SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BA=6, DA=9, DC=7, ∠BAF=32°, ∠CBF=40°, ∠FAD=20°, quadrilateral ADCB is a ▱. Find the measure of ∠BDC.
问题：如图所示，BA=6，DA=9，DC=7，∠BAF=32°，∠CBF=40°，∠FAD=20°，ADCB是▱。求∠BDC的大小。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(FC,CB,BF)
Shape(FB,BA,AF)
Collinear(DFB)
Collinear(AFC)
Conditions (CDL):
Equal(LengthOfLine(BA),6)
Equal(LengthOfLine(DA),9)
Equal(LengthOfLine(DC),7)
Equal(MeasureOfAngle(BAF),32)
Equal(MeasureOfAngle(CBF),40)
Equal(MeasureOfAngle(FAD),20)
Parallelogram(ADCB)
Goal: Value(MeasureOfAngle(BDC))
Answer: 88
Theorem sequence:
angle_addition(1,BAF,FAD)
parallelogram_property_opposite_angle_equal(1,ADCB)
triangle_property_angle_sum(1,CBD)
Problem: As shown in the diagram, the diameter of circle A is 10, the diameter of circle B is 20, the diameter of ⊙C is 14, the center of ⊙A is A, B is the center of circle B, C is the center of circle C. Find the length of line BY.
问题：如图所示，圆A的直径为10，圆B的直径为20，圆C的直径为14，⊙A的圆心为A，⊙B的圆心为B，圆C的圆心为C。求直线BY的长度。
Construction (CDL):
Shape(BEA,BAF,AEF)
Shape(BEA,AD,ADE)
Shape(DA,BAF,AFD)
Shape(BD,ADE,BGE,CGY,YB)
Shape(YB,BD,AFD,BFH,CYH)
Shape(YC,BCG,CGY)
Shape(CY,CYH,BHC)
Shape(BHC,BCG,CHG)
Collinear(ADBYC)
Cocircular(A,EFD)
Cocircular(B,HCGEAF)
Cocircular(C,YHG)
Conditions (CDL):
Equal(DiameterOfCircle(A),10)
Equal(DiameterOfCircle(B),20)
Equal(DiameterOfCircle(C),14)
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
Goal: Value(LengthOfLine(BY))
Answer: 3
Theorem sequence:
line_addition(1,BY,YC)
radius_of_circle_property_length_equal(1,BC,B)
radius_of_circle_property_length_equal(1,CY,C)
circle_property_length_of_radius_and_diameter(1,B)
circle_property_length_of_radius_and_diameter(1,C)
Problem: As shown in the diagram, ∠EBF=26°, the measure of ⌒ACD is 89, the measure of ⌒AEF is x. Find the value of x.
问题：如图所示，∠EBF=26°，⌒ACD的角度为89，弧AEF的角度为x。求x的值。
Construction (CDL):
Shape(FC,ACD,DE,AEF)
Shape(CF,AFC)
Shape(ED,ADE)
Shape(EB,BF,AEF)
Collinear(CFB)
Collinear(DEB)
Cocircular(A,CDEF)
Conditions (CDL):
Equal(MeasureOfAngle(EBF),26)
Equal(MeasureOfArc(ACD),89)
Equal(MeasureOfArc(AEF),x)
Goal: Value(x)
Answer: 37
Theorem sequence:
circle_property_circular_power_segment_and_segment_angle(2,BFC,BED,A)
Problem: As shown in the diagram, AB=4*x-17, CD=2*x-1, ∠BCD=4*y-19°, ∠CBA=3*y+3°, CA and DB are opposite sides of the ▱ ACDB. Find the value of x.
问题：如图所示，AB=4*x-17，CD=2*x-1，∠BCD=4*y-19°，∠CBA=3*y+3°，CA和DB是▱ACDB的一组对边。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Shape(BC,CD,DB)
Conditions (CDL):
Equal(LengthOfLine(AB),4*x-17)
Equal(LengthOfLine(CD),2*x-1)
Equal(MeasureOfAngle(BCD),4*y-19)
Equal(MeasureOfAngle(CBA),3*y+3)
Parallelogram(ACDB)
Goal: Value(x)
Answer: 8
Theorem sequence:
parallelogram_property_opposite_line_equal(1,CDBA)
Problem: As shown in the diagram, AB=9, AC=x, ∠ACB=58°, ∠CBA=35°. Find the value of x.
