SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠MPG=89°, the measure of arc PNE is 66, P is the center of ⊙P, the diameter of circle P is GN. Find the measure of ∠GME.
问题：如图所示，∠MPG=89°，⌒PNE的角度为66，P是⊙P的圆心，⊙P的直径为GN。求∠GME的大小。
Construction (CDL):
Shape(GE,PEG)
Shape(MG,PGM)
Shape(PMN,NM)
Shape(PNE,EN)
Shape(GM,MP,PG)
Shape(PM,MB,BP)
Shape(BM,MN,NB)
Shape(BN,NE,EB)
Shape(GP,PE,EG)
Shape(EP,PB,BE)
Collinear(GPBN)
Collinear(MBE)
Cocircular(P,GMNE)
Conditions (CDL):
Equal(MeasureOfAngle(MPG),89)
Equal(MeasureOfArc(PNE),66)
IsCentreOfCircle(P,P)
IsDiameterOfCircle(GN,P)
Goal: Value(MeasureOfAngle(GME))
Answer: 57
Theorem sequence:
diameter_of_circle_property_right_angle(1,GMN,P)
arc_property_circumference_angle_external(1,PNE,M)
angle_addition(1,GME,EMN)
Problem: As shown in the diagram, AF=1/2*y+20, CF=3*y, CG=20-3*x, FC=AF, HG=2*x-5, FG∥AH. Find the value of x.
问题：如图所示，AF=1/2*y+20，CF=3*y，CG=20-3*x，FC=AF，HG=2*x-5，FG平行于AH。求x的值。
Construction (CDL):
Shape(CF,FG,GC)
Shape(FA,AH,HG,GF)
Collinear(CFA)
Collinear(CGH)
Conditions (CDL):
Equal(LengthOfLine(AF),1/2*y+20)
Equal(LengthOfLine(CF),3*y)
Equal(LengthOfLine(CG),20-3*x)
Equal(LengthOfLine(FC),LengthOfLine(AF))
Equal(LengthOfLine(HG),2*x-5)
ParallelBetweenLine(FG,AH)
Goal: Value(x)
Answer: 5
Theorem sequence:
parallel_property_corresponding_angle(1,FG,AH,C)
line_addition(1,AF,FC)
line_addition(1,HG,GC)
similar_triangle_judgment_aa(1,GCF,HCA)
similar_triangle_property_line_ratio(1,GCF,HCA)
similar_triangle_property_line_ratio(1,FGC,AHC)
Problem: As shown in the diagram, AB=2*x+9, CD=3*x-7, the measure of ⌒EAB is 76, the measure of ⌒EDC is 284, E is the center of circle E. Find the value of x.
问题：如图所示，AB=2*x+9，CD=3*x-7，弧EAB的角度为76，⌒EDC的角度为284，E是⊙E的圆心。求x的值。
Construction (CDL):
Shape(EAB,BA)
Shape(AB,EBC,CE,ED,EDA)
Shape(EC,CD,DE)
Shape(ECD,DC)
Cocircular(E,BCDA)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x+9)
Equal(LengthOfLine(CD),3*x-7)
Equal(MeasureOfArc(EAB),76)
Equal(MeasureOfArc(EDC),284)
IsCentreOfCircle(E,E)
Goal: Value(x)
Answer: 16
Theorem sequence:
arc_property_center_angle(1,EDC,E)
arc_property_center_angle(1,ECD,E)
round_angle(1,CED,DEC)
congruent_arc_judgment_measure_equal(1,EAB,ECD)
congruent_arc_property_chord_equal(1,EAB,ECD)
Problem: As shown in the diagram, AE=13, CE=x, ∠AEC=18°, ∠CAE=141°. Find the value of x.
