SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠ILR=2*y+8°, ∠JME=z°, ∠KIP=4*x+6°, ∠RLP=142°, HK∥BR, JN is parallel to HK. Find the value of z.
问题：如图所示，∠ILR=2*y+8°，∠JME=z°，∠KIP=4*x+6°，∠RLP=142°，HK平行于BR，JN∥HK。求z的值。
Construction (CDL):
Shape(JM,ME)
Shape(EM,MN)
Shape(NM,MP)
Shape(PM,MJ)
Shape(HI,IE)
Shape(EI,IK)
Shape(KI,IP)
Shape(PI,IH)
Shape(BL,LE)
Shape(EL,LR)
Shape(RL,LP)
Shape(PL,LB)
Collinear(JSCMN)
Collinear(HFIAK)
Collinear(BQLGODR)
Collinear(PLIME)
Conditions (CDL):
Equal(MeasureOfAngle(ILR),2*y+8)
Equal(MeasureOfAngle(JME),z)
Equal(MeasureOfAngle(KIP),4*x+6)
Equal(MeasureOfAngle(RLP),142)
ParallelBetweenLine(HK,BR)
ParallelBetweenLine(JN,HK)
Goal: Value(z)
Answer: 142
Theorem sequence:
adjacent_complementary_angle(1,ELR,RLP)
parallel_property_collinear_extend(3,HK,BR,I)
parallel_property_collinear_extend(3,RB,KI,L)
parallel_property_collinear_extend(3,JN,HK,M)
parallel_property_collinear_extend(3,KH,NM,I)
parallel_property_corresponding_angle(1,IK,LR,E)
parallel_property_corresponding_angle(1,MN,IK,E)
adjacent_complementary_angle(1,JME,EMN)
Problem: As shown in the diagram, QS=8, TQ=8, ∠RQS=43°, QS is perpendicular to RS, RT⊥QT. Find the measure of ∠TQR.
问题：如图所示，QS=8，TQ=8，∠RQS=43°，QS⊥RS，RT⊥QT。求∠TQR的大小。
Construction (CDL):
Shape(SR,RQ,QS)
Shape(QR,RT,TQ)
Conditions (CDL):
Equal(LengthOfLine(QS),8)
Equal(LengthOfLine(TQ),8)
Equal(MeasureOfAngle(RQS),43)
PerpendicularBetweenLine(QS,RS)
PerpendicularBetweenLine(RT,QT)
Goal: Value(MeasureOfAngle(TQR))
Answer: 43
Theorem sequence:
triangle_property_angle_sum(1,SRQ)
sine_theorem(1,QSR)
sine_theorem(1,QRT)
triangle_property_angle_sum(1,QRT)
Problem: As shown in the diagram, BA=3*y-8, BC=2*x+7, DA=13, DC=10, ∠ADF=59°, ∠BFC=49°, ∠FBA=20°, quadrilateral ADCB is a parallelogram. Find the value of y.
问题：如图所示，BA=3*y-8，BC=2*x+7，DA=13，DC=10，∠ADF=59°，∠BFC=49°，∠FBA=20°，四边形ADCB是平行四边形。求y的值。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(FC,CB,BF)
Shape(BA,AF,FB)
Collinear(AFC)
Collinear(DFB)
Conditions (CDL):
Equal(LengthOfLine(BA),3*y-8)
Equal(LengthOfLine(BC),2*x+7)
Equal(LengthOfLine(DA),13)
Equal(LengthOfLine(DC),10)
Equal(MeasureOfAngle(ADF),59)
Equal(MeasureOfAngle(BFC),49)
Equal(MeasureOfAngle(FBA),20)
Parallelogram(ADCB)
Goal: Value(y)
Answer: 6
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BADC)
Problem: As shown in the diagram, ∠ADE=68°, ∠CBD=28°, ∠ECB=82°. Find the measure of ∠EAD.
问题：如图所示，∠ADE=68°，∠CBD=28°，∠ECB=82°。求∠EAD的大小。
Construction (CDL):
Shape(CB,BE,EC)
Shape(EA,AD,DE)
Collinear(CEA)
Collinear(BED)
Conditions (CDL):
Equal(MeasureOfAngle(ADE),68)
Equal(MeasureOfAngle(CBD),28)
Equal(MeasureOfAngle(ECB),82)
Goal: Value(MeasureOfAngle(EAD))
Answer: 42
Theorem sequence:
triangle_property_angle_sum(1,CBE)
triangle_property_angle_sum(1,EAD)
vertical_angle(1,BEC,DEA)
Problem: As shown in the diagram, AB=y, AC=11, BC=x, ∠ADB=60°, ∠BCA=60°, ∠CAB=30°, DB⊥AB. Find the value of x.
