SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=20, BC=x, CA=10, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=20，BC=x，CA=10，BC垂直于AC。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CA),10)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 10*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠LPM=65°, P is the center of ⊙P, LP is perpendicular to NP, NP is perpendicular to JP. Find the measure of ⌒PKN.
问题：如图所示，∠LPM=65°，P是⊙P的圆心，LP垂直于NP，NP垂直于JP。求⌒PKN的角度。
Construction (CDL):
Shape(PK,PKJ,JP)
Shape(PJ,PJN,NP)
Shape(PN,PNM,MP)
Shape(PM,PML,LP)
Shape(PL,PLK,KP)
Collinear(JPL)
Collinear(KPM)
Cocircular(P,KJNML)
Conditions (CDL):
Equal(MeasureOfAngle(LPM),65)
IsCentreOfCircle(P,P)
PerpendicularBetweenLine(LP,NP)
PerpendicularBetweenLine(NP,JP)
Goal: Value(MeasureOfArc(PKN))
Answer: 155
Theorem sequence:
vertical_angle(1,JPK,LPM)
angle_addition(1,NPJ,JPK)
arc_property_center_angle(1,PKN,P)
Problem: As shown in the diagram, CG=13, FC=12, the incenter of △BAC is G, CF is perpendicular to GF, GD⊥CD, GE⊥AE. Find the length of line EG.
问题：如图所示，CG=13，FC=12，三角形BAC的内心为G，CF⊥GF，GD⊥CD，GE⊥AE。求直线EG的长度。
Construction (CDL):
Shape(BE,EG,GB)
Shape(EA,AG,GE)
Shape(AD,DG,GA)
Shape(GD,DC,CG)
Shape(GC,CF,FG)
Shape(BG,GF,FB)
Collinear(BEA)
Collinear(ADC)
Collinear(CFB)
Conditions (CDL):
Equal(LengthOfLine(CG),13)
Equal(LengthOfLine(FC),12)
IsIncenterOfTriangle(G,BAC)
PerpendicularBetweenLine(CF,GF)
PerpendicularBetweenLine(GD,CD)
PerpendicularBetweenLine(GE,AE)
Goal: Value(LengthOfLine(EG))
Answer: 5
Theorem sequence:
adjacent_complementary_angle(1,BEG,GEA)
adjacent_complementary_angle(1,CFG,GFB)
mirror_congruent_triangle_judgment_aas(3,GFB,GBE)
mirror_congruent_triangle_property_line_equal(1,BGF,BEG)
right_triangle_judgment_angle(1,CFG)
right_triangle_property_pythagorean(1,CFG)
Problem: As shown in the diagram, GH=KJ, ∠JLG=83°, ∠KLH=2*x-27°, L is the center of circle L. Find the value of x.
问题：如图所示，GH=KJ，∠JLG=83°，∠KLH=2*x-27°，圆L的圆心为L。求x的值。
Construction (CDL):
Shape(LKG,GL,LK)
Shape(LG,GC,CL)
Shape(CG,LGJ,JC)
Shape(LC,CB,BD,DL)
Shape(BC,CJ,JB)
Shape(BJ,LJH,HB)
Shape(DB,BH,HD)
Shape(KL,LD,DK)
Shape(KD,DH,LHK)
Collinear(KDBJ)
Collinear(LCJ)
Collinear(LDH)
Collinear(GCBH)
Cocircular(L,KGJH)
Conditions (CDL):
Equal(LengthOfLine(GH),LengthOfLine(KJ))
Equal(MeasureOfAngle(JLG),83)
Equal(MeasureOfAngle(KLH),2*x-27)
IsCentreOfCircle(L,L)
Goal: Value(x)
Answer: 55
Theorem sequence:
congruent_arc_judgment_chord_equal(1,LGH,LJK)
congruent_arc_property_length_equal(1,LGH,LJK)
arc_addition_length(1,LGJ,LJH)
arc_addition_length(1,LJH,LHK)
congruent_arc_judgment_length_equal(1,LHK,LGJ)
congruent_arc_property_measure_equal(1,LHK,LGJ)
arc_property_center_angle(1,LGJ,L)
arc_property_center_angle(1,LHK,L)
Problem: As shown in the diagram, ∠TYZ=52°, ∠ZYX=38°, TY⊥XY, TZ⊥WZ, XW is perpendicular to TW, YX⊥WX. Find the measure of ∠YWT.