问题：如图所示，AB=9，AC=x，∠ACB=58°，∠CBA=35°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),9)
Equal(LengthOfLine(AC),x)
Equal(MeasureOfAngle(ACB),58)
Equal(MeasureOfAngle(CBA),35)
Goal: Value(x)
Answer: 9*sin(7*pi/36)/sin(29*pi/90)
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, AB=3, AB=CN, AC=BN, ∠ANB=62°, AC is perpendicular to NC, NB is perpendicular to AB. Find the perimeter of ACNB.
问题：如图所示，AB=3，AB=CN，AC=BN，∠ANB=62°，AC⊥NC，NB垂直于AB。求四边形ACNB的周长。
Construction (CDL):
Shape(AC,CN,NA)
Shape(AN,NB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AB),LengthOfLine(CN))
Equal(LengthOfLine(AC),LengthOfLine(BN))
Equal(MeasureOfAngle(ANB),62)
PerpendicularBetweenLine(AC,NC)
PerpendicularBetweenLine(NB,AB)
Goal: Value(PerimeterOfQuadrilateral(ACNB))
Answer: 6*tan(7*pi/45)+6
Theorem sequence:
triangle_property_angle_sum(1,ANB)
sine_theorem(1,BAN)
quadrilateral_perimeter_formula(1,ACNB)
Problem: As shown in the diagram, ∠HMJ=79°, ∠KML=77°. Find the measure of ∠JMK.
问题：如图所示，∠HMJ=79°，∠KML=77°。求∠JMK的大小。
Construction (CDL):
Shape(HM,MJ,MJH)
Shape(LM,MH,MHL)
Shape(KM,ML,MLK)
Shape(JM,MK,MKJ)
Collinear(HMK)
Collinear(LMJ)
Cocircular(M,HLKJ)
Conditions (CDL):
Equal(MeasureOfAngle(HMJ),79)
Equal(MeasureOfAngle(KML),77)
Goal: Value(MeasureOfAngle(JMK))
Answer: 103
Theorem sequence:
adjacent_complementary_angle(1,JMK,KML)
Problem: As shown in the diagram, ∠ACD=x°, ∠BCA=130°, ∠DCB=60°. Find the value of x.
问题：如图所示，∠ACD=x°，∠BCA=130°，∠DCB=60°。求x的值。
Construction (CDL):
Shape(DC,CB,CBD)
Shape(AC,CD,CDA)
Shape(BC,CA,CAB)
Cocircular(C,DAB)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),x)
Equal(MeasureOfAngle(BCA),130)
Equal(MeasureOfAngle(DCB),60)
Goal: Value(x)
Answer: 170
Theorem sequence:
angle_addition(1,DCB,BCA)
round_angle(1,DCA,ACD)
Problem: As shown in the diagram, ∠MPG=89°, the measure of arc PNE is 66, the center of circle P is P. Find the measure of ∠MNG.
问题：如图所示，∠MPG=89°，弧PNE的角度为66，P是圆P的圆心。求∠MNG的大小。
Construction (CDL):
Shape(GE,PEG)
Shape(MG,PGM)
Shape(NM,PMN)
Shape(EN,PNE)
Shape(GP,PB,BE,EG)
Shape(GM,MP,PG)
Shape(PM,MB,BP)
Shape(BM,MN,NB)
Shape(BN,NE,EB)
Collinear(GPBN)
Collinear(MBE)
Cocircular(P,GMNE)
Conditions (CDL):
Equal(MeasureOfAngle(MPG),89)
Equal(MeasureOfArc(PNE),66)
IsCentreOfCircle(P,P)
Goal: Value(MeasureOfAngle(MNG))
Answer: 89/2
Theorem sequence:
adjacent_complementary_angle(1,NPM,MPG)
triangle_property_angle_sum(1,PMN)
radius_of_circle_property_length_equal(1,PM,P)
radius_of_circle_property_length_equal(1,PN,P)
isosceles_triangle_judgment_line_equal(1,PMN)
isosceles_triangle_property_angle_equal(1,PMN)
Problem: As shown in the diagram, AB=18, AC=9, BC=x, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=18，AC=9，BC=x，BC垂直于AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(AC),9)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 9*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, the area of quadrilateral BDAC is 177, AD=11, AE=13, BC=x, AE⊥BE, BC and DA are the parallel sides of trapezoid BDAC. Find the value of x.
