SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠ADG=36°, ∠AGE=104°, ∠FEC=40°, GB⊥CB. Find the measure of ∠GAD.
问题：如图所示，∠ADG=36°，∠AGE=104°，∠FEC=40°，GB垂直于CB。求∠GAD的大小。
Construction (CDL):
Shape(AD,DG,GA)
Shape(GB,BE,FG)
Shape(FE,EC)
Shape(EB,BC,CE)
Shape(AG,GE)
Shape(GE,EF)
Shape(BG,GD)
Collinear(DGEC)
Collinear(AGB)
Collinear(BEF)
Conditions (CDL):
Equal(MeasureOfAngle(ADG),36)
Equal(MeasureOfAngle(AGE),104)
Equal(MeasureOfAngle(FEC),40)
PerpendicularBetweenLine(GB,CB)
Goal: Value(MeasureOfAngle(GAD))
Answer: 68
Theorem sequence:
adjacent_complementary_angle(1,DGA,AGE)
triangle_property_angle_sum(1,DGA)
Problem: As shown in the diagram, AB=20, AD=18, CE=16, AB and DC are opposite sides of the ▱ BADC, CE⊥DE. Find the perimeter of BADC.
问题：如图所示，AB=20，AD=18，CE=16，AB和DC是平行四边形BADC的一组对边，CE垂直于DE。求四边形BADC的周长。
Construction (CDL):
Shape(BA,AE,EC,CB)
Shape(CE,ED,DC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(AD),18)
Equal(LengthOfLine(CE),16)
Parallelogram(BADC)
PerpendicularBetweenLine(CE,DE)
Goal: Value(PerimeterOfQuadrilateral(BADC))
Answer: 76
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ADCB)
parallelogram_property_opposite_line_equal(1,BADC)
quadrilateral_perimeter_formula(1,BADC)
Problem: As shown in the diagram, BC=6, the center of ⊙A is A. Find the area of the ⊙A.
问题：如图所示，BC=6，⊙A的圆心为A。求⊙A的面积。
Construction (CDL):
Shape(BA,AC,ACB)
Shape(CA,AB,ABC)
Collinear(BAC)
Cocircular(A,CB)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
IsCentreOfCircle(A,A)
Goal: Value(AreaOfCircle(A))
Answer: 9*pi
Theorem sequence:
line_addition(1,BA,AC)
radius_of_circle_property_length_equal(1,AB,A)
radius_of_circle_property_length_equal(1,AC,A)
circle_area_formula(1,A)
Problem: As shown in the diagram, AC=20, BA=x, CB=y, ∠ACB=60°, CB⊥AB. Find the value of x.
问题：如图所示，AC=20，BA=x，CB=y，∠ACB=60°，CB垂直于AB。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BA),x)
Equal(LengthOfLine(CB),y)
Equal(MeasureOfAngle(ACB),60)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 10*sqrt(3)
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, JM=6, KL=6, ∠JML=80°, JK is parallel to ML. Find the measure of ∠LKJ.
问题：如图所示，JM=6，KL=6，∠JML=80°，JK∥ML。求∠LKJ的大小。
Construction (CDL):
Shape(JM,ML,LK,KJ)
Conditions (CDL):
Equal(LengthOfLine(JM),6)
Equal(LengthOfLine(KL),6)
Equal(MeasureOfAngle(JML),80)
ParallelBetweenLine(JK,ML)
Goal: Value(MeasureOfAngle(LKJ))
Answer: 100
Theorem sequence:
trapezoid_judgment_parallel(1,JMLK)
isosceles_trapezoid_judgment_line_equal(1,JMLK)
parallel_property_ipsilateral_internal_angle(1,JK,ML)
isosceles_trapezoid_property_angle_equal(1,LKJM)
Problem: As shown in the diagram, AB=AC, BC=7/2, ∠BCA=45°. Find the perimeter of △ABC.
问题：如图所示，AB=AC，BC=7/2，∠BCA=45°。求三角形ABC的周长。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(BC),7/2)
Equal(MeasureOfAngle(BCA),45)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 7/2+7*sqrt(2)/2
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
cosine_theorem(1,BCA)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, ∠CAD=150°, ∠DAB=85°, ∠EAC=x°, the center of ⊙A is A, BA⊥EA. Find the value of x.
