SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=12, CD=8, DA=CB, ∠ADE=30°, BF is perpendicular to CF, DE⊥AE, quadrilateral CBAD is a trapezoid. Find the perimeter of quadrilateral ADCB.
问题：如图所示，AB=12，CD=8，DA=CB，∠ADE=30°，BF垂直于CF，DE⊥AE，四边形CBAD是梯形。求四边形ADCB的周长。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DC,CF,FE)
Shape(FC,CB,BF)
Collinear(AEFB)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(CD),8)
Equal(LengthOfLine(DA),LengthOfLine(CB))
Equal(MeasureOfAngle(ADE),30)
PerpendicularBetweenLine(BF,CF)
PerpendicularBetweenLine(DE,AE)
Trapezoid(CBAD)
Goal: Value(PerimeterOfQuadrilateral(ADCB))
Answer: 28
Theorem sequence:
parallel_property_collinear_extend(3,AB,DC,E)
parallel_property_collinear_extend(3,EB,DC,F)
adjacent_complementary_angle(1,BFC,CFA)
parallel_judgment_corresponding_angle(2,FC,ED,A)
parallelogram_judgment_parallel_and_parallel(1,DCFE)
parallelogram_property_opposite_line_equal(1,FEDC)
isosceles_trapezoid_judgment_line_equal(1,CBAD)
isosceles_trapezoid_property_angle_equal(1,CBAD)
mirror_congruent_triangle_judgment_aas(3,DEA,CBF)
mirror_congruent_triangle_property_line_equal(1,DEA,CBF)
line_addition(1,AE,EF)
line_addition(1,AF,FB)
sine_theorem(1,ADE)
quadrilateral_perimeter_formula(1,ADCB)
Problem: As shown in the diagram, AB=24, AX=x, BC=7, C is the center of ⊙C, AB is the tangent to circle C. Find the value of x.
问题：如图所示，AB=24，AX=x，BC=7，⊙C的圆心为C，圆O的切线为AB。求x的值。
Construction (CDL):
Shape(CX,CXB,BC)
Shape(CB,CBX,XC)
Shape(AX,CBX,BA)
Collinear(CXA)
Cocircular(C,BX)
Conditions (CDL):
Equal(LengthOfLine(AB),24)
Equal(LengthOfLine(AX),x)
Equal(LengthOfLine(BC),7)
IsCentreOfCircle(C,C)
IsTangentOfCircle(AB,C)
Goal: Value(x)
Answer: 18
Theorem sequence:
radius_of_circle_property_length_equal(1,CB,C)
radius_of_circle_property_length_equal(1,CX,C)
tangent_of_circle_property_perpendicular(1,AB,C,C)
line_addition(1,CX,XA)
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, AB=60, DE=48, C is the center of circle C, the diameter of ⊙C is AB, DF is perpendicular to CF. Find the length of line CF.
问题：如图所示，AB=60，DE=48，圆C的圆心为C，AB是圆C的直径，DF垂直于CF。求直线CF的长度。
Construction (CDL):
Shape(EF,FA,CAE)
Shape(AF,FD,CDA)
Shape(FC,CD,DF)
Shape(FE,CEB,BF)
Shape(CB,CBD,DC)
Collinear(AFCB)
Collinear(EFD)
Cocircular(C,AEBD)
Conditions (CDL):
Equal(LengthOfLine(AB),60)
Equal(LengthOfLine(DE),48)
IsCentreOfCircle(C,C)
IsDiameterOfCircle(AB,C)
PerpendicularBetweenLine(DF,CF)
Goal: Value(LengthOfLine(CF))
Answer: 18
Theorem sequence:
diameter_of_circle_property_length_equal(1,AB,C)
circle_property_length_of_radius_and_diameter(1,C)
radius_of_circle_property_length_equal(1,CD,C)
circle_property_chord_perpendicular_bisect_chord(1,C,CF,DE)
line_addition(1,EF,FD)
right_triangle_judgment_angle(1,DFC)
right_triangle_property_pythagorean(1,DFC)
Problem: As shown in the diagram, the measure of arc GAB is 102, the center of circle G is G, the tangent to ⊙G is CB. Find the measure of ∠ABC.