问题：如图所示，∠TYZ=52°，∠ZYX=38°，TY垂直于XY，TZ垂直于WZ，XW垂直于TW，YX⊥WX。求∠YWT的大小。
Construction (CDL):
Shape(TY,YZ,ZT)
Shape(TZ,ZW,WT)
Shape(WZ,ZY,YX,XW)
Collinear(YZW)
Conditions (CDL):
Equal(MeasureOfAngle(TYZ),52)
Equal(MeasureOfAngle(ZYX),38)
PerpendicularBetweenLine(TY,XY)
PerpendicularBetweenLine(TZ,WZ)
PerpendicularBetweenLine(XW,TW)
PerpendicularBetweenLine(YX,WX)
Goal: Value(MeasureOfAngle(YWT))
Answer: 38
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,XY,WT)
parallel_property_alternate_interior_angle(2,XY,WT)
Problem: As shown in the diagram, AB=6, AF=8, BC=x, CD=y, DE=2*y-3, FE=x+10/3, BF∥CD, CB is parallel to DF. Find the length of line BC.
问题：如图所示，AB=6，AF=8，BC=x，CD=y，DE=2*y-3，FE=x+10/3，BF平行于CD，CB∥DF。求直线BC的长度。
Construction (CDL):
Shape(AB,BF,FA)
Shape(BC,CD,DF,FB)
Shape(FD,DE,EF)
Collinear(ABC)
Collinear(CDE)
Collinear(EFA)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AF),8)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CD),y)
Equal(LengthOfLine(DE),2*y-3)
Equal(LengthOfLine(FE),x+10/3)
ParallelBetweenLine(BF,CD)
ParallelBetweenLine(CB,DF)
Goal: Value(LengthOfLine(BC))
Answer: 10
Theorem sequence:
line_addition(1,AB,BC)
line_addition(1,EF,FA)
parallel_property_corresponding_angle(1,BF,CD,A)
similar_triangle_judgment_aa(1,FAB,EAC)
similar_triangle_property_line_ratio(1,FAB,EAC)
similar_triangle_property_line_ratio(1,BFA,CEA)
Problem: As shown in the diagram, ∠ADC=2*y+16°, ∠BAD=58°, ∠BCD=2*x°, ∠CBA=3*y+4°, O is the center of circle O. Find the measure of ∠ADC.
问题：如图所示，∠ADC=2*y+16°，∠BAD=58°，∠BCD=2*x°，∠CBA=3*y+4°，O是圆O的圆心。求∠ADC的大小。
Construction (CDL):
Shape(OBA,AB)
Shape(OAD,DA)
Shape(ODC,CD)
Shape(OCB,BC)
Shape(BA,AO,OC,CB)
Shape(OA,AD,DC,CO)
Cocircular(O,BADC)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),2*y+16)
Equal(MeasureOfAngle(BAD),58)
Equal(MeasureOfAngle(BCD),2*x)
Equal(MeasureOfAngle(CBA),3*y+4)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(ADC))
Answer: 80
Theorem sequence:
arc_property_circumference_angle_external(1,OCA,D)
arc_property_circumference_angle_internal(1,OCA,B)
Problem: As shown in the diagram, FH=8, FH=FL, FJ=17, the center of ⊙F is F, JH⊥FH. Find the length of line JG.
问题：如图所示，FH=8，FH=FL，FJ=17，F是圆F的圆心，JH⊥FH。求直线JG的长度。
Construction (CDL):
Shape(JH,HF,FJ)
Shape(FH,HG,GF)
Shape(FJG,GJ)
Shape(FG,FGJ,JF)
Collinear(JHG)
Cocircular(F,JG)
Conditions (CDL):
Equal(LengthOfLine(FH),8)
Equal(LengthOfLine(FH),LengthOfLine(FL))
Equal(LengthOfLine(FJ),17)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(JH,FH)
Goal: Value(LengthOfLine(JG))
Answer: 30
Theorem sequence:
adjacent_complementary_angle(1,JHF,FHG)
right_triangle_judgment_angle(1,JHF)
right_triangle_judgment_angle(1,FHG)
radius_of_circle_property_length_equal(1,FJ,F)
radius_of_circle_property_length_equal(1,FG,F)
right_triangle_property_pythagorean(1,JHF)
right_triangle_property_pythagorean(1,FHG)
line_addition(1,JH,HG)
Problem: As shown in the diagram, AD=10, BC=15, BD=CD, EG=x, FG=HG, FH=12, ∠ABC=∠GHE, ∠EFG=∠BCA. Find the value of x.
