SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, the area of triangle CED is 336, DF=24, DF is perpendicular to EF. Find the length of line CE.
问题：如图所示，三角形CED的面积为336，DF=24，DF⊥EF。求直线CE的长度。
Construction (CDL):
Shape(CF,FD,DC)
Shape(FE,ED,DF)
Collinear(CFE)
Conditions (CDL):
Equal(AreaOfTriangle(CED),336)
Equal(LengthOfLine(DF),24)
PerpendicularBetweenLine(DF,EF)
Goal: Value(LengthOfLine(CE))
Answer: 28
Theorem sequence:
adjacent_complementary_angle(1,CFD,DFE)
altitude_of_triangle_judgment(1,DF,DCE)
triangle_area_formula_common(1,DCE)
Problem: As shown in the diagram, AB=16, AD=8, BD=x, ∠ABD=y°, BD⊥AD. Find the value of y.
问题：如图所示，AB=16，AD=8，BD=x，∠ABD=y°，BD⊥AD。求y的值。
Construction (CDL):
Shape(CB,BA,AC)
Shape(AB,BD,DA)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BD),x)
Equal(MeasureOfAngle(ABD),y)
PerpendicularBetweenLine(BD,AD)
Goal: Value(y)
Answer: 30
Theorem sequence:
sine_theorem(1,ABD)
Problem: As shown in the diagram, FA=AE, HB=14, A is the center of circle A, the center of ⊙E is E, the center of circle F is F. Find the area of circle A minus the sum of the areas of circles F and E.
问题：如图所示，FA=AE，HB=14，⊙A的圆心为A，E是⊙E的圆心，F是⊙F的圆心。求圆A的面积减去圆F和E的面积和。
Construction (CDL):
Shape(BF,FA,FAB)
Shape(AF,FB,FBA)
Shape(AE,EH,EHA)
Shape(HE,EA,EAH)
Shape(EHA,AHB,FAB)
Shape(ABH,EAH,FBA)
Collinear(BFAEH)
Cocircular(F,AB)
Cocircular(E,AH)
Cocircular(A,BH)
Conditions (CDL):
Equal(LengthOfLine(FA),LengthOfLine(AE))
Equal(LengthOfLine(HB),14)
IsCentreOfCircle(A,A)
IsCentreOfCircle(E,E)
IsCentreOfCircle(F,F)
Goal: Value(Sub(AreaOfCircle(A),Add(AreaOfCircle(F),AreaOfCircle(E))))
Answer: 49*pi/2
Theorem sequence:
radius_of_circle_property_length_equal(1,FB,F)
radius_of_circle_property_length_equal(1,FA,F)
radius_of_circle_property_length_equal(1,EA,E)
radius_of_circle_property_length_equal(1,EH,E)
radius_of_circle_property_length_equal(1,AB,A)
line_addition(1,BF,FA)
line_addition(1,BA,AE)
line_addition(1,BE,EH)
circle_area_formula(1,A)
circle_area_formula(1,F)
circle_area_formula(1,E)
Problem: As shown in the diagram, BA=y, BD=16, CB=z, CD=8, DA=x, AD is perpendicular to BD, CB is perpendicular to AB. Find the value of x.
问题：如图所示，BA=y，BD=16，CB=z，CD=8，DA=x，AD⊥BD，CB⊥AB。求x的值。
Construction (CDL):
Shape(CB,BD,DC)
Shape(DB,BA,AD)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(BA),y)
Equal(LengthOfLine(BD),16)
Equal(LengthOfLine(CB),z)
Equal(LengthOfLine(CD),8)
Equal(LengthOfLine(DA),x)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 32
Theorem sequence:
angle_addition(1,CBD,DBA)
triangle_property_angle_sum(1,CBD)
triangle_property_angle_sum(1,DBA)
triangle_property_angle_sum(1,CBA)
similar_triangle_judgment_aa(1,CBD,BAD)
similar_triangle_property_line_ratio(1,BDC,ADB)
similar_triangle_property_line_ratio(1,CBD,BAD)
Problem: As shown in the diagram, BA=y, BC=z, DA=x, DB=8, ∠BAC=45°, ∠BDA=60°, AC is perpendicular to BC, DA is perpendicular to BA. Find the value of y.