问题：如图所示，AB=y，AC=11，BC=x，∠ADB=60°，∠BCA=60°，∠CAB=30°，DB⊥AB。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Shape(BA,AD,DB)
Collinear(CBD)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),11)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ADB),60)
Equal(MeasureOfAngle(BCA),60)
Equal(MeasureOfAngle(CAB),30)
PerpendicularBetweenLine(DB,AB)
Goal: Value(x)
Answer: 11/2
Theorem sequence:
adjacent_complementary_angle(1,DBA,ABC)
sine_theorem(1,CAB)
Problem: As shown in the diagram, AC=9, AD=4, AE=x, BE=8, E is the center of ⊙E. Find the value of x.
问题：如图所示，AC=9，AD=4，AE=x，BE=8，E是⊙E的圆心。求x的值。
Construction (CDL):
Shape(FA,AD,EDF)
Shape(CA,AF,EFC)
Shape(BE,EA,AC,ECB)
Shape(DA,AE,EB,EBD)
Collinear(FAEB)
Collinear(CAD)
Cocircular(E,FCBD)
Conditions (CDL):
Equal(LengthOfLine(AC),9)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(AE),x)
Equal(LengthOfLine(BE),8)
IsCentreOfCircle(E,E)
Goal: Value(x)
Answer: 2*sqrt(7)
Theorem sequence:
radius_of_circle_property_length_equal(1,EB,E)
radius_of_circle_property_length_equal(1,EF,E)
line_addition(1,AE,EB)
line_addition(1,FA,AE)
circle_property_circular_power_chord_and_chord(1,CAD,FAB,E)
Problem: As shown in the diagram, CB=23, DE=5, ∠EDB=45°, AC and DB are opposite sides of the parallelogram CADB, BE is perpendicular to DE. Find the area of quadrilateral CADB.
问题：如图所示，CB=23，DE=5，∠EDB=45°，AC和DB是平行四边形CADB的一组对边，BE⊥DE。求CADB的面积。
Construction (CDL):
Shape(BC,CA,AE,EB)
Shape(BE,ED,DB)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(CB),23)
Equal(LengthOfLine(DE),5)
Equal(MeasureOfAngle(EDB),45)
Parallelogram(CADB)
PerpendicularBetweenLine(BE,DE)
Goal: Value(AreaOfQuadrilateral(CADB))
Answer: 115
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BCAD)
triangle_property_angle_sum(1,BED)
sine_theorem(1,DBE)
parallelogram_area_formula_sine(1,ADBC)
Problem: As shown in the diagram, ∠CEB=40°, ∠DEF=y°, ∠EFD=2*x°, ∠FDE=∠EFD. Find the value of y.
问题：如图所示，∠CEB=40°，∠DEF=y°，∠EFD=2*x°，∠FDE=∠EFD。求y的值。
Construction (CDL):
Shape(FD,DE,EF)
Shape(EB,BC,CE)
Collinear(DEC)
Collinear(FEB)
Conditions (CDL):
Equal(MeasureOfAngle(CEB),40)
Equal(MeasureOfAngle(DEF),y)
Equal(MeasureOfAngle(EFD),2*x)
Equal(MeasureOfAngle(FDE),MeasureOfAngle(EFD))
Goal: Value(y)
Answer: 40
Theorem sequence:
vertical_angle(1,DEF,CEB)
Problem: As shown in the diagram, AC=22, BD=31, CB=15, O is the center of circle O, AD is the diameter of circle O, CEDA is a rectangle. Find the sum of the area of triangle CBE, the area of quadrilateral ACED and the area of the sector ODA.