问题：如图所示，AD=10，BC=15，BD=CD，EG=x，FG=HG，FH=12，∠ABC=∠GHE，∠EFG=∠BCA。求x的值。
Construction (CDL):
Shape(EF,FG,GE)
Shape(EG,GH,HE)
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(FGH)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(BD),LengthOfLine(CD))
Equal(LengthOfLine(EG),x)
Equal(LengthOfLine(FG),LengthOfLine(HG))
Equal(LengthOfLine(FH),12)
Equal(MeasureOfAngle(ABC),MeasureOfAngle(GHE))
Equal(MeasureOfAngle(EFG),MeasureOfAngle(BCA))
Goal: Value(x)
Answer: 8
Theorem sequence:
line_addition(1,FG,GH)
line_addition(1,BD,DC)
mirror_similar_triangle_judgment_aa(1,EFH,ABC)
mirror_similar_triangle_property_line_ratio(1,EFH,ABC)
mirror_similar_triangle_property_line_ratio(1,FHE,CAB)
mirror_similar_triangle_judgment_sas(1,HEG,BDA)
mirror_similar_triangle_property_line_ratio(1,HEG,BDA)
mirror_similar_triangle_property_line_ratio(1,EGH,ABD)
Problem: As shown in the diagram, DF=x, DG=15, FG=19, ∠DGF=125°. Find the value of x.
问题：如图所示，DF=x，DG=15，FG=19，∠DGF=125°。求x的值。
Construction (CDL):
Shape(DG,GF,FD)
Conditions (CDL):
Equal(LengthOfLine(DF),x)
Equal(LengthOfLine(DG),15)
Equal(LengthOfLine(FG),19)
Equal(MeasureOfAngle(DGF),125)
Goal: Value(x)
Answer: sqrt(570*cos(11*pi/36)+586)
Theorem sequence:
cosine_theorem(1,GFD)
Problem: As shown in the diagram, PQ=25, RS=10, RT=x, SQ=5, ST∥QP, TR⊥SR. Find the length of line PT.
问题：如图所示，PQ=25，RS=10，RT=x，SQ=5，ST平行于QP，TR垂直于SR。求直线PT的长度。
Construction (CDL):
Shape(PT,TS,SQ,QP)
Shape(TR,RS,ST)
Collinear(PTR)
Collinear(RSQ)
Conditions (CDL):
Equal(LengthOfLine(PQ),25)
Equal(LengthOfLine(RS),10)
Equal(LengthOfLine(RT),x)
Equal(LengthOfLine(SQ),5)
ParallelBetweenLine(ST,QP)
PerpendicularBetweenLine(TR,SR)
Goal: Value(LengthOfLine(PT))
Answer: 20/3
Theorem sequence:
line_addition(1,PT,TR)
line_addition(1,RS,SQ)
parallel_property_corresponding_angle(1,ST,QP,R)
parallel_property_corresponding_angle(2,PQ,TS,R)
similar_triangle_judgment_aa(1,RQP,RST)
right_triangle_judgment_angle(1,PRQ)
right_triangle_property_pythagorean(1,PRQ)
similar_triangle_property_line_ratio(1,PRQ,TRS)
similar_triangle_property_line_ratio(1,QPR,STR)
Problem: As shown in the diagram, BC=8, CE=14, the center of ⊙B is B, CD⊥BD. Find the length of line BD.
问题：如图所示，BC=8，CE=14，B是圆B的圆心，CD⊥BD。求直线BD的长度。
Construction (CDL):
Shape(CB,BA,BAC)
Shape(BC,CD,DB)
Shape(DC,BCF,FD)
Shape(ED,DF,BFE)
Shape(AB,BD,DE,BEA)
Collinear(CDE)
Collinear(ABDF)
Cocircular(B,ACFE)
Conditions (CDL):
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(CE),14)
IsCentreOfCircle(B,B)
PerpendicularBetweenLine(CD,BD)
Goal: Value(LengthOfLine(BD))
Answer: sqrt(15)
Theorem sequence:
right_triangle_judgment_angle(1,CDB)
circle_property_chord_perpendicular_bisect_chord(1,B,BD,CE)
line_addition(1,CD,DE)
right_triangle_property_pythagorean(1,CDB)
Problem: As shown in the diagram, KP=3, XJ=8, P is the centroid of triangle XYZ. Find the length of line YJ.
