SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BF=14, DB=21, EC=24, FH=8, BD is parallel to CE, CB∥AD, FA is parallel to HP, FH⊥PH, PE⊥AE. Find the sum of the area of BCPD and the area of triangle APE.
问题：如图所示，BF=14，DB=21，EC=24，FH=8，BD∥CE，CB平行于AD，FA∥HP，FH垂直于PH，PE⊥AE。求四边形BCPD的面积与三角形APE的面积之和。
Construction (CDL):
Shape(BG,GF,FB)
Shape(GC,CH,HF,FG)
Shape(AF,FH,HP,PA)
Shape(AD,DB,BF,FA)
Shape(AP,PE,EA)
Collinear(BGC)
Collinear(BFH)
Collinear(CHPE)
Collinear(DAP)
Conditions (CDL):
Equal(LengthOfLine(BF),14)
Equal(LengthOfLine(DB),21)
Equal(LengthOfLine(EC),24)
Equal(LengthOfLine(FH),8)
ParallelBetweenLine(BD,CE)
ParallelBetweenLine(CB,AD)
ParallelBetweenLine(FA,HP)
PerpendicularBetweenLine(FH,PH)
PerpendicularBetweenLine(PE,AE)
Goal: Value(Add(AreaOfQuadrilateral(BCPD),AreaOfTriangle(APE)))
Answer: 474
Theorem sequence:
line_addition(1,BF,FH)
line_addition(1,CP,PE)
parallel_property_collinear_extend(3,EC,DB,P)
parallel_property_collinear_extend(2,DA,BC,P)
parallel_property_collinear_extend(1,PH,AF,E)
adjacent_complementary_angle(1,CHF,FHP)
perpendicular_judgment_angle(1,HE,AE)
perpendicular_judgment_angle(1,FH,EH)
parallel_judgment_per_per(1,HF,EA)
parallelogram_judgment_parallel_and_parallel(1,BCPD)
parallelogram_judgment_parallel_and_parallel(1,AFHE)
parallelogram_property_opposite_line_equal(1,FHEA)
parallelogram_property_opposite_line_equal(1,DBCP)
altitude_of_quadrilateral_judgment_left_vertex(1,BH,BCPD)
parallelogram_area_formula_common(1,BCPD)
triangle_area_formula_sine(1,EAP)
Problem: As shown in the diagram, CB=40, DB=38, SD=28, ACBS is a parallelogram, BD is perpendicular to SD. Find the perimeter of ACBS.
问题：如图所示，CB=40，DB=38，SD=28，AS和CB是平行四边形ACBS的一组对边，BD⊥SD。求ACBS的周长。
Construction (CDL):
Shape(AC,CB,BS,SA)
Shape(SB,BD,DS)
Collinear(ASD)
Conditions (CDL):
Equal(LengthOfLine(CB),40)
Equal(LengthOfLine(DB),38)
Equal(LengthOfLine(SD),28)
Parallelogram(ACBS)
PerpendicularBetweenLine(BD,SD)
Goal: Value(PerimeterOfQuadrilateral(ACBS))
Answer: 80+4*sqrt(557)
Theorem sequence:
right_triangle_judgment_angle(1,BDS)
right_triangle_property_pythagorean(1,BDS)
parallelogram_property_opposite_line_equal(1,ACBS)
parallelogram_property_opposite_line_equal(1,CBSA)
quadrilateral_perimeter_formula(1,ACBS)
Problem: As shown in the diagram, AN=5, BC=12, BD=8, CN=15, BC is perpendicular to EC, DB⊥CB, DE is perpendicular to NE, EN is perpendicular to AN. Find the sum of the area of BCED and the area of DENA.
