SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AN=19, BF=10, CE=2, FN=20, AE⊥FE, EF⊥NF, ACBN is a trapezoid. Find the perimeter of quadrilateral ACBN.
问题：如图所示，AN=19，BF=10，CE=2，FN=20，AE垂直于FE，EF垂直于NF，ACBN是梯形。求四边形ACBN的周长。
Construction (CDL):
Shape(AC,CE,EA)
Shape(AE,EF,FN,NA)
Shape(NF,FB,BN)
Collinear(CEFB)
Conditions (CDL):
Equal(LengthOfLine(AN),19)
Equal(LengthOfLine(BF),10)
Equal(LengthOfLine(CE),2)
Equal(LengthOfLine(FN),20)
PerpendicularBetweenLine(AE,FE)
PerpendicularBetweenLine(EF,NF)
Trapezoid(ACBN)
Goal: Value(PerimeterOfQuadrilateral(ACBN))
Answer: 2*sqrt(101)+10*sqrt(5)+50
Theorem sequence:
adjacent_complementary_angle(1,CEA,AEF)
adjacent_complementary_angle(1,CFN,NFB)
parallel_judgment_per_per(1,EA,FN)
parallel_property_collinear_extend(3,BC,NA,F)
parallel_property_collinear_extend(3,FC,NA,E)
parallelogram_judgment_parallel_and_parallel(1,AEFN)
parallelogram_property_opposite_line_equal(1,AEFN)
parallelogram_property_opposite_line_equal(1,EFNA)
right_triangle_judgment_angle(1,CEA)
right_triangle_judgment_angle(1,NFB)
right_triangle_property_pythagorean(1,CEA)
right_triangle_property_pythagorean(1,NFB)
line_addition(1,CE,EF)
line_addition(1,CF,FB)
quadrilateral_perimeter_formula(1,ACBN)
Problem: As shown in the diagram, ∠GKM=62°, ∠HBF=38°, NO∥MH. Find the measure of ∠ICO.
问题：如图所示，∠GKM=62°，∠HBF=38°，NO∥MH。求∠ICO的大小。
Construction (CDL):
Shape(NC,CJ)
Shape(JC,CI)
Shape(IC,CO)
Shape(OC,CB)
Shape(CB,BH)
Shape(HB,BF)
Shape(FB,BK)
Shape(BK,KG)
Shape(GK,KM)
Shape(MK,KC)
Shape(KC,CN)
Shape(CK,KB,BC)
Collinear(JCBF)
Collinear(ICKG)
Collinear(NCO)
Collinear(MKBH)
Conditions (CDL):
Equal(MeasureOfAngle(GKM),62)
Equal(MeasureOfAngle(HBF),38)
ParallelBetweenLine(NO,MH)
Goal: Value(MeasureOfAngle(ICO))
Answer: 62
Theorem sequence:
parallel_property_collinear_extend(3,NO,MH,C)
parallel_property_collinear_extend(3,HM,CN,K)
parallel_property_corresponding_angle(1,KM,CN,G)
vertical_angle(1,ICO,KCN)
Problem: As shown in the diagram, EF=9, GE=13, GF=x, ∠FEG=48°. Find the value of x.
问题：如图所示，EF=9，GE=13，GF=x，∠FEG=48°。求x的值。
Construction (CDL):
Shape(EG,GF,FE)
Conditions (CDL):
Equal(LengthOfLine(EF),9)
Equal(LengthOfLine(GE),13)
Equal(LengthOfLine(GF),x)
Equal(MeasureOfAngle(FEG),48)
Goal: Value(x)
Answer: 2**(3/4)*sqrt(-234*sqrt(3)*sqrt(sqrt(5)+5)+117*sqrt(10)+883*sqrt(2))/4
Theorem sequence:
cosine_theorem(1,EGF)
Problem: As shown in the diagram, AB=6, the center of ⊙B is B. Find the circumference of the circle B.
问题：如图所示，AB=6，B是圆B的圆心。求⊙B的周长。
Construction (CDL):
Shape(AB)
Shape(BAC,BCA)
Cocircular(B,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
IsCentreOfCircle(B,B)
Goal: Value(PerimeterOfCircle(B))
Answer: 12*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,BA,B)
circle_perimeter_formula(1,B)
Problem: As shown in the diagram, NQ=8, PR=12, VX=4*x+2, WY=7*x-1, NQ is the median of triangle NPR, VX is the median of △ VWY, △PRN is similar to △WYV.. Find the value of x.