问题：如图所示，BDAC的面积为177，AD=11，AE=13，BC=x，AE⊥BE，BDAC是梯形。求x的值。
Construction (CDL):
Shape(BD,DA,AE,EB)
Shape(EA,AC,CE)
Collinear(BEC)
Conditions (CDL):
Equal(AreaOfQuadrilateral(BDAC),177)
Equal(LengthOfLine(AD),11)
Equal(LengthOfLine(AE),13)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(AE,BE)
Trapezoid(BDAC)
Goal: Value(x)
Answer: 211/13
Theorem sequence:
adjacent_complementary_angle(1,CEA,AEB)
altitude_of_quadrilateral_judgment_left_vertex(2,AE,ACBD)
trapezoid_area_formula(1,ACBD)
Problem: As shown in the diagram, AB=18, AD=y, BC=x, BD=z, ∠ABC=45°, ∠ADB=60°, BA is perpendicular to DA, BC is perpendicular to AC. Find the value of z.
问题：如图所示，AB=18，AD=y，BC=x，BD=z，∠ABC=45°，∠ADB=60°，BA⊥DA，BC⊥AC。求z的值。
Construction (CDL):
Shape(BA,AD,DB)
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(BD),z)
Equal(MeasureOfAngle(ABC),45)
Equal(MeasureOfAngle(ADB),60)
PerpendicularBetweenLine(BA,DA)
PerpendicularBetweenLine(BC,AC)
Goal: Value(z)
Answer: 12*sqrt(3)
Theorem sequence:
sine_theorem(1,BAD)
Problem: As shown in the diagram, RP=y+4, RS=27, TP=2*y-5, TQ=5*x, ∠PQT=95°, ∠RQP=33°, ∠TSP=3*z°, QR and TS are opposite sides of the ▱ QTSR. Find the value of y.
问题：如图所示，RP=y+4，RS=27，TP=2*y-5，TQ=5*x，∠PQT=95°，∠RQP=33°，∠TSP=3*z°，QR和TS是平行四边形QTSR的一组对边。求y的值。
Construction (CDL):
Shape(QT,TP,PQ)
Shape(PT,TS,SP)
Shape(PS,SR,RP)
Shape(PR,RQ,QP)
Collinear(QPS)
Collinear(TPR)
Conditions (CDL):
Equal(LengthOfLine(RP),y+4)
Equal(LengthOfLine(RS),27)
Equal(LengthOfLine(TP),2*y-5)
Equal(LengthOfLine(TQ),5*x)
Equal(MeasureOfAngle(PQT),95)
Equal(MeasureOfAngle(RQP),33)
Equal(MeasureOfAngle(TSP),3*z)
Parallelogram(QTSR)
Goal: Value(y)
Answer: 9
Theorem sequence:
parallelogram_property_diagonal_bisection(1,TSRQ,P)
Problem: As shown in the diagram, AB=6, AC=4, DC=2, CD is perpendicular to AD, CD⊥BD. Find the perimeter of triangle CBA.
问题：如图所示，AB=6，AC=4，DC=2，CD⊥AD，CD垂直于BD。求△CBA的周长。
Construction (CDL):
Shape(CB,BD,DC)
Shape(CD,DA,AC)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(DC),2)
PerpendicularBetweenLine(CD,AD)
PerpendicularBetweenLine(CD,BD)
Goal: Value(PerimeterOfTriangle(CBA))
Answer: 2*sqrt(13-6*sqrt(3))+10
Theorem sequence:
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,CDA)
line_addition(1,BD,DA)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,CDA)
triangle_perimeter_formula(1,BAC)
Problem: As shown in the diagram, ∠BCD=2*x-20°, ∠CDE=x°, ∠DEA=2*x+10°, AB is perpendicular to CB, EA⊥BA. Find the measure of ∠ABC.
问题：如图所示，∠BCD=2*x-20°，∠CDE=x°，∠DEA=2*x+10°，AB⊥CB，EA⊥BA。求∠ABC的大小。
Construction (CDL):
Shape(AB,BC,CD,DE,EA)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),2*x-20)
Equal(MeasureOfAngle(CDE),x)
Equal(MeasureOfAngle(DEA),2*x+10)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(EA,BA)
Goal: Value(MeasureOfAngle(ABC))
Answer: 90
Theorem sequence:
Problem: As shown in the diagram, AB=10, AC=x, BC=y, ∠CBA=45°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=10，AC=x，BC=y，∠CBA=45°，AC⊥BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 5*sqrt(2)
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, ∠CFD=x+36°, ∠DEC=2*y°, ∠ECF=78°, ∠FDE=110°, CE is parallel to FD. Find the value of x.