问题：如图所示，AN=5，BC=12，BD=8，CN=15，BC垂直于EC，DB垂直于CB，DE垂直于NE，EN⊥AN。求BCED的面积与DENA的面积之和。
Construction (CDL):
Shape(DB,BC,CE,ED)
Shape(DE,EN,NA,AD)
Collinear(CEN)
Conditions (CDL):
Equal(LengthOfLine(AN),5)
Equal(LengthOfLine(BC),12)
Equal(LengthOfLine(BD),8)
Equal(LengthOfLine(CN),15)
PerpendicularBetweenLine(BC,EC)
PerpendicularBetweenLine(DB,CB)
PerpendicularBetweenLine(DE,NE)
PerpendicularBetweenLine(EN,AN)
Goal: Value(Add(AreaOfQuadrilateral(BCED),AreaOfQuadrilateral(DENA)))
Answer: 311/2
Theorem sequence:
line_addition(1,CE,EN)
adjacent_complementary_angle(1,CED,DEN)
perpendicular_judgment_angle(1,CE,DE)
parallel_judgment_per_per(1,ED,NA)
parallel_judgment_per_per(1,CB,ED)
parallel_judgment_per_per(1,BD,CE)
parallelogram_judgment_parallel_and_parallel(1,BCED)
parallelogram_property_opposite_line_equal(1,BCED)
parallelogram_property_opposite_line_equal(1,DBCE)
trapezoid_judgment_parallel(1,ENAD)
right_trapezoid_judgment_right_angle(1,ENAD)
right_trapezoid_area_formular(1,ENAD)
parallelogram_area_formula_sine(1,BCED)
Problem: As shown in the diagram, DE=14+2*x, GF=4*x-4*3+6, DG⊥GG, DGFE is a rectangle. Find the length of line GF.
问题：如图所示，DE=14+2*x，GF=4*x-4*3+6，DG垂直于GG，DGFE是长方形。求直线GF的长度。
Construction (CDL):
Shape(DG,GB,BD)
Shape(ED,DB,BE)
Shape(BG,GF,FB)
Shape(EB,BF,FE)
Collinear(DBF)
Collinear(GBE)
Conditions (CDL):
Equal(LengthOfLine(DE),14+2*x)
Equal(LengthOfLine(GF),4*x-4*3+6)
PerpendicularBetweenLine(DG,GF)
Rectangle(DGFE)
Goal: Value(LengthOfLine(GF))
Answer: 34
Theorem sequence:
parallelogram_property_opposite_line_equal(1,EDGF)
Problem: As shown in the diagram, RS=3*x+2, SR=RT, TR=2*x+4. Find the value of x.
问题：如图所示，RS=3*x+2，SR=RT，TR=2*x+4。求x的值。
Construction (CDL):
Shape(RT,TS,SR)
Conditions (CDL):
Equal(LengthOfLine(RS),3*x+2)
Equal(LengthOfLine(SR),LengthOfLine(RT))
Equal(LengthOfLine(TR),2*x+4)
Goal: Value(x)
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, ∠ABF=32°, EB is perpendicular to CB. Find the measure of ∠FBE.
问题：如图所示，∠ABF=32°，EB垂直于CB。求∠FBE的大小。
Construction (CDL):
Shape(FB,BE)
Shape(EB,BC)
Shape(AB,BE)
Shape(DB,DA)
Shape(CB,BD)
Collinear(FBC)
Collinear(EBD)
Conditions (CDL):
Equal(MeasureOfAngle(ABF),32)
PerpendicularBetweenLine(EB,CB)
Goal: Value(MeasureOfAngle(FBE))
Answer: 90
Theorem sequence:
adjacent_complementary_angle(1,FBE,EBC)
Problem: As shown in the diagram, ∠DCB=120°, BADC is a rhombus. Find the measure of ∠CAD.
问题：如图所示，∠DCB=120°，四边形BADC是菱形。求∠CAD的大小。
Construction (CDL):
Shape(CB,BA,AC)
Shape(CA,AD,DC)
Conditions (CDL):
Equal(MeasureOfAngle(DCB),120)
Rhombus(BADC)
Goal: Value(MeasureOfAngle(CAD))
Answer: 60
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,DA,CB)
isosceles_triangle_judgment_line_equal(1,DCA)
isosceles_triangle_property_angle_equal(1,DCA)
triangle_property_angle_sum(1,DCA)
Problem: As shown in the diagram, ST=8, ∠RST=72°, the center of ⊙S is S. Find the area of the sector STR.
问题：如图所示，ST=8，∠RST=72°，⊙S的圆心为S。求扇形STR的面积。
Construction (CDL):
Shape(RS,ST,STR)
Cocircular(S,TR)
Conditions (CDL):
Equal(LengthOfLine(ST),8)
Equal(MeasureOfAngle(RST),72)
IsCentreOfCircle(S,S)
Goal: Value(AreaOfSector(STR))
Answer: 64*pi/5
Theorem sequence:
radius_of_circle_property_length_equal(1,ST,S)
arc_property_center_angle(1,STR,S)
sector_area_formula(1,STR)
Problem: As shown in the diagram, AC=12, BA=8, D is the center of ⊙D, the diameter of ⊙D is BC. Find the circumference of the ⊙D.