问题：如图所示，NQ=8，PR=12，VX=4*x+2，WY=7*x-1，NQ是△NPR的中线，VX是三角形VWY的中线，三角形PRN相似于三角形WYV。求x的值。
Construction (CDL):
Shape(NP,PQ,QN)
Shape(NQ,QR,RN)
Shape(VW,WX,XV)
Shape(VX,XY,YV)
Collinear(PQR)
Collinear(WXY)
Conditions (CDL):
Equal(LengthOfLine(NQ),8)
Equal(LengthOfLine(PR),12)
Equal(LengthOfLine(VX),4*x+2)
Equal(LengthOfLine(WY),7*x-1)
IsMedianOfTriangle(NQ,NPR)
IsMedianOfTriangle(VX,VWY)
SimilarBetweenTriangle(PRN,WYV)
Goal: Value(x)
Answer: 4
Theorem sequence:
line_addition(1,PQ,QR)
line_addition(1,WX,XY)
similar_triangle_property_line_ratio(1,NPR,VWY)
similar_triangle_property_line_ratio(1,RNP,YVW)
similar_triangle_property_angle_equal(1,PRN,WYV)
similar_triangle_judgment_sas(1,PQN,WXV)
similar_triangle_property_line_ratio(1,PQN,WXV)
similar_triangle_property_line_ratio(1,NPQ,VWX)
Problem: As shown in the diagram, CY=y, DY=3/5*y+2, XF=20-5*x, XF=XJ, XJ=2*x+6, YC=YD. Find the value of x.
问题：如图所示，CY=y，DY=3/5*y+2，XF=20-5*x，XF=XJ，XJ=2*x+6，YC=YD。求x的值。
Construction (CDL):
Shape(FX,XY,YC,CF)
Shape(XJ,JD,DY,YX)
Collinear(FXJ)
Collinear(CYD)
Conditions (CDL):
Equal(LengthOfLine(CY),y)
Equal(LengthOfLine(DY),3/5*y+2)
Equal(LengthOfLine(XF),20-5*x)
Equal(LengthOfLine(XF),LengthOfLine(XJ))
Equal(LengthOfLine(XJ),2*x+6)
Equal(LengthOfLine(YC),LengthOfLine(YD))
Goal: Value(x)
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, BE=5, CF=y, FD=2, B is the center of circle B, BF⊥AF. Find the value of y.
问题：如图所示，BE=5，CF=y，FD=2，圆B的圆心为B，BF⊥AF。求y的值。
Construction (CDL):
Shape(BCD,DF,FC)
Shape(FD,BDA,AF)
Shape(CF,FB,BC)
Shape(BE,BEC,CB)
Shape(BF,FA,BAE,EB)
Collinear(DFBE)
Collinear(CFA)
Cocircular(B,CDAE)
Conditions (CDL):
Equal(LengthOfLine(BE),5)
Equal(LengthOfLine(CF),y)
Equal(LengthOfLine(FD),2)
IsCentreOfCircle(B,B)
PerpendicularBetweenLine(BF,AF)
Goal: Value(y)
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,CFB,BFA)
radius_of_circle_property_length_equal(1,BC,B)
radius_of_circle_property_length_equal(1,BD,B)
radius_of_circle_property_length_equal(1,BE,B)
right_triangle_judgment_angle(1,CFB)
line_addition(1,DF,FB)
right_triangle_property_pythagorean(1,CFB)
Problem: As shown in the diagram, AF=y, DA=8, DC=z, DF=x, FE=6, ∠DAF=30°, ∠EBC=45°, DC∥FE, AF⊥DF, CE is perpendicular to BE. Find the perimeter of quadrilateral ABCD.