问题：如图所示，BA=y，BC=z，DA=x，DB=8，∠BAC=45°，∠BDA=60°，AC垂直于BC，DA⊥BA。求y的值。
Construction (CDL):
Shape(BD,DA,AB)
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(BA),y)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(DA),x)
Equal(LengthOfLine(DB),8)
Equal(MeasureOfAngle(BAC),45)
Equal(MeasureOfAngle(BDA),60)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(DA,BA)
Goal: Value(y)
Answer: 4*sqrt(3)
Theorem sequence:
sine_theorem(1,BDA)
Problem: As shown in the diagram, AP=6*r+4, PH=22+3*r, ∠AHW=8*q+17°, ∠HWP=10+q°, WP is the perpendicular bisector of AH. Find the value of q.
问题：如图所示，AP=6*r+4，PH=22+3*r，∠AHW=8*q+17°，∠HWP=10+q°，WP垂直平分AB。求q的值。
Construction (CDL):
Shape(HW,WX,XH)
Shape(XW,WQ,QX)
Shape(HX,XP,PH)
Shape(PX,XQ,QA,AP)
Collinear(HXQ)
Collinear(WXP)
Collinear(WQA)
Collinear(HPA)
Conditions (CDL):
Equal(LengthOfLine(AP),6*r+4)
Equal(LengthOfLine(PH),22+3*r)
Equal(MeasureOfAngle(AHW),8*q+17)
Equal(MeasureOfAngle(HWP),10+q)
IsPerpendicularBisectorOfLine(WP,AH)
Goal: Value(q)
Answer: 7
Theorem sequence:
triangle_property_angle_sum(1,WPH)
Problem: As shown in the diagram, AD=3*x+5, DC=5*x-1, BD is the median of △ BAC. Find the length of line AC.
问题：如图所示，AD=3*x+5，DC=5*x-1，BD是三角形BAC的中线。求直线AC的长度。
Construction (CDL):
Shape(BA,AD,DB)
Shape(BD,DC,CB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AD),3*x+5)
Equal(LengthOfLine(DC),5*x-1)
IsMedianOfTriangle(BD,BAC)
Goal: Value(LengthOfLine(AC))
Answer: 28
Theorem sequence:
line_addition(1,AD,DC)
Problem: As shown in the diagram, AB=20, AC=x, CB=z, DA=10, DB=y, CA⊥BA, DB⊥CB. Find the value of y.
问题：如图所示，AB=20，AC=x，CB=z，DA=10，DB=y，CA⊥BA，DB垂直于CB。求y的值。
Construction (CDL):
Shape(AD,DB,BA)
Shape(CA,AB,BC)
Collinear(DAC)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(CB),z)
Equal(LengthOfLine(DA),10)
Equal(LengthOfLine(DB),y)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(DB,CB)
Goal: Value(y)
Answer: 10*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,CAB,BAD)
right_triangle_judgment_angle(1,BAD)
right_triangle_property_pythagorean(1,BAD)
Problem: As shown in the diagram, DC=46, EH=38, the midsegment of quadrilateral CGFD is HE, the midsegment of quadrilateral CHED is WX, CG and FD are the non-parallel sides (legs) of trapezoid CGFD. Find the length of line WX.
问题：如图所示，DC=46，EH=38，四边形CGFD的中位线为HE，CHED的中位线为WX，CG和FD是梯形CGFD的腰。求直线WX的长度。
Construction (CDL):
Shape(CW,WX,XD,DC)
Shape(WH,HE,EX,XW)
Shape(HG,GF,FE,EH)
Collinear(CWHG)
Collinear(DXEF)
Conditions (CDL):
Equal(LengthOfLine(DC),46)
Equal(LengthOfLine(EH),38)
IsMidsegmentOfQuadrilateral(HE,CGFD)
IsMidsegmentOfQuadrilateral(WX,CHED)
Trapezoid(CGFD)
Goal: Value(LengthOfLine(WX))
Answer: 42
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,WX,CHED)
Problem: As shown in the diagram, BO=15, CO=20, ∠DCO=60°, quadrilateral ACOB is a parallelogram, AD⊥OD. Find the area of ACOB.
问题：如图所示，BO=15，CO=20，∠DCO=60°，CA和OB是▱ACOB的一组对边，AD⊥OD。求ACOB的面积。
Construction (CDL):
Shape(CO,OD,DC)
Shape(OB,BA,AD,DO)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(BO),15)
Equal(LengthOfLine(CO),20)
Equal(MeasureOfAngle(DCO),60)
Parallelogram(ACOB)
PerpendicularBetweenLine(AD,OD)
Goal: Value(AreaOfQuadrilateral(ACOB))
Answer: 150*sqrt(3)
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ACOB)
parallelogram_area_formula_sine(1,ACOB)
Problem: As shown in the diagram, BD=24, CO=23, DO=7, BA and OC are opposite sides of the parallelogram BOCA, BD⊥OD. Find the perimeter of ABOC.