问题：如图所示，AC=12，BA=8，⊙D的圆心为D，BC是⊙D的直径。求⊙D的周长。
Construction (CDL):
Shape(CD,DB,DBC)
Shape(DC,CA,AB,BD)
Shape(AC,DCA)
Shape(BA,DAB)
Collinear(CDB)
Cocircular(D,CAB)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(BA),8)
IsCentreOfCircle(D,D)
IsDiameterOfCircle(BC,D)
Goal: Value(PerimeterOfCircle(D))
Answer: 4*sqrt(13)*pi
Theorem sequence:
diameter_of_circle_property_right_angle(1,CAB,D)
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
line_addition(1,CD,DB)
radius_of_circle_property_length_equal(1,DC,D)
radius_of_circle_property_length_equal(1,DB,D)
circle_perimeter_formula(1,D)
Problem: As shown in the diagram, MN=3*x-4, MQ=4*w-3, NQ=15, PN=2*y+5, PQ=11.1, RM=17, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, RM and PN are opposite sides of the parallelogram MRPN. Find the value of w.
问题：如图所示，MN=3*x-4，MQ=4*w-3，NQ=15，PN=2*y+5，PQ=11.1，RM=17，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，MRPN是▱。求w的值。
Construction (CDL):
Shape(MR,RQ,QM)
Shape(NM,MQ,QN)
Shape(QR,RP,PQ)
Shape(NQ,QP,PN)
Collinear(MQP)
Collinear(RQN)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(MQ),4*w-3)
Equal(LengthOfLine(NQ),15)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),11.1)
Equal(LengthOfLine(RM),17)
Equal(LengthOfLine(RP),20)
Equal(LengthOfLine(RQ),3*z-3)
Equal(MeasureOfAngle(MRQ),38)
Equal(MeasureOfAngle(NQP),83)
Equal(MeasureOfAngle(QNM),33)
Parallelogram(MRPN)
Goal: Value(w)
Answer: 3.525000
Theorem sequence:
parallelogram_property_diagonal_bisection(1,MRPN,Q)
Problem: As shown in the diagram, AB=x, AC=20, BC=22, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=x，AC=20，BC=22，BC垂直于AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BC),22)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 2*sqrt(221)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠DAF=4*x+5°, ∠FAB=9*x+20°, ∠FBC=4*y+4°, ∠FDA=y**2-1°, ABCD is a rectangle. Find the value of x.
问题：如图所示，∠DAF=4*x+5°，∠FAB=9*x+20°，∠FBC=4*y+4°，∠FDA=y**2-1°，四边形ABCD是矩形。求x的值。
Construction (CDL):
Shape(DA,AF,FD)
Shape(AB,BF,FA)
Shape(FB,BC,CF)
Shape(DF,FC,CD)
Collinear(AFC)
Collinear(BFD)
Conditions (CDL):
Equal(MeasureOfAngle(DAF),4*x+5)
Equal(MeasureOfAngle(FAB),9*x+20)
Equal(MeasureOfAngle(FBC),4*y+4)
Equal(MeasureOfAngle(FDA),y**2-1)
Rectangle(ABCD)
Goal: Value(x)
Answer: 5
Theorem sequence:
angle_addition(1,DAF,FAB)
Problem: As shown in the diagram, ∠ABC=4*x°, ∠BCA=3*x°, ∠CAB=5*x°. Find the value of x.
问题：如图所示，∠ABC=4*x°，∠BCA=3*x°，∠CAB=5*x°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),4*x)
Equal(MeasureOfAngle(BCA),3*x)
Equal(MeasureOfAngle(CAB),5*x)
Goal: Value(x)
Answer: 15
Theorem sequence:
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, QT=11, the center of circle Q is Q. Find the length of line QU.
问题：如图所示，QT=11，Q是⊙Q的圆心。求直线QU的长度。
Construction (CDL):
Shape(TQ,QV,QVT)
Shape(VQ,QU,QUV)
Shape(UQ,QT,QTU)
Collinear(TQU)
Cocircular(Q,TUV)
Conditions (CDL):
Equal(LengthOfLine(QT),11)
IsCentreOfCircle(Q,Q)
Goal: Value(LengthOfLine(QU))
Answer: 11
Theorem sequence:
radius_of_circle_property_length_equal(1,QT,Q)
radius_of_circle_property_length_equal(1,QU,Q)
Problem: As shown in the diagram, AC=2*x+2, AD=y+10, CB=2*y+5, DB=x+5, BC and DA are opposite sides of the parallelogram BDAC. Find the value of y.