问题：如图所示，KP=3，XJ=8，△XYZ的重心为O。求直线YJ的长度。
Construction (CDL):
Shape(XJ,JP,PX)
Shape(JY,YP,PJ)
Shape(PY,YK,KP)
Shape(PK,KZ,ZP)
Shape(PZ,ZL,LP)
Shape(PL,LX,XP)
Collinear(XJY)
Collinear(ZLX)
Collinear(YKZ)
Collinear(XPK)
Collinear(YPL)
Collinear(JPZ)
Conditions (CDL):
Equal(LengthOfLine(KP),3)
Equal(LengthOfLine(XJ),8)
IsCentroidOfTriangle(P,XYZ)
Goal: Value(LengthOfLine(YJ))
Answer: 8
Theorem sequence:
centroid_of_triangle_property_intersection(1,P,ZXY,J)
Problem: As shown in the diagram, QP=17, RP=8, RQ=15, QR is perpendicular to PR. Find the value of cos(RP).
问题：如图所示，QP=17，RP=8，RQ=15，QR垂直于PR。求cos(RP)的值。
Construction (CDL):
Shape(QR,RP,PQ)
Conditions (CDL):
Equal(LengthOfLine(QP),17)
Equal(LengthOfLine(RP),8)
Equal(LengthOfLine(RQ),15)
PerpendicularBetweenLine(QR,PR)
Goal: Value(Cos(MeasureOfAngle(RPQ)))
Answer: 8/17
Theorem sequence:
cosine_theorem(1,PQR)
Problem: As shown in the diagram, AC=16, CD=23, DB and CA are opposite sides of the ▱ BDCA. Find the perimeter of BDCA.
问题：如图所示，AC=16，CD=23，四边形BDCA是▱。求BDCA的周长。
Construction (CDL):
Shape(BD,DC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),16)
Equal(LengthOfLine(CD),23)
Parallelogram(BDCA)
Goal: Value(PerimeterOfQuadrilateral(BDCA))
Answer: 78
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BDCA)
parallelogram_property_opposite_line_equal(1,DCAB)
quadrilateral_perimeter_formula(1,BDCA)
Problem: As shown in the diagram, AC=y, AD=3, BC=z, CD=x, DB=12, AC⊥BC, CD is perpendicular to AD. Find the value of z.
问题：如图所示，AC=y，AD=3，BC=z，CD=x，DB=12，AC⊥BC，CD⊥AD。求z的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(DC,CB,BD)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AD),3)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(CD),x)
Equal(LengthOfLine(DB),12)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(CD,AD)
Goal: Value(z)
Answer: 6*sqrt(5)
Theorem sequence:
line_addition(1,AD,DB)
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, ML=6, ∠LKJ=109°, JK and ML are opposite sides of the ▱ JMLK. Find the measure of ∠JML.
问题：如图所示，ML=6，∠LKJ=109°，JMLK是平行四边形。求∠JML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(JML))
Answer: 109
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,MLKJ)
Problem: As shown in the diagram, ∠JML=105°, ∠KJM=75°, ∠MLK=2*x-1°, KL and JM are opposite sides of the parallelogram KJML. Find the value of x.
问题：如图所示，∠JML=105°，∠KJM=75°，∠MLK=2*x-1°，KL和JM是平行四边形KJML的一组对边。求x的值。
Construction (CDL):
Shape(JM,ML,LK,KJ)
Conditions (CDL):
Equal(MeasureOfAngle(JML),105)
Equal(MeasureOfAngle(KJM),75)
Equal(MeasureOfAngle(MLK),2*x-1)
Parallelogram(KJML)
Goal: Value(x)
Answer: 38
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,JMLK)
Problem: As shown in the diagram, DE=3, EC=8, FC=3, the radius of circle A is 1, the radius of circle B is 1, x=Add(AreaOfSector(A, y=Add(x, the center of ⊙A is A, the center of ⊙B is B, GH is perpendicular to CH, DECF is a rectangle. Find the value of y.