问题：如图所示，AF=y，DA=8，DC=z，DF=x，FE=6，∠DAF=30°，∠EBC=45°，DC平行于FE，AF垂直于DF，CE⊥BE。求四边形ABCD的周长。
Construction (CDL):
Shape(DA,AF,FD)
Shape(DF,FE,EC,CD)
Shape(CE,EB,BC)
Collinear(AFEB)
Conditions (CDL):
Equal(LengthOfLine(AF),y)
Equal(LengthOfLine(DA),8)
Equal(LengthOfLine(DC),z)
Equal(LengthOfLine(DF),x)
Equal(LengthOfLine(FE),6)
Equal(MeasureOfAngle(DAF),30)
Equal(MeasureOfAngle(EBC),45)
ParallelBetweenLine(DC,FE)
PerpendicularBetweenLine(AF,DF)
PerpendicularBetweenLine(CE,BE)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 4*sqrt(2)+4*sqrt(3)+24
Theorem sequence:
adjacent_complementary_angle(1,AFD,DFE)
adjacent_complementary_angle(1,AEC,CEB)
perpendicular_judgment_angle(1,DF,EF)
perpendicular_judgment_angle(1,FE,CE)
parallel_judgment_per_per(1,FD,EC)
parallelogram_judgment_parallel_and_parallel(1,DFEC)
parallelogram_property_opposite_line_equal(1,DFEC)
parallelogram_property_opposite_line_equal(1,FECD)
triangle_property_angle_sum(1,DAF)
triangle_property_angle_sum(1,CEB)
sine_theorem(1,DAF)
sine_theorem(1,AFD)
sine_theorem(1,CEB)
sine_theorem(1,EBC)
line_addition(1,AF,FE)
line_addition(1,AE,EB)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, AB=26, AC=24, CB=10, BC is perpendicular to AC. Find the value of cos(CA).
问题：如图所示，AB=26，AC=24，CB=10，BC垂直于AC。求cos(CA)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AC),24)
Equal(LengthOfLine(CB),10)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Cos(MeasureOfAngle(CAB)))
Answer: 12/13
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, the perimeter of quadrilateral DCIA is 200, the perimeter of IGHB is 150, AI⊥BI, quadrilateral ABEF is a square, DCIA is a square, IGHB is a square. Find the perimeter of ABEF.
问题：如图所示，四边形DCIA的周长为200，IGHB的周长为150，AI垂直于BI，ABEF是正方形，DCIA是正方形，IGHB是正方形。求四边形ABEF的周长。
Construction (CDL):
Shape(DC,CI,IA,AD)
Shape(AI,IB,BA)
Shape(AB,BE,EF,FA)
Shape(IG,GH,HB,BI)
Collinear(CIB)
Collinear(AIG)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(DCIA),200)
Equal(PerimeterOfQuadrilateral(IGHB),150)
PerpendicularBetweenLine(AI,BI)
Square(ABEF)
Square(DCIA)
Square(IGHB)
Goal: Value(PerimeterOfQuadrilateral(ABEF))
Answer: 250
Theorem sequence:
quadrilateral_perimeter_formula(1,DCIA)
quadrilateral_perimeter_formula(1,ABEF)
quadrilateral_perimeter_formula(1,IGHB)
right_triangle_judgment_angle(1,AIB)
right_triangle_property_pythagorean(1,AIB)
Problem: As shown in the diagram, MP=10, ∠MPQ=8*x°, ∠QLM=2*x**2-10°, LMPQ is a rhombus. Find the measure of ∠PQL.
问题：如图所示，MP=10，∠MPQ=8*x°，∠QLM=2*x**2-10°，LMPQ是菱形。求∠PQL的大小。
Construction (CDL):
Shape(QL,LR,RQ)
Shape(RL,LM,MR)
Shape(QR,RP,PQ)
Shape(RM,MP,PR)
Collinear(QRM)
Collinear(LRP)
Conditions (CDL):
Equal(LengthOfLine(MP),10)
Equal(MeasureOfAngle(MPQ),8*x)
Equal(MeasureOfAngle(QLM),2*x**2-10)
Rhombus(LMPQ)
Goal: Value(MeasureOfAngle(PQL))
Answer: 140
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,LMPQ)
parallel_property_ipsilateral_internal_angle(1,PM,QL)
Problem: As shown in the diagram, BO=8, ∠BOA=120°, O is the center of circle O. Find the length of ⌒OAB.
问题：如图所示，BO=8，∠BOA=120°，圆O的圆心为O。求⌒OAB的长度。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(AO,OB,OBA)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(BO),8)
Equal(MeasureOfAngle(BOA),120)
IsCentreOfCircle(O,O)
Goal: Value(LengthOfArc(OAB))
Answer: 16*pi/3
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
arc_property_center_angle(1,OAB,O)
arc_length_formula(1,OAB)
Problem: As shown in the diagram, AD=3*x+11, BD=5*x-8, BD=AD, BE=2*y-1, CE=4*y-7, AC is parallel to DE, DE is parallel to AC. Find the value of y.