问题：如图所示，AC=2*x+2，AD=y+10，CB=2*y+5，DB=x+5，BDAC是平行四边形。求y的值。
Construction (CDL):
Shape(BD,DA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),2*x+2)
Equal(LengthOfLine(AD),y+10)
Equal(LengthOfLine(CB),2*y+5)
Equal(LengthOfLine(DB),x+5)
Parallelogram(BDAC)
Goal: Value(y)
Answer: 5
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DACB)
Problem: As shown in the diagram, EF=60, FD=65, A is the center of circle A, FD is the diameter of circle A. Find the length of line ED.
问题：如图所示，EF=60，FD=65，圆A的圆心为A，FD是圆A的直径。求直线ED的长度。
Construction (CDL):
Shape(ED,DA,AF,FE)
Shape(DE,AED)
Shape(EF,AFE)
Shape(DA，AF,AFD)
Collinear(DAF)
Cocircular(A,EDF)
Conditions (CDL):
Equal(LengthOfLine(EF),60)
Equal(LengthOfLine(FD),65)
IsCentreOfCircle(A,A)
IsDiameterOfCircle(FD,A)
Goal: Value(LengthOfLine(ED))
Answer: 25
Theorem sequence:
diameter_of_circle_property_right_angle(1,FED,A)
right_triangle_judgment_angle(1,FED)
right_triangle_property_pythagorean(1,FED)
Problem: As shown in the diagram, AB=3*x, CD=9*x, ST=15, S bisects segment AC, T is the midpoint of segment BD, quadrilateral ACDB is a trapezoid. Find the value of x.
问题：如图所示，AB=3*x，CD=9*x，ST=15，S是线段AC的中点，T平分线段BD，ACDB是梯形。求x的值。
Construction (CDL):
Shape(AS,ST,TB,BA)
Shape(SC,CD,DT,TS)
Collinear(ASC)
Collinear(BTD)
Conditions (CDL):
Equal(LengthOfLine(AB),3*x)
Equal(LengthOfLine(CD),9*x)
Equal(LengthOfLine(ST),15)
IsMidpointOfLine(S,AC)
IsMidpointOfLine(T,BD)
Trapezoid(ACDB)
Goal: Value(x)
Answer: 5/2
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,ST,ACDB)
midsegment_of_quadrilateral_property_length(1,ST,ACDB)
Problem: As shown in the diagram, ∠AGB=30°, G is the center of ⊙G, CG is perpendicular to DG. Find the measure of arc GFD.
问题：如图所示，∠AGB=30°，圆G的圆心为G，CG⊥DG。求弧GFD的角度。
Construction (CDL):
Shape(AG,GB,GBA)
Shape(BG,GC,GCB)
Shape(CG,GD,GDC)
Shape(DG,GF,GFD)
Shape(FG,GA,GAF)
Collinear(AGD)
Collinear(BGF)
Cocircular(G,CBAFD)
Conditions (CDL):
Equal(MeasureOfAngle(AGB),30)
IsCentreOfCircle(G,G)
PerpendicularBetweenLine(CG,DG)
Goal: Value(MeasureOfArc(GFD))
Answer: 30
Theorem sequence:
vertical_angle(1,AGB,DGF)
arc_property_center_angle(1,GFD,G)
Problem: As shown in the diagram, SB=5, ∠TBS=126°, the measure of arc BQR is 108, the center of circle B is B. Find the measure of arc BST.
问题：如图所示，SB=5，∠TBS=126°，弧BQR的角度为108，⊙B的圆心为B。求弧BST的角度。
Construction (CDL):
Shape(QB,BT,BTQ)
Shape(TB,BS,BST)
Shape(SB,BR,BRS)
Shape(RB,BQ,BQR)
Cocircular(B,QRST)
Conditions (CDL):
Equal(LengthOfLine(SB),5)
Equal(MeasureOfAngle(TBS),126)
Equal(MeasureOfArc(BQR),108)
IsCentreOfCircle(B,B)
Goal: Value(MeasureOfArc(BST))
Answer: 126
Theorem sequence:
arc_property_center_angle(1,BST,B)
Problem: As shown in the diagram, AB=x, AC=y, AD=z, BD=8, CD=3, BD⊥AD, CA is perpendicular to BA. Find the value of y.