问题：如图所示，BD=24，CO=23，DO=7，四边形BOCA是▱，BD垂直于OD。求ABOC的周长。
Construction (CDL):
Shape(BD,DO,OB)
Shape(BO,OC,CA,AB)
Collinear(DOC)
Conditions (CDL):
Equal(LengthOfLine(BD),24)
Equal(LengthOfLine(CO),23)
Equal(LengthOfLine(DO),7)
Parallelogram(BOCA)
PerpendicularBetweenLine(BD,OD)
Goal: Value(PerimeterOfQuadrilateral(ABOC))
Answer: 96
Theorem sequence:
right_triangle_judgment_angle(1,BDO)
right_triangle_property_pythagorean(1,BDO)
parallelogram_property_opposite_line_equal(1,BOCA)
parallelogram_property_opposite_line_equal(1,ABOC)
quadrilateral_perimeter_formula(1,BOCA)
Problem: As shown in the diagram, JK=8, JS=10, ∠KPL=28°, ∠SLK=24.5°, the center of the inscribed circle of △PLJ is S, JK is perpendicular to SK, SM⊥MM, SQ⊥PQ. Find the measure of ∠QPM.
问题：如图所示，JK=8，JS=10，∠KPL=28°，∠SLK=24.5°，S是△PLJ内切圆的圆心，JK垂直于SK，SM垂直于MM，SQ垂直于PQ。求∠QPM的大小。
Construction (CDL):
Shape(JQ,QS,SJ)
Shape(QP,PS,SQ)
Shape(SP,PM,MS)
Shape(SM,ML,LS)
Shape(SL,LK,KS)
Shape(JS,SK,KJ)
Collinear(PQJ)
Collinear(PSK)
Collinear(PML)
Collinear(QSL)
Collinear(JSM)
Collinear(JKL)
Conditions (CDL):
Equal(LengthOfLine(JK),8)
Equal(LengthOfLine(JS),10)
Equal(MeasureOfAngle(KPL),28)
Equal(MeasureOfAngle(SLK),24.5)
IsIncenterOfTriangle(S,PLJ)
PerpendicularBetweenLine(JK,SK)
PerpendicularBetweenLine(SM,MP)
PerpendicularBetweenLine(SQ,PQ)
Goal: Value(MeasureOfAngle(QPM))
Answer: 56
Theorem sequence:
angle_addition(1,QPS,SPM)
Problem: As shown in the diagram, TR=18, WT=12, ∠TWV=60°, TWVR is a ▱, TA is perpendicular to VA. Find the perimeter of TWVR.
问题：如图所示，TR=18，WT=12，∠TWV=60°，四边形TWVR是平行四边形，TA垂直于VA。求TWVR的周长。
Construction (CDL):
Shape(TA,AV,VR,RT)
Shape(TW,WA,AT)
Collinear(WAV)
Conditions (CDL):
Equal(LengthOfLine(TR),18)
Equal(LengthOfLine(WT),12)
Equal(MeasureOfAngle(TWV),60)
Parallelogram(TWVR)
PerpendicularBetweenLine(TA,VA)
Goal: Value(PerimeterOfQuadrilateral(TWVR))
Answer: 60
Theorem sequence:
parallelogram_property_opposite_line_equal(1,TWVR)
parallelogram_property_opposite_line_equal(1,WVRT)
quadrilateral_perimeter_formula(1,TWVR)
Problem: As shown in the diagram, FG=9*x-6, FG=GH, FH=17, GH=7*x+4. Find the value of x.
问题：如图所示，FG=9*x-6，FG=GH，FH=17，GH=7*x+4。求x的值。
Construction (CDL):
Shape(FG,GH,HF)
Conditions (CDL):
Equal(LengthOfLine(FG),9*x-6)
Equal(LengthOfLine(FG),LengthOfLine(GH))
Equal(LengthOfLine(FH),17)
Equal(LengthOfLine(GH),7*x+4)
Goal: Value(x)
Answer: 5
Theorem sequence:
Problem: As shown in the diagram, AB=6*sqrt(3), BC=y, CA=x, ∠BCA=60°, CA⊥BA. Find the value of y.