问题：如图所示，DE=3，EC=8，FC=3，圆A的半径为1，圆B的半径为1，x=Add(AreaOfSector(A，y=Add(x，A是⊙A的圆心，圆B的圆心为B，GH垂直于CH，DECF是矩形。求y的值。
Construction (CDL):
Shape(AD,ADE,EA)
Shape(DA,AE,EC,CB,BF,FD)
Shape(FB,BC,BCF)
Collinear(DAE)
Collinear(FBC)
Cocircular(A,DE)
Cocircular(B,FC)
Conditions (CDL):
Equal(LengthOfLine(DE),3)
Equal(LengthOfLine(EC),8)
Equal(LengthOfLine(FC),3)
Equal(RadiusOfCircle(A),1)
Equal(RadiusOfCircle(B),1)
Equal(x,Add(AreaOfSector(A,DE),AreaOfSector(B,CF)))
Equal(y,Add(x,AreaOfQuadrilateral(DECF)))
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
PerpendicularBetweenLine(GH,CH)
Rectangle(DECF)
Goal: Value(y)
Answer: pi+24
Theorem sequence:
flat_angle(1,EAD)
flat_angle(1,FBC)
arc_property_center_angle(1,ADE,A)
arc_property_center_angle(1,BCF,B)
sector_area_formula(1,ADE)
sector_area_formula(1,BCF)
parallelogram_area_formula_sine(1,DECF)
Problem: As shown in the diagram, ED=30, W is the center of circle W, WF⊥DF. Find the length of line DF.
问题：如图所示，ED=30，W是圆W的圆心，WF⊥DF。求直线DF的长度。
Construction (CDL):
Shape(WED,DE)
Shape(WF,FD,WDE,EF,FW)
Collinear(EFD)
Cocircular(W,ED)
Conditions (CDL):
Equal(LengthOfLine(ED),30)
IsCentreOfCircle(W,W)
PerpendicularBetweenLine(WF,DF)
Goal: Value(LengthOfLine(DF))
Answer: 15
Theorem sequence:
adjacent_complementary_angle(1,EFW,WFD)
circle_property_chord_perpendicular_bisect_chord(1,W,WF,ED)
line_addition(1,EF,FD)
Problem: As shown in the diagram, AB=5, ∠BCA=59°, CA is perpendicular to BA. Find the area of triangle ABC.
问题：如图所示，AB=5，∠BCA=59°，CA⊥BA。求△ABC的面积。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(MeasureOfAngle(BCA),59)
PerpendicularBetweenLine(CA,BA)
Goal: Value(AreaOfTriangle(ABC))
Answer: 25*tan(31*pi/180)/2
Theorem sequence:
triangle_property_angle_sum(1,ABC)
triangle_area_formula_sine(1,ABC)
triangle_area_formula_sine(1,BCA)
triangle_area_formula_sine(1,CAB)
Problem: As shown in the diagram, △WYX is congruent to △WXZ, YX=19, ZX=3*y+7, ∠YXW=20*y+10°, WX⊥ZX. Find the value of y.
问题：如图所示，△WYX与△WXZ是全等三角形，YX=19，ZX=3*y+7，∠YXW=20*y+10°，WX⊥ZX。求y的值。
Construction (CDL):
Shape(WY,YX,XW)
Shape(WX,XZ,ZW)
Collinear(YXZ)
Conditions (CDL):
CongruentBetweenTriangle(WYX,WXZ)
Equal(LengthOfLine(YX),19)
Equal(LengthOfLine(ZX),3*y+7)
Equal(MeasureOfAngle(YXW),20*y+10)
PerpendicularBetweenLine(WX,ZX)
Goal: Value(y)
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,YXW,WXZ)
Problem: As shown in the diagram, JL=9*x-5, KJ=6*x+7, KN=LN, JN⊥LN. Find the length of line JK.
问题：如图所示，JL=9*x-5，KJ=6*x+7，KN=LN，JN垂直于LN。求直线JK的长度。
Construction (CDL):
Shape(JK,KN,NJ)
Shape(JN,NL,LJ)
Collinear(KNL)
Conditions (CDL):
Equal(LengthOfLine(JL),9*x-5)
Equal(LengthOfLine(KJ),6*x+7)
Equal(LengthOfLine(KN),LengthOfLine(LN))
PerpendicularBetweenLine(JN,LN)
Goal: Value(LengthOfLine(JK))
Answer: 31
Theorem sequence:
adjacent_complementary_angle(1,KNJ,JNL)
mirror_congruent_triangle_judgment_sas(1,NJK,NLJ)
mirror_congruent_triangle_property_line_equal(1,NJK,NLJ)
Problem: As shown in the diagram, MN=39, MP=36, NP=15, MP is perpendicular to NP. Find the value of sin(PN).