问题：如图所示，⌒GAB的角度为102，G是圆G的圆心，CB是⊙O的切线。求∠ABC的大小。
Construction (CDL):
Shape(GA,AB,BG)
Shape(GB,GBA,AG)
Shape(BA,GAB)
Shape(GB,BC)
Cocircular(G,AB)
Conditions (CDL):
Equal(MeasureOfArc(GAB),102)
IsCentreOfCircle(G,G)
IsTangentOfCircle(CB,G)
Goal: Value(MeasureOfAngle(ABC))
Answer: 129
Theorem sequence:
arc_property_center_angle(1,GAB,G)
tangent_of_circle_property_perpendicular(1,CB,G,G)
radius_of_circle_property_length_equal(1,GA,G)
radius_of_circle_property_length_equal(1,GB,G)
isosceles_triangle_judgment_line_equal(1,GAB)
isosceles_triangle_property_angle_equal(1,GAB)
triangle_property_angle_sum(1,GAB)
angle_addition(1,ABG,GBC)
Problem: As shown in the diagram, ∠AFD=3*x-9°, ∠ECP=57°, CP∥FA. Find the value of x.
问题：如图所示，∠AFD=3*x-9°，∠ECP=57°，CP平行于FA。求x的值。
Construction (CDL):
Shape(GC,CE)
Shape(EC,CP)
Shape(FC,CG)
Shape(PC,CF)
Shape(BF,FC)
Shape(CF,FA)
Shape(DF,FB)
Shape(AF,FD)
Collinear(GCP)
Collinear(BFA)
Collinear(ECFD)
Conditions (CDL):
Equal(MeasureOfAngle(AFD),3*x-9)
Equal(MeasureOfAngle(ECP),57)
ParallelBetweenLine(CP,FA)
Goal: Value(x)
Answer: 44
Theorem sequence:
parallel_property_corresponding_angle(1,CP,FA,E)
adjacent_complementary_angle(1,CFA,AFD)
Problem: As shown in the diagram, NR=3, PR=x, QR=10, RM=6. Find the value of x.
问题：如图所示，NR=3，PR=x，QR=10，RM=6。求x的值。
Construction (CDL):
Shape(MR,RN,ANM)
Shape(RM,AMQ,QR)
Shape(RA,AQP,PR)
Shape(NR,RP,APN)
Collinear(NRQ)
Collinear(MRP)
Cocircular(A,MQPN)
Conditions (CDL):
Equal(LengthOfLine(NR),3)
Equal(LengthOfLine(PR),x)
Equal(LengthOfLine(QR),10)
Equal(LengthOfLine(RM),6)
Goal: Value(x)
Answer: 5
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,PRM,NRQ,A)
Problem: As shown in the diagram, AP=3*x-1, PC=x+9, DP is perpendicular to AP, quadrilateral ADCB is a rhombus. Find the length of line AC.
问题：如图所示，AP=3*x-1，PC=x+9，DP垂直于AP，ADCB是菱形。求直线AC的长度。
Construction (CDL):
Shape(AD,DP,PA)
Shape(AP,PB,BA)
Shape(PC,CB,BP)
Shape(PD,DC,CP)
Collinear(APC)
Collinear(BPD)
Conditions (CDL):
Equal(LengthOfLine(AP),3*x-1)
Equal(LengthOfLine(PC),x+9)
PerpendicularBetweenLine(DP,AP)
Rhombus(ADCB)
Goal: Value(LengthOfLine(AC))
Answer: 28
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ADCB,P)
line_addition(1,AP,PC)
Problem: As shown in the diagram, BA=6, DA=9, FC=6.86, ∠BAF=32°, ∠CBF=40.1°, ∠FAD=20°, ADCB is a ▱. Find the measure of ∠DCB.
问题：如图所示，BA=6，DA=9，FC=6.86，∠BAF=32°，∠CBF=40.1°，∠FAD=20°，DA和CB是平行四边形ADCB的一组对边。求∠DCB的大小。
Construction (CDL):
Shape(DC,CF,FD)
Shape(DF,FA,AD)
Shape(FB,BA,AF)
Shape(FC,CB,BF)
Collinear(DFB)
Collinear(AFC)
Conditions (CDL):
Equal(LengthOfLine(BA),6)
Equal(LengthOfLine(DA),9)
Equal(LengthOfLine(FC),6.86)
Equal(MeasureOfAngle(BAF),32)
Equal(MeasureOfAngle(CBF),40.1)
Equal(MeasureOfAngle(FAD),20)
Parallelogram(ADCB)
Goal: Value(MeasureOfAngle(DCB))
Answer: 52
Theorem sequence:
angle_addition(1,BAF,FAD)
parallelogram_property_opposite_angle_equal(1,ADCB)
Problem: As shown in the diagram, AB=6, BC=y, YA=x, ∠BCA=45°, AY⊥CY, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AB=6，BC=y，YA=x，∠BCA=45°，AY垂直于CY，CA垂直于BA。求x的值。
Construction (CDL):
Shape(AB,BY,YA)
Shape(AY,YC,CA)
Collinear(BYC)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(YA),x)
Equal(MeasureOfAngle(BCA),45)
PerpendicularBetweenLine(AY,CY)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 3*sqrt(2)
Theorem sequence:
adjacent_complementary_angle(1,BYA,AYC)
triangle_property_angle_sum(1,ABC)
sine_theorem(1,ABY)
Problem: As shown in the diagram, ∠AFB=8*x-6°, ∠AFC=90°, ∠BFC=14*x+8°. Find the value of x.