问题：如图所示，∠CAD=150°，∠DAB=85°，∠EAC=x°，A是圆A的圆心，BA垂直于EA。求x的值。
Construction (CDL):
Shape(CA,AD,ADC)
Shape(EA,AC,ACE)
Shape(BA,AE,AEB)
Shape(DA,AB,ABD)
Cocircular(A,CEBD)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),150)
Equal(MeasureOfAngle(DAB),85)
Equal(MeasureOfAngle(EAC),x)
IsCentreOfCircle(A,A)
PerpendicularBetweenLine(BA,EA)
Goal: Value(x)
Answer: 35
Theorem sequence:
angle_addition(1,CAD,DAB)
angle_addition(1,CAB,BAE)
round_angle(1,CAE,EAC)
Problem: As shown in the diagram, JP=18, MQ=9, Q is the incenter of triangle JNL, LK is perpendicular to QK, QM⊥LM, QP is perpendicular to NP. Find the length of line JQ.
问题：如图所示，JP=18，MQ=9，三角形JNL的内切圆圆心为Q，LK⊥QK，QM垂直于LM，QP⊥NP。求直线JQ的长度。
Construction (CDL):
Shape(LK,KQ,QL)
Shape(KJ,JQ,QK)
Shape(QJ,JP,PQ)
Shape(QP,PN,NQ)
Shape(MQ,QN,NM)
Shape(QM,ML,LQ)
Collinear(LKJ)
Collinear(JPN)
Collinear(NML)
Conditions (CDL):
Equal(LengthOfLine(JP),18)
Equal(LengthOfLine(MQ),9)
IsIncenterOfTriangle(Q,JNL)
PerpendicularBetweenLine(LK,QK)
PerpendicularBetweenLine(QM,LM)
PerpendicularBetweenLine(QP,NP)
Goal: Value(LengthOfLine(JQ))
Answer: 9*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,LKQ,QKJ)
adjacent_complementary_angle(1,JPQ,QPN)
mirror_congruent_triangle_judgment_aas(1,QLK,QML)
mirror_congruent_triangle_judgment_aas(1,QJP,QKJ)
mirror_congruent_triangle_property_line_equal(1,LKQ,LQM)
mirror_congruent_triangle_property_line_equal(1,QJP,QKJ)
right_triangle_judgment_angle(1,QKJ)
right_triangle_property_pythagorean(1,QKJ)
Problem: As shown in the diagram, AD=10, BA=x-2, BC=5, ED=4, EB is parallel to DC. Find the value of x.
问题：如图所示，AD=10，BA=x-2，BC=5，ED=4，EB∥DC。求x的值。
Construction (CDL):
Shape(AE,EB,BA)
Shape(ED,DC,CB,BE)
Collinear(AED)
Collinear(ABC)
Conditions (CDL):
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BA),x-2)
Equal(LengthOfLine(BC),5)
Equal(LengthOfLine(ED),4)
ParallelBetweenLine(EB,DC)
Goal: Value(x)
Answer: 19/2
Theorem sequence:
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
line_addition(1,AE,ED)
line_addition(1,AB,BC)
similar_triangle_property_line_ratio(1,BAE,CAD)
similar_triangle_property_line_ratio(1,EBA,DCA)
Problem: As shown in the diagram, AD=8, BC=z, CA=y, DB=25/2, DC=x, BC is perpendicular to AC, CD is perpendicular to BD. Find the value of y.
问题：如图所示，AD=8，BC=z，CA=y，DB=25/2，DC=x，BC垂直于AC，CD⊥BD。求y的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(CA),y)
Equal(LengthOfLine(DB),25/2)
Equal(LengthOfLine(DC),x)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(CD,BD)
Goal: Value(y)
Answer: 2*sqrt(41)
Theorem sequence:
line_addition(1,AD,DB)
adjacent_complementary_angle(1,ADC,CDB)
right_triangle_judgment_angle(1,ADC)
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,ADC)
right_triangle_property_pythagorean(1,CDB)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠BFC=72°, ∠CFD=4*a-1°, ∠DFA=2*a-5°, the center of circle F is F. Find the measure of ∠AFB.
问题：如图所示，∠BFC=72°，∠CFD=4*a-1°，∠DFA=2*a-5°，F是圆F的圆心。求∠AFB的大小。
Construction (CDL):
Shape(BF,FC,FCB)
Shape(CF,FD,FDC)
Shape(DF,FA,FAD)
Shape(AF,FB,FBA)
Collinear(AFC)
Cocircular(F,BADC)
Conditions (CDL):
Equal(MeasureOfAngle(BFC),72)
Equal(MeasureOfAngle(CFD),4*a-1)
Equal(MeasureOfAngle(DFA),2*a-5)
IsCentreOfCircle(F,F)
Goal: Value(MeasureOfAngle(AFB))
Answer: 108
Theorem sequence:
adjacent_complementary_angle(1,AFB,BFC)
Problem: As shown in the diagram, ∠AGB=120°, ∠BDC=50°, ∠DCF=78°, ∠FAE=56°. Find the measure of ∠GFB.