问题：如图所示，AD=3*x+11，BD=5*x-8，BD=AD，BE=2*y-1，CE=4*y-7，AC∥DE，DE∥AC。求y的值。
Construction (CDL):
Shape(AC,CE,ED,DA)
Shape(DE,EB,BD)
Collinear(ADB)
Collinear(CEB)
Conditions (CDL):
Equal(LengthOfLine(AD),3*x+11)
Equal(LengthOfLine(BD),5*x-8)
Equal(LengthOfLine(BD),LengthOfLine(AD))
Equal(LengthOfLine(BE),2*y-1)
Equal(LengthOfLine(CE),4*y-7)
ParallelBetweenLine(AC,DE)
ParallelBetweenLine(DE,AC)
Goal: Value(y)
Answer: 3
Theorem sequence:
parallel_property_corresponding_angle(2,CA,ED,B)
similar_triangle_judgment_aa(1,DEB,ACB)
line_addition(1,AD,DB)
line_addition(1,CE,EB)
similar_triangle_property_line_ratio(1,DEB,ACB)
similar_triangle_property_line_ratio(1,EBD,CBA)
Problem: As shown in the diagram, RP=RT, RQ=RS, ∠RQS=80°. Find the measure of ∠RPT.
问题：如图所示，RP=RT，RQ=RS，∠RQS=80°。求∠RPT的大小。
Construction (CDL):
Shape(RQ,QS,SR)
Shape(QP,PU,UQ)
Shape(QU,US,SQ)
Shape(SU,UT,TS)
Collinear(PQR)
Collinear(PUT)
Collinear(RST)
Conditions (CDL):
Equal(LengthOfLine(RP),LengthOfLine(RT))
Equal(LengthOfLine(RQ),LengthOfLine(RS))
Equal(MeasureOfAngle(RQS),80)
Goal: Value(MeasureOfAngle(RPT))
Answer: 80
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,RQS)
isosceles_triangle_property_angle_equal(1,RQS)
isosceles_triangle_judgment_line_equal(1,RPT)
isosceles_triangle_property_angle_equal(1,RPT)
triangle_property_angle_sum(1,RQS)
triangle_property_angle_sum(1,RPT)
Problem: As shown in the diagram, AB=10, PA=11, QA=9, P is the center of ⊙P, the center of ⊙Q is Q, AH is perpendicular to QH, BH⊥PH, PH⊥AH, QH is perpendicular to BH. Find the length of line PQ.
问题：如图所示，AB=10，PA=11，QA=9，⊙P的圆心为P，Q是圆Q的圆心，AH⊥QH，BH垂直于PH，PH⊥AH，QH垂直于BH。求直线PQ的长度。
Construction (CDL):
Shape(PA,PAB,BP)
Shape(AP,PC,QAC)
Shape(PB,QCB,CP)
Shape(QAC,CH,HA)
Shape(QCB,BH,HC)
Shape(AH,HF,PFA)
Shape(HB,PBF,FH)
Shape(PFA,FQ,QA)
Shape(PBF,BQ,QF)
Shape(AQ,QB,QBA)
Collinear(PCHFQ)
Collinear(AHB)
Cocircular(P,BFA)
Cocircular(Q,ACB)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(PA),11)
Equal(LengthOfLine(QA),9)
IsCentreOfCircle(P,P)
IsCentreOfCircle(Q,Q)
PerpendicularBetweenLine(AH,QH)
PerpendicularBetweenLine(BH,PH)
PerpendicularBetweenLine(PH,AH)
PerpendicularBetweenLine(QH,BH)
Goal: Value(LengthOfLine(PQ))
Answer: 2*sqrt(14)+4*sqrt(6)
Theorem sequence:
radius_of_circle_property_length_equal(1,PB,P)
radius_of_circle_property_length_equal(1,PA,P)
radius_of_circle_property_length_equal(1,QA,Q)
radius_of_circle_property_length_equal(1,QB,Q)
cosine_theorem(1,PBA)
cosine_theorem(1,QAB)
altitude_of_triangle_judgment(1,PH,PBA)
altitude_of_triangle_judgment(1,QH,QAB)
triangle_area_formula_common(1,PBA)
triangle_area_formula_common(1,QAB)
triangle_area_formula_sine(1,PBA)
triangle_area_formula_sine(1,QAB)
line_addition(1,PH,HQ)
Problem: As shown in the diagram, AB=CD, DA=12, DA=BC, ∠ADB=28°, ∠DBA=40°. Find the area of quadrilateral ADCB.