问题：如图所示，MN=39，MP=36，NP=15，MP⊥NP。求sin(PN)的值。
Construction (CDL):
Shape(PN,NM,MP)
Conditions (CDL):
Equal(LengthOfLine(MN),39)
Equal(LengthOfLine(MP),36)
Equal(LengthOfLine(NP),15)
PerpendicularBetweenLine(MP,NP)
Goal: Value(Sin(MeasureOfAngle(PNM)))
Answer: 12/13
Theorem sequence:
sine_theorem(1,MPN)
Problem: As shown in the diagram, AB=12, AC=5, BA is perpendicular to CA. Find the length of line CB.
问题：如图所示，AB=12，AC=5，BA⊥CA。求直线CB的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),5)
PerpendicularBetweenLine(BA,CA)
Goal: Value(LengthOfLine(CB))
Answer: 13
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, AM=10, CB=6, MB=y, the center of ⊙P is P, the tangent to ⊙P is MA. Find the value of y.
问题：如图所示，AM=10，CB=6，MB=y，P是⊙P的圆心，圆O的切线为MA。求y的值。
Construction (CDL):
Shape(AM,MB,PAB)
Shape(CB,PBC)
Shape(BC,PCB)
Collinear(MBC)
Cocircular(P,ABC)
Conditions (CDL):
Equal(LengthOfLine(AM),10)
Equal(LengthOfLine(CB),6)
Equal(LengthOfLine(MB),y)
IsCentreOfCircle(P,P)
IsTangentOfCircle(MA,P)
Goal: Value(y)
Answer: -3+sqrt(109)
Theorem sequence:
line_addition(1,MB,BC)
circle_property_circular_power_tangent_and_segment_line(1,MA,MBC,P)
Problem: As shown in the diagram, ∠BAC=2*x+1°, ∠BDC=3*y+10°, ∠CAD=5*x+5°, ∠DBA=y*y°, ADCB is a rectangle. Find the value of x.
问题：如图所示，∠BAC=2*x+1°，∠BDC=3*y+10°，∠CAD=5*x+5°，∠DBA=y*y°，ADCB是长方形。求x的值。
Construction (CDL):
Shape(AD,DE,EA)
Shape(BA,AE,EB)
Shape(ED,DC,CE)
Shape(BE,EC,CB)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),2*x+1)
Equal(MeasureOfAngle(BDC),3*y+10)
Equal(MeasureOfAngle(CAD),5*x+5)
Equal(MeasureOfAngle(DBA),y*y)
Rectangle(ADCB)
Goal: Value(x)
Answer: 12
Theorem sequence:
angle_addition(1,BAE,EAD)
Problem: As shown in the diagram, LJ=ML, ∠LKM=37°, KM is perpendicular to LM, LJ is perpendicular to KJ. Find the measure of ∠JKL.
问题：如图所示，LJ=ML，∠LKM=37°，KM⊥LM，LJ垂直于KJ。求∠JKL的大小。
Construction (CDL):
Shape(JK,KL,LJ)
Shape(KM,ML,LK)
Conditions (CDL):
Equal(LengthOfLine(LJ),LengthOfLine(ML))
Equal(MeasureOfAngle(LKM),37)
PerpendicularBetweenLine(KM,LM)
PerpendicularBetweenLine(LJ,KJ)
Goal: Value(MeasureOfAngle(JKL))
Answer: 37
Theorem sequence:
mirror_congruent_triangle_judgment_hl(2,LJK,LKM)
mirror_congruent_triangle_property_angle_equal(1,KLJ,KML)
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 y.
问题：如图所示，JM=5，KM=20，LJ=y，LK=z，LM=x，JL垂直于KL，KM垂直于LM。求y的值。
Construction (CDL):
Shape(KM,ML,LK)
Shape(MJ,JL,LM)
Collinear(KMJ)
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(y)
Answer: 5*sqrt(5)
Theorem sequence:
line_addition(1,KM,MJ)
adjacent_complementary_angle(1,KML,LMJ)
right_triangle_judgment_angle(1,KML)
right_triangle_judgment_angle(1,LMJ)
right_triangle_judgment_angle(1,JLK)
right_triangle_property_pythagorean(1,KML)
right_triangle_property_pythagorean(1,LMJ)
right_triangle_property_pythagorean(1,JLK)
Problem: As shown in the diagram, ∠WZX=24°, ∠XWZ=23°, ∠XZY=y°, ∠YXZ=105°, ∠ZXW=z°, ∠ZYX=∠XZY, ∠ZYX=x°. Find the value of x.