问题：如图所示，∠AGB=120°，∠BDC=50°，∠DCF=78°，∠FAE=56°。求∠GFB的大小。
Construction (CDL):
Shape(DC,CB,BD)
Shape(FB,BG,GF)
Shape(GA,AE,EG)
Shape(DB,BF)
Shape(CB,BG)
Shape(AG,GB)
Shape(FG,GE)
Shape(HE,EA)
Collinear(DBGEH)
Collinear(CBF)
Collinear(AGF)
Conditions (CDL):
Equal(MeasureOfAngle(AGB),120)
Equal(MeasureOfAngle(BDC),50)
Equal(MeasureOfAngle(DCF),78)
Equal(MeasureOfAngle(FAE),56)
Goal: Value(MeasureOfAngle(GFB))
Answer: 68
Theorem sequence:
adjacent_complementary_angle(1,AGB,BGF)
triangle_property_angle_sum(1,DCB)
vertical_angle(1,CBD,FBG)
triangle_property_angle_sum(1,FBG)
Problem: As shown in the diagram, FL=24, HJ=48, the measure of ⌒MHP is 65, M is the center of circle M, JN is perpendicular to PN, KL is perpendicular to GL. Find the measure of ∠JMN.
问题：如图所示，FL=24，HJ=48，⌒MHP的角度为65，⊙M的圆心为M，JN垂直于PN，KL垂直于GL。求∠JMN的大小。
Construction (CDL):
Shape(FL,LK,MKF)
Shape(KL,LG,MGK)
Shape(HM,ML,LF,MFH)
Shape(GL,LM,MJ,MJG)
Shape(MH,HN,NM)
Shape(MN,NJ,JM)
Shape(PN,NH,MHP)
Shape(JN,NP,MPJ)
Collinear(HNJ)
Collinear(FLG)
Collinear(KLM)
Collinear(MNP)
Cocircular(M,HPJGKF)
Conditions (CDL):
Equal(LengthOfLine(FL),24)
Equal(LengthOfLine(HJ),48)
Equal(MeasureOfArc(MHP),65)
IsCentreOfCircle(M,M)
PerpendicularBetweenLine(JN,PN)
PerpendicularBetweenLine(KL,GL)
Goal: Value(MeasureOfAngle(JMN))
Answer: 65
Theorem sequence:
arc_property_center_angle(1,MHP,M)
radius_of_circle_property_length_equal(1,MH,M)
radius_of_circle_property_length_equal(1,MJ,M)
isosceles_triangle_judgment_line_equal(1,MHJ)
isosceles_triangle_property_angle_equal(1,MHJ)
adjacent_complementary_angle(1,JNP,PNH)
adjacent_complementary_angle(1,MNJ,JNP)
adjacent_complementary_angle(1,PNH,HNM)
mirror_congruent_triangle_judgment_aas(1,MHN,MNJ)
mirror_congruent_triangle_property_angle_equal(1,MHN,MNJ)
Problem: As shown in the diagram, ∠ADG=36°, ∠AGF=104°, ∠EFC=40°, GB⊥CB. Find the measure of ∠FGB.
问题：如图所示，∠ADG=36°，∠AGF=104°，∠EFC=40°，GB垂直于CB。求∠FGB的大小。
Construction (CDL):
Shape(AD,DG,GA)
Shape(GB,BF,FG)
Shape(FB,BC,CF)
Shape(AG,GF)
Shape(BG,GD)
Shape(GF,FE)
Shape(EF,FC)
Collinear(BFE)
Collinear(AGB)
Collinear(DGFC)
Conditions (CDL):
Equal(MeasureOfAngle(ADG),36)
Equal(MeasureOfAngle(AGF),104)
Equal(MeasureOfAngle(EFC),40)
PerpendicularBetweenLine(GB,CB)
Goal: Value(MeasureOfAngle(FGB))
Answer: 76
Theorem sequence:
adjacent_complementary_angle(1,AGF,FGB)
Problem: As shown in the diagram, ∠ADC=x°, ∠BAD=3*x°, CB⊥AB, DC is perpendicular to BC. Find the measure of ∠BAD.