问题：如图所示，∠AFB=8*x-6°，∠AFC=90°，∠BFC=14*x+8°。求x的值。
Construction (CDL):
Shape(BF,FC)
Shape(CF,FD)
Shape(DF,FE)
Shape(AF,FB)
Collinear(BFE)
Conditions (CDL):
Equal(MeasureOfAngle(AFB),8*x-6)
Equal(MeasureOfAngle(AFC),90)
Equal(MeasureOfAngle(BFC),14*x+8)
Goal: Value(x)
Answer: 4
Theorem sequence:
angle_addition(1,AFB,BFC)
Problem: As shown in the diagram, HG=12, ∠FGH=60°, ∠HFG=60°. Find the length of line FH.
问题：如图所示，HG=12，∠FGH=60°，∠HFG=60°。求直线FH的长度。
Construction (CDL):
Shape(FG,GH,HF)
Conditions (CDL):
Equal(LengthOfLine(HG),12)
Equal(MeasureOfAngle(FGH),60)
Equal(MeasureOfAngle(HFG),60)
Goal: Value(LengthOfLine(FH))
Answer: 12
Theorem sequence:
isosceles_triangle_judgment_angle_equal(1,HFG)
Problem: As shown in the diagram, ∠HEL=55°, ∠IHF=110°, KC is parallel to GF. Find the measure of ∠GEA.
问题：如图所示，∠HEL=55°，∠IHF=110°，KC平行于GF。求∠GEA的大小。
Construction (CDL):
Shape(DI,IC)
Shape(CI,IH)
Shape(IH,HF)
Shape(FH,HB)
Shape(AI,ID)
Shape(HI,IA)
Shape(EH,HI)
Shape(BH,HE)
Shape(JA,AI)
Shape(IA,AE)
Shape(AE,EH)
Shape(HE,EL)
Shape(KA,AJ)
Shape(EA,AK)
Shape(GE,EA)
Shape(LE,EG)
Collinear(CIAK)
Collinear(FHEG)
Collinear(DIHB)
Collinear(JAEL)
Conditions (CDL):
Equal(MeasureOfAngle(HEL),55)
Equal(MeasureOfAngle(IHF),110)
ParallelBetweenLine(KC,GF)
Goal: Value(MeasureOfAngle(GEA))
Answer: 55
Theorem sequence:
vertical_angle(1,HEL,GEA)
Problem: As shown in the diagram, AB=34, BC=x, ∠CBA=31°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=34，BC=x，∠CBA=31°，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),34)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(CBA),31)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 34*sin(59*pi/180)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,BAC)
Problem: As shown in the diagram, ∠ABC=30°, ∠BCA=x°, ∠CAB=x°. Find the value of x.
问题：如图所示，∠ABC=30°，∠BCA=x°，∠CAB=x°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),30)
Equal(MeasureOfAngle(BCA),x)
Equal(MeasureOfAngle(CAB),x)
Goal: Value(x)
Answer: 75
Theorem sequence:
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, AC=20, BE=12, DE=15, DF=10, DF⊥AF, FEBC is a rectangle. Find the sum of the area of triangle DFA and the area of quadrilateral FEBC.
问题：如图所示，AC=20，BE=12，DE=15，DF=10，DF垂直于AF，FEBC是矩形。求△DFA的面积与四边形FEBC的面积之和。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FE,EB,BC,CF)
Collinear(FDE)
Collinear(AFC)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BE),12)
Equal(LengthOfLine(DE),15)
Equal(LengthOfLine(DF),10)
PerpendicularBetweenLine(DF,AF)
Rectangle(FEBC)
Goal: Value(Add(AreaOfTriangle(DFA),AreaOfQuadrilateral(FEBC)))
Answer: 340
Theorem sequence:
parallelogram_property_opposite_line_equal(1,CFEB)
line_addition(1,AF,FC)
triangle_area_formula_sine(1,FAD)
line_addition(1,FD,DE)
parallelogram_area_formula_sine(1,CFEB)
Problem: As shown in the diagram, MN=3*x-4, NQ=15, PN=2*y+5, PQ=11.1, RM=18, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, MRPN is a parallelogram. Find the value of z.