问题：如图所示，AB=6*sqrt(3)，BC=y，CA=x，∠BCA=60°，CA垂直于BA。求y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),6*sqrt(3))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CA),x)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 12
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=x+2, BC=8, DC=y-3, EH=10, FE=15, HG=5, ∠ABC=∠FEH, ∠BCD=∠EHG, ∠CDA=∠HGF, ∠DAB=∠GFE, quadrilateral ABCD is similar to quadrilateral FEHG. Find the value of y.
问题：如图所示，AB=x+2，BC=8，DC=y-3，EH=10，FE=15，HG=5，∠ABC=∠FEH，∠BCD=∠EHG，∠CDA=∠HGF，∠DAB=∠GFE，ABCD与FEHG相似。求y的值。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(FE,EH,HG,GF)
Conditions (CDL):
Equal(LengthOfLine(AB),x+2)
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(DC),y-3)
Equal(LengthOfLine(EH),10)
Equal(LengthOfLine(FE),15)
Equal(LengthOfLine(HG),5)
Equal(MeasureOfAngle(ABC),MeasureOfAngle(FEH))
Equal(MeasureOfAngle(BCD),MeasureOfAngle(EHG))
Equal(MeasureOfAngle(CDA),MeasureOfAngle(HGF))
Equal(MeasureOfAngle(DAB),MeasureOfAngle(GFE))
SimilarBetweenQuadrilateral(ABCD,FEHG)
Goal: Value(y)
Answer: 7
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,BCDA,EHGF)
similar_quadrilateral_property_line_ratio(1,CDAB,HGFE)
Problem: As shown in the diagram, ∠ABE=x+10°, ∠BDG=31°, ∠CAF=x-11°, ∠GCH=2*x-42°. Find the value of x.
问题：如图所示，∠ABE=x+10°，∠BDG=31°，∠CAF=x-11°，∠GCH=2*x-42°。求x的值。
Construction (CDL):
Shape(CA,AF)
Shape(AB,BE)
Shape(BD,DG)
Shape(DC,CH)
Shape(AC,CD,DB,BA)
Collinear(FAB)
Collinear(DBE)
Collinear(CDG)
Collinear(ACH)
Conditions (CDL):
Equal(MeasureOfAngle(ABE),x+10)
Equal(MeasureOfAngle(BDG),31)
Equal(MeasureOfAngle(CAF),x-11)
Equal(MeasureOfAngle(GCH),2*x-42)
Goal: Value(x)
Answer: 93
Theorem sequence:
adjacent_complementary_angle(1,BAC,CAF)
adjacent_complementary_angle(1,ACG,GCH)
adjacent_complementary_angle(1,CDE,EDG)
adjacent_complementary_angle(1,DBA,ABE)
quadrilateral_property_angle_sum(1,ACDB)
Problem: As shown in the diagram, ∠CAB=42°, ∠CFG=77°, △CDE is an equilateral △, △CAB is an isosceles △, ∠CFG and ∠FGC are the base angles of the isosceles triangle CFG. Find the measure of ∠BCE.
问题：如图所示，∠CAB=42°，∠CFG=77°，△CDE为等边△，CA和CB是等腰△CAB的腰，∠CFG和∠FGC是等腰△CFG的底角。求∠BCE的大小。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DF,FC)
Shape(CF,FG,GC)
Shape(CG,GE,EC)
Shape(CE,EB,BC)
Collinear(ADFGEB)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),42)
Equal(MeasureOfAngle(CFG),77)
EquilateralTriangle(CDE)
IsoscelesTriangle(CAB)
IsoscelesTriangle(CFG)
Goal: Value(MeasureOfAngle(BCE))
Answer: 18
Theorem sequence:
equilateral_triangle_property_angle(1,ECD)
isosceles_triangle_property_angle_equal(1,CAB)
adjacent_complementary_angle(1,GEC,CEB)
triangle_property_angle_sum(1,CEB)
Problem: As shown in the diagram, ∠PML=∠NMP, quadrilateral LJMK is a parallelogram, quadrilateral LPNM is a rectangle. Find the measure of ∠PJL.