问题：如图所示，∠WZX=24°，∠XWZ=23°，∠XZY=y°，∠YXZ=105°，∠ZXW=z°，∠ZYX=∠XZY，∠ZYX=x°。求x的值。
Construction (CDL):
Shape(YX,XZ,ZY)
Shape(XW,WZ,ZX)
Conditions (CDL):
Equal(MeasureOfAngle(WZX),24)
Equal(MeasureOfAngle(XWZ),23)
Equal(MeasureOfAngle(XZY),y)
Equal(MeasureOfAngle(YXZ),105)
Equal(MeasureOfAngle(ZXW),z)
Equal(MeasureOfAngle(ZYX),MeasureOfAngle(XZY))
Equal(MeasureOfAngle(ZYX),x)
Goal: Value(x)
Answer: 75/2
Theorem sequence:
triangle_property_angle_sum(1,YXZ)
Problem: As shown in the diagram, AB=z, AC=y, AD=36, BD=6*x, CD=x, BD⊥AD, CA⊥BA. Find the value of x.
问题：如图所示，AB=z，AC=y，AD=36，BD=6*x，CD=x，BD垂直于AD，CA垂直于BA。求x的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AD),36)
Equal(LengthOfLine(BD),6*x)
Equal(LengthOfLine(CD),x)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 6*sqrt(6)
Theorem sequence:
line_addition(1,BD,DC)
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,CAB)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,ADC)
right_triangle_property_pythagorean(1,CAB)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,ADC)
Problem: As shown in the diagram, AC=b, BA=c, BC=a, ∠ABC=60°, ∠CAB=30°, c=5, BC⊥AC. Find the value of b.
问题：如图所示，AC=b，BA=c，BC=a，∠ABC=60°，∠CAB=30°，c=5，BC⊥AC。求b的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(BA),c)
Equal(LengthOfLine(BC),a)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),30)
Equal(c,5)
PerpendicularBetweenLine(BC,AC)
Goal: Value(b)
Answer: 5*sqrt(3)/2
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, ∠AEN=64°, IL∥RN, WJ is parallel to AK. Find the measure of ∠EZD.
问题：如图所示，∠AEN=64°，IL平行于RN，WJ∥AK。求∠EZD的大小。
Construction (CDL):
Shape(AE,EN)
Shape(EZ,ZD)
Collinear(IDZQL)
Collinear(RHEN)
Collinear(WDHJ)
Collinear(AZEK)
Conditions (CDL):
Equal(MeasureOfAngle(AEN),64)
ParallelBetweenLine(IL,RN)
ParallelBetweenLine(WJ,AK)
Goal: Value(MeasureOfAngle(EZD))
Answer: 64
Theorem sequence:
parallel_property_collinear_extend(3,IL,RN,Z)
parallel_property_collinear_extend(3,NR,ZI,E)
parallel_property_alternate_interior_angle(2,NE,ZI)
Problem: As shown in the diagram, BC=10, BD=14, DC=x, BC is perpendicular to DC. Find the value of x.
问题：如图所示，BC=10，BD=14，DC=x，BC垂直于DC。求x的值。
Construction (CDL):
Shape(DA,AB,BD)
Shape(DB,BC,CD)
Conditions (CDL):
Equal(LengthOfLine(BC),10)
Equal(LengthOfLine(BD),14)
Equal(LengthOfLine(DC),x)
PerpendicularBetweenLine(BC,DC)
Goal: Value(x)
Answer: 4*sqrt(6)
Theorem sequence:
right_triangle_judgment_angle(1,BCD)
right_triangle_property_pythagorean(1,BCD)
Problem: As shown in the diagram, CB=26, CE=24, DB=22, CBDA is a parallelogram, BE⊥CE. Find the area of quadrilateral CBDA.