问题：如图所示，AE=13，CE=x，∠AEC=18°，∠CAE=141°。求x的值。
Construction (CDL):
Shape(EC,CA,AE)
Conditions (CDL):
Equal(LengthOfLine(AE),13)
Equal(LengthOfLine(CE),x)
Equal(MeasureOfAngle(AEC),18)
Equal(MeasureOfAngle(CAE),141)
Goal: Value(x)
Answer: (-13*sqrt(5*sqrt(5)+25)-13*sqrt(30)-13*sqrt(3*sqrt(5)+15)-13*sqrt(10)-13*sqrt(6)-13*sqrt(2)+13*sqrt(sqrt(5)+5)+13*sqrt(15*sqrt(5)+75))/(-sqrt(15)*sqrt(sqrt(5)+5)-sqrt(5)*sqrt(sqrt(5)+5)-sqrt(10)-sqrt(2)+sqrt(6)+sqrt(sqrt(5)+5)+sqrt(3)*sqrt(sqrt(5)+5)+sqrt(30))
Theorem sequence:
triangle_property_angle_sum(1,ECA)
sine_theorem(1,ECA)
Problem: As shown in the diagram, ∠AXB=7*x+10°, ∠BXC=90°, ∠CXA=3*x°. Find the value of x.
问题：如图所示，∠AXB=7*x+10°，∠BXC=90°，∠CXA=3*x°。求x的值。
Construction (CDL):
Shape(CX,XA,XAC)
Shape(XC,XCB,BX)
Shape(XB,XBA,AX)
Cocircular(X,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(AXB),7*x+10)
Equal(MeasureOfAngle(BXC),90)
Equal(MeasureOfAngle(CXA),3*x)
Goal: Value(x)
Answer: 26
Theorem sequence:
angle_addition(1,BXC,CXA)
round_angle(1,BXA,AXB)
Problem: As shown in the diagram, DF=9, DG=d, DH=11, HE=15, DG is the tangent to ⊙C. Find the value of d.
问题：如图所示，DF=9，DG=d，DH=11，HE=15，⊙O的切线为DG。求d的值。
Construction (CDL):
Shape(AF,CFA)
Shape(CAE,EH,CHF,FA)
Shape(CEH,HE)
Shape(AD,DM)
Shape(ED,DA)
Collinear(AFD)
Collinear(EHD)
Collinear(MGD)
Cocircular(C,AEHFG)
Conditions (CDL):
Equal(LengthOfLine(DF),9)
Equal(LengthOfLine(DG),d)
Equal(LengthOfLine(DH),11)
Equal(LengthOfLine(HE),15)
IsTangentOfCircle(DG,C)
Goal: Value(d)
Answer: sqrt(286)
Theorem sequence:
circle_property_circular_power_tangent_and_segment_line(1,DG,DHE,C)
line_addition(1,EH,HD)
Problem: As shown in the diagram, ∠BDC=40°, the measure of arc AED is 98. Find the measure of ∠DCB.
问题：如图所示，∠BDC=40°，弧AED的角度为98。求∠DCB的大小。
Construction (CDL):
Shape(ADC,CD)
Shape(CB,BD,DC)
Shape(ACF,FB,BC)
Shape(BF,AFE,EB)
Shape(DB,BE,AED)
Collinear(DABF)
Collinear(CBE)
Cocircular(A,CFED)
Conditions (CDL):
Equal(MeasureOfAngle(BDC),40)
Equal(MeasureOfArc(AED),98)
Goal: Value(MeasureOfAngle(DCB))
Answer: 49
Theorem sequence:
arc_property_circumference_angle_external(1,AED,C)
Problem: As shown in the diagram, ∠AFC=70°, ∠EBF=35°, AD⊥ED, EB is perpendicular to CB. Find the measure of ∠BCF.
问题：如图所示，∠AFC=70°，∠EBF=35°，AD垂直于ED，EB⊥CB。求∠BCF的大小。
Construction (CDL):
Shape(DE,EF,FA,AD)
Shape(EB,BF,FE)
Shape(FB,BC,CF)
Shape(AF,FC)
Collinear(DEB)
Collinear(AFB)
Collinear(EFC)
Conditions (CDL):
Equal(MeasureOfAngle(AFC),70)
Equal(MeasureOfAngle(EBF),35)
PerpendicularBetweenLine(AD,ED)
PerpendicularBetweenLine(EB,CB)
Goal: Value(MeasureOfAngle(BCF))
Answer: 15
Theorem sequence:
vertical_angle(1,BFE,AFC)
triangle_property_angle_sum(1,EBF)
triangle_property_angle_sum(1,EBC)
Problem: As shown in the diagram, AC=5*sqrt(26), AD=25, BC=sqrt(26), CD=5, DB=1, ∠BCD=y°, ∠DCA=x°, AD is perpendicular to CD, BC⊥AC. Find the value of sin(CA).