问题：如图所示，AB=x，AC=y，AD=z，BD=8，CD=3，BD垂直于AD，CA垂直于BA。求y的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AD),z)
Equal(LengthOfLine(BD),8)
Equal(LengthOfLine(CD),3)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: sqrt(33)
Theorem sequence:
line_addition(1,BD,DC)
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,ADC)
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,ADC)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, AC=3*x-7, BD=2*x+8, AD and CB are the non-parallel sides of isosceles trapezoid ADCB, AB and DC are the parallel sides of trapezoid ADCB. Find the value of x.
问题：如图所示，AC=3*x-7，BD=2*x+8，AD和CB是等腰梯形ADCB的腰，ADCB是梯形。求x的值。
Construction (CDL):
Shape(PA,AD,DP)
Shape(PD,DC,CP)
Shape(BP,PC,CB)
Shape(BA,AP,PB)
Collinear(APC)
Collinear(DPB)
Conditions (CDL):
Equal(LengthOfLine(AC),3*x-7)
Equal(LengthOfLine(BD),2*x+8)
IsoscelesTrapezoid(ADCB)
Trapezoid(ADCB)
Goal: Value(x)
Answer: 15
Theorem sequence:
isosceles_trapezoid_property_diagonal_equal(1,ADCB)
Problem: As shown in the diagram, AB=19, AC=x, BC=y, ∠CBA=45°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=19，AC=x，BC=y，∠CBA=45°，AC⊥BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),19)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 19*sqrt(2)/2
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, DE=8, EH=8, DE⊥HE, EH⊥KH, HK⊥DK, KD is perpendicular to ED. Find the area of quadrilateral DEHK.
问题：如图所示，DE=8，EH=8，DE⊥HE，EH垂直于KH，HK垂直于DK，KD垂直于ED。求四边形DEHK的面积。
Construction (CDL):
Shape(HK,KD,DE,EH)
Conditions (CDL):
Equal(LengthOfLine(DE),8)
Equal(LengthOfLine(EH),8)
PerpendicularBetweenLine(DE,HE)
PerpendicularBetweenLine(EH,KH)
PerpendicularBetweenLine(HK,DK)
PerpendicularBetweenLine(KD,ED)
Goal: Value(AreaOfQuadrilateral(DEHK))
Answer: 64
Theorem sequence:
parallel_judgment_per_per(1,KH,DE)
parallel_judgment_per_per(1,DK,EH)
parallelogram_judgment_parallel_and_parallel(1,DEHK)
parallelogram_area_formula_sine(1,DEHK)
Problem: As shown in the diagram, KL=NL, NM=ML, ∠JLK=25°, ∠KLN=18°, ∠NKJ=130°, ∠NLM=20°, △LNK is an isosceles △, triangleMNL is an isosceles triangle. Find the measure of ∠NKL.
问题：如图所示，KL=NL，NM=ML，∠JLK=25°，∠KLN=18°，∠NKJ=130°，∠NLM=20°，三角形LNK为等腰三角形，MN和ML是等腰三角形MNL的腰。求∠NKL的大小。
Construction (CDL):
Shape(MN,NL,LM)
Shape(NK,KL,LN)
Shape(KJ,JL,LK)
Conditions (CDL):
Equal(LengthOfLine(KL),LengthOfLine(NL))
Equal(LengthOfLine(NM),LengthOfLine(ML))
Equal(MeasureOfAngle(JLK),25)
Equal(MeasureOfAngle(KLN),18)
Equal(MeasureOfAngle(NKJ),130)
Equal(MeasureOfAngle(NLM),20)
IsoscelesTriangle(LNK)
IsoscelesTriangle(MNL)
Goal: Value(MeasureOfAngle(NKL))
Answer: 81
Theorem sequence:
isosceles_triangle_property_angle_equal(1,LNK)
triangle_property_angle_sum(1,NKL)
Problem: As shown in the diagram, AB=7, AI=7, the center of ⊙C is C, O is the center of ⊙O, the diameter of ⊙O is DB, the diameter of ⊙C is ID, AB is the tangent to circle O, AI is the tangent to ⊙C, IA⊥BA. Find the sum of the area of quadrilateral ABDI, half the area of circle C, and half the area of circle O.