问题：如图所示，AC=22，BD=31，CB=15，⊙O的圆心为O，⊙O的直径为AD，CEDA是长方形。求△CBE的面积、四边形ACED的面积与扇形ODA的面积之和。
Construction (CDL):
Shape(AC,CE,ED,DO,OA)
Shape(CB,BE,EC)
Shape(AO,OD,ODA)
Collinear(AOD)
Collinear(BED)
Cocircular(O,DA)
Conditions (CDL):
Equal(LengthOfLine(AC),22)
Equal(LengthOfLine(BD),31)
Equal(LengthOfLine(CB),15)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AD,O)
Rectangle(CEDA)
Goal: Value(Add(AreaOfTriangle(CBE),AreaOfQuadrilateral(ACED),AreaOfSector(ODA)))
Answer: 18*pi+318
Theorem sequence:
parallelogram_property_opposite_line_equal(1,CEDA)
parallelogram_property_opposite_line_equal(1,EDAC)
line_addition(1,BE,ED)
adjacent_complementary_angle(1,BEC,CED)
right_triangle_judgment_angle(1,BEC)
right_triangle_property_pythagorean(1,BEC)
triangle_area_formula_sine(1,ECB)
parallelogram_area_formula_sine(1,EDAC)
diameter_of_circle_property_length_equal(1,AD,O)
circle_property_length_of_radius_and_diameter(1,O)
flat_angle(1,AOD)
arc_property_center_angle(1,ODA,O)
sector_area_formula(1,ODA)
Problem: As shown in the diagram, AB=y, AC=8, BC=6, SA=z, SB=x, BA⊥SA, SC⊥AC. Find the value of x.
问题：如图所示，AB=y，AC=8，BC=6，SA=z，SB=x，BA垂直于SA，SC垂直于AC。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Shape(CA,AS,SC)
Collinear(BCS)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(SA),z)
Equal(LengthOfLine(SB),x)
PerpendicularBetweenLine(BA,SA)
PerpendicularBetweenLine(SC,AC)
Goal: Value(x)
Answer: 50/3
Theorem sequence:
line_addition(1,BC,CS)
adjacent_complementary_angle(1,SCA,ACB)
mirror_similar_triangle_judgment_aa(1,ASC,BAS)
mirror_similar_triangle_property_angle_equal(1,ASC,BAS)
similar_triangle_judgment_aa(1,ACB,SCA)
similar_triangle_property_line_ratio(1,ACB,SCA)
similar_triangle_property_line_ratio(1,BAC,ASC)
Problem: As shown in the diagram, AC=x, BC=16, ∠CBA=45°, BA⊥CA. Find the value of x.
问题：如图所示，AC=x，BC=16，∠CBA=45°，BA垂直于CA。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),16)
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 8*sqrt(2)
Theorem sequence:
sine_theorem(1,CBA)
Problem: As shown in the diagram, AD=48, AH=15, HB=30, the center of ⊙K is K, HA is perpendicular to DA, KH is perpendicular to AH. Find the sum of the area of quadrilateral HADB and half the area of circle K.
问题：如图所示，AD=48，AH=15，HB=30，K是⊙K的圆心，HA垂直于DA，KH⊥AH。求四边形HADB面积和1/2圆K面积之和。
Construction (CDL):
Shape(HB,KBH)
Shape(HA,AD,DB,BK,KH)
Collinear(HKB)
Cocircular(K,BH)
Conditions (CDL):
Equal(LengthOfLine(AD),48)
Equal(LengthOfLine(AH),15)
Equal(LengthOfLine(HB),30)
IsCentreOfCircle(K,K)
PerpendicularBetweenLine(HA,DA)
PerpendicularBetweenLine(KH,AH)
Goal: Value(Add(AreaOfQuadrilateral(HADB),Mul(1/2,AreaOfCircle(K))))
Answer: 225*pi/2+585
Theorem sequence:
line_addition(1,HK,KB)
perpendicular_judgment_angle(1,BH,AH)
parallel_judgment_per_per(1,HB,AD)
trapezoid_judgment_parallel(1,HADB)
right_trapezoid_judgment_right_angle(1,HADB)
right_trapezoid_area_formular(1,HADB)
radius_of_circle_property_length_equal(1,KH,K)
radius_of_circle_property_length_equal(1,KB,K)
circle_area_formula(1,K)
Problem: As shown in the diagram, PQ=37, RQ=58, RP is perpendicular to QP. Find the measure of ∠QRP.