问题：如图所示，AC=5*sqrt(26)，AD=25，BC=sqrt(26)，CD=5，DB=1，∠BCD=y°，∠DCA=x°，AD⊥CD，BC⊥AC。求sin(CA)的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),5*sqrt(26))
Equal(LengthOfLine(AD),25)
Equal(LengthOfLine(BC),sqrt(26))
Equal(LengthOfLine(CD),5)
Equal(LengthOfLine(DB),1)
Equal(MeasureOfAngle(BCD),y)
Equal(MeasureOfAngle(DCA),x)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Sin(MeasureOfAngle(CAD)))
Answer: sqrt(26)/26
Theorem sequence:
cosine_theorem(1,CDB)
angle_addition(1,BCD,DCA)
triangle_property_angle_sum(1,ADC)
Problem: As shown in the diagram, KJ=2, LK=4, LP=2, MP=3, PQ=6, RS=6, KQ is parallel to JR, LP is parallel to HS, LP is parallel to JR, LP is parallel to KQ. Find the length of line QR.
问题：如图所示，KJ=2，LK=4，LP=2，MP=3，PQ=6，RS=6，KQ∥JR，LP∥HS，LP平行于JR，LP平行于KQ。求直线QR的长度。
Construction (CDL):
Shape(HS,SR,RJ,JH)
Shape(JR,RQ,QK,KJ)
Shape(KQ,QP,PL,LK)
Shape(LP,PM,ML)
Collinear(SRQPM)
Collinear(HJKLM)
Conditions (CDL):
Equal(LengthOfLine(KJ),2)
Equal(LengthOfLine(LK),4)
Equal(LengthOfLine(LP),2)
Equal(LengthOfLine(MP),3)
Equal(LengthOfLine(PQ),6)
Equal(LengthOfLine(RS),6)
ParallelBetweenLine(KQ,JR)
ParallelBetweenLine(LP,HS)
ParallelBetweenLine(LP,JR)
ParallelBetweenLine(LP,KQ)
Goal: Value(LengthOfLine(QR))
Answer: 3
Theorem sequence:
parallel_property_corresponding_angle(1,LP,KQ,M)
parallel_property_corresponding_angle(1,KQ,JR,M)
similar_triangle_judgment_aa(1,PML,QMK)
similar_triangle_judgment_aa(1,QMK,RMJ)
line_addition(1,ML,LK)
line_addition(1,MP,PQ)
similar_triangle_property_line_ratio(1,LPM,KQM)
similar_triangle_property_line_ratio(1,PML,QMK)
line_addition(1,MK,KJ)
line_addition(1,MQ,QR)
similar_triangle_property_line_ratio(1,QMK,RMJ)
similar_triangle_property_line_ratio(1,KQM,JRM)
Problem: As shown in the diagram, AC=DiameterOfCircle(F), ∠BFC=72°, ∠CFD=4*a-1°, ∠DFA=2*a-5°. Find the measure of ∠DFB.
问题：如图所示，AC=DiameterOfCircle(F)，∠BFC=72°，∠CFD=4*a-1°，∠DFA=2*a-5°。求∠DFB的大小。
Construction (CDL):
Shape(BF,FC,FCB)
Shape(AF,FB,FBA)
Shape(FA,FAD,DF)
Shape(FD,FDC,CF)
Collinear(AFC)
Cocircular(F,ADCB)
Conditions (CDL):
Equal(LengthOfLine(AC),DiameterOfCircle(F))
Equal(MeasureOfAngle(BFC),72)
Equal(MeasureOfAngle(CFD),4*a-1)
Equal(MeasureOfAngle(DFA),2*a-5)
Goal: Value(MeasureOfAngle(DFB))
Answer: 165
Theorem sequence:
flat_angle(1,AFC)
round_angle(1,AFC,CFA)
angle_addition(1,CFD,DFA)
angle_addition(1,BFC,CFD)
round_angle(1,DFB,BFD)
Problem: As shown in the diagram, AB=y, AC=x, BC=7*sqrt(2), ∠CBA=45°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=y，AC=x，BC=7*sqrt(2)，∠CBA=45°，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),7*sqrt(2))
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 7*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,ACB)
sine_theorem(1,BAC)
Problem: As shown in the diagram, YZ=11, ZY=RadiusOfCircle(Y), ∠XYZ=55°, the center of circle Y is Y. Find the area of the sector YXZ.