问题：如图所示，∠PML=∠NMP，四边形LJMK是平行四边形，LPNM是矩形。求∠PJL的大小。
Construction (CDL):
Shape(PN,NJ,JP)
Shape(JN,NM,MJ)
Shape(JM,ML,LJ)
Shape(JL,LP,PJ)
Shape(MK,KL,LM)
Collinear(PJM)
Collinear(NJL)
Conditions (CDL):
Equal(MeasureOfAngle(PML),MeasureOfAngle(NMP))
Parallelogram(LJMK)
Rectangle(LPNM)
Goal: Value(MeasureOfAngle(PJL))
Answer: 90
Theorem sequence:
angle_addition(1,NMJ,JML)
rectangle_property_diagonal_equal(1,LPNM)
parallelogram_property_diagonal_bisection(1,LPNM,J)
parallelogram_property_diagonal_bisection(1,PNML,J)
line_addition(1,PJ,JM)
line_addition(1,LJ,JN)
isosceles_triangle_judgment_line_equal(1,JML)
isosceles_triangle_property_angle_equal(1,JML)
triangle_property_angle_sum(1,JML)
adjacent_complementary_angle(1,PJL,LJM)
Problem: As shown in the diagram, ∠ABD=45°, ∠BCD=70°, ∠DAB=40°. Find the measure of ∠CDB.
问题：如图所示，∠ABD=45°，∠BCD=70°，∠DAB=40°。求∠CDB的大小。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),45)
Equal(MeasureOfAngle(BCD),70)
Equal(MeasureOfAngle(DAB),40)
Goal: Value(MeasureOfAngle(CDB))
Answer: 85
Theorem sequence:
triangle_property_angle_sum(1,BDA)
adjacent_complementary_angle(1,CDB,BDA)
Problem: As shown in the diagram, ∠ADG=36°, ∠AGF=104°, ∠EFC=40°, AB is perpendicular to CB. Find the measure of ∠FBC.
问题：如图所示，∠ADG=36°，∠AGF=104°，∠EFC=40°，AB⊥CB。求∠FBC的大小。
Construction (CDL):
Shape(AD,DG,GA)
Shape(GB,BF,FG)
Shape(FB,BC,CF)
Shape(EF,FC)
Shape(AG,GF)
Collinear(DGFC)
Collinear(AGB)
Collinear(BFE)
Conditions (CDL):
Equal(MeasureOfAngle(ADG),36)
Equal(MeasureOfAngle(AGF),104)
Equal(MeasureOfAngle(EFC),40)
PerpendicularBetweenLine(AB,CB)
Goal: Value(MeasureOfAngle(FBC))
Answer: 26
Theorem sequence:
adjacent_complementary_angle(1,AGC,CGB)
adjacent_complementary_angle(1,EFC,CFB)
triangle_property_angle_sum(1,GBC)
triangle_property_angle_sum(1,FBC)
Problem: As shown in the diagram, SU=16, the center of circle R is R, the diameter of circle R is SU. Find the length of line RT.
问题：如图所示，SU=16，R是圆R的圆心，SU是⊙R的直径。求直线RT的长度。
Construction (CDL):
Shape(SR,RT,RTS)
Shape(TR,RU,RUT)
Shape(VU,US,RSV)
Shape(UV,RVU)
Collinear(SRU)
Cocircular(R,SVUT)
Conditions (CDL):
Equal(LengthOfLine(SU),16)
IsCentreOfCircle(R,R)
IsDiameterOfCircle(SU,R)
Goal: Value(LengthOfLine(RT))
Answer: 8
Theorem sequence:
diameter_of_circle_property_length_equal(1,SU,R)
radius_of_circle_property_length_equal(1,RT,R)
circle_property_length_of_radius_and_diameter(1,R)
Problem: As shown in the diagram, KJ=15, ∠LKJ=164°, the center of ⊙K is K. Find the area of the sector KJL.
问题：如图所示，KJ=15，∠LKJ=164°，⊙K的圆心为K。求扇形KJL的面积。
Construction (CDL):
Shape(JK,KL,KLJ)
Shape(LK,KJ,KJL)
Cocircular(K,JL)
Conditions (CDL):
Equal(LengthOfLine(KJ),15)
Equal(MeasureOfAngle(LKJ),164)
IsCentreOfCircle(K,K)
Goal: Value(AreaOfSector(KJL))
Answer: 205*pi/2
Theorem sequence:
radius_of_circle_property_length_equal(1,KJ,K)
sector_area_formula(1,KJL)
arc_property_center_angle(1,KJL,K)
Problem: As shown in the diagram, JL=3*y-2, MJ=4, ML=6*x-3, SQ=5, ST=3, TQ=2, ∠LMJ=∠STQ, ∠MJL=∠TQS, △JLM is similar to △QST.. Find the value of y.