问题：如图所示，AB=7，AI=7，⊙C的圆心为C，O是⊙O的圆心，⊙O的直径为DB，ID是圆C的直径，圆O的切线为AB，AI是圆O的切线，IA垂直于BA。求四边形ABDI的面积、1/2圆C面积和1/2圆O面积之和。
Construction (CDL):
Shape(IC,CD,CDI)
Shape(DO,OB,OBD)
Shape(DC,CI,IA,AB,BO,OD)
Collinear(DOB)
Collinear(DCI)
Cocircular(C,ID)
Cocircular(O,DB)
Conditions (CDL):
Equal(LengthOfLine(AB),7)
Equal(LengthOfLine(AI),7)
IsCentreOfCircle(C,C)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(DB,O)
IsDiameterOfCircle(ID,C)
IsTangentOfCircle(AB,O)
IsTangentOfCircle(AI,C)
PerpendicularBetweenLine(IA,BA)
Goal: Value(Add(AreaOfQuadrilateral(ABDI),Mul(1/2,AreaOfCircle(C)),Mul(1/2,AreaOfCircle(O))))
Answer: 49*pi/4+49
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AI,C,C)
tangent_of_circle_property_perpendicular(2,AB,O,O)
perpendicular_judgment_angle(1,DI,AI)
perpendicular_judgment_angle(1,AB,DB)
parallel_judgment_per_per(1,ID,AB)
parallel_judgment_per_per(1,AI,BD)
parallelogram_judgment_parallel_and_parallel(1,ABDI)
parallelogram_area_formula_sine(1,ABDI)
parallelogram_property_opposite_line_equal(1,ABDI)
parallelogram_property_opposite_line_equal(1,BDIA)
diameter_of_circle_property_length_equal(1,ID,C)
diameter_of_circle_property_length_equal(1,DB,O)
circle_property_length_of_radius_and_diameter(1,C)
circle_property_length_of_radius_and_diameter(1,O)
circle_area_formula(1,C)
circle_area_formula(1,O)
Problem: As shown in the diagram, AC=2*x+9, BD=4*x-5, ∠CAE=42°, ∠DCE=4*y°, ∠EAB=23°, ∠EBD=83°, quadrilateral ABDC is a parallelogram. Find the value of x.
问题：如图所示，AC=2*x+9，BD=4*x-5，∠CAE=42°，∠DCE=4*y°，∠EAB=23°，∠EBD=83°，四边形ABDC是平行四边形。求x的值。
Construction (CDL):
Shape(CA,AE,EC)
Shape(DC,CE,ED)
Shape(EA,AB,BE)
Shape(DE,EB,BD)
Collinear(CEB)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AC),2*x+9)
Equal(LengthOfLine(BD),4*x-5)
Equal(MeasureOfAngle(CAE),42)
Equal(MeasureOfAngle(DCE),4*y)
Equal(MeasureOfAngle(EAB),23)
Equal(MeasureOfAngle(EBD),83)
Parallelogram(ABDC)
Goal: Value(x)
Answer: 7
Theorem sequence:
parallelogram_property_opposite_line_equal(1,CABD)
Problem: As shown in the diagram, ∠AEF=148°, ∠AEF=∠ZXY, ∠FAE=18°, ∠XYZ=3*x+y°, ∠XYZ=∠EFA, ∠YZX=5*x-y°. Find the value of y.
问题：如图所示，∠AEF=148°，∠AEF=∠ZXY，∠FAE=18°，∠XYZ=3*x+y°，∠XYZ=∠EFA，∠YZX=5*x-y°。求y的值。
Construction (CDL):
Shape(AE,EF,FA)
Shape(XY,YZ,ZX)
Conditions (CDL):
Equal(MeasureOfAngle(AEF),148)
Equal(MeasureOfAngle(AEF),MeasureOfAngle(ZXY))
Equal(MeasureOfAngle(FAE),18)
Equal(MeasureOfAngle(XYZ),3*x+y)
Equal(MeasureOfAngle(XYZ),MeasureOfAngle(EFA))
Equal(MeasureOfAngle(YZX),5*x-y)
Goal: Value(y)
Answer: 2
Theorem sequence:
triangle_property_angle_sum(1,AEF)
triangle_property_angle_sum(1,XYZ)
Problem: As shown in the diagram, AE=28, BE=16, DE=20, LD=15, AC⊥DC, DE is perpendicular to AE, EA⊥CA, LD is perpendicular to CD. Find the sum of the area of △LBE, the area of △LDC and the area of quadrilateral CDEA.