问题：如图所示，YZ=11，ZY=RadiusOfCircle(Y)，∠XYZ=55°，Y是⊙Y的圆心。求扇形YXZ的面积。
Construction (CDL):
Shape(XY,YZ,YZX)
Shape(YX,YXZ,ZY)
Cocircular(Y,XZ)
Conditions (CDL):
Equal(LengthOfLine(YZ),11)
Equal(LengthOfLine(ZY),RadiusOfCircle(Y))
Equal(MeasureOfAngle(XYZ),55)
IsCentreOfCircle(Y,Y)
Goal: Value(AreaOfSector(YXZ))
Answer: 7381*pi/72
Theorem sequence:
round_angle(1,XYZ,ZYX)
arc_property_center_angle(1,YXZ,Y)
sector_area_formula(1,YXZ)
Problem: As shown in the diagram, ∠JKL=y°, ∠LJK=4*x-13°, ∠WZY=44°, ∠YWZ=71°, Triangle JKL is mirror similar to triangle WZY.. Find the value of y.
问题：如图所示，∠JKL=y°，∠LJK=4*x-13°，∠WZY=44°，∠YWZ=71°，三角形JKL镜像相似于三角形WZY。求y的值。
Construction (CDL):
Shape(JK,KL,LJ)
Shape(WZ,ZY,YW)
Conditions (CDL):
Equal(MeasureOfAngle(JKL),y)
Equal(MeasureOfAngle(LJK),4*x-13)
Equal(MeasureOfAngle(WZY),44)
Equal(MeasureOfAngle(YWZ),71)
MirrorSimilarBetweenTriangle(JKL,WZY)
Goal: Value(y)
Answer: 65
Theorem sequence:
triangle_property_angle_sum(1,WZY)
mirror_similar_triangle_property_angle_equal(1,KLJ,YWZ)
Problem: As shown in the diagram, AB=x, BC=18, ∠ACB=135°, ∠BAC=20°. Find the value of x.
问题：如图所示，AB=x，BC=18，∠ACB=135°，∠BAC=20°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),18)
Equal(MeasureOfAngle(ACB),135)
Equal(MeasureOfAngle(BAC),20)
Goal: Value(x)
Answer: 9*sqrt(2)/sin(pi/9)
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, PR=12, quadrilateral MPON is a rhombus. Find the length of line RN.
问题：如图所示，PR=12，四边形MPON是菱形。求直线RN的长度。
Construction (CDL):
Shape(MP,PR,RM)
Shape(MR,RN,NM)
Shape(RP,PO,OR)
Shape(RO,ON,NR)
Collinear(MRO)
Collinear(PRN)
Conditions (CDL):
Equal(LengthOfLine(PR),12)
Rhombus(MPON)
Goal: Value(LengthOfLine(RN))
Answer: 12
Theorem sequence:
parallelogram_property_diagonal_bisection(1,PONM,R)
Problem: As shown in the diagram, ∠PXR=x°, ∠QPX=4*x°, ∠RQP=2*x°, ∠XRQ=5*x°. Find the measure of ∠RQP.
问题：如图所示，∠PXR=x°，∠QPX=4*x°，∠RQP=2*x°，∠XRQ=5*x°。求∠RQP的大小。
Construction (CDL):
Shape(XR,RQ,QP,PX)
Conditions (CDL):
Equal(MeasureOfAngle(PXR),x)
Equal(MeasureOfAngle(QPX),4*x)
Equal(MeasureOfAngle(RQP),2*x)
Equal(MeasureOfAngle(XRQ),5*x)
Goal: Value(MeasureOfAngle(RQP))
Answer: 60
Theorem sequence:
quadrilateral_property_angle_sum(1,XRQP)
Problem: As shown in the diagram, AB=x, AC=12, AD=y, BD=z, DC=4, AD is perpendicular to CD, CA is perpendicular to BA. Find the value of y.
问题：如图所示，AB=x，AC=12，AD=y，BD=z，DC=4，AD垂直于CD，CA垂直于BA。求y的值。
Construction (CDL):
Shape(BD,DA,AB)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BD),z)
Equal(LengthOfLine(DC),4)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 8*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,ADC)
right_triangle_property_pythagorean(1,ADC)
Problem: As shown in the diagram, DC=9, FA=12, FD=x, VZ=3*y-1, YW=6, YZ=10, quadrilateral AFDC is similar to quadrilateral VZYW. Find the value of y.