问题：如图所示，AB=CD，DA=12，DA=BC，∠ADB=28°，∠DBA=40°。求四边形ADCB的面积。
Construction (CDL):
Shape(AD,DB,BA)
Shape(BD,DC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(CD))
Equal(LengthOfLine(DA),12)
Equal(LengthOfLine(DA),LengthOfLine(BC))
Equal(MeasureOfAngle(ADB),28)
Equal(MeasureOfAngle(DBA),40)
Goal: Value(AreaOfQuadrilateral(ADCB))
Answer: 144*sin(7*pi/45)*sin(17*pi/45)/sin(2*pi/9)
Theorem sequence:
sine_theorem(1,ADB)
parallelogram_judgment_equal_and_equal(1,ADCB)
parallel_property_alternate_interior_angle(2,AB,DC)
angle_addition(1,ADB,BDC)
parallelogram_area_formula_sine(1,ADCB)
Problem: As shown in the diagram, BD=10, BE=7, CE=2*x, the center of circle A is A, the tangent to ⊙A is BD. Find the value of x.
问题：如图所示，BD=10，BE=7，CE=2*x，A是圆A的圆心，圆O的切线为BD。求x的值。
Construction (CDL):
Shape(A)
Shape(AED,CE)
Shape(EC,ACD,ADE)
Shape(ADE,DB,BE)
Collinear(CEB)
Cocircular(A,CDE)
Conditions (CDL):
Equal(LengthOfLine(BD),10)
Equal(LengthOfLine(BE),7)
Equal(LengthOfLine(CE),2*x)
IsCentreOfCircle(A,A)
IsTangentOfCircle(BD,A)
Goal: Value(x)
Answer: 51/14
Theorem sequence:
line_addition(1,CE,EB)
circle_property_circular_power_tangent_and_segment_line(1,BD,BEC,A)
Problem: As shown in the diagram, RQ=4*x, RS=2*x+1, SQ=6*x-1, △QSR is an equilateral △. Find the value of x.
问题：如图所示，RQ=4*x，RS=2*x+1，SQ=6*x-1，三角形QSR为等边三角形。求x的值。
Construction (CDL):
Shape(QS,SR,RQ)
Conditions (CDL):
Equal(LengthOfLine(RQ),4*x)
Equal(LengthOfLine(RS),2*x+1)
Equal(LengthOfLine(SQ),6*x-1)
EquilateralTriangle(QSR)
Goal: Value(x)
Answer: 1/2
Theorem sequence:
Problem: As shown in the diagram, ∠CEA=x°, the measure of arc DAC is 90, the measure of ⌒DGH is 30. Find the value of x.
问题：如图所示，∠CEA=x°，弧DAC的角度为90，⌒DGH的角度为30。求x的值。
Construction (CDL):
Shape(DHA,AH)
Shape(HA,DAC,CG,DGH)
Shape(D)
Shape(GC,DCG)
Shape(EH,DGH,GE)
Collinear(AHE)
Collinear(CGE)
Cocircular(D,ACGH)
Conditions (CDL):
Equal(MeasureOfAngle(CEA),x)
Equal(MeasureOfArc(DAC),90)
Equal(MeasureOfArc(DGH),30)
Goal: Value(x)
Answer: 30
Theorem sequence:
circle_property_circular_power_segment_and_segment_angle(1,EGC,EHA,D)
Problem: As shown in the diagram, AB=z, AC=5, BC=15, AY is perpendicular to BY, BA is perpendicular to CA. Find the length of line BY.
问题：如图所示，AB=z，AC=5，BC=15，AY⊥BY，BA垂直于CA。求直线BY的长度。
Construction (CDL):
Shape(AC,CY,YA)
Shape(AY,YB,BA)
Collinear(CYB)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),15)
PerpendicularBetweenLine(AY,BY)
PerpendicularBetweenLine(BA,CA)
Goal: Value(LengthOfLine(BY))
Answer: 40/3
Theorem sequence:
adjacent_complementary_angle(1,CYA,AYB)
right_triangle_judgment_angle(1,CYA)
right_triangle_judgment_angle(1,AYB)
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,CYA)
right_triangle_property_pythagorean(1,AYB)
right_triangle_property_pythagorean(1,BAC)
line_addition(1,CY,YB)
Problem: As shown in the diagram, ∠ARQ=7*x-1°, ∠QPA=3*x+1°, A is the center of ⊙A, the diameter of ⊙A is RP. Find the measure of ∠ARQ.