问题：如图所示，CB=26，CE=24，DB=22，CA和BD是▱CBDA的一组对边，BE垂直于CE。求四边形CBDA的面积。
Construction (CDL):
Shape(CB,BE,EC)
Shape(CE,ED,DA,AC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(CB),26)
Equal(LengthOfLine(CE),24)
Equal(LengthOfLine(DB),22)
Parallelogram(CBDA)
PerpendicularBetweenLine(BE,CE)
Goal: Value(AreaOfQuadrilateral(CBDA))
Answer: 528
Theorem sequence:
altitude_of_quadrilateral_judgment_left_vertex(1,CE,CBDA)
parallelogram_area_formula_common(1,CBDA)
Problem: As shown in the diagram, ∠AOF=∠FOE, ∠EOD=∠DOB, ∠FOE=∠EOD. Find the measure of ∠AOF.
问题：如图所示，∠AOF=∠FOE，∠EOD=∠DOB，∠FOE=∠EOD。求∠AOF的大小。
Construction (CDL):
Shape(OFA,AO,OF)
Shape(OEF,FO,OE)
Shape(ODE,EO,OD)
Shape(OBD,DO,OB)
Collinear(AOB)
Cocircular(O,AFEDB)
Conditions (CDL):
Equal(MeasureOfAngle(AOF),MeasureOfAngle(FOE))
Equal(MeasureOfAngle(EOD),MeasureOfAngle(DOB))
Equal(MeasureOfAngle(FOE),MeasureOfAngle(EOD))
Goal: Value(MeasureOfAngle(AOF))
Answer: 45
Theorem sequence:
flat_angle(1,AOB)
angle_addition(1,AOF,FOB)
angle_addition(1,FOE,EOB)
angle_addition(1,EOD,DOB)
Problem: As shown in the diagram, JK=JL, JK=KL, JK=x+7, JL=4*x-8, LK=JL. Find the length of line KL.
问题：如图所示，JK=JL，JK=KL，JK=x+7，JL=4*x-8，LK=JL。求直线KL的长度。
Construction (CDL):
Shape(KJ,JL,LK)
Conditions (CDL):
Equal(LengthOfLine(JK),LengthOfLine(JL))
Equal(LengthOfLine(JK),LengthOfLine(KL))
Equal(LengthOfLine(JK),x+7)
Equal(LengthOfLine(JL),4*x-8)
Equal(LengthOfLine(LK),LengthOfLine(JL))
Goal: Value(LengthOfLine(KL))
Answer: 12
Theorem sequence:
Problem: As shown in the diagram, QR=x+3, QT=2*x-6, RS=7, TS=7, TS⊥QS. Find the length of line TQ.
问题：如图所示，QR=x+3，QT=2*x-6，RS=7，TS=7，TS垂直于QS。求直线TQ的长度。
Construction (CDL):
Shape(RQ,QS,SR)
Shape(SQ,QT,TS)
Collinear(RST)
Conditions (CDL):
Equal(LengthOfLine(QR),x+3)
Equal(LengthOfLine(QT),2*x-6)
Equal(LengthOfLine(RS),7)
Equal(LengthOfLine(TS),7)
PerpendicularBetweenLine(TS,QS)
Goal: Value(LengthOfLine(TQ))
Answer: 12
Theorem sequence:
adjacent_complementary_angle(1,TSQ,QSR)
mirror_congruent_triangle_judgment_sas(1,SRQ,SQT)
mirror_congruent_triangle_property_line_equal(1,SRQ,SQT)
Problem: As shown in the diagram, HF=21, HG=x+3, JH=x-8, KH=7, GJ is parallel to FK. Find the value of x.
问题：如图所示，HF=21，HG=x+3，JH=x-8，KH=7，GJ∥FK。求x的值。
Construction (CDL):
Shape(HG,GJ,JH)
Shape(GF,FK,KJ,JG)
Collinear(HGF)
Collinear(KJH)
Conditions (CDL):
Equal(LengthOfLine(HF),21)
Equal(LengthOfLine(HG),x+3)
Equal(LengthOfLine(JH),x-8)
Equal(LengthOfLine(KH),7)
ParallelBetweenLine(GJ,FK)
Goal: Value(x)
Answer: 27/2
Theorem sequence:
parallel_property_corresponding_angle(1,GJ,FK,H)
parallel_property_corresponding_angle(2,KF,JG,H)
similar_triangle_judgment_aa(1,HGJ,HFK)
similar_triangle_property_line_ratio(1,JHG,KHF)
similar_triangle_property_line_ratio(1,GJH,FKH)