问题：如图所示，∠CFD=x+36°，∠DEC=2*y°，∠ECF=78°，∠FDE=110°，CE平行于FD。求x的值。
Construction (CDL):
Shape(EC,CF,FD,DE)
Conditions (CDL):
Equal(MeasureOfAngle(CFD),x+36)
Equal(MeasureOfAngle(DEC),2*y)
Equal(MeasureOfAngle(ECF),78)
Equal(MeasureOfAngle(FDE),110)
ParallelBetweenLine(CE,FD)
Goal: Value(x)
Answer: 66
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,CE,FD)
Problem: As shown in the diagram, AB=6, AC=6*sqrt(2), BC=6, ∠BCA=x°. Find the value of x.
问题：如图所示，AB=6，AC=6*sqrt(2)，BC=6，∠BCA=x°。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AC),6*sqrt(2))
Equal(LengthOfLine(BC),6)
Equal(MeasureOfAngle(BCA),x)
Goal: Value(x)
Answer: 45
Theorem sequence:
cosine_theorem(1,CAB)
Problem: As shown in the diagram, ∠GFH=130°, HF⊥JF. Find the measure of ∠JFG.
问题：如图所示，∠GFH=130°，HF垂直于JF。求∠JFG的大小。
Construction (CDL):
Shape(FG,FGJ,JF)
Shape(FJ,FJH,HF)
Shape(FH,FHG,GF)
Cocircular(F,GJH)
Conditions (CDL):
Equal(MeasureOfAngle(GFH),130)
PerpendicularBetweenLine(HF,JF)
Goal: Value(MeasureOfAngle(JFG))
Answer: 140
Theorem sequence:
angle_addition(1,GFH,HFJ)
round_angle(1,GFJ,JFG)
Problem: As shown in the diagram, AE=x, BE=2/3*x, CE=4*y, DE=3*y+4, AC and BD are opposite sides of the ▱ CABD. Find the value of y.
问题：如图所示，AE=x，BE=2/3*x，CE=4*y，DE=3*y+4，四边形CABD是▱。求y的值。
Construction (CDL):
Shape(CA,AE,EC)
Shape(EA,AB,BE)
Shape(EB,BD,DE)
Shape(ED,DC,CE)
Collinear(AED)
Collinear(CEB)
Conditions (CDL):
Equal(LengthOfLine(AE),x)
Equal(LengthOfLine(BE),2/3*x)
Equal(LengthOfLine(CE),4*y)
Equal(LengthOfLine(DE),3*y+4)
Parallelogram(CABD)
Goal: Value(y)
Answer: 4/3
Theorem sequence:
parallelogram_property_diagonal_bisection(1,CABD,E)
parallelogram_property_diagonal_bisection(1,ABDC,E)
Problem: As shown in the diagram, AE=4, BE=3, DA and DC are one pair of adjacent sides of the kite BADC. Find the length of line AB.
问题：如图所示，AE=4，BE=3，DA和DC是风筝形BADC的一组临边。求直线AB的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DC,CE,ED)
Shape(AE,EB,BA)
Shape(EC,CB,BE)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(AE),4)
Equal(LengthOfLine(BE),3)
Kite(BADC)
Goal: Value(LengthOfLine(AB))
Answer: 5
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,BADC,E)
right_triangle_judgment_angle(1,AEB)
right_triangle_property_pythagorean(1,AEB)
Problem: As shown in the diagram, AB=y, AC=5, BC=x, ∠BAC=60°, AC⊥BC. Find the value of y.
问题：如图所示，AB=y，AC=5，BC=x，∠BAC=60°，AC⊥BC。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(BAC),60)
PerpendicularBetweenLine(AC,BC)
Goal: Value(y)
Answer: 10
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,ACB)
Problem: As shown in the diagram, ∠AGE=55°, ∠BGA=x°, ∠GBH=y°, BD∥GA. Find the value of y.
问题：如图所示，∠AGE=55°，∠BGA=x°，∠GBH=y°，BD平行于GA。求y的值。
Construction (CDL):
Shape(HB,BF)
Shape(CG,GB)
Shape(GB,BH)
Shape(EG,GC)
Shape(AG,GE)
Shape(DB,BG)
Shape(BG,GA)
Shape(FB,BD)
Collinear(HBD)
Collinear(CGA)
Collinear(FBGE)
Conditions (CDL):
Equal(MeasureOfAngle(AGE),55)
Equal(MeasureOfAngle(BGA),x)
Equal(MeasureOfAngle(GBH),y)
ParallelBetweenLine(BD,GA)
Goal: Value(y)
Answer: 125
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,BD,GA)
adjacent_complementary_angle(1,BGA,AGE)
adjacent_complementary_angle(1,DBG,GBH)
Problem: As shown in the diagram, ∠LCR=3*x+5°, ∠MCN=60°, ∠RCM=x-1°. Find the measure of ∠LCR.