问题：如图所示，JL=3*y-2，MJ=4，ML=6*x-3，SQ=5，ST=3，TQ=2，∠LMJ=∠STQ，∠MJL=∠TQS，三角形JLM与三角形QST是相似三角形。求y的值。
Construction (CDL):
Shape(JL,LM,MJ)
Shape(QS,ST,TQ)
Conditions (CDL):
Equal(LengthOfLine(JL),3*y-2)
Equal(LengthOfLine(MJ),4)
Equal(LengthOfLine(ML),6*x-3)
Equal(LengthOfLine(SQ),5)
Equal(LengthOfLine(ST),3)
Equal(LengthOfLine(TQ),2)
Equal(MeasureOfAngle(LMJ),MeasureOfAngle(STQ))
Equal(MeasureOfAngle(MJL),MeasureOfAngle(TQS))
SimilarBetweenTriangle(JLM,QST)
Goal: Value(y)
Answer: 4
Theorem sequence:
similar_triangle_property_line_ratio(1,MJL,TQS)
similar_triangle_property_line_ratio(1,LMJ,STQ)
Problem: As shown in the diagram, AB=20, AC=x, BC=y, ∠CAB=60°, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=20，AC=x，BC=y，∠CAB=60°，BC垂直于AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(CAB),60)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 10
Theorem sequence:
triangle_property_angle_sum(1,BCA)
sine_theorem(1,ABC)
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 z.
问题：如图所示，JM=5，KM=20，LJ=y，LK=z，LM=x，JL⊥KL，KM垂直于LM。求z的值。
Construction (CDL):
Shape(JL,LM,MJ)
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(z)
Answer: 10*sqrt(5)
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,LKM,JLK)
line_addition(1,JM,MK)
mirror_similar_triangle_property_line_ratio(1,LKM,JLK)
mirror_similar_triangle_property_line_ratio(1,MLK,LKJ)
Problem: As shown in the diagram, AB=x, AC=20, BC=28, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=x，AC=20，BC=28，BC⊥AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BC),28)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 4*sqrt(74)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, GN=13+6*y, HP=3*x-9, KN=17+5*y, PQ=4*x-22, PQ=HP, PN is parallel to HK, QG∥PN. Find the value of y.
问题：如图所示，GN=13+6*y，HP=3*x-9，KN=17+5*y，PQ=4*x-22，PQ=HP，PN平行于HK，QG∥PN。求y的值。
Construction (CDL):
Shape(KN,NP,PH,HK)
Shape(NG,GQ,QP,PN)
Collinear(KNG)
Collinear(HPQ)
Conditions (CDL):
Equal(LengthOfLine(GN),13+6*y)
Equal(LengthOfLine(HP),3*x-9)
Equal(LengthOfLine(KN),17+5*y)
Equal(LengthOfLine(PQ),4*x-22)
Equal(LengthOfLine(PQ),LengthOfLine(HP))
ParallelBetweenLine(PN,HK)
ParallelBetweenLine(QG,PN)
Goal: Value(y)
Answer: 4
Theorem sequence:
parallel_judgment_par_par(1,QG,PN,HK)
trapezoid_judgment_parallel(1,QHKG)
midsegment_of_quadrilateral_judgment_parallel(3,NP,KGQH)
Problem: As shown in the diagram, MP=10, ∠MPQ=8*x°, ∠QLM=2*x**2-10°, quadrilateral LMPQ is a rhombus. Find the measure of ∠MQL.
问题：如图所示，MP=10，∠MPQ=8*x°，∠QLM=2*x**2-10°，LMPQ是菱形。求∠MQL的大小。
Construction (CDL):
Shape(LR,RQ,QL)
Shape(LM,MR,RL)
Shape(MP,PR,RM)
Shape(QR,RP,PQ)
Collinear(LRP)
Collinear(QRM)
Conditions (CDL):
Equal(LengthOfLine(MP),10)
Equal(MeasureOfAngle(MPQ),8*x)
Equal(MeasureOfAngle(QLM),2*x**2-10)
Rhombus(LMPQ)
Goal: Value(MeasureOfAngle(MQL))
Answer: 70
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,LMPQ)
isosceles_triangle_judgment_line_equal(1,LMQ)
isosceles_triangle_property_angle_equal(1,LMQ)
triangle_property_angle_sum(1,LMQ)
Problem: As shown in the diagram, the area of ABLT is 147, the area of quadrilateral EGHD is 27, BC=14, EF=x, BC⊥LC, EF is perpendicular to GF, EGHD is similar to BLTA, △EFG is similar to △BCL.. Find the value of x.