问题：如图所示，MN=3*x-4，NQ=15，PN=2*y+5，PQ=11.1，RM=18，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，四边形MRPN是平行四边形。求z的值。
Construction (CDL):
Shape(MR,RQ,QM)
Shape(QR,RP,PQ)
Shape(QP,PN,NQ)
Shape(MQ,QN,NM)
Collinear(MQP)
Collinear(RQN)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(NQ),15)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),11.1)
Equal(LengthOfLine(RM),18)
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(z)
Answer: 6
Theorem sequence:
parallelogram_property_diagonal_bisection(1,NMRP,Q)
Problem: As shown in the diagram, AC=CL, AL=3*sqrt(2), LB=6, ALBT is a kite, LC⊥AC. Find the area of quadrilateral ALBT.
问题：如图所示，AC=CL，AL=3*sqrt(2)，LB=6，BL和BT是风筝形ALBT的一组临边，LC⊥AC。求四边形ALBT的面积。
Construction (CDL):
Shape(AL,LC,CA)
Shape(BC,CL,LB)
Shape(BT,TC,CB)
Shape(CT,TA,AC)
Collinear(LCT)
Collinear(BCA)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(CL))
Equal(LengthOfLine(AL),3*sqrt(2))
Equal(LengthOfLine(LB),6)
Kite(ALBT)
PerpendicularBetweenLine(LC,AC)
Goal: Value(AreaOfQuadrilateral(ALBT))
Answer: 9+9*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,LCA)
right_triangle_property_pythagorean(1,LCA)
adjacent_complementary_angle(1,BCL,LCA)
right_triangle_judgment_angle(1,BCL)
right_triangle_property_pythagorean(1,BCL)
kite_property_diagonal_perpendicular_bisection(1,ALBT,C)
line_addition(1,BC,CA)
line_addition(1,LC,CT)
kite_area_formula_diagonal(1,ALBT)
Problem: As shown in the diagram, CA=y, DA=3, DB=8, DC=x, AC is perpendicular to BC, BD⊥CD. Find the value of x.
问题：如图所示，CA=y，DA=3，DB=8，DC=x，AC⊥BC，BD垂直于CD。求x的值。
Construction (CDL):
Shape(BD,DC,CB)
Shape(CD,DA,AC)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(CA),y)
Equal(LengthOfLine(DA),3)
Equal(LengthOfLine(DB),8)
Equal(LengthOfLine(DC),x)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BD,CD)
Goal: Value(x)
Answer: 2*sqrt(6)
Theorem sequence:
adjacent_complementary_angle(1,BDC,CDA)
line_addition(1,BD,DA)
mirror_similar_triangle_judgment_aa(1,BAC,CDA)
mirror_similar_triangle_property_line_ratio(1,BAC,CDA)
mirror_similar_triangle_property_line_ratio(1,CBA,DAC)
right_triangle_judgment_angle(1,CDA)
right_triangle_property_pythagorean(1,CDA)
Problem: As shown in the diagram, the area of quadrilateral SRUT is 675, RB=15, TB=15, TB⊥UB, quadrilateral SRUT is a rhombus. Find the length of line SU.
问题：如图所示，四边形SRUT的面积为675，RB=15，TB=15，TB垂直于UB，SRUT是菱形。求直线SU的长度。
Construction (CDL):
Shape(RU,UB,BR)
Shape(BU,UT,TB)
Shape(BT,TS,SB)
Shape(RB,BS,SR)
Collinear(RBT)
Collinear(SBU)
Conditions (CDL):
Equal(AreaOfQuadrilateral(SRUT),675)
Equal(LengthOfLine(RB),15)
Equal(LengthOfLine(TB),15)
PerpendicularBetweenLine(TB,UB)
Rhombus(SRUT)
Goal: Value(LengthOfLine(SU))
Answer: 45
Theorem sequence:
line_addition(1,RB,BT)
kite_area_formula_diagonal(1,SRUT)
Problem: As shown in the diagram, AB=6*sqrt(2), AC=AB, the center of circle I is I, BC is the diameter of circle I, CA⊥BA. Find the circumference of the circle I.