问题：如图所示，∠LCR=3*x+5°，∠MCN=60°，∠RCM=x-1°。求∠LCR的大小。
Construction (CDL):
Shape(CR,CRL,LC)
Shape(CM,CMR,RC)
Shape(CN,CNM,MC)
Shape(CL,CLN,NC)
Collinear(LCM)
Cocircular(C,LNMR)
Conditions (CDL):
Equal(MeasureOfAngle(LCR),3*x+5)
Equal(MeasureOfAngle(MCN),60)
Equal(MeasureOfAngle(RCM),x-1)
Goal: Value(MeasureOfAngle(LCR))
Answer: 137
Theorem sequence:
adjacent_complementary_angle(1,LCR,RCM)
Problem: As shown in the diagram, CE=7, ED=x, the center of circle C is C, DB is the tangent to ⊙C, CE is perpendicular to DE, DB⊥CB, ED⊥BD. Find the value of x.
问题：如图所示，CE=7，ED=x，⊙C的圆心为C，DB是⊙O的切线，CE垂直于DE，DB⊥CB，ED⊥BD。求x的值。
Construction (CDL):
Shape(ED,DB,CEB)
Shape(CE,CEB,BC)
Shape(CB,CBE,EC)
Cocircular(C,BE)
Conditions (CDL):
Equal(LengthOfLine(CE),7)
Equal(LengthOfLine(ED),x)
IsCentreOfCircle(C,C)
IsTangentOfCircle(DB,C,B)
PerpendicularBetweenLine(CE,DE)
PerpendicularBetweenLine(DB,CB)
PerpendicularBetweenLine(ED,BD)
Goal: Value(x)
Answer: 7
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,DE,BC)
parallel_judgment_ipsilateral_internal_angle(1,EC,DB)
parallelogram_judgment_parallel_and_parallel(1,EDBC)
parallelogram_property_opposite_line_equal(1,EDBC)
radius_of_circle_property_length_equal(1,CE,C)
radius_of_circle_property_length_equal(1,CB,C)
Problem: As shown in the diagram, the height of quadrilateral BCAD is 14, AC=8, DB=4, EF is perpendicular to FF, BD and CA are the parallel sides of trapezoid BCAD. Find the area of BCAD.
问题：如图所示，四边形BCAD的高为14，AC=8，DB=4，EF⊥FF，四边形BCAD是梯形。求四边形BCAD的面积。
Construction (CDL):
Shape(BC,CA,AD,DB)
Conditions (CDL):
Equal(HeightOfQuadrilateral(BCAD),14)
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(DB),4)
PerpendicularBetweenLine(EF,FA)
Trapezoid(BCAD)
Goal: Value(AreaOfQuadrilateral(BCAD))
Answer: 84
Theorem sequence:
trapezoid_area_formula(1,BCAD)
Problem: As shown in the diagram, AB=35, CD=19, the perimeter of quadrilateral DBAC is 74, DBAC is a isosceles trapezoid, CF⊥AF, DE⊥AE. Find the area of DBAC.
问题：如图所示，AB=35，CD=19，四边形DBAC的周长为74，四边形DBAC是等腰梯形，CF垂直于AF，DE垂直于AE。求DBAC的面积。
Construction (CDL):
Shape(DB,BE,ED)
Shape(DE,EF,FC,CD)
Shape(CF,FA,AC)
Collinear(BEFA)
Conditions (CDL):
Equal(LengthOfLine(AB),35)
Equal(LengthOfLine(CD),19)
Equal(PerimeterOfQuadrilateral(DBAC),74)
IsoscelesTrapezoid(DBAC)
PerpendicularBetweenLine(CF,AF)
PerpendicularBetweenLine(DE,AE)
Goal: Value(AreaOfQuadrilateral(DBAC))
Answer: 162
Theorem sequence:
quadrilateral_perimeter_formula(1,DBAC)
adjacent_complementary_angle(1,BED,DEF)
adjacent_complementary_angle(1,BFC,CFA)
altitude_of_quadrilateral_judgment_left_vertex(2,DE,DBAC)
altitude_of_quadrilateral_judgment_right_vertex(2,CF,DBAC)
parallel_judgment_ipsilateral_internal_angle(1,DC,EF)
parallel_judgment_ipsilateral_internal_angle(1,ED,FC)
parallelogram_judgment_parallel_and_parallel(1,DEFC)
parallelogram_property_opposite_line_equal(1,EFCD)
mirror_congruent_triangle_judgment_hl(1,BED,ACF)
mirror_congruent_triangle_property_line_equal(1,DBE,CFA)
line_addition(1,BE,EF)
line_addition(1,BF,FA)
right_triangle_judgment_angle(1,BED)
right_triangle_property_pythagorean(1,BED)
trapezoid_area_formula(1,DBAC)
Problem: As shown in the diagram, ∠RST=63°, the center of ⊙A is A. Find the measure of ⌒ATR.