问题：如图所示，DC=9，FA=12，FD=x，VZ=3*y-1，YW=6，YZ=10，四边形AFDC相似于四边形VZYW。求y的值。
Construction (CDL):
Shape(AF,FD,DC,CA)
Shape(YW,WV,VZ,ZY)
Conditions (CDL):
Equal(LengthOfLine(DC),9)
Equal(LengthOfLine(FA),12)
Equal(LengthOfLine(FD),x)
Equal(LengthOfLine(VZ),3*y-1)
Equal(LengthOfLine(YW),6)
Equal(LengthOfLine(YZ),10)
SimilarBetweenQuadrilateral(AFDC,VZYW)
Goal: Value(y)
Answer: 3
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,AFDC,VZYW)
similar_quadrilateral_property_line_ratio(1,DCAF,YWVZ)
Problem: As shown in the diagram, AB=36, AC=20, BC=x, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AB=36，AC=20，BC=x，CA垂直于BA。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),36)
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 4*sqrt(106)
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, CD=6, DF=x, ∠ACD=45°, ∠BED=45°, ∠EAD=45°, ∠FBD=45°, CD⊥ED, DA⊥CA, DB⊥EB, DE⊥AE, DF⊥BF. Find the value of x.
问题：如图所示，CD=6，DF=x，∠ACD=45°，∠BED=45°，∠EAD=45°，∠FBD=45°，CD垂直于ED，DA⊥CA，DB垂直于EB，DE垂直于AE，DF垂直于BF。求x的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DE,EA)
Shape(ED,DB,BE)
Shape(BD,DF,FB)
Conditions (CDL):
Equal(LengthOfLine(CD),6)
Equal(LengthOfLine(DF),x)
Equal(MeasureOfAngle(ACD),45)
Equal(MeasureOfAngle(BED),45)
Equal(MeasureOfAngle(EAD),45)
Equal(MeasureOfAngle(FBD),45)
PerpendicularBetweenLine(CD,ED)
PerpendicularBetweenLine(DA,CA)
PerpendicularBetweenLine(DB,EB)
PerpendicularBetweenLine(DE,AE)
PerpendicularBetweenLine(DF,BF)
Goal: Value(x)
Answer: 3/2
Theorem sequence:
sine_theorem(1,DAC)
sine_theorem(1,DEA)
sine_theorem(1,DBE)
sine_theorem(1,DFB)
Problem: As shown in the diagram, AF=4, Sub(AreaOfCircle(F)=AreaOfTriangle(ADC)), triangle ADC is an equilateral triangle, the center of ⊙F is F. Find the value of x.
问题：如图所示，AF=4，Sub(AreaOfCircle(F)=AreaOfTriangle(ADC))，三角形ADC为等边三角形，F是⊙F的圆心。求x的值。
Construction (CDL):
Shape(FAD,DA)
Shape(FDC,CD)
Shape(AC,FCA)
Shape(FA,AD,DF)
Shape(FD,DC,CA,AF)
Cocircular(F,ADC)
Conditions (CDL):
Equal(LengthOfLine(AF),4)
Equal(Sub(AreaOfCircle(F),AreaOfTriangle(ADC)),x)
EquilateralTriangle(ADC)
IsCentreOfCircle(F,F)
Goal: Value(x)
Answer: -12*sqrt(3)+16*pi
Theorem sequence:
equilateral_triangle_property_angle(1,ADC)
equilateral_triangle_property_angle(1,CAD)
arc_property_circumference_angle_external(1,FAD,C)
arc_property_center_angle(1,FAD,F)
radius_of_circle_property_length_equal(1,FA,F)
radius_of_circle_property_length_equal(1,FD,F)
cosine_theorem(1,FAD)
triangle_area_formula_sine(1,ADC)
circle_area_formula(1,F)
Problem: As shown in the diagram, CB=12, CB=AB, ∠BAC=44°. Find the length of line AB.
问题：如图所示，CB=12，CB=AB，∠BAC=44°。求直线AB的长度。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(CB),12)
Equal(LengthOfLine(CB),LengthOfLine(AB))
Equal(MeasureOfAngle(BAC),44)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
Problem: As shown in the diagram, AB=75, AC=x, CB=40, CB⊥AB. Find the value of x.