问题：如图所示，PQ=37，RQ=58，RP⊥QP。求∠QRP的大小。
Construction (CDL):
Shape(PQ,QR,RP)
Conditions (CDL):
Equal(LengthOfLine(PQ),37)
Equal(LengthOfLine(RQ),58)
PerpendicularBetweenLine(RP,QP)
Goal: Value(MeasureOfAngle(QRP))
Answer: 180*asin(37/58)/pi
Theorem sequence:
sine_theorem(1,QRP)
Problem: As shown in the diagram, AB=50, AC=14, BC=x, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=50，AC=14，BC=x，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),50)
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 48
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, ∠WXH=130°, ∠YZI=20°, Mul(LengthOfLine(IJ)=LengthOfLine(YJ)). Find the measure of ∠GHJ.
问题：如图所示，∠WXH=130°，∠YZI=20°，Mul(LengthOfLine(IJ)=LengthOfLine(YJ))。求∠GHJ的大小。
Construction (CDL):
Shape(HX,XY,YI,IH)
Shape(IY,YZ,ZI)
Shape(JH,HI,IJ)
Shape(GH,HJ)
Shape(WX,XJ)
Collinear(JHX)
Collinear(JIY)
Collinear(GHIZ)
Collinear(WXYZ)
Conditions (CDL):
Equal(MeasureOfAngle(WXH),130)
Equal(MeasureOfAngle(YZI),20)
Equal(Mul(LengthOfLine(IJ),LengthOfLine(YJ)),Mul(LengthOfLine(XJ),LengthOfLine(HJ)))
Goal: Value(MeasureOfAngle(GHJ))
Answer: 110
Theorem sequence:
flat_angle(1,WXY)
angle_addition(1,WXH,HXY)
mirror_similar_triangle_judgment_sas(1,JHI,JXY)
mirror_similar_triangle_property_angle_equal(1,HIJ,YJX)
mirror_similar_triangle_property_angle_equal(1,IJH,XYJ)
vertical_angle(1,HIJ,ZIY)
triangle_property_angle_sum(1,IYZ)
flat_angle(1,XYZ)
angle_addition(1,XYI,IYZ)
flat_angle(1,GHI)
angle_addition(1,GHJ,JHI)
Problem: As shown in the diagram, ∠BGC=20*x°, ∠DGB=100+20*x°. Find the measure of ∠BGC.
问题：如图所示，∠BGC=20*x°，∠DGB=100+20*x°。求∠BGC的大小。
Construction (CDL):
Shape(BG,GC)
Shape(CG,GE)
Shape(EG,GD)
Shape(EG,GD)
Collinear(CGD)
Collinear(BGE)
Conditions (CDL):
Equal(MeasureOfAngle(BGC),20*x)
Equal(MeasureOfAngle(DGB),100+20*x)
Goal: Value(MeasureOfAngle(BGC))
Answer: 40
Theorem sequence:
adjacent_complementary_angle(1,DGB,BGC)
Problem: As shown in the diagram, AB=y, AC=10, BC=x, ∠BAC=45°, CB⊥AB. Find the value of y.
问题：如图所示，AB=y，AC=10，BC=x，∠BAC=45°，CB垂直于AB。求y的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(BAC),45)
PerpendicularBetweenLine(CB,AB)
Goal: Value(y)
Answer: 5*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,CBA)
sine_theorem(1,ACB)
Problem: As shown in the diagram, AD=x, CA=8, CD=17, the center of circle C is C, the tangent to ⊙C is DA. Find the value of x.
问题：如图所示，AD=x，CA=8，CD=17，圆C的圆心为C，圆O的切线为DA。求x的值。
Construction (CDL):
Shape(AC,CE,CEA)
Shape(CEA,ED,DA)
Shape(CAE,EC,CA)
Collinear(CED)
Cocircular(C,EA)
Conditions (CDL):
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(CA),8)
Equal(LengthOfLine(CD),17)
IsCentreOfCircle(C,C)
IsTangentOfCircle(DA,C)
Goal: Value(x)
Answer: 15
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DA,C,C)
right_triangle_judgment_angle(1,DAC)
right_triangle_property_pythagorean(1,DAC)
Problem: As shown in the diagram, AB=5, AC=15, OB=20, OC=8, OC is perpendicular to AC. Find the perimeter of △OAB.
问题：如图所示，AB=5，AC=15，OB=20，OC=8，OC垂直于AC。求三角形OAB的周长。
Construction (CDL):
Shape(OC,CA,AO)
Shape(OA,AB,BO)
Collinear(CAB)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),15)
Equal(LengthOfLine(OB),20)
Equal(LengthOfLine(OC),8)
PerpendicularBetweenLine(OC,AC)
Goal: Value(PerimeterOfTriangle(OAB))
Answer: 42
Theorem sequence:
right_triangle_judgment_angle(1,OCA)
right_triangle_property_pythagorean(1,OCA)
triangle_perimeter_formula(1,OAB)
Problem: As shown in the diagram, AB=c, AC=b, BC=a, ∠ABC=60°, ∠CAB=30°, c=8, BC is perpendicular to AC. Find the value of a.