问题：如图所示，∠ADC=x°，∠BAD=3*x°，CB⊥AB，DC垂直于BC。求∠BAD的大小。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),x)
Equal(MeasureOfAngle(BAD),3*x)
PerpendicularBetweenLine(CB,AB)
PerpendicularBetweenLine(DC,BC)
Goal: Value(MeasureOfAngle(BAD))
Answer: 135
Theorem sequence:
quadrilateral_property_angle_sum(1,ADCB)
Problem: As shown in the diagram, CD=12, DE=8, DF=5, EF=4, triangle BCD is similar to triangle FDE.. Find the perimeter of △BCD.
问题：如图所示，CD=12，DE=8，DF=5，EF=4，△BCD与△FDE是相似三角形。求△BCD的周长。
Construction (CDL):
Shape(BC,CD,DB)
Shape(FD,DE,EF)
Shape(BD,DF,FA,AB)
Collinear(ABC)
Collinear(CDE)
Collinear(EFA)
Conditions (CDL):
Equal(LengthOfLine(CD),12)
Equal(LengthOfLine(DE),8)
Equal(LengthOfLine(DF),5)
Equal(LengthOfLine(EF),4)
SimilarBetweenTriangle(BCD,FDE)
Goal: Value(PerimeterOfTriangle(BCD))
Answer: 51/2
Theorem sequence:
similar_triangle_property_line_ratio(1,BCD,FDE)
similar_triangle_property_line_ratio(1,CDB,DEF)
similar_triangle_property_line_ratio(1,DBC,EFD)
triangle_perimeter_formula(1,BCD)
Problem: As shown in the diagram, ML=6, ∠LKJ=109°, JMLK is a ▱. Find the measure of ∠MLK.
问题：如图所示，ML=6，∠LKJ=109°，MJ和LK是平行四边形JMLK的一组对边。求∠MLK的大小。
Construction (CDL):
Shape(JM,ML,LK,KJ)
Conditions (CDL):
Equal(LengthOfLine(ML),6)
Equal(MeasureOfAngle(LKJ),109)
Parallelogram(JMLK)
Goal: Value(MeasureOfAngle(MLK))
Answer: 71
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,MLKJ)
parallel_property_ipsilateral_internal_angle(1,JK,ML)
parallelogram_property_opposite_angle_equal(1,JMLK)
Problem: As shown in the diagram, XW=2, the measure of ⌒XYW is 52, X is the center of circle X. Find the area of the sector XYW.
问题：如图所示，XW=2，⌒XYW的角度为52，X是⊙X的圆心。求扇形XYW的面积。
Construction (CDL):
Shape(WX,XY,XYW)
Shape(YX,XW,XWY)
Cocircular(X,YW)
Conditions (CDL):
Equal(LengthOfLine(XW),2)
Equal(MeasureOfArc(XYW),52)
IsCentreOfCircle(X,X)
Goal: Value(AreaOfSector(XYW))
Answer: 26*pi/45
Theorem sequence:
radius_of_circle_property_length_equal(1,XW,X)
sector_area_formula(1,XYW)
Problem: As shown in the diagram, QT=2*x, QU=14, SR=27, ST=17, the center of ⊙W is W, QT is the tangent to circle W, the tangent to circle W is QU, RU is the tangent to circle W, RV is the tangent to ⊙W, ST is the tangent to ⊙W, the tangent to ⊙W is SV. Find the value of x.
问题：如图所示，QT=2*x，QU=14，SR=27，ST=17，W是⊙W的圆心，QT是⊙O的切线，QU是圆O的切线，⊙O的切线为RU，RV是圆O的切线，圆O的切线为ST，SV是⊙O的切线。求x的值。
Construction (CDL):
Shape(VR,RU,WVU)
Shape(TS,SV,WTV)
Shape(UW,WT,WTU)
Shape(AW,WU,WUA)
Shape(TW,WA,WAT)
Shape(UQ,QA,WUA)
Shape(AQ,QT,WAT)
Collinear(QTS)
Collinear(RUQ)
Collinear(RVS)
Collinear(QAW)
Cocircular(W,UATV)
Conditions (CDL):
Equal(LengthOfLine(QT),2*x)
Equal(LengthOfLine(QU),14)
Equal(LengthOfLine(SR),27)
Equal(LengthOfLine(ST),17)
IsCentreOfCircle(W,W)
IsTangentOfCircle(QT,W)
IsTangentOfCircle(QU,W)
IsTangentOfCircle(RU,W)
IsTangentOfCircle(RV,W)
IsTangentOfCircle(ST,W)
IsTangentOfCircle(SV,W)
Goal: Value(x)
Answer: 7
Theorem sequence:
tangent_of_circle_property_perpendicular(1,QU,W,W)
tangent_of_circle_property_perpendicular(2,QT,W,W)
radius_of_circle_property_length_equal(1,WU,W)
radius_of_circle_property_length_equal(1,WT,W)
mirror_congruent_triangle_judgment_hl(2,WUQ,WQT)
mirror_congruent_triangle_property_line_equal(1,WUQ,WQT)
Problem: As shown in the diagram, ∠EDA=5*x-3°, ∠GFA=3*x+7°, DG⊥FG, DGFE is a rectangle. Find the measure of ∠EDA.