问题：如图所示，∠ARQ=7*x-1°，∠QPA=3*x+1°，A是圆A的圆心，RP是⊙A的直径。求∠ARQ的大小。
Construction (CDL):
Shape(AQP,PQ)
Shape(QP,PA,AR,RQ)
Shape(RA,AP,APR)
Shape(ARQ,QR)
Collinear(PAR)
Cocircular(A,PRQ)
Conditions (CDL):
Equal(MeasureOfAngle(ARQ),7*x-1)
Equal(MeasureOfAngle(QPA),3*x+1)
IsCentreOfCircle(A,A)
IsDiameterOfCircle(RP,A)
Goal: Value(MeasureOfAngle(ARQ))
Answer: 62
Theorem sequence:
diameter_of_circle_property_right_angle(1,RQP,A)
triangle_property_angle_sum(1,QPR)
Problem: As shown in the diagram, ∠PQR=101°, ∠QRS=x°, ∠RSP=y°, RQ and SP are opposite sides of the parallelogram QRSP. Find the value of y.
问题：如图所示，∠PQR=101°，∠QRS=x°，∠RSP=y°，QRSP是▱。求y的值。
Construction (CDL):
Shape(QR,RS,SP,PQ)
Conditions (CDL):
Equal(MeasureOfAngle(PQR),101)
Equal(MeasureOfAngle(QRS),x)
Equal(MeasureOfAngle(RSP),y)
Parallelogram(QRSP)
Goal: Value(y)
Answer: 101
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,QRSP)
Problem: As shown in the diagram, AF=2, BC=5/2, BG=1, CE=7/2, EI=3, FG=3/2, JE=1, AD is parallel to EI, BG is parallel to FA, CB∥FG, DI is parallel to FA, FA∥CE, FG∥EI, BC is perpendicular to KC, KE is perpendicular to JE. Find the sum of the area of quadrilateral CKGB, the area of KEJF and the area of quadrilateral AJID.
问题：如图所示，AF=2，BC=5/2，BG=1，CE=7/2，EI=3，FG=3/2，JE=1，AD∥EI，BG平行于FA，CB∥FG，DI∥FA，FA∥CE，FG平行于EI，BC垂直于KC，KE⊥JE。求CKGB的面积、KEJF的面积与四边形AJID的面积之和。
Construction (CDL):
Shape(BC,CK,KF,FG,GB)
Shape(FK,KE,EJ,JA,AF)
Shape(AJ,JI,ID,DA)
Collinear(KFG)
Collinear(CKE)
Collinear(FAJ)
Collinear(EJI)
Conditions (CDL):
Equal(LengthOfLine(AF),2)
Equal(LengthOfLine(BC),5/2)
Equal(LengthOfLine(BG),1)
Equal(LengthOfLine(CE),7/2)
Equal(LengthOfLine(EI),3)
Equal(LengthOfLine(FG),3/2)
Equal(LengthOfLine(JE),1)
ParallelBetweenLine(AD,EI)
ParallelBetweenLine(BG,FA)
ParallelBetweenLine(CB,FG)
ParallelBetweenLine(DI,FA)
ParallelBetweenLine(FA,CE)
ParallelBetweenLine(FG,EI)
PerpendicularBetweenLine(BC,KC)
PerpendicularBetweenLine(KE,JE)
Goal: Value(Add(AreaOfQuadrilateral(CKGB),AreaOfQuadrilateral(KEJF),AreaOfQuadrilateral(AJID)))
Answer: 6
Theorem sequence:
line_addition(1,CK,KE)
line_addition(1,FA,AJ)
line_addition(1,EJ,JI)
parallel_property_collinear_extend(2,GF,BC,K)
parallel_property_collinear_extend(3,IE,DA,J)
parallel_property_collinear_extend(1,AF,ID,J)
parallel_property_collinear_extend(2,FA,CE,J)
parallel_property_collinear_extend(3,EC,JF,K)
parallel_judgment_par_par(1,BG,FA,CE)
parallel_property_collinear_extend(3,EC,GB,K)
parallel_property_collinear_extend(1,FG,EI,K)
parallel_property_collinear_extend(3,IE,FK,J)
parallelogram_judgment_parallel_and_parallel(1,AJID)
parallelogram_judgment_parallel_and_parallel(1,BCKG)
parallelogram_judgment_parallel_and_parallel(1,KEJF)
parallelogram_property_opposite_line_equal(1,CKGB)
parallelogram_property_opposite_line_equal(1,KEJF)
parallel_property_corresponding_angle(2,EC,JF,I)
parallelogram_area_formula_sine(1,KEJF)
parallelogram_area_formula_sine(1,BCKG)
parallelogram_area_formula_sine(1,AJID)
Problem: As shown in the diagram, ∠LCR=3*x+5°, ∠MCN=60°, ∠RCM=x-1°, C is the center of ⊙C. Find the measure of ∠RCN.