问题：如图所示，ABLT的面积为147，四边形EGHD的面积为27，BC=14，EF=x，BC垂直于LC，EF⊥GF，四边形EGHD相似于四边形BLTA，三角形EFG相似于三角形BCL。求x的值。
Construction (CDL):
Shape(EF,FG,GE)
Shape(EG,GH,HD,DE)
Shape(BC,CL,LB)
Shape(BL,LT,TA,AB)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ABLT),147)
Equal(AreaOfQuadrilateral(EGHD),27)
Equal(LengthOfLine(BC),14)
Equal(LengthOfLine(EF),x)
PerpendicularBetweenLine(BC,LC)
PerpendicularBetweenLine(EF,GF)
SimilarBetweenQuadrilateral(EGHD,BLTA)
SimilarBetweenTriangle(EFG,BCL)
Goal: Value(x)
Answer: 6
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,EGHD,BLTA)
similar_quadrilateral_property_area_square_ratio(1,EGHD,BLTA)
similar_triangle_property_line_ratio(1,FGE,CLB)
similar_triangle_property_line_ratio(1,GEF,LBC)
Problem: As shown in the diagram, ∠BGF=36°, ∠BIG=x°, ∠DIB=31°, ∠IGB=y°, ID is parallel to GF. Find the measure of ∠GBI.
问题：如图所示，∠BGF=36°，∠BIG=x°，∠DIB=31°，∠IGB=y°，ID∥GF。求∠GBI的大小。
Construction (CDL):
Shape(IG,GB,BI)
Shape(BG,GF)
Shape(DI,IB)
Shape(DI,IG)
Shape(IG,GF)
Conditions (CDL):
Equal(MeasureOfAngle(BGF),36)
Equal(MeasureOfAngle(BIG),x)
Equal(MeasureOfAngle(DIB),31)
Equal(MeasureOfAngle(IGB),y)
ParallelBetweenLine(ID,GF)
Goal: Value(MeasureOfAngle(GBI))
Answer: 67
Theorem sequence:
angle_addition(1,IGB,BGF)
angle_addition(1,DIB,BIG)
parallel_property_ipsilateral_internal_angle(1,ID,GF)
triangle_property_angle_sum(1,BIG)
Problem: As shown in the diagram, QS=3*x+10, RQ=5*x, RQ=SQ, RS=6*x-5, RS=RQ. Find the length of line RS.
问题：如图所示，QS=3*x+10，RQ=5*x，RQ=SQ，RS=6*x-5，RS=RQ。求直线RS的长度。
Construction (CDL):
Shape(QS,SR,RQ)
Conditions (CDL):
Equal(LengthOfLine(QS),3*x+10)
Equal(LengthOfLine(RQ),5*x)
Equal(LengthOfLine(RQ),LengthOfLine(SQ))
Equal(LengthOfLine(RS),6*x-5)
Equal(LengthOfLine(RS),LengthOfLine(RQ))
Goal: Value(LengthOfLine(RS))
Answer: 25
Theorem sequence:
Problem: As shown in the diagram, BE=7*sqrt(3), ∠CAB=30°, ∠EBC=60°, AE is perpendicular to CE, BC⊥AC. Find the length of line CA.
问题：如图所示，BE=7*sqrt(3)，∠CAB=30°，∠EBC=60°，AE⊥CE，BC⊥AC。求直线CA的长度。
Construction (CDL):
Shape(BC,CE,EB)
Shape(EC,CA,AE)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(BE),7*sqrt(3))
Equal(MeasureOfAngle(CAB),30)
Equal(MeasureOfAngle(EBC),60)
PerpendicularBetweenLine(AE,CE)
PerpendicularBetweenLine(BC,AC)
Goal: Value(LengthOfLine(CA))
Answer: 42
Theorem sequence:
line_addition(1,BE,EA)
triangle_property_angle_sum(1,AEC)
sine_theorem(1,AEC)
sine_theorem(1,ABC)
Problem: As shown in the diagram, BC=6, CA=8, BA⊥SA, SC is perpendicular to AC. Find the length of line AB.
问题：如图所示，BC=6，CA=8，BA垂直于SA，SC⊥AC。求直线AB的长度。
Construction (CDL):
Shape(AC,CB,BA)
Shape(CA,AS,SC)
Collinear(SCB)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(CA),8)
PerpendicularBetweenLine(BA,SA)
PerpendicularBetweenLine(SC,AC)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,SCA,ACB)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, WF=14, WX=x, XF=10. Find the ratio of the length of line FX to the length of line FW.