问题：如图所示，∠EDA=5*x-3°，∠GFA=3*x+7°，DG⊥FG，四边形DGFE是长方形。求∠EDA的大小。
Construction (CDL):
Shape(DG,GA,AD)
Shape(AG,GF,FA)
Shape(EA,AF,FE)
Shape(DA,AE,ED)
Collinear(DAF)
Collinear(GAE)
Conditions (CDL):
Equal(MeasureOfAngle(EDA),5*x-3)
Equal(MeasureOfAngle(GFA),3*x+7)
PerpendicularBetweenLine(DG,FG)
Rectangle(DGFE)
Goal: Value(MeasureOfAngle(EDA))
Answer: 22
Theorem sequence:
parallel_property_alternate_interior_angle(1,DE,GF)
Problem: As shown in the diagram, AE=m-2, BA=20, DA=4, EC=m+4, ED∥CB. Find the value of m.
问题：如图所示，AE=m-2，BA=20，DA=4，EC=m+4，ED∥CB。求m的值。
Construction (CDL):
Shape(DA,AE,ED)
Shape(DE,EC,CB,BD)
Collinear(BDA)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AE),m-2)
Equal(LengthOfLine(BA),20)
Equal(LengthOfLine(DA),4)
Equal(LengthOfLine(EC),m+4)
ParallelBetweenLine(ED,CB)
Goal: Value(m)
Answer: 4
Theorem sequence:
line_addition(1,AE,EC)
parallel_property_corresponding_angle(1,ED,CB,A)
similar_triangle_judgment_aa(1,DAE,BAC)
similar_triangle_property_line_ratio(1,DAE,BAC)
similar_triangle_property_line_ratio(1,EDA,CBA)
Problem: As shown in the diagram, AB=2*x+3, BC=5*x, ADCB is a rhombus. Find the length of line AD.
问题：如图所示，AB=2*x+3，BC=5*x，四边形ADCB是菱形。求直线AD的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DC,CE)
Shape(EC,CB,BE)
Shape(AE,EB,BA)
Collinear(AEC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x+3)
Equal(LengthOfLine(BC),5*x)
Rhombus(ADCB)
Goal: Value(LengthOfLine(AD))
Answer: 5
Theorem sequence:
Problem: As shown in the diagram, NZ=9, XM=4, XN=6, YZ∥MN. Find the length of line XY.
问题：如图所示，NZ=9，XM=4，XN=6，YZ∥MN。求直线XY的长度。
Construction (CDL):
Shape(XN,NM,MX)
Shape(MN,NZ,ZY,YM)
Collinear(XMY)
Collinear(XNZ)
Conditions (CDL):
Equal(LengthOfLine(NZ),9)
Equal(LengthOfLine(XM),4)
Equal(LengthOfLine(XN),6)
ParallelBetweenLine(YZ,MN)
Goal: Value(LengthOfLine(XY))
Answer: 10
Theorem sequence:
line_addition(1,XN,NZ)
parallel_property_corresponding_angle(1,NM,ZY,X)
similar_triangle_judgment_aa(1,MXN,YXZ)
similar_triangle_property_line_ratio(1,MXN,YXZ)
similar_triangle_property_line_ratio(1,NMX,ZYX)
Problem: As shown in the diagram, ∠BEC=80°, ∠DEA=40°, ∠EAD=60°, ∠EBA=70°, ∠ECB=60°. Prove that DA∥EB.
问题：如图所示，∠BEC=80°，∠DEA=40°，∠EAD=60°，∠EBA=70°，∠ECB=60°。求证DA∥EB。
Construction (CDL):
Shape(AD,DE,EA)
Shape(AE,EB,BA)
Shape(BE,EC,CB)
Collinear(DEC)
Conditions (CDL):
Equal(MeasureOfAngle(BEC),80)
Equal(MeasureOfAngle(DEA),40)
Equal(MeasureOfAngle(EAD),60)
Equal(MeasureOfAngle(EBA),70)
Equal(MeasureOfAngle(ECB),60)
Goal: Relation(ParallelBetweenLine(DA,EB))
Answer: ('D','A','E','B')
Theorem sequence:
angle_addition(1,DEA,AEB)
adjacent_complementary_angle(1,DEB,BEC)
parallel_judgment_alternate_interior_angle(2,DA,EB)
Problem: As shown in the diagram, ∠GKM=62°, ∠HBF=38°, CO∥BH. Find the measure of ∠NCJ.