问题：如图所示，AB=c，AC=b，BC=a，∠ABC=60°，∠CAB=30°，c=8，BC⊥AC。求a的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),c)
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(BC),a)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),30)
Equal(c,8)
PerpendicularBetweenLine(BC,AC)
Goal: Value(a)
Answer: 4
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, ∠ABO=p°, ∠AOC=n°, ∠DAO=m°, m=120, n=130. Find the value of p.
问题：如图所示，∠ABO=p°，∠AOC=n°，∠DAO=m°，m=120，n=130。求p的值。
Construction (CDL):
Shape(BO,OA,AB)
Shape(AO,OC)
Shape(DA,AO)
Collinear(COB)
Collinear(BAD)
Conditions (CDL):
Equal(MeasureOfAngle(ABO),p)
Equal(MeasureOfAngle(AOC),n)
Equal(MeasureOfAngle(DAO),m)
Equal(m,120)
Equal(n,130)
Goal: Value(p)
Answer: 70
Theorem sequence:
adjacent_complementary_angle(1,BOA,AOC)
adjacent_complementary_angle(1,DAO,OAB)
triangle_property_angle_sum(1,BOA)
Problem: As shown in the diagram, NM=9, NP=10, QM=8, QP=7, WX=4, quadrilateral WZYX is similar to quadrilateral QPNM. Find The ratio of similar quadrilaterals WZYX and QPNM.
问题：如图所示，NM=9，NP=10，QM=8，QP=7，WX=4，WZYX与QPNM相似。求四边形WZYX和四边形QPNM的相似比。
Construction (CDL):
Shape(WZ,ZY,YX,XW)
Shape(NM,MQ,QP,PN)
Conditions (CDL):
Equal(LengthOfLine(NM),9)
Equal(LengthOfLine(NP),10)
Equal(LengthOfLine(QM),8)
Equal(LengthOfLine(QP),7)
Equal(LengthOfLine(WX),4)
SimilarBetweenQuadrilateral(WZYX,QPNM)
Goal: Value(RatioOfSimilarQuadrilateral(WZYX,QPNM))
Answer: 1/2
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,XWZY,MQPN)
Problem: As shown in the diagram, AB=x+5, AC=3*x+1, BD=12, CF=15, BC∥DF. Find the value of x.
问题：如图所示，AB=x+5，AC=3*x+1，BD=12，CF=15，BC平行于DF。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Shape(BD,DF,FC,CB)
Collinear(ABD)
Collinear(ACF)
Conditions (CDL):
Equal(LengthOfLine(AB),x+5)
Equal(LengthOfLine(AC),3*x+1)
Equal(LengthOfLine(BD),12)
Equal(LengthOfLine(CF),15)
ParallelBetweenLine(BC,DF)
Goal: Value(x)
Answer: 3
Theorem sequence:
line_addition(1,AB,BD)
line_addition(1,AC,CF)
parallel_property_corresponding_angle(1,BC,DF,A)
similar_triangle_judgment_aa(1,CAB,FAD)
similar_triangle_property_line_ratio(1,CAB,FAD)
similar_triangle_property_line_ratio(1,BCA,DFA)
Problem: As shown in the diagram, AB=6, BC=y, YA=x, ∠YCA=45°, AY⊥CY, CA is perpendicular to BA. Find the value of y.
问题：如图所示，AB=6，BC=y，YA=x，∠YCA=45°，AY垂直于CY，CA⊥BA。求y的值。
Construction (CDL):
Shape(AB,BY,YA)
Shape(AY,YC,CA)
Collinear(BYC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(YA),x)
Equal(MeasureOfAngle(YCA),45)
PerpendicularBetweenLine(AY,CY)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 6*sqrt(2)
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, ∠HLI=59°, L is the center of ⊙L, IL is perpendicular to JL. Find the measure of arc LIH.