问题：如图所示，∠RST=63°，A是⊙A的圆心。求⌒ATR的角度。
Construction (CDL):
Shape(SR,ARS)
Shape(TS,AST)
Shape(AT,ATR,RA)
Shape(ST,TA,AR,RS)
Cocircular(A,RST)
Conditions (CDL):
Equal(MeasureOfAngle(RST),63)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfArc(ATR))
Answer: 126
Theorem sequence:
arc_property_circumference_angle_external(1,ATR,S)
Problem: As shown in the diagram, ∠BPY=3*x°, the measure of arc PAC is 2*x+15, the measure of arc PCY is 3*x-3, P is the center of circle P. Find the measure of ⌒PCY.
问题：如图所示，∠BPY=3*x°，弧PAC的角度为2*x+15，弧PCY的角度为3*x-3，P是圆P的圆心。求⌒PCY的角度。
Construction (CDL):
Shape(BP,PY,PYB)
Shape(PC,PCY,YP)
Shape(PA,PAC,CP)
Shape(PX,PXA,AP)
Shape(PB,PBX,XP)
Collinear(BPA)
Cocircular(P,BXACY)
Conditions (CDL):
Equal(MeasureOfAngle(BPY),3*x)
Equal(MeasureOfArc(PAC),2*x+15)
Equal(MeasureOfArc(PCY),3*x-3)
IsCentreOfCircle(P,P)
Goal: Value(MeasureOfArc(PCY))
Answer: 60
Theorem sequence:
arc_property_center_angle(1,PCY,P)
arc_property_center_angle(1,PAC,P)
angle_addition(1,BPY,YPC)
adjacent_complementary_angle(1,BPC,CPA)
Problem: As shown in the diagram, ∠FYD=4*y+10°, EH is parallel to BY, YH is perpendicular to EH. Find the value of y.
问题：如图所示，∠FYD=4*y+10°，EH平行于BY，YH垂直于EH。求y的值。
Construction (CDL):
Shape(EH,HJ)
Shape(JH,HA)
Shape(YH,HE)
Shape(BY,YH)
Shape(DY,YB)
Shape(FY,YD)
Shape(HY,YF)
Shape(AH,HY)
Collinear(JHYD)
Collinear(EHA)
Collinear(BYF)
Conditions (CDL):
Equal(MeasureOfAngle(FYD),4*y+10)
ParallelBetweenLine(EH,BY)
PerpendicularBetweenLine(YH,EH)
Goal: Value(y)
Answer: 20
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,YB,HE)
vertical_angle(1,BYH,FYD)
Problem: As shown in the diagram, ∠CBD=12°, the measure of arc GEB is 28. Find the measure of ∠BKE.
问题：如图所示，∠CBD=12°，⌒GEB的角度为28。求∠BKE的大小。
Construction (CDL):
Shape(GCE,EK,KC)
Shape(GEB,BK,KE)
Shape(GBD,DB)
Shape(KB,BD,DK)
Shape(GDC,CK,KD)
Collinear(EKD)
Collinear(CKB)
Cocircular(G,EBDC)
Conditions (CDL):
Equal(MeasureOfAngle(CBD),12)
Equal(MeasureOfArc(GEB),28)
Goal: Value(MeasureOfAngle(BKE))
Answer: 26
Theorem sequence:
arc_property_circumference_angle_external(1,GEB,D)
triangle_property_angle_sum(1,KBD)
adjacent_complementary_angle(1,DKB,BKE)
Problem: As shown in the diagram, BC=32, BC=AB, BD=y, CD=x, ∠CAB=54°, BD⊥AD, CD is perpendicular to BD. Find the value of y.