问题：如图所示，∠GKM=62°，∠HBF=38°，CO∥BH。求∠NCJ的大小。
Construction (CDL):
Shape(NC,CJ)
Shape(JC,CI)
Shape(IC,CO)
Shape(KC,CN)
Shape(KC,CB)
Shape(BC,CO)
Shape(MK,KC)
Shape(CK,KB)
Shape(KB,BC)
Shape(CB,BH)
Shape(GK,KM)
Shape(BK,KG)
Shape(FB,BK)
Shape(HB,BF)
Collinear(NCO)
Collinear(MKBH)
Collinear(JCBF)
Collinear(ICKG)
Conditions (CDL):
Equal(MeasureOfAngle(GKM),62)
Equal(MeasureOfAngle(HBF),38)
ParallelBetweenLine(CO,BH)
Goal: Value(MeasureOfAngle(NCJ))
Answer: 38
Theorem sequence:
parallel_property_corresponding_angle(2,CO,BH,F)
vertical_angle(1,NCJ,OCB)
Problem: As shown in the diagram, AC=x, BA=y, BD=16, CB=z, DC=8, AD⊥BD, CB⊥AB. Find the value of y.
问题：如图所示，AC=x，BA=y，BD=16，CB=z，DC=8，AD⊥BD，CB⊥AB。求y的值。
Construction (CDL):
Shape(CB,BD,DC)
Shape(DB,BA,AD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BA),y)
Equal(LengthOfLine(BD),16)
Equal(LengthOfLine(CB),z)
Equal(LengthOfLine(DC),8)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(CB,AB)
Goal: Value(y)
Answer: 16*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,ADB,BDC)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,ADB)
right_triangle_judgment_angle(1,CBA)
line_addition(1,AD,DC)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,ADB)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, JK=2*b+3, JM=3, LK=21, LM=45, ∠MLR=30°, ∠RLK=70°, JMLK is a parallelogram. Find the measure of ∠LKJ.
问题：如图所示，JK=2*b+3，JM=3，LK=21，LM=45，∠MLR=30°，∠RLK=70°，JMLK是▱。求∠LKJ的大小。
Construction (CDL):
Shape(JM,MR,RJ)
Shape(RM,ML,LR)
Shape(RL,LK,KR)
Shape(JR,RK,KJ)
Collinear(JRL)
Collinear(MRK)
Conditions (CDL):
Equal(LengthOfLine(JK),2*b+3)
Equal(LengthOfLine(JM),3)
Equal(LengthOfLine(LK),21)
Equal(LengthOfLine(LM),45)
Equal(MeasureOfAngle(MLR),30)
Equal(MeasureOfAngle(RLK),70)
Parallelogram(JMLK)
Goal: Value(MeasureOfAngle(LKJ))
Answer: 80
Theorem sequence:
angle_addition(1,MLR,RLK)
parallel_property_ipsilateral_internal_angle(1,LM,KJ)
Problem: As shown in the diagram, AE=8, AF=x, BE=8, CF=6, C is the center of ⊙C, AE is the tangent to circle C. Find the value of x.
问题：如图所示，AE=8，AF=x，BE=8，CF=6，⊙C的圆心为C，⊙O的切线为AE。求x的值。
Construction (CDL):
Shape(DC,CE,CED)
Shape(EC,CF,CFE)
Shape(FC,CD,CDF)
Shape(FA,AE,CFE)
Shape(EB,BD,CED)
Collinear(AEB)
Collinear(CFA)
Collinear(CDB)
Cocircular(C,FED)
Conditions (CDL):
Equal(LengthOfLine(AE),8)
Equal(LengthOfLine(AF),x)
Equal(LengthOfLine(BE),8)
Equal(LengthOfLine(CF),6)
IsCentreOfCircle(C,C)
IsTangentOfCircle(AE,C)
Goal: Value(x)
Answer: 4
Theorem sequence:
radius_of_circle_property_length_equal(1,CF,C)
radius_of_circle_property_length_equal(1,CE,C)
line_addition(1,CF,FA)
tangent_of_circle_property_perpendicular(2,AE,C,C)
right_triangle_judgment_angle(1,AEC)
right_triangle_property_pythagorean(1,AEC)
Problem: As shown in the diagram, ∠BED=3*x°, ∠DCB=5*x-12°, the center of ⊙B is B, the diameter of circle B is CE. Find the value of x.