问题：如图所示，∠HLI=59°，⊙L的圆心为L，IL垂直于JL。求⌒LIH的角度。
Construction (CDL):
Shape(HL,LI,LIH)
Shape(GL,LH,LHG)
Shape(KL,LG,LGK)
Shape(JL,LK,LKJ)
Shape(IL,LJ,LJI)
Collinear(HLK)
Collinear(ILG)
Cocircular(L,GKJIH)
Conditions (CDL):
Equal(MeasureOfAngle(HLI),59)
IsCentreOfCircle(L,L)
PerpendicularBetweenLine(IL,JL)
Goal: Value(MeasureOfArc(LIH))
Answer: 59
Theorem sequence:
arc_property_center_angle(1,LIH,L)
Problem: As shown in the diagram, JM=5, KM=20, LJ=y, LK=z, LM=x, JL is perpendicular to KL, KM⊥LM. Find the value of x.
问题：如图所示，JM=5，KM=20，LJ=y，LK=z，LM=x，JL垂直于KL，KM⊥LM。求x的值。
Construction (CDL):
Shape(MJ,JL,LM)
Shape(ML,LK,KM)
Collinear(JMK)
Conditions (CDL):
Equal(LengthOfLine(JM),5)
Equal(LengthOfLine(KM),20)
Equal(LengthOfLine(LJ),y)
Equal(LengthOfLine(LK),z)
Equal(LengthOfLine(LM),x)
PerpendicularBetweenLine(JL,KL)
PerpendicularBetweenLine(KM,LM)
Goal: Value(x)
Answer: 10
Theorem sequence:
line_addition(1,JM,MK)
adjacent_complementary_angle(1,KML,LMJ)
right_triangle_judgment_angle(1,LMJ)
right_triangle_judgment_angle(1,KML)
right_triangle_judgment_angle(1,JLK)
right_triangle_property_pythagorean(1,LMJ)
right_triangle_property_pythagorean(1,KML)
right_triangle_property_pythagorean(1,JLK)
Problem: As shown in the diagram, AB=4, AC=3, CA⊥BA. Find the length of line BC.
问题：如图所示，AB=4，AC=3，CA垂直于BA。求直线BC的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(AC),3)
PerpendicularBetweenLine(CA,BA)
Goal: Value(LengthOfLine(BC))
Answer: 5
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, AC=8, AD=11, BE=2, BF=3, CH=6, EG=2, FD=5, HG=1, BE⊥GE, BN is perpendicular to DN, CA is perpendicular to MA, DF⊥BF, GH is perpendicular to CH, GM⊥NM, HC⊥AC, ND is perpendicular to FD. Find the sum of the area of quadrilateral HCAM, the area of quadrilateral GMNE and the area of quadrilateral FBND.
问题：如图所示，AC=8，AD=11，BE=2，BF=3，CH=6，EG=2，FD=5，HG=1，BE⊥GE，BN⊥DN，CA垂直于MA，DF⊥BF，GH⊥CH，GM⊥NM，HC垂直于AC，ND⊥FD。求HCAM的面积、四边形GMNE的面积与四边形FBND的面积之和。
Construction (CDL):
Shape(HC,CA,AM,MG,GH)
Shape(EG,GM,MN,NB,BE)
Shape(BN,ND,DF,FB)
Collinear(HGM)
Collinear(EBN)
Collinear(AMND)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(AD),11)
Equal(LengthOfLine(BE),2)
Equal(LengthOfLine(BF),3)
Equal(LengthOfLine(CH),6)
Equal(LengthOfLine(EG),2)
Equal(LengthOfLine(FD),5)
Equal(LengthOfLine(HG),1)
PerpendicularBetweenLine(BE,GE)
PerpendicularBetweenLine(BN,DN)
PerpendicularBetweenLine(CA,MA)
PerpendicularBetweenLine(DF,BF)
PerpendicularBetweenLine(GH,CH)
PerpendicularBetweenLine(GM,NM)
PerpendicularBetweenLine(HC,AC)
PerpendicularBetweenLine(ND,FD)
Goal: Value(Add(AreaOfQuadrilateral(HCAM),AreaOfQuadrilateral(GMNE),AreaOfQuadrilateral(FBND)))
Answer: 77
Theorem sequence:
line_addition(1,EB,BN)
parallel_judgment_ipsilateral_internal_angle(1,NB,DF)
parallel_judgment_ipsilateral_internal_angle(1,DN,FB)
parallelogram_judgment_parallel_and_parallel(1,BNDF)
parallelogram_area_formula_sine(1,DFBN)
parallelogram_property_opposite_line_equal(1,BNDF)
adjacent_complementary_angle(1,ANB,BND)
parallel_judgment_ipsilateral_internal_angle(1,NM,EG)
parallel_judgment_ipsilateral_internal_angle(1,MG,NE)
parallelogram_judgment_parallel_and_parallel(1,GMNE)
parallelogram_area_formula_sine(1,NEGM)
parallel_judgment_ipsilateral_internal_angle(1,CH,AM)
parallel_judgment_ipsilateral_internal_angle(1,HG,CA)
parallel_property_collinear_extend(2,HG,CA,M)
parallelogram_judgment_parallel_and_parallel(1,CAMH)
parallelogram_area_formula_sine(1,HCAM)
Problem: As shown in the diagram, AD=5, quadrilateral ADBC is a square. Find the circumference of the circle E.