问题：如图所示，∠LCR=3*x+5°，∠MCN=60°，∠RCM=x-1°，圆C的圆心为C。求∠RCN的大小。
Construction (CDL):
Shape(CMR,RC,CM)
Shape(CRL,LC,CR)
Shape(CLN,NC,CL)
Shape(CNM,MC,CN)
Collinear(LCM)
Cocircular(C,RLNM)
Conditions (CDL):
Equal(MeasureOfAngle(LCR),3*x+5)
Equal(MeasureOfAngle(MCN),60)
Equal(MeasureOfAngle(RCM),x-1)
IsCentreOfCircle(C,C)
Goal: Value(MeasureOfAngle(RCN))
Answer: 103
Theorem sequence:
angle_addition(1,RCM,MCN)
adjacent_complementary_angle(1,LCR,RCM)
Problem: As shown in the diagram, AB=8, BC=10, BD=y, CA=6, CD=x, BD⊥AD, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AB=8，BC=10，BD=y，CA=6，CD=x，BD垂直于AD，CA⊥BA。求x的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(BC),10)
Equal(LengthOfLine(BD),y)
Equal(LengthOfLine(CA),6)
Equal(LengthOfLine(CD),x)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 18/5
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,ADC)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,ADC)
line_addition(1,BD,DC)
Problem: As shown in the diagram, ∠CAD=42°, ∠CFB=77°, triangle DEC is an equilateral triangle, △CAB is an isosceles △, ∠FCG and ∠CGF are the base angles of the isosceles triangle FCG. Find the measure of ∠DCA.
问题：如图所示，∠CAD=42°，∠CFB=77°，△DEC为等边△，CA和CB是等腰三角形CAB的腰，∠FCG和∠CGF是等腰三角形FCG的底角。求∠DCA的大小。
Construction (CDL):
Shape(AD,DC,CA)
Shape(CD,DF,FC)
Shape(CF,FG,GC)
Shape(CG,GE,EC)
Shape(CE,EB,BC)
Collinear(ADFGEB)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),42)
Equal(MeasureOfAngle(CFB),77)
EquilateralTriangle(DEC)
IsoscelesTriangle(CAB)
IsoscelesTriangle(FCG)
Goal: Value(MeasureOfAngle(DCA))
Answer: 18
Theorem sequence:
equilateral_triangle_property_angle(1,DEC)
adjacent_complementary_angle(1,ADC,CDF)
triangle_property_angle_sum(1,ADC)
Problem: As shown in the diagram, AB=9, AC=x, AD=4, BC=y, CD=z, CA⊥DA, DC⊥BC. Find the value of y.
问题：如图所示，AB=9，AC=x，AD=4，BC=y，CD=z，CA⊥DA，DC⊥BC。求y的值。
Construction (CDL):
Shape(BA,AC,CB)
Shape(CA,AD,DC)
Collinear(BAD)
Conditions (CDL):
Equal(LengthOfLine(AB),9)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CD),z)
PerpendicularBetweenLine(CA,DA)
PerpendicularBetweenLine(DC,BC)
Goal: Value(y)
Answer: 3*sqrt(13)
Theorem sequence:
adjacent_complementary_angle(1,BAC,CAD)
right_triangle_judgment_angle(1,BAC)
right_triangle_judgment_angle(1,CAD)
right_triangle_judgment_angle(1,DCB)
right_triangle_property_pythagorean(1,BAC)
right_triangle_property_pythagorean(1,CAD)
right_triangle_property_pythagorean(1,DCB)
line_addition(1,BA,AD)
Problem: As shown in the diagram, AK=10, BL=14, JA=25, BA is parallel to LK. Find the length of line JB.
问题：如图所示，AK=10，BL=14，JA=25，BA∥LK。求直线JB的长度。
Construction (CDL):
Shape(AJ,JB,BA)
Shape(AB,BL,LK,KA)
Collinear(JAK)
Collinear(JBL)
Conditions (CDL):
Equal(LengthOfLine(AK),10)
Equal(LengthOfLine(BL),14)
Equal(LengthOfLine(JA),25)
ParallelBetweenLine(BA,LK)
Goal: Value(LengthOfLine(JB))
Answer: 35
Theorem sequence:
parallel_property_corresponding_angle(1,BA,LK,J)
similar_triangle_judgment_aa(1,AJB,KJL)
line_addition(1,JB,BL)
line_addition(1,JA,AK)
similar_triangle_property_line_ratio(1,AJB,KJL)
similar_triangle_property_line_ratio(1,BAJ,LKJ)
Problem: As shown in the diagram, PR=15, ∠QPR=120°, P is the center of ⊙P. Find the length of ⌒PRQ.