问题：如图所示，∠BED=3*x°，∠DCB=5*x-12°，⊙B的圆心为B，⊙B的直径为CE。求x的值。
Construction (CDL):
Shape(DC,CB,BE,ED)
Shape(EB,BC,BCE)
Shape(DE,BED)
Shape(DC,BDC)
Collinear(CBE)
Cocircular(B,CED)
Conditions (CDL):
Equal(MeasureOfAngle(BED),3*x)
Equal(MeasureOfAngle(DCB),5*x-12)
IsCentreOfCircle(B,B)
IsDiameterOfCircle(CE,B)
Goal: Value(x)
Answer: 51/4
Theorem sequence:
diameter_of_circle_property_right_angle(1,EDC,B)
triangle_property_angle_sum(1,DCE)
Problem: As shown in the diagram, QS=6*x-1, RQ=4*x, SR=2*x+1, △RQS is an equilateral △. Find the length of line SR.
问题：如图所示，QS=6*x-1，RQ=4*x，SR=2*x+1，三角形RQS为等边三角形。求直线SR的长度。
Construction (CDL):
Shape(RQ,QS,SR)
Conditions (CDL):
Equal(LengthOfLine(QS),6*x-1)
Equal(LengthOfLine(RQ),4*x)
Equal(LengthOfLine(SR),2*x+1)
EquilateralTriangle(RQS)
Goal: Value(LengthOfLine(SR))
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, AB=8*sqrt(3), BC=y, CA=x, ∠BCA=60°, CA⊥BA. Find the value of x.
问题：如图所示，AB=8*sqrt(3)，BC=y，CA=x，∠BCA=60°，CA垂直于BA。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),8*sqrt(3))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CA),x)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 8
Theorem sequence:
cosine_theorem(1,CAB)
cosine_theorem(1,ABC)
Problem: As shown in the diagram, SV=x, UV=8, VR=4, VT=4. Find the value of x.
问题：如图所示，SV=x，UV=8，VR=4，VT=4。求x的值。
Construction (CDL):
Shape(UV,VR,ARU)
Shape(TV,VU,AUT)
Shape(SV,VT,ATS)
Shape(RV,VS,ASR)
Collinear(RVT)
Collinear(SVU)
Cocircular(A,UTSR)
Conditions (CDL):
Equal(LengthOfLine(SV),x)
Equal(LengthOfLine(UV),8)
Equal(LengthOfLine(VR),4)
Equal(LengthOfLine(VT),4)
Goal: Value(x)
Answer: 2
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,UVS,TVR,A)
Problem: As shown in the diagram, AC=4, AD=5, BC=4, DE=6, AD is perpendicular to ED, DE⊥BE, EB⊥AB. Find the area of quadrilateral BADE minus the area of △BAC.
问题：如图所示，AC=4，AD=5，BC=4，DE=6，AD垂直于ED，DE垂直于BE，EB垂直于AB。求四边形BADE的面积减去三角形BAC的面积。
Construction (CDL):
Shape(BC,CA,AD,DE,EB)
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(AD),5)
Equal(LengthOfLine(BC),4)
Equal(LengthOfLine(DE),6)
PerpendicularBetweenLine(AD,ED)
PerpendicularBetweenLine(DE,BE)
PerpendicularBetweenLine(EB,AB)
Goal: Value(Sub(AreaOfQuadrilateral(BADE),AreaOfTriangle(BAC)))
Answer: 30-3*sqrt(7)
Theorem sequence:
quadrilateral_property_angle_sum(1,BADE)
parallel_judgment_ipsilateral_internal_angle(1,BE,AD)
parallel_judgment_ipsilateral_internal_angle(1,ED,BA)
parallelogram_judgment_parallel_and_parallel(1,BADE)
parallelogram_area_formula_sine(1,ADEB)
parallelogram_property_opposite_line_equal(1,DEBA)
cosine_theorem(1,CBA)
triangle_area_formula_sine(1,CBA)
Problem: As shown in the diagram, AB=3, A is the center of ⊙A. Find the area of the ⊙A.