问题：如图所示，AD=5，四边形ADBC是正方形。求⊙E的周长。
Construction (CDL):
Shape(AD,DB,BA)
Shape(AB,BC,CA)
Shape(DA,EAD)
Shape(BD,EDB)
Shape(CB,EBC)
Shape(AC,ECA)
Collinear(AEB)
Cocircular(E,ADBC)
Conditions (CDL):
Equal(LengthOfLine(AD),5)
Square(ADBC)
Goal: Value(PerimeterOfCircle(E))
Answer: 5*sqrt(2)*pi
Theorem sequence:
right_triangle_judgment_angle(1,ADB)
right_triangle_property_pythagorean(1,ADB)
diameter_of_circle_judgment_right_angle(1,ADB,E)
diameter_of_circle_property_length_equal(1,AB,E)
circle_property_length_of_radius_and_diameter(1,E)
circle_perimeter_formula(1,E)
Problem: As shown in the diagram, AC=21, AD=6*y+1/2*x, CB=2*x+4*y, DB=3*x+3*y, quadrilateral CADB is a parallelogram. Find the value of y.
问题：如图所示，AC=21，AD=6*y+1/2*x，CB=2*x+4*y，DB=3*x+3*y，CADB是平行四边形。求y的值。
Construction (CDL):
Shape(BC,CA,AD,DB)
Conditions (CDL):
Equal(LengthOfLine(AC),21)
Equal(LengthOfLine(AD),6*y+1/2*x)
Equal(LengthOfLine(CB),2*x+4*y)
Equal(LengthOfLine(DB),3*x+3*y)
Parallelogram(CADB)
Goal: Value(y)
Answer: 3
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BCAD)
parallelogram_property_opposite_line_equal(1,CADB)
Problem: As shown in the diagram, ∠PZW=x-9°, ∠WXP=x-11°, ZW is parallel to YX, WX⊥WX, quadrilateral ZWXY is a rectangle. Find the measure of ∠ZXY.
问题：如图所示，∠PZW=x-9°，∠WXP=x-11°，ZW∥YX，WX⊥WX，ZWXY是矩形。求∠ZXY的大小。
Construction (CDL):
Shape(ZW,WP,PZ)
Shape(PW,WX,XP)
Shape(PX,XY,YP)
Shape(ZP,PY,YZ)
Collinear(WPY)
Collinear(ZPX)
Conditions (CDL):
Equal(MeasureOfAngle(PZW),x-9)
Equal(MeasureOfAngle(WXP),x-11)
ParallelBetweenLine(ZW,YX)
PerpendicularBetweenLine(WX,WZ)
Rectangle(ZWXY)
Goal: Value(MeasureOfAngle(ZXY))
Answer: 46
Theorem sequence:
triangle_property_angle_sum(1,ZWX)
parallel_property_alternate_interior_angle(2,WZ,XY)
Problem: As shown in the diagram, KN=9, LN=16, PM=Mul(LengthOfLine(KP),2), LK∥RP, KN⊥MN, RM⊥PM. Find the length of line PR.