问题：如图所示，AE=28，BE=16，DE=20，LD=15，AC垂直于DC，DE垂直于AE，EA⊥CA，LD垂直于CD。求△LBE的面积、△LDC的面积与CDEA的面积之和。
Construction (CDL):
Shape(LB,BE,ED,DL)
Shape(LD,DC,CL)
Shape(CD,DE,EA,AC)
Collinear(LDE)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(AE),28)
Equal(LengthOfLine(BE),16)
Equal(LengthOfLine(DE),20)
Equal(LengthOfLine(LD),15)
PerpendicularBetweenLine(AC,DC)
PerpendicularBetweenLine(DE,AE)
PerpendicularBetweenLine(EA,CA)
PerpendicularBetweenLine(LD,CD)
Goal: Value(Add(AreaOfTriangle(LBE),AreaOfTriangle(LDC),AreaOfQuadrilateral(CDEA)))
Answer: 1050
Theorem sequence:
line_addition(1,LD,DE)
adjacent_complementary_angle(1,BED,DEA)
parallel_judgment_per_per(1,ED,AC)
parallel_judgment_per_per(1,AE,CD)
parallelogram_judgment_parallel_and_parallel(1,ACDE)
parallelogram_property_opposite_line_equal(1,ACDE)
parallelogram_property_opposite_line_equal(1,CDEA)
parallelogram_area_formula_sine(1,EACD)
triangle_area_formula_sine(1,ELB)
triangle_area_formula_sine(1,DCL)
Problem: As shown in the diagram, AD=21, AE=16, BF=7, DC=9, EB=6, CD⊥AD, DA⊥GA, EB is perpendicular to FB, FC⊥DC, GE is perpendicular to BE. Find the sum of the area of quadrilateral AGCD and the area of GEBF.
问题：如图所示，AD=21，AE=16，BF=7，DC=9，EB=6，CD⊥AD，DA⊥GA，EB垂直于FB，FC⊥DC，GE⊥BE。求AGCD的面积与四边形GEBF的面积之和。
Construction (CDL):
Shape(AG,GF,FA)
Shape(AF,FC,CD,DA)
Shape(GE,EB,BF,FG)
Collinear(GFC)
Collinear(AGE)
Conditions (CDL):
Equal(LengthOfLine(AD),21)
Equal(LengthOfLine(AE),16)
Equal(LengthOfLine(BF),7)
Equal(LengthOfLine(DC),9)
Equal(LengthOfLine(EB),6)
PerpendicularBetweenLine(CD,AD)
PerpendicularBetweenLine(DA,GA)
PerpendicularBetweenLine(EB,FB)
PerpendicularBetweenLine(FC,DC)
PerpendicularBetweenLine(GE,BE)
Goal: Value(Add(AreaOfQuadrilateral(AGCD),AreaOfQuadrilateral(GEBF)))
Answer: 231
Theorem sequence:
parallel_judgment_per_per(1,CF,DA)
parallel_judgment_per_per(1,DC,AG)
parallel_judgment_per_per(1,EG,BF)
parallel_property_collinear_extend(2,CF,DA,G)
parallelogram_judgment_parallel_and_parallel(1,AGCD)
parallelogram_area_formula_sine(1,CDAG)
rectangle_judgment_right_angle(1,CDAG)
adjacent_complementary_angle(1,AGF,FGE)
perpendicular_judgment_angle(1,FG,EG)
parallel_judgment_per_per(1,GF,EB)
parallelogram_judgment_parallel_and_parallel(1,GEBF)
parallelogram_area_formula_sine(1,EBFG)
Problem: As shown in the diagram, MK=20, MP=16, YQ=x, YX=15, ∠MLP=∠YZQ, ∠PKM=∠QXY, LP⊥MP, ZQ is perpendicular to YQ. Find the value of x.
问题：如图所示，MK=20，MP=16，YQ=x，YX=15，∠MLP=∠YZQ，∠PKM=∠QXY，LP⊥MP，ZQ垂直于YQ。求x的值。
Construction (CDL):
Shape(KM,MP,PK)
Shape(PM,ML,LP)
Shape(XY,YQ,QX)
Shape(YZ,ZQ,QY)
Collinear(KPL)
Collinear(ZQX)
Conditions (CDL):
Equal(LengthOfLine(MK),20)
Equal(LengthOfLine(MP),16)
Equal(LengthOfLine(YQ),x)
Equal(LengthOfLine(YX),15)
Equal(MeasureOfAngle(MLP),MeasureOfAngle(YZQ))
Equal(MeasureOfAngle(PKM),MeasureOfAngle(QXY))
PerpendicularBetweenLine(LP,MP)
PerpendicularBetweenLine(ZQ,YQ)
Goal: Value(x)
Answer: 12
Theorem sequence:
adjacent_complementary_angle(1,LPM,MPK)
adjacent_complementary_angle(1,ZQY,YQX)
similar_triangle_judgment_aa(1,MPK,YQX)
similar_triangle_property_line_ratio(1,PKM,QXY)
similar_triangle_property_line_ratio(1,KMP,XYQ)
Problem: As shown in the diagram, ∠GHJ=2*x°, ∠MGI=54°, JF∥DM. Find the value of x.