问题：如图所示，AB=3，A是圆A的圆心。求圆A的面积。
Construction (CDL):
Shape(AB)
Cocircular(A,B)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
IsCentreOfCircle(A,A)
Goal: Value(AreaOfCircle(A))
Answer: 9*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,AB,A)
circle_area_formula(1,A)
Problem: As shown in the diagram, JK=2*b+3, JM=3*a, LK=21, LM=45, ∠MLR=30°, ∠RLK=70°, JK and ML are opposite sides of the parallelogram JMLK. Find the measure of ∠KJR.
问题：如图所示，JK=2*b+3，JM=3*a，LK=21，LM=45，∠MLR=30°，∠RLK=70°，JMLK是平行四边形。求∠KJR的大小。
Construction (CDL):
Shape(JM,MR,RJ)
Shape(RM,ML,LR)
Shape(RL,LK,KR)
Shape(JR,RK,KJ)
Collinear(JRL)
Collinear(MRK)
Conditions (CDL):
Equal(LengthOfLine(JK),2*b+3)
Equal(LengthOfLine(JM),3*a)
Equal(LengthOfLine(LK),21)
Equal(LengthOfLine(LM),45)
Equal(MeasureOfAngle(MLR),30)
Equal(MeasureOfAngle(RLK),70)
Parallelogram(JMLK)
Goal: Value(MeasureOfAngle(KJR))
Answer: 30
Theorem sequence:
parallel_property_alternate_interior_angle(1,JK,ML)
Problem: As shown in the diagram, BX=13, XA=5, XA is perpendicular to BA. Find the value of tan(BX).
问题：如图所示，BX=13，XA=5，XA⊥BA。求tan(BX)的值。
Construction (CDL):
Shape(XA,AB,BX)
Conditions (CDL):
Equal(LengthOfLine(BX),13)
Equal(LengthOfLine(XA),5)
PerpendicularBetweenLine(XA,BA)
Goal: Value(Tan(MeasureOfAngle(BXA)))
Answer: 12/5
Theorem sequence:
right_triangle_judgment_angle(1,XAB)
right_triangle_property_pythagorean(1,XAB)
cosine_theorem(1,XAB)
Problem: As shown in the diagram, the area of quadrilateral BADC is 2000, BC=x+10, BE=x, AB and DC are opposite sides of the ▱ BADC, BE is perpendicular to DE. Find the value of x.
问题：如图所示，四边形BADC的面积为2000，BC=x+10，BE=x，BC和AD是▱BADC的一组对边，BE⊥DE。求x的值。
Construction (CDL):
Shape(BA,AE,EB)
Shape(BE,ED,DC,CB)
Collinear(AED)
Conditions (CDL):
Equal(AreaOfQuadrilateral(BADC),2000)
Equal(LengthOfLine(BC),x+10)
Equal(LengthOfLine(BE),x)
Parallelogram(BADC)
PerpendicularBetweenLine(BE,DE)
Goal: Value(x)
Answer: 40
Theorem sequence:
adjacent_complementary_angle(1,AEB,BED)
altitude_of_quadrilateral_judgment_left_vertex(1,BE,BADC)
parallelogram_property_opposite_line_equal(1,ADCB)
parallelogram_area_formula_common(1,BADC)
Problem: As shown in the diagram, ∠HGK=13°, EH is perpendicular to GH, EHGF is a rectangle. Find the measure of ∠KGF.
问题：如图所示，∠HGK=13°，EH垂直于GH，EHGF是矩形。求∠KGF的大小。
Construction (CDL):
Shape(EH,HK,KE)
Shape(KH,HG,GK)
Shape(KG,GF,FK)
Shape(EK,KF,FE)
Collinear(EKG)
Collinear(HKF)
Conditions (CDL):
Equal(MeasureOfAngle(HGK),13)
PerpendicularBetweenLine(EH,GH)
Rectangle(EHGF)
Goal: Value(MeasureOfAngle(KGF))
Answer: 77
Theorem sequence:
angle_addition(1,HGK,KGF)
Problem: As shown in the diagram, AB=2*y+19, DC=3*y+13, ∠ADC=115°, ∠CBA=2*x+41°, CD and BA are opposite sides of the ▱ DCBA. Find the value of x.
问题：如图所示，AB=2*y+19，DC=3*y+13，∠ADC=115°，∠CBA=2*x+41°，DA和CB是▱DCBA的一组对边。求x的值。
Construction (CDL):
Shape(DC,CB,BA,AD)
Conditions (CDL):
Equal(LengthOfLine(AB),2*y+19)
Equal(LengthOfLine(DC),3*y+13)
Equal(MeasureOfAngle(ADC),115)
Equal(MeasureOfAngle(CBA),2*x+41)
Parallelogram(DCBA)
Goal: Value(x)
Answer: 37
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,DCBA)