如图所示，OC=OE，∠BAO=70°，AB平行于OD。求∠OCE的大小。
Construction (CDL):
Shape(BA,AO)
Shape(AO,OD)
Shape(DO,OE)
Shape(EO,OC,CE)
Shape(CO,OA)
Collinear(AOE)
Collinear(DOC)
Conditions (CDL):
Equal(LengthOfLine(OC),LengthOfLine(OE))
Equal(MeasureOfAngle(BAO),70)
ParallelBetweenLine(AB,OD)
Goal: Value(MeasureOfAngle(OCE))
Answer: 35
Theorem sequence:
parallel_property_corresponding_angle(2,AB,OD,E)
adjacent_complementary_angle(1,DOE,EOC)
isosceles_triangle_judgment_line_equal(1,OCE)
isosceles_triangle_property_angle_equal(1,OCE)
triangle_property_angle_sum(1,EOC)
Problem: As shown in the diagram, AB=6, AB=AC, AE is the median of △ ABC, ED is the median of △ EAB. Find the length of line DE.
问题：如图所示，AB=6，AB=AC，AE是三角形ABC的中线，ED是△EAB的中线。求直线DE的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(AE,EC,CA)
Shape(DB,BE,ED)
Collinear(ADB)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AB),LengthOfLine(AC))
IsMedianOfTriangle(AE,ABC)
IsMedianOfTriangle(ED,EAB)
Goal: Value(LengthOfLine(DE))
Answer: 3
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_line_coincidence(2,ABC,E)
right_triangle_judgment_angle(1,BEA)
right_triangle_property_length_of_median(1,BEA,D)
Problem: As shown in the diagram, AC=BC, ∠EAC=118°, AE∥CD. Find the measure of ∠ABC.
问题：如图所示，AC=BC，∠EAC=118°，AE∥CD。求∠ABC的大小。
Construction (CDL):
Shape(AB,BC,CA)
Shape(EA,AC)
Shape(AC,CD)
Collinear(BCD)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(BC))
Equal(MeasureOfAngle(EAC),118)
ParallelBetweenLine(AE,CD)
Goal: Value(MeasureOfAngle(ABC))
Answer: 31
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AE,CD)
adjacent_complementary_angle(1,BCA,ACD)
isosceles_triangle_judgment_line_equal(1,CAB)
isosceles_triangle_property_angle_equal(1,CAB)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, AB=10, AC=18, AM=2*x, AM=AN, CN=3*x. Find the value of x.
问题：如图所示，AB=10，AC=18，AM=2*x，AM=AN，CN=3*x。求x的值。
Construction (CDL):
Shape(AM,MN,NA)
Shape(MB,BC,CN,NM)
Collinear(AMB)
Collinear(ANC)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),18)
Equal(LengthOfLine(AM),2*x)
Equal(LengthOfLine(AM),LengthOfLine(AN))
Equal(LengthOfLine(CN),3*x)
Goal: Value(x)
Answer: 18/5
Theorem sequence:
line_addition(1,AN,NC)