问题：如图所示，∠GHJ=2*x°，∠MGI=54°，JF∥DM。求x的值。
Construction (CDL):
Shape(JH,HB)
Shape(BH,HF)
Shape(FH,HG)
Shape(GH,HJ)
Shape(DG,GH)
Shape(HG,GM)
Shape(MG,GI)
Shape(IG,GD)
Collinear(BHGI)
Collinear(JHF)
Collinear(DGM)
Conditions (CDL):
Equal(MeasureOfAngle(GHJ),2*x)
Equal(MeasureOfAngle(MGI),54)
ParallelBetweenLine(JF,DM)
Goal: Value(x)
Answer: 63
Theorem sequence:
vertical_angle(1,DGH,MGI)
parallel_property_collinear_extend(3,JF,DM,H)
parallel_property_collinear_extend(3,MD,HJ,G)
parallel_property_ipsilateral_internal_angle(1,GD,HJ)
Problem: As shown in the diagram, SU=20, YW=20, the measure of arc RYX is 45, R is the center of circle R, RV is perpendicular to SV, YZ⊥RZ. Find the length of line WZ.
问题：如图所示，SU=20，YW=20，弧RYX的角度为45，圆R的圆心为R，RV垂直于SV，YZ⊥RZ。求直线WZ的长度。
Construction (CDL):
Shape(SV,VT,RTS)
Shape(TV,VU,SUT)
Shape(YZ,ZR,RV,VS,RSY)
Shape(UV,VR,RZ,ZW,RWU)
Shape(XZ,ZY,RYX)
Shape(WZ,ZX,RXW)
Collinear(YZW)
Collinear(RZX)
Collinear(RVT)
Collinear(SVU)
Cocircular(R,SYXWUT)
Conditions (CDL):
Equal(LengthOfLine(SU),20)
Equal(LengthOfLine(YW),20)
Equal(MeasureOfArc(RYX),45)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(RV,SV)
PerpendicularBetweenLine(YZ,RZ)
Goal: Value(LengthOfLine(WZ))
Answer: 10
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,R,RZ,YW)
line_addition(1,YZ,ZW)
Problem: As shown in the diagram, AB=12, AC=y, AD=AC, ∠ACB=60°, AB is perpendicular to DB. Find the value of y.
问题：如图所示，AB=12，AC=y，AD=AC，∠ACB=60°，AB垂直于DB。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Shape(AB,BD,DA)
Collinear(CBD)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AD),LengthOfLine(AC))
Equal(MeasureOfAngle(ACB),60)
PerpendicularBetweenLine(AB,DB)
Goal: Value(y)
Answer: 8*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,CBA,ABD)
sine_theorem(1,ACB)
Problem: As shown in the diagram, TV=13, VW=2, TW⊥UW, VU is perpendicular to TU. Find the length of line VU.
问题：如图所示，TV=13，VW=2，TW⊥UW，VU垂直于TU。求直线VU的长度。
Construction (CDL):
Shape(TW,WU,UT)
Shape(WV,VU,UW)
Collinear(VWT)
Conditions (CDL):
Equal(LengthOfLine(TV),13)
Equal(LengthOfLine(VW),2)
PerpendicularBetweenLine(TW,UW)
PerpendicularBetweenLine(VU,TU)
Goal: Value(LengthOfLine(VU))
Answer: sqrt(26)
Theorem sequence:
line_addition(1,VW,WT)
adjacent_complementary_angle(1,TWU,UWV)
right_triangle_judgment_angle(1,TWU)
right_triangle_judgment_angle(1,UWV)
right_triangle_judgment_angle(1,VUT)
right_triangle_property_pythagorean(1,TWU)
right_triangle_property_pythagorean(1,UWV)
right_triangle_property_pythagorean(1,VUT)