问题：如图所示，AB=75，AC=x，CB=40，CB垂直于AB。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),75)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(CB),40)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 85
Theorem sequence:
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, AB=30, CA=16, the center of circle D is D, the diameter of circle D is BC, BA is perpendicular to CA. Find the circumference of the circle D.
问题：如图所示，AB=30，CA=16，⊙D的圆心为D，圆D的直径为BC，BA⊥CA。求⊙D的周长。
Construction (CDL):
Shape(AB,DBA)
Shape(CA,DAC)
Shape(BD,DC,DCB)
Shape(BA,AC,CD,DB)
Collinear(BDC)
Cocircular(D,BAC)
Conditions (CDL):
Equal(LengthOfLine(AB),30)
Equal(LengthOfLine(CA),16)
IsCentreOfCircle(D,D)
IsDiameterOfCircle(BC,D)
PerpendicularBetweenLine(BA,CA)
Goal: Value(PerimeterOfCircle(D))
Answer: 34*pi
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
diameter_of_circle_property_length_equal(1,BC,D)
circle_property_length_of_radius_and_diameter(1,D)
circle_perimeter_formula(1,D)
Problem: As shown in the diagram, HK=5, HN=2, KL=x, ML=4, the tangent to circle O is HJ, the tangent to circle O is HN, the tangent to ⊙O is KJ, the tangent to ⊙O is KL, ML is the tangent to ⊙O, the tangent to ⊙O is MN. Find the value of x.
问题：如图所示，HK=5，HN=2，KL=x，ML=4，⊙O的切线为HJ，圆O的切线为HN，圆O的切线为KJ，圆O的切线为KL，ML是⊙O的切线，⊙O的切线为MN。求x的值。
Construction (CDL):
Shape(KJ,OJL,LK)
Shape(JH,HN,ONJ)
Shape(NM,ML,OLN)
Shape(OLJ,OJN,ONL)
Collinear(KJH)
Collinear(HNM)
Collinear(KLM)
Cocircular(O,JNL)
Conditions (CDL):
Equal(LengthOfLine(HK),5)
Equal(LengthOfLine(HN),2)
Equal(LengthOfLine(KL),x)
Equal(LengthOfLine(ML),4)
IsTangentOfCircle(HJ,O)
IsTangentOfCircle(HN,O)
IsTangentOfCircle(KJ,O)
IsTangentOfCircle(KL,O)
IsTangentOfCircle(ML,O)
IsTangentOfCircle(MN,O)
Goal: Value(x)
Answer: 3
Theorem sequence:
tangent_of_circle_property_length_equal(1,KJ,KL,O)
tangent_of_circle_property_length_equal(1,HJ,HN,O)
line_addition(1,KJ,JH)
Problem: As shown in the diagram, AB=26, AD=12, DG=5, EF=8, GF=14, ∠AGF=108°, quadrilateral AGFE is similar to quadrilateral ADCB. Find the length of line DC.
问题：如图所示，AB=26，AD=12，DG=5，EF=8，GF=14，∠AGF=108°，四边形AGFE相似于四边形ADCB。求直线DC的长度。
Construction (CDL):
Shape(AG,GF,FE,EA)
Shape(GD,DC,CB,BE,EF,FG)
Collinear(AGD)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DG),5)
Equal(LengthOfLine(EF),8)
Equal(LengthOfLine(GF),14)
Equal(MeasureOfAngle(AGF),108)
SimilarBetweenQuadrilateral(AGFE,ADCB)
Goal: Value(LengthOfLine(DC))
Answer: 24
Theorem sequence:
line_addition(1,AG,GD)
similar_quadrilateral_property_line_ratio(1,AGFE,ADCB)
similar_quadrilateral_property_line_ratio(1,GFEA,DCBA)
Problem: As shown in the diagram, ∠ABD=32°, ∠CAD=62°, ∠CDS=125°. Find the measure of ∠BDA.
问题：如图所示，∠ABD=32°，∠CAD=62°，∠CDS=125°。求∠BDA的大小。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Shape(DS)
Collinear(CDB)
Collinear(ADS)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),32)
Equal(MeasureOfAngle(CAD),62)
Equal(MeasureOfAngle(CDS),125)
Goal: Value(MeasureOfAngle(BDA))
Answer: 125
Theorem sequence:
vertical_angle(1,CDS,BDA)
Problem: As shown in the diagram, AC=x, AD=4, BC=13, BD=6, ∠DCA=∠BCD. Find the value of x.