问题：如图所示，AB=6*sqrt(2)，AC=AB，圆I的圆心为I，圆I的直径为BC，CA垂直于BA。求⊙I的周长。
Construction (CDL):
Shape(BA,IAB)
Shape(AB,BC,CA)
Shape(AC,ICA)
Shape(CB,IBC)
Collinear(BIC)
Cocircular(I,ABC)
Conditions (CDL):
Equal(LengthOfLine(AB),6*sqrt(2))
Equal(LengthOfLine(AC),LengthOfLine(AB))
IsCentreOfCircle(I,I)
IsDiameterOfCircle(BC,I)
PerpendicularBetweenLine(CA,BA)
Goal: Value(PerimeterOfCircle(I))
Answer: 12*pi
Theorem sequence:
cosine_theorem(1,ABC)
diameter_of_circle_property_length_equal(1,BC,I)
circle_property_length_of_radius_and_diameter(1,I)
circle_perimeter_formula(1,I)
Problem: As shown in the diagram, SU=20, YW=20, the measure of arc RYX is 45, R is the center of circle R, RZ is perpendicular to WZ, UV is perpendicular to RV. Find the length of line SV.
问题：如图所示，SU=20，YW=20，弧RYX的角度为45，R是圆R的圆心，RZ⊥WZ，UV⊥RV。求直线SV的长度。
Construction (CDL):
Shape(RV,VS,RSY,YZ,ZR)
Shape(ZY,RYX,XZ)
Shape(ZX,RXW,WZ)
Shape(RZ,ZW,RWU,UV,VR)
Shape(TV,VU,RUT)
Shape(SV,VT,RTS)
Collinear(SVU)
Collinear(RVT)
Collinear(RZX)
Collinear(YZW)
Cocircular(R,YXWUTS)
Conditions (CDL):
Equal(LengthOfLine(SU),20)
Equal(LengthOfLine(YW),20)
Equal(MeasureOfArc(RYX),45)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(RZ,WZ)
PerpendicularBetweenLine(UV,RV)
Goal: Value(LengthOfLine(SV))
Answer: 10
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,R,RV,US)
line_addition(1,SV,VU)
Problem: As shown in the diagram, CB=14, ∠CBA=60°, BA is perpendicular to CA. Find the length of line AC.
问题：如图所示，CB=14，∠CBA=60°，BA垂直于CA。求直线AC的长度。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(CB),14)
Equal(MeasureOfAngle(CBA),60)
PerpendicularBetweenLine(BA,CA)
Goal: Value(LengthOfLine(AC))
Answer: 7*sqrt(3)
Theorem sequence:
sine_theorem(1,CBA)
Problem: As shown in the diagram, ∠CXG=x-8°, ∠DEA=120°, XG∥EL. Find the value of x.
问题：如图所示，∠CXG=x-8°，∠DEA=120°，XG∥EL。求x的值。
Construction (CDL):
Shape(FX,XC)
Shape(CX,XG)
Shape(EX,XF)
Shape(GX,XE)
Shape(AE,EX)
Shape(XE,EL)
Shape(AE,ED)
Shape(LE,ED)
Collinear(CXED)
Collinear(FXG)
Collinear(AEL)
Conditions (CDL):
Equal(MeasureOfAngle(CXG),x-8)
Equal(MeasureOfAngle(DEA),120)
ParallelBetweenLine(XG,EL)
Goal: Value(x)
Answer: 128
Theorem sequence:
parallel_property_corresponding_angle(1,XG,EL,C)
vertical_angle(1,XEL,DEA)
Problem: As shown in the diagram, AB=12, EB=9, ∠DBA=55°, ADCB is a rhombus. Find the length of line AE.
问题：如图所示，AB=12，EB=9，∠DBA=55°，四边形ADCB是菱形。求直线AE的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(AE,EB,BA)
Shape(ED,DC,CE)
Shape(BE,EC,CB)
Collinear(AEC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(EB),9)
Equal(MeasureOfAngle(DBA),55)
Rhombus(ADCB)
Goal: Value(LengthOfLine(AE))
Answer: 3*sqrt(25-24*cos(11*pi/36))
Theorem sequence:
cosine_theorem(1,BAE)
Problem: As shown in the diagram, JL=y+4/5, LC=2*y-11/5, MI=12-3*x, TM=10-2*x, TM=MI, JT is parallel to LM, LM is parallel to CI, quadrilateral JLMT is similar to quadrilateral JCIT. Find the value of y.