问题：如图所示，BC=32，BC=AB，BD=y，CD=x，∠CAB=54°，BD⊥AD，CD垂直于BD。求y的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(BC),32)
Equal(LengthOfLine(BC),LengthOfLine(AB))
Equal(LengthOfLine(BD),y)
Equal(LengthOfLine(CD),x)
Equal(MeasureOfAngle(CAB),54)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CD,BD)
Goal: Value(y)
Answer: 8+8*sqrt(5)
Theorem sequence:
sine_theorem(1,BDA)
Problem: As shown in the diagram, DE=7, EX=24, the center of ⊙D is D, DA is perpendicular to XA, XE is perpendicular to DE. Find the length of line AX.
问题：如图所示，DE=7，EX=24，D是圆D的圆心，DA垂直于XA，XE垂直于DE。求直线AX的长度。
Construction (CDL):
Shape(AX,XQ,DAQ)
Shape(DAQ,QD,DA)
Shape(DT,DTA,AD)
Shape(QX,XE,DQE)
Shape(DQ,DQE,ED)
Shape(DE,DET,TD)
Collinear(XQDT)
Cocircular(D,AQET)
Conditions (CDL):
Equal(LengthOfLine(DE),7)
Equal(LengthOfLine(EX),24)
IsCentreOfCircle(D,D)
PerpendicularBetweenLine(DA,XA)
PerpendicularBetweenLine(XE,DE)
Goal: Value(LengthOfLine(AX))
Answer: 24
Theorem sequence:
radius_of_circle_property_length_equal(1,DA,D)
radius_of_circle_property_length_equal(1,DE,D)
right_triangle_judgment_angle(1,XED)
right_triangle_judgment_angle(1,DAX)
right_triangle_property_pythagorean(1,DAX)
right_triangle_property_pythagorean(1,XED)
Problem: As shown in the diagram, ∠CBI=84°, ∠HBD=x°, ∠IBH=16°, DB is perpendicular to CB. Find the value of x.
问题：如图所示，∠CBI=84°，∠HBD=x°，∠IBH=16°，DB⊥CB。求x的值。
Construction (CDL):
Shape(BC,BCD,DB)
Shape(BD,BDH,HB)
Shape(BH,BHI,IB)
Shape(BI,BIC,CB)
Cocircular(B,DHIC)
Conditions (CDL):
Equal(MeasureOfAngle(CBI),84)
Equal(MeasureOfAngle(HBD),x)
Equal(MeasureOfAngle(IBH),16)
PerpendicularBetweenLine(DB,CB)
Goal: Value(x)
Answer: 170
Theorem sequence:
angle_addition(1,DBC,CBI)
angle_addition(1,IBH,HBD)
round_angle(1,DBI,IBD)
Problem: As shown in the diagram, ∠AQT=32°, ∠QSR=5*x+4°, ∠QTR=6*x-2°. Find the measure of ∠SRT.
问题：如图所示，∠AQT=32°，∠QSR=5*x+4°，∠QTR=6*x-2°。求∠SRT的大小。
Construction (CDL):
Shape(CSR,RS)
Shape(AS,SR,RA)
Shape(AR,CRQ,QA)
Shape(AQ,QT,TA)
Shape(CQT,TQ)
Shape(AT,CTS,SA)
Collinear(RAT)
Collinear(QAS)
Cocircular(C,RQTS)
Conditions (CDL):
Equal(MeasureOfAngle(AQT),32)
Equal(MeasureOfAngle(QSR),5*x+4)
Equal(MeasureOfAngle(QTR),6*x-2)
Goal: Value(MeasureOfAngle(SRT))
Answer: 32
Theorem sequence:
arc_property_circumference_angle_external(1,CTS,R)
arc_property_circumference_angle_external(1,CTS,Q)
Problem: As shown in the diagram, WZ=4, XW=3, XZ=b, XW is perpendicular to ZW, quadrilateral XWZY is a rectangle. Find the length of line YW.