问题：如图所示，AC=x，AD=4，BC=13，BD=6，∠DCA=∠BCD。求x的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(BC),13)
Equal(LengthOfLine(BD),6)
Equal(MeasureOfAngle(DCA),MeasureOfAngle(BCD))
Goal: Value(x)
Answer: 26/3
Theorem sequence:
bisector_of_angle_judgment_angle_equal(1,CD,BCA)
bisector_of_angle_property_line_ratio(1,CD,BCA)
Problem: As shown in the diagram, the diameter of circle K is equal to the length of line GJ, ∠GKH=122°. Find the measure of ∠HKG.
问题：如图所示，⊙K的直径与直线GJ的长度相等，∠GKH=122°。求∠HKG的大小。
Construction (CDL):
Shape(KHG,GK,KH)
Shape(KGJ,JK,KG)
Shape(KJ,KJH,HK)
Collinear(GKJ)
Cocircular(K,HGJ)
Conditions (CDL):
Equal(DiameterOfCircle(K),LengthOfLine(GJ))
Equal(MeasureOfAngle(GKH),122)
Goal: Value(MeasureOfAngle(HKG))
Answer: 238
Theorem sequence:
round_angle(1,GKH,HKG)
Problem: As shown in the diagram, AC=10, AD=13, ∠CBD=45°, CA and BD are opposite sides of the ▱ ACBD. Find the perimeter of ACBD.
问题：如图所示，AC=10，AD=13，∠CBD=45°，AD和CB是▱ACBD的一组对边。求ACBD的周长。
Construction (CDL):
Shape(AC,CB,BD,DA)
Conditions (CDL):
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(AD),13)
Equal(MeasureOfAngle(CBD),45)
Parallelogram(ACBD)
Goal: Value(PerimeterOfQuadrilateral(ACBD))
Answer: 46
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ACBD)
parallelogram_property_opposite_line_equal(1,CBDA)
quadrilateral_perimeter_formula(1,ACBD)
Problem: As shown in the diagram, ∠AGB=120°, ∠BCD=50°, ∠CDB=78°, ∠GAF=56°. Find the measure of ∠FGA.
问题：如图所示，∠AGB=120°，∠BCD=50°，∠CDB=78°，∠GAF=56°。求∠FGA的大小。
Construction (CDL):
Shape(CD,DB,BC)
Shape(BG,GE,EB)
Shape(GA,AF,FG)
Shape(FH)
Shape(AG,GB)
Collinear(CBGFH)
Collinear(DBE)
Collinear(EGA)
Conditions (CDL):
Equal(MeasureOfAngle(AGB),120)
Equal(MeasureOfAngle(BCD),50)
Equal(MeasureOfAngle(CDB),78)
Equal(MeasureOfAngle(GAF),56)
Goal: Value(MeasureOfAngle(FGA))
Answer: 60
Theorem sequence:
adjacent_complementary_angle(1,FGA,AGB)
Problem: As shown in the diagram, AC=AB, BA=5*sqrt(2), BC=y, BE=x, AE⊥BE, BC⊥DC, CA⊥BA, CD is perpendicular to AD, EB⊥CB. Find the value of y.
问题：如图所示，AC=AB，BA=5*sqrt(2)，BC=y，BE=x，AE⊥BE，BC⊥DC，CA垂直于BA，CD垂直于AD，EB⊥CB。求y的值。
Construction (CDL):
Shape(CD,DA,AC)
Shape(CA,AB,BC)
Shape(AE,EB,BA)
Collinear(DAE)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(AB))
Equal(LengthOfLine(BA),5*sqrt(2))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BE),x)
PerpendicularBetweenLine(AE,BE)
PerpendicularBetweenLine(BC,DC)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(CD,AD)
PerpendicularBetweenLine(EB,CB)
Goal: Value(y)
Answer: 10
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, DC=46, EH=38, the midsegment of CGFD is HE, the midsegment of quadrilateral HGFE is YZ, CGFD is a trapezoid. Find the length of line YZ.