问题：如图所示，KN=9，LN=16，PM=Mul(LengthOfLine(KP),2)，LK平行于RP，KN⊥MN，RM垂直于PM。求直线PR的长度。
Construction (CDL):
Shape(LR,RQ,QN,NL)
Shape(RM,MQ,QR)
Shape(NQ,QP,PK,KN)
Shape(QM,MP,PQ)
Collinear(LRM)
Collinear(KPM)
Collinear(LNK)
Collinear(RQP)
Collinear(NQM)
Conditions (CDL):
Equal(LengthOfLine(KN),9)
Equal(LengthOfLine(LN),16)
Equal(LengthOfLine(PM),Mul(LengthOfLine(KP),2))
ParallelBetweenLine(LK,RP)
PerpendicularBetweenLine(KN,MN)
PerpendicularBetweenLine(RM,PM)
Goal: Value(LengthOfLine(PR))
Answer: 50/3
Theorem sequence:
parallel_property_corresponding_angle(2,LK,RP,M)
similar_triangle_judgment_aa(1,KLM,PRM)
similar_triangle_property_line_ratio(1,LMK,RMP)
similar_triangle_property_line_ratio(1,MKL,MPR)
line_addition(1,KP,PM)
line_addition(1,LN,NK)
Problem: As shown in the diagram, AB=18, AC=x, AD=z, BD=y, ∠BAC=45°, ∠BDA=60°, AB⊥DB, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=18，AC=x，AD=z，BD=y，∠BAC=45°，∠BDA=60°，AB⊥DB，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),z)
Equal(LengthOfLine(BD),y)
Equal(MeasureOfAngle(BAC),45)
Equal(MeasureOfAngle(BDA),60)
PerpendicularBetweenLine(AB,DB)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 9*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,ACB)
Problem: As shown in the diagram, CD=33, CE=25, DE=x, DE is perpendicular to CE. Find the value of x.
问题：如图所示，CD=33，CE=25，DE=x，DE垂直于CE。求x的值。
Construction (CDL):
Shape(DA,AE,ED)
Shape(AB,BE,EA)
Shape(BC,CE,EB)
Shape(CD,DE,EC)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(CD),33)
Equal(LengthOfLine(CE),25)
Equal(LengthOfLine(DE),x)
PerpendicularBetweenLine(DE,CE)
Goal: Value(x)
Answer: 4*sqrt(29)
Theorem sequence:
right_triangle_judgment_angle(1,DEC)
right_triangle_property_pythagorean(1,DEC)
Problem: As shown in the diagram, WX=7, ∠BZC=41°, ∠XZD=41°, XW⊥DW, XY⊥ZY. Find the length of line XY.
问题：如图所示，WX=7，∠BZC=41°，∠XZD=41°，XW⊥DW，XY⊥ZY。求直线XY的长度。
Construction (CDL):
Shape(XY,YZ,ZX)
Shape(XZ,ZW,WX)
Collinear(BYZ)
Collinear(CXZ)
Collinear(DWZ)
Conditions (CDL):
Equal(LengthOfLine(WX),7)
Equal(MeasureOfAngle(BZC),41)
Equal(MeasureOfAngle(XZD),41)
PerpendicularBetweenLine(XW,DW)
PerpendicularBetweenLine(XY,ZY)
Goal: Value(LengthOfLine(XY))
Answer: 7
Theorem sequence:
adjacent_complementary_angle(1,ZWX,XWD)
sine_theorem(1,XZW)
sine_theorem(1,XYZ)
Problem: As shown in the diagram, AZ=y, QZ=z, RQ=12, RS=10, RZ=x, ∠AQR=30°, ∠SPA=45°, SR∥AZ, PA is perpendicular to SA, RZ⊥QZ. Find the value of z.
问题：如图所示，AZ=y，QZ=z，RQ=12，RS=10，RZ=x，∠AQR=30°，∠SPA=45°，SR∥AZ，PA⊥SA，RZ垂直于QZ。求z的值。
Construction (CDL):
Shape(SP,PA,AS)
Shape(SA,AZ,ZR,RS)
Shape(ZQ,QR,RZ)
Collinear(PAZQ)
Conditions (CDL):
Equal(LengthOfLine(AZ),y)
Equal(LengthOfLine(QZ),z)
Equal(LengthOfLine(RQ),12)
Equal(LengthOfLine(RS),10)
Equal(LengthOfLine(RZ),x)
Equal(MeasureOfAngle(AQR),30)
Equal(MeasureOfAngle(SPA),45)
ParallelBetweenLine(SR,AZ)
PerpendicularBetweenLine(PA,SA)
PerpendicularBetweenLine(RZ,QZ)
Goal: Value(z)
Answer: 6*sqrt(3)
Theorem sequence:
cosine_theorem(1,ZQR)
cosine_theorem(1,QRZ)