问题：如图所示，WZ=4，XW=3，XZ=b，XW⊥ZW，XWZY是矩形。求直线YW的长度。
Construction (CDL):
Shape(AY,YX,XA)
Shape(AX,XW,WA)
Shape(AW,WZ,ZA)
Shape(AZ,ZY,YA)
Collinear(XAZ)
Collinear(WAY)
Conditions (CDL):
Equal(LengthOfLine(WZ),4)
Equal(LengthOfLine(XW),3)
Equal(LengthOfLine(XZ),b)
PerpendicularBetweenLine(XW,ZW)
Rectangle(XWZY)
Goal: Value(LengthOfLine(YW))
Answer: 5
Theorem sequence:
right_triangle_judgment_angle(1,XWZ)
right_triangle_property_pythagorean(1,XWZ)
rectangle_property_diagonal_equal(1,XWZY)
Problem: As shown in the diagram, AB=5*y, DC=y+8, ∠CDA=4*x°, ∠DAB=2*x-6°, quadrilateral DABC is a parallelogram. Find the value of y.
问题：如图所示，AB=5*y，DC=y+8，∠CDA=4*x°，∠DAB=2*x-6°，四边形DABC是平行四边形。求y的值。
Construction (CDL):
Shape(DA,AB,BC,CD)
Conditions (CDL):
Equal(LengthOfLine(AB),5*y)
Equal(LengthOfLine(DC),y+8)
Equal(MeasureOfAngle(CDA),4*x)
Equal(MeasureOfAngle(DAB),2*x-6)
Parallelogram(DABC)
Goal: Value(y)
Answer: 2
Theorem sequence:
parallelogram_property_opposite_line_equal(1,CDAB)
Problem: As shown in the diagram, BE=16, BF=x, CF=12, DE=16, C is the center of circle C, the tangent to ⊙C is BE, DE is the tangent to circle C. Find the value of x.
问题：如图所示，BE=16，BF=x，CF=12，DE=16，C是⊙C的圆心，BE是⊙O的切线，DE是⊙O的切线。求x的值。
Construction (CDL):
Shape(BE,CFE,FB)
Shape(FC,CA,CAF)
Shape(CEA,ED,DA)
Shape(CF,CFE,EC)
Shape(CE,CEA,AC)
Collinear(BFC)
Collinear(CAD)
Collinear(BED)
Cocircular(C,FEA)
Conditions (CDL):
Equal(LengthOfLine(BE),16)
Equal(LengthOfLine(BF),x)
Equal(LengthOfLine(CF),12)
Equal(LengthOfLine(DE),16)
IsCentreOfCircle(C,C)
IsTangentOfCircle(BE,C)
IsTangentOfCircle(DE,C)
Goal: Value(x)
Answer: 8
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BE,C,C)
radius_of_circle_property_length_equal(1,CF,C)
radius_of_circle_property_length_equal(1,CE,C)
line_addition(1,CF,FB)
right_triangle_judgment_angle(1,BEC)
right_triangle_property_pythagorean(1,BEC)
Problem: As shown in the diagram, BG=7, DC=14, the center of ⊙B is B, quadrilateral ADCE is a square. Find the area of ADCE minus the area of the circle B.
问题：如图所示，BG=7，DC=14，⊙B的圆心为B，四边形ADCE是正方形。求ADCE的面积减去⊙B的面积。
Construction (CDL):
Shape(AF,BIF,IA)
Shape(FD,DG,BFG)
Shape(GC,CH,BGH)
Shape(HE,EI,BHI)
Shape(BIF,BFG,BGH,BHI)
Shape(BG)
Collinear(AFD)
Collinear(DGC)
Collinear(CHE)
Collinear(AIE)
Cocircular(B,FGHI)
Conditions (CDL):
Equal(LengthOfLine(BG),7)
Equal(LengthOfLine(DC),14)
IsCentreOfCircle(B,B)
Square(ADCE)
Goal: Value(Sub(AreaOfQuadrilateral(ADCE),AreaOfCircle(B)))
Answer: 196-49*pi
Theorem sequence:
parallelogram_area_formula_sine(1,ADCE)
radius_of_circle_property_length_equal(1,BG,B)
circle_area_formula(1,B)
Problem: As shown in the diagram, ∠BCD=2*x+4°, ∠ECB=2*x-4°. Find the measure of ∠ECB.
问题：如图所示，∠BCD=2*x+4°，∠ECB=2*x-4°。求∠ECB的大小。
Construction (CDL):
Shape(EC,CB)
Shape(BC,CD)
Shape(DC,CA)
Shape(AC,CE)
Collinear(BCA)
Collinear(ECD)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),2*x+4)
Equal(MeasureOfAngle(ECB),2*x-4)
Goal: Value(MeasureOfAngle(ECB))
Answer: 86
Theorem sequence:
adjacent_complementary_angle(1,ECB,BCD)