问题：如图所示，JL=y+4/5，LC=2*y-11/5，MI=12-3*x，TM=10-2*x，TM=MI，JT∥LM，LM∥CI，四边形JLMT相似于四边形JCIT。求y的值。
Construction (CDL):
Shape(NJ,JR)
Shape(RJ,JT)
Shape(NT,TQ)
Shape(QT,TA)
Shape(LJ,JN)
Shape(JL,LM,MT,TJ)
Shape(AT,TM)
Shape(VL,LJ)
Shape(TM,MS)
Shape(CL,LV)
Shape(SM,MJ)
Shape(ZC,CL)
Shape(MI,IO)
Shape(LC,CI,IM,ML)
Shape(UC,CZ)
Shape(JC,CU)
Shape(PI,IC)
Shape(OI,IP)
Collinear(RJLCU)
Collinear(QTMIP)
Collinear(NJTA)
Collinear(VLMS)
Collinear(ZCIO)
Conditions (CDL):
Equal(LengthOfLine(JL),y+4/5)
Equal(LengthOfLine(LC),2*y-11/5)
Equal(LengthOfLine(MI),12-3*x)
Equal(LengthOfLine(TM),10-2*x)
Equal(LengthOfLine(TM),LengthOfLine(MI))
ParallelBetweenLine(JT,LM)
ParallelBetweenLine(LM,CI)
SimilarBetweenQuadrilateral(JLMT,JCIT)
Goal: Value(y)
Answer: 3
Theorem sequence:
line_addition(1,JL,LC)
line_addition(1,TM,MI)
similar_quadrilateral_property_line_ratio(1,JLMT,JCIT)
similar_quadrilateral_property_line_ratio(1,MTJL,ITJC)
Problem: As shown in the diagram, AB=40, AL=27, LN=36, AN and BC are opposite sides of the ▱ NABC, NL⊥AL. Find the perimeter of quadrilateral NABC.
问题：如图所示，AB=40，AL=27，LN=36，NC和AB是▱NABC的一组对边，NL⊥AL。求四边形NABC的周长。
Construction (CDL):
Shape(NA,AB,BC,CN)
Shape(NL,LA,AN)
Collinear(LAB)
Conditions (CDL):
Equal(LengthOfLine(AB),40)
Equal(LengthOfLine(AL),27)
Equal(LengthOfLine(LN),36)
Parallelogram(NABC)
PerpendicularBetweenLine(NL,AL)
Goal: Value(PerimeterOfQuadrilateral(NABC))
Answer: 170
Theorem sequence:
right_triangle_judgment_angle(1,NLA)
right_triangle_property_pythagorean(1,NLA)
parallelogram_property_opposite_line_equal(1,NABC)
parallelogram_property_opposite_line_equal(1,ABCN)
quadrilateral_perimeter_formula(1,NABC)
Problem: As shown in the diagram, ∠CFG=3*x+3°, ∠JGD=4*x+4°, ∠KBF=2*x+2°, ∠ODB=x+1°. Find the value of x.
问题：如图所示，∠CFG=3*x+3°，∠JGD=4*x+4°，∠KBF=2*x+2°，∠ODB=x+1°。求x的值。
Construction (CDL):
Shape(JG,GD)
Shape(OD,DB)
Shape(CF,FG)
Shape(KB,BF)
Shape(GF,FB,BD,DG)
Collinear(JGF)
Collinear(CFB)
Collinear(KBD)
Collinear(ODG)
Conditions (CDL):
Equal(MeasureOfAngle(CFG),3*x+3)
Equal(MeasureOfAngle(JGD),4*x+4)
Equal(MeasureOfAngle(KBF),2*x+2)
Equal(MeasureOfAngle(ODB),x+1)
Goal: Value(x)
Answer: 35
Theorem sequence:
adjacent_complementary_angle(1,JGD,DGF)
adjacent_complementary_angle(1,CFG,GFB)
adjacent_complementary_angle(1,KBF,FBD)
adjacent_complementary_angle(1,ODB,BDG)
quadrilateral_property_angle_sum(1,GFBD)
Problem: As shown in the diagram, AE=3, BD=x, CB=12, CF=6, EF=y, EH=2, the center of circle D is D, CB is the tangent to circle D. Find the value of x.