问题：如图所示，DC=46，EH=38，CGFD的中位线为HE，HGFE的中位线为YZ，四边形CGFD是梯形。求直线YZ的长度。
Construction (CDL):
Shape(CH,HE,ED,DC)
Shape(HY,YZ,ZE,EH)
Shape(YG,GF,FZ,ZY)
Collinear(CHYG)
Collinear(DEZF)
Conditions (CDL):
Equal(LengthOfLine(DC),46)
Equal(LengthOfLine(EH),38)
IsMidsegmentOfQuadrilateral(HE,CGFD)
IsMidsegmentOfQuadrilateral(YZ,HGFE)
Trapezoid(CGFD)
Goal: Value(LengthOfLine(YZ))
Answer: 34
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,HE,CGFD)
midsegment_of_quadrilateral_property_length(1,YZ,HGFE)
Problem: As shown in the diagram, WY=WX, ∠WYX=4*x+20°, ∠XWY=52°, ∠YXW=6*y-2°. Find the value of y.
问题：如图所示，WY=WX，∠WYX=4*x+20°，∠XWY=52°，∠YXW=6*y-2°。求y的值。
Construction (CDL):
Shape(WY,YX,XW)
Conditions (CDL):
Equal(LengthOfLine(WY),LengthOfLine(WX))
Equal(MeasureOfAngle(WYX),4*x+20)
Equal(MeasureOfAngle(XWY),52)
Equal(MeasureOfAngle(YXW),6*y-2)
Goal: Value(y)
Answer: 11
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,WYX)
isosceles_triangle_property_angle_equal(1,WYX)
triangle_property_angle_sum(1,WYX)
Problem: As shown in the diagram, CA=AD, CM=4.5, MD=4.5, ∠AMC=4*x+1°, ∠DMA=5*x-5°. Find the value of x.
问题：如图所示，CA=AD，CM=4.5，MD=4.5，∠AMC=4*x+1°，∠DMA=5*x-5°。求x的值。
Construction (CDL):
Shape(MC,CA,AM)
Shape(MA,AD,DM)
Conditions (CDL):
Equal(LengthOfLine(CA),LengthOfLine(AD))
Equal(LengthOfLine(CM),4.5)
Equal(LengthOfLine(MD),4.5)
Equal(MeasureOfAngle(AMC),4*x+1)
Equal(MeasureOfAngle(DMA),5*x-5)
Goal: Value(x)
Answer: 6
Theorem sequence:
mirror_congruent_triangle_judgment_sss(1,MCA,MAD)
mirror_congruent_triangle_property_angle_equal(1,MCA,MAD)
Problem: As shown in the diagram, AC=x, BA=8/9, BC=4/9, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AC=x，BA=8/9，BC=4/9，BC垂直于AC。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BA),8/9)
Equal(LengthOfLine(BC),4/9)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 4*sqrt(3)/9
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, AB=17, AC=AD, AC=BC, BC=x, BD=AC, ∠ABD=y°, BD⊥AD. Find the value of x.
问题：如图所示，AB=17，AC=AD，AC=BC，BC=x，BD=AC，∠ABD=y°，BD⊥AD。求x的值。
Construction (CDL):
Shape(BD,DA,AB)
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),17)
Equal(LengthOfLine(AC),LengthOfLine(AD))
Equal(LengthOfLine(AC),LengthOfLine(BC))
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(BD),LengthOfLine(AC))
Equal(MeasureOfAngle(ABD),y)
PerpendicularBetweenLine(BD,AD)
Goal: Value(x)
Answer: 17*sqrt(2)/2
Theorem sequence:
right_triangle_judgment_angle(1,BDA)
right_triangle_property_pythagorean(1,BDA)
Problem: As shown in the diagram, AB=12, CB=x, CD=12, CD is the tangent to circle E. Find the value of x.
问题：如图所示，AB=12，CB=x，CD=12，CD是圆O的切线。求x的值。
Construction (CDL):
Shape(AE,EB,EBA)
Shape(BE,EA,EAD,EDB)
Shape(CB,EBD,DC)
Collinear(AEBC)
Cocircular(E,ADB)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(CB),x)
Equal(LengthOfLine(CD),12)
IsTangentOfCircle(CD,E)
Goal: Value(x)
Answer: -6+6*sqrt(5)
Theorem sequence:
line_addition(1,AB,BC)
circle_property_circular_power_tangent_and_segment_line(1,CD,CBA,E)