问题：如图所示，PR=15，∠QPR=120°，⊙P的圆心为P。求⌒PRQ的长度。
Construction (CDL):
Shape(PQR,RP,PQ)
Shape(PRQ,QP,PR)
Cocircular(P,QR)
Conditions (CDL):
Equal(LengthOfLine(PR),15)
Equal(MeasureOfAngle(QPR),120)
IsCentreOfCircle(P,P)
Goal: Value(LengthOfArc(PRQ))
Answer: 10*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,PR,P)
arc_property_center_angle(1,PRQ,P)
arc_length_formula(1,PRQ)
Problem: As shown in the diagram, SR=5, TR=3, TS=4, RT is perpendicular to ST. Find the value of tan(TS).
问题：如图所示，SR=5，TR=3，TS=4，RT垂直于ST。求tan(TS)的值。
Construction (CDL):
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(SR),5)
Equal(LengthOfLine(TR),3)
Equal(LengthOfLine(TS),4)
PerpendicularBetweenLine(RT,ST)
Goal: Value(Tan(MeasureOfAngle(TSR)))
Answer: 3/4
Theorem sequence:
cosine_theorem(1,SRT)
Problem: As shown in the diagram, AB=y, CD=9, CE=x, DE=7, XA=20, XB=14, triangle DCE is similar to triangle BAX.. Find the value of y.
问题：如图所示，AB=y，CD=9，CE=x，DE=7，XA=20，XB=14，△DCE相似于△BAX。求y的值。
Construction (CDL):
Shape(ED,DC,CE)
Shape(XB,BA,AX)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(CD),9)
Equal(LengthOfLine(CE),x)
Equal(LengthOfLine(DE),7)
Equal(LengthOfLine(XA),20)
Equal(LengthOfLine(XB),14)
SimilarBetweenTriangle(DCE,BAX)
Goal: Value(y)
Answer: 18
Theorem sequence:
similar_triangle_property_line_ratio(1,EDC,XBA)
similar_triangle_property_line_ratio(1,CED,AXB)
Problem: As shown in the diagram, BD=5, CD=9, BA is perpendicular to CA, CD⊥AD. Find the length of line AD.
问题：如图所示，BD=5，CD=9，BA垂直于CA，CD垂直于AD。求直线AD的长度。
Construction (CDL):
Shape(BA,AD,DB)
Shape(DA,AC,CD)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(BD),5)
Equal(LengthOfLine(CD),9)
PerpendicularBetweenLine(BA,CA)
PerpendicularBetweenLine(CD,AD)
Goal: Value(LengthOfLine(AD))
Answer: 3*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,CDA,ADB)
right_triangle_judgment_angle(1,ADB)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,ADB)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,BAC)
line_addition(1,BD,DC)
Problem: As shown in the diagram, AB=36, CD=12, DE=x, AF⊥EF, BA⊥FA, DC⊥BC, EB⊥AB, ED is perpendicular to CD, FE is perpendicular to BE, quadrilateral AFEB is similar to quadrilateral CBED. Find the value of x.
问题：如图所示，AB=36，CD=12，DE=x，AF⊥EF，BA⊥FA，DC垂直于BC，EB垂直于AB，ED垂直于CD，FE垂直于BE，AFEB相似于CBED。求x的值。
Construction (CDL):
Shape(AF,FE,EB,BA)
Shape(BE,ED,DC,CB)
Collinear(FED)
Collinear(ABC)
Conditions (CDL):
Equal(LengthOfLine(AB),36)
Equal(LengthOfLine(CD),12)
Equal(LengthOfLine(DE),x)
PerpendicularBetweenLine(AF,EF)
PerpendicularBetweenLine(BA,FA)
PerpendicularBetweenLine(DC,BC)
PerpendicularBetweenLine(EB,AB)
PerpendicularBetweenLine(ED,CD)
PerpendicularBetweenLine(FE,BE)
SimilarBetweenQuadrilateral(AFEB,CBED)
Goal: Value(x)
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,FEB,BED)
adjacent_complementary_angle(1,CBE,EBA)
parallelogram_judgment_angle_and_angle(1,AFEB)
parallelogram_judgment_angle_and_angle(1,BEDC)
parallelogram_property_opposite_line_equal(1,FEBA)
parallelogram_property_opposite_line_equal(1,BEDC)
similar_quadrilateral_property_line_ratio(1,FEBA,BEDC)
similar_quadrilateral_property_line_ratio(1,EBAF,EDCB)