SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=x, AD=8, BD=z, CB=y, CD=3, AD⊥BD, CB is perpendicular to AB. Find the value of x.
问题：如图所示，AB=x，AD=8，BD=z，CB=y，CD=3，AD⊥BD，CB垂直于AB。求x的值。
Construction (CDL):
Shape(AD,DB,BA)
Shape(DC,CB,BD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BD),z)
Equal(LengthOfLine(CB),y)
Equal(LengthOfLine(CD),3)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 2*sqrt(22)
Theorem sequence:
adjacent_complementary_angle(1,ADB,BDC)
mirror_similar_triangle_judgment_aa(1,BDC,ACB)
line_addition(1,CD,DA)
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,CBA)
mirror_similar_triangle_property_line_ratio(1,BDC,ACB)
mirror_similar_triangle_property_line_ratio(1,DCB,BAC)
Problem: As shown in the diagram, ∠QRS=x°, ∠RST=2*x+7°, ∠STQ=x°, ∠TQR=2*x+5°. Find the measure of ∠RST.
问题：如图所示，∠QRS=x°，∠RST=2*x+7°，∠STQ=x°，∠TQR=2*x+5°。求∠RST的大小。
Construction (CDL):
Shape(QR,RS,ST,TQ)
Conditions (CDL):
Equal(MeasureOfAngle(QRS),x)
Equal(MeasureOfAngle(RST),2*x+7)
Equal(MeasureOfAngle(STQ),x)
Equal(MeasureOfAngle(TQR),2*x+5)
Goal: Value(MeasureOfAngle(RST))
Answer: 123
Theorem sequence:
quadrilateral_property_angle_sum(1,QRST)
Problem: As shown in the diagram, AB=24, AC=3*x-1, BD=18, CD=2*x+1, ∠BAD=∠DAC. Find the value of x.
问题：如图所示，AB=24，AC=3*x-1，BD=18，CD=2*x+1，∠BAD=∠DAC。求x的值。
Construction (CDL):
Shape(BA,AD,DB)
Shape(AC,CD,DA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),24)
Equal(LengthOfLine(AC),3*x-1)
Equal(LengthOfLine(BD),18)
Equal(LengthOfLine(CD),2*x+1)
Equal(MeasureOfAngle(BAD),MeasureOfAngle(DAC))
Goal: Value(x)
Answer: 7
Theorem sequence:
bisector_of_angle_judgment_angle_equal(1,AD,BAC)
bisector_of_angle_property_line_ratio(1,AD,BAC)
Problem: As shown in the diagram, ∠ABC=64°, AD∥BC. Find the measure of ∠DAB.
问题：如图所示，∠ABC=64°，AD∥BC。求∠DAB的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),64)
ParallelBetweenLine(AD,BC)
Goal: Value(MeasureOfAngle(DAB))
Answer: 116
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AD,BC)
Problem: As shown in the diagram, AD=12, BD=3, ∠BCA=90°, AD⊥CD. Find the length of line AC.
问题：如图所示，AD=12，BD=3，∠BCA=90°，AD垂直于CD。求直线AC的长度。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(BD),3)
Equal(MeasureOfAngle(BCA),90)
PerpendicularBetweenLine(AD,CD)
Goal: Value(LengthOfLine(AC))
Answer: 6*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,ADC,CDB)
line_addition(1,AD,DB)
mirror_similar_triangle_judgment_aa(1,ABC,CDB)
right_triangle_judgment_angle(1,ADC)
right_triangle_property_pythagorean(1,ADC)
mirror_similar_triangle_property_line_ratio(1,ABC,CDB)
mirror_similar_triangle_property_line_ratio(1,BCA,BCD)
mirror_similar_triangle_property_line_ratio(1,CAB,DBC)
Problem: As shown in the diagram, XY=2, the radius of circle W is 4, the radius of ⊙Z is 7, the center of circle W is W, the center of ⊙Z is Z. Find the length of line YZ.
问题：如图所示，XY=2，⊙W的半径为4，⊙Z的半径为7，圆W的圆心为W，Z是圆Z的圆心。求直线YZ的长度。
Construction (CDL):
Shape(WI,WIB,ZXB,XW)
Shape(IW,WX,ZAX,WAI)
Shape(ZAX,XY,WYA)
Shape(YX,ZXB,WBY)
Shape(YZ,ZC,ZCA,WAY)
Shape(ZY,WYB,ZBC,CZ)
Collinear(IWXYZC)
Cocircular(W,IBYA)
Cocircular(Z,AXBC)
Conditions (CDL):
Equal(LengthOfLine(XY),2)
Equal(RadiusOfCircle(W),4)
Equal(RadiusOfCircle(Z),7)
IsCentreOfCircle(W,W)
IsCentreOfCircle(Z,Z)
Goal: Value(LengthOfLine(YZ))
Answer: 5
Theorem sequence:
radius_of_circle_property_length_equal(1,ZX,Z)
line_addition(1,XY,YZ)
Problem: As shown in the diagram, AC=20, DB=2*x+1, FB=2*x-1, FC=12, ∠ACF=22°, ∠FBD=22°, CF⊥DF. Find the length of line DB.
问题：如图所示，AC=20，DB=2*x+1，FB=2*x-1，FC=12，∠ACF=22°，∠FBD=22°，CF⊥DF。求直线DB的长度。
Construction (CDL):
Shape(AC,CF,FA)
Shape(DF,FB,BD)
Collinear(ADF)
Collinear(CFB)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(DB),2*x+1)
Equal(LengthOfLine(FB),2*x-1)
Equal(LengthOfLine(FC),12)
Equal(MeasureOfAngle(ACF),22)
Equal(MeasureOfAngle(FBD),22)
PerpendicularBetweenLine(CF,DF)
Goal: Value(LengthOfLine(DB))
Answer: 5
Theorem sequence:
adjacent_complementary_angle(1,CFA,AFB)
mirror_similar_triangle_judgment_aa(1,ACF,DFB)
mirror_similar_triangle_property_line_ratio(1,ACF,DFB)
mirror_similar_triangle_property_line_ratio(1,FAC,FBD)
Problem: As shown in the diagram, ∠ACD=50°, ∠CDA=70°, ∠DBA=21°. Find the measure of ∠ADB.
问题：如图所示，∠ACD=50°，∠CDA=70°，∠DBA=21°。求∠ADB的大小。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),50)
Equal(MeasureOfAngle(CDA),70)
Equal(MeasureOfAngle(DBA),21)