问题：如图所示，WF=14，WX=x，XF=10。求直线FX的长度与直线FW的长度之比。
Construction (CDL):
Shape(XF,FB,BX)
Shape(BF,FC,CB)
Shape(CY,YR,RB,BC)
Shape(WX,XB,BR,RW)
Collinear(FXW)
Collinear(FBR)
Collinear(FCY)
Conditions (CDL):
Equal(LengthOfLine(WF),14)
Equal(LengthOfLine(WX),x)
Equal(LengthOfLine(XF),10)
Goal: Value(Div(LengthOfLine(FX),LengthOfLine(FW)))
Answer: 5/7
Theorem sequence:
Problem: As shown in the diagram, ∠HJP=57°, ∠JHL=x°, HP is the midsegment of triangle MLK, JH is the midsegment of triangle LKM, PJ is the midsegment of △ KML. Find the value of x.
问题：如图所示，∠HJP=57°，∠JHL=x°，HP是三角形MLK的中位线，JH是三角形LKM的中位线，PJ是三角形KML的中位线。求x的值。
Construction (CDL):
Shape(KP,PJ,JK)
Shape(PM,MH,HP)
Shape(JP,PH,HJ)
Shape(JH,HL,LJ)
Collinear(KJL)
Collinear(KPM)
Collinear(LHM)
Conditions (CDL):
Equal(MeasureOfAngle(HJP),57)
Equal(MeasureOfAngle(JHL),x)
IsMidsegmentOfTriangle(HP,MLK)
IsMidsegmentOfTriangle(JH,LKM)
IsMidsegmentOfTriangle(PJ,KML)
Goal: Value(x)
Answer: 57
Theorem sequence:
midsegment_of_triangle_property_parallel(1,PJ,KML)
parallel_property_collinear_extend(3,LM,JP,H)
parallel_property_alternate_interior_angle(2,LH,JP)
Problem: As shown in the diagram, AD=9, BD=3, BF=x, CF=5. Find the value of x.
问题：如图所示，AD=9，BD=3，BF=x，CF=5。求x的值。
Construction (CDL):
Shape(CF,EFC)
Shape(AD,EDF,FC,ECA)
Shape(EAD,DA)
Shape(DB,BF,EDF)
Collinear(AEDB)
Collinear(CFB)
Cocircular(E,ADFC)
Conditions (CDL):
Equal(LengthOfLine(AD),9)
Equal(LengthOfLine(BD),3)
Equal(LengthOfLine(BF),x)
Equal(LengthOfLine(CF),5)
Goal: Value(x)
Answer: 4
Theorem sequence:
circle_property_circular_power_segment_and_segment_line(1,BDA,BFC,E)
line_addition(1,AD,DB)
line_addition(1,CF,FB)
Problem: As shown in the diagram, ∠GHM=3*y-2°, ∠HMK=68°, KM is parallel to GH. Find the value of y.
问题：如图所示，∠GHM=3*y-2°，∠HMK=68°，KM∥GH。求y的值。
Construction (CDL):
Shape(FH,HE)
Shape(EH,HG)
Shape(MH,HF)
Shape(GH,HM)
Shape(DM,MH)
Shape(HM,MK)
Shape(PM,MD)
Shape(KM,MP)
Collinear(EHMP)
Collinear(GHF)
Collinear(KMD)
Conditions (CDL):
Equal(MeasureOfAngle(GHM),3*y-2)
Equal(MeasureOfAngle(HMK),68)
ParallelBetweenLine(KM,GH)
Goal: Value(y)
Answer: 38
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,HG,MK)
Problem: As shown in the diagram, ∠BIC=57°, DA is parallel to KB, GC∥LJ. Find the measure of ∠CIF.
问题：如图所示，∠BIC=57°，DA平行于KB，GC平行于LJ。求∠CIF的大小。
Construction (CDL):
Shape(GE,EA)
Shape(AE,EI)
Shape(EI,IB)
Shape(BI,IC)
Shape(HE,EG)
Shape(LH,HE)
Shape(EH,HF,FI,IE)
Shape(IF,FJ)
Shape(CI,IF)
Shape(DH,HL)
Shape(FH,HD)
Shape(KF,FH)
Shape(JF,FK)
Shape(BI,IF)
Collinear(AEHD)
Collinear(BIFK)
Collinear(GEIC)
Collinear(LHFJ)
Conditions (CDL):
Equal(MeasureOfAngle(BIC),57)
ParallelBetweenLine(DA,KB)
ParallelBetweenLine(GC,LJ)
Goal: Value(MeasureOfAngle(CIF))
Answer: 123
Theorem sequence:
adjacent_complementary_angle(1,BIC,CIF)