问题：如图所示，AE=3，BD=x，CB=12，CF=6，EF=y，EH=2，圆D的圆心为D，CB是圆O的切线。求x的值。
Construction (CDL):
Shape(AE,EB,DBA)
Shape(EA,DAH,HE)
Shape(EH,DHF,FE)
Shape(BE,EF,DFB)
Shape(FC,CB,DFB)
Collinear(BDEH)
Collinear(CFEA)
Cocircular(D,AHFB)
Conditions (CDL):
Equal(LengthOfLine(AE),3)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CB),12)
Equal(LengthOfLine(CF),6)
Equal(LengthOfLine(EF),y)
Equal(LengthOfLine(EH),2)
IsCentreOfCircle(D,D)
IsTangentOfCircle(CB,D)
Goal: Value(x)
Answer: 49/4
Theorem sequence:
line_addition(1,BD,DH)
line_addition(1,BE,EH)
line_addition(1,CF,FA)
line_addition(1,FE,EA)
circle_property_circular_power_tangent_and_segment_line(1,CB,CFA,D)
circle_property_circular_power_chord_and_chord(1,AEF,BEH,D)
radius_of_circle_property_length_equal(1,DB,D)
radius_of_circle_property_length_equal(1,DH,D)
Problem: As shown in the diagram, ∠TYZ=52°, ∠ZYX=38°, TY is perpendicular to XY, XW⊥TW, YX is perpendicular to WX, YZ⊥TZ. Find the measure of ∠XWY.
问题：如图所示，∠TYZ=52°，∠ZYX=38°，TY垂直于XY，XW垂直于TW，YX⊥WX，YZ垂直于TZ。求∠XWY的大小。
Construction (CDL):
Shape(TY,YZ,ZT)
Shape(TZ,ZW,WT)
Shape(WY,YX,XW)
Collinear(YZW)
Conditions (CDL):
Equal(MeasureOfAngle(TYZ),52)
Equal(MeasureOfAngle(ZYX),38)
PerpendicularBetweenLine(TY,XY)
PerpendicularBetweenLine(XW,TW)
PerpendicularBetweenLine(YX,WX)
PerpendicularBetweenLine(YZ,TZ)
Goal: Value(MeasureOfAngle(XWY))
Answer: 52
Theorem sequence:
triangle_property_angle_sum(1,WYX)
Problem: As shown in the diagram, ∠FAD=2*x°, ∠HAE=3*x°, HA is perpendicular to EA. Find the value of x.
问题：如图所示，∠FAD=2*x°，∠HAE=3*x°，HA⊥EA。求x的值。
Construction (CDL):
Shape(HA,AE)
Shape(DA,AH)
Shape(FA,AD)
Shape(EA,AF)
Collinear(EAD)
Conditions (CDL):
Equal(MeasureOfAngle(FAD),2*x)
Equal(MeasureOfAngle(HAE),3*x)
PerpendicularBetweenLine(HA,EA)
Goal: Value(x)
Answer: 30
Theorem sequence:
Problem: As shown in the diagram, ∠CAE=38°, ABDC is a rectangle. Find the measure of ∠CBD.
问题：如图所示，∠CAE=38°，四边形ABDC是长方形。求∠CBD的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,EC,CA)
Shape(EB,BD,DE)
Shape(ED,DC,CE)
Collinear(AED)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(CAE),38)
Rectangle(ABDC)
Goal: Value(MeasureOfAngle(CBD))
Answer: 38
Theorem sequence:
parallel_property_alternate_interior_angle(1,AC,BD)
rectangle_property_diagonal_equal(1,ABDC)
parallelogram_property_diagonal_bisection(1,ABDC,E)
parallelogram_property_diagonal_bisection(1,BDCA,E)
line_addition(1,AE,ED)
line_addition(1,BE,EC)
isosceles_triangle_judgment_line_equal(1,EBD)
isosceles_triangle_property_angle_equal(1,EBD)
Problem: As shown in the diagram, DG=9, DK=16, FK=12, K is the centroid of △CDF. Find the length of line FG.
问题：如图所示，DG=9，DK=16，FK=12，K是三角形CDF的重心。求直线FG的长度。
Construction (CDL):
Shape(CG,GK,KC)
Shape(GD,DK,KG)
Shape(KD,DJ,JK)
Shape(KJ,JF,FK)
Shape(KF,FH,HK)
Shape(KH,HC,CK)
Collinear(CGD)
Collinear(DJF)
Collinear(FHC)
Collinear(GKF)
Collinear(HKD)
Collinear(CKJ)
Conditions (CDL):
Equal(LengthOfLine(DG),9)
Equal(LengthOfLine(DK),16)
Equal(LengthOfLine(FK),12)
IsCentroidOfTriangle(K,CDF)
Goal: Value(LengthOfLine(FG))
Answer: 18
Theorem sequence:
centroid_of_triangle_property_line_ratio(1,K,FCD,G)
line_addition(1,FK,KG)
Problem: As shown in the diagram, ∠CFA=38°, ∠GEC=52°, DF⊥DF. Find the measure of ∠DFI.
问题：如图所示，∠CFA=38°，∠GEC=52°，DF垂直于DF。求∠DFI的大小。
Construction (CDL):
Shape(IF,FB)
Shape(BF,FE)
Shape(FE,EG)
Shape(GE,EC)
Shape(DF,FI)
Shape(FD,DE,EF)
Shape(CE,ED)
Shape(HD,DF)
Shape(AD,DH)
Shape(ED,DA)
Collinear(BFDA)
Collinear(HDEC)
Collinear(IFEC)
Conditions (CDL):
Equal(MeasureOfAngle(CFA),38)
Equal(MeasureOfAngle(GEC),52)
PerpendicularBetweenLine(DF,DE)
Goal: Value(MeasureOfAngle(DFI))
Answer: 142
Theorem sequence:
adjacent_complementary_angle(1,EFD,DFI)
Problem: As shown in the diagram, AB=x-2, AD=10, CB=5, ED=4, EB is parallel to DC. Find the value of x.
问题：如图所示，AB=x-2，AD=10，CB=5，ED=4，EB∥DC。求x的值。
Construction (CDL):
Shape(AE,EB,BA)
Shape(ED,DC,CB,BE)
Collinear(AED)
Collinear(ABC)
Conditions (CDL):
Equal(LengthOfLine(AB),x-2)
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(CB),5)
Equal(LengthOfLine(ED),4)
ParallelBetweenLine(EB,DC)
Goal: Value(x)
Answer: 19/2
Theorem sequence:
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
line_addition(1,AE,ED)
line_addition(1,AB,BC)
similar_triangle_property_line_ratio(1,BAE,CAD)
similar_triangle_property_line_ratio(1,EBA,DCA)
Problem: As shown in the diagram, AB=10, the measure of ⌒OBA is 60, the radius of circle O is 10, the center of circle O is O, BX⊥OX. Find the measure of ⌒OYA.
问题：如图所示，AB=10，⌒OBA的角度为60，⊙O的半径为10，圆O的圆心为O，BX垂直于OX。求弧OYA的角度。
Construction (CDL):
Shape(AX,XY,OYA)
Shape(YX,XB,OBY)
Shape(BA,OAB)
Collinear(AXB)
Collinear(YXO)
Cocircular(O,BYA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(MeasureOfArc(OBA),60)
Equal(RadiusOfCircle(O),10)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BX,OX)
Goal: Value(MeasureOfArc(OYA))
Answer: 30
Theorem sequence:
circle_property_chord_perpendicular_bisect_arc(1,OBA,OXY)
congruent_arc_judgment_length_equal(1,OBY,OYA)
congruent_arc_property_measure_equal(1,OBY,OYA)
arc_addition_measure(1,OBY,OYA)
Problem: As shown in the diagram, ∠WXY=4*x+15°, ∠ZXV=2*x+65°. Find the measure of ∠ZXV.
问题：如图所示，∠WXY=4*x+15°，∠ZXV=2*x+65°。求∠ZXV的大小。
Construction (CDL):
Shape(YX,XZ,XZY)
Shape(XY,XYW,WX)
Shape(XW,WXV,VX)
Collinear(ZXW)
Collinear(YXV)
Conditions (CDL):
Equal(MeasureOfAngle(WXY),4*x+15)
Equal(MeasureOfAngle(ZXV),2*x+65)
Goal: Value(MeasureOfAngle(ZXV))
Answer: 115
Theorem sequence:
vertical_angle(1,ZXV,WXY)
Problem: As shown in the diagram, AD=6, AE=x, BE=5, AD is the tangent to circle C. Find the value of x.
问题：如图所示，AD=6，AE=x，BE=5，圆O的切线为AD。求x的值。
Construction (CDL):
Shape(BE,CEB)
Shape(EA,AD,CED)
Shape(EB,CBE)
Collinear(BEA)
Cocircular(C,BED)
Conditions (CDL):
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(AE),x)
Equal(LengthOfLine(BE),5)
IsTangentOfCircle(AD,C)
Goal: Value(x)
Answer: 4
Theorem sequence:
line_addition(1,BE,EA)
circle_property_circular_power_tangent_and_segment_line(1,AD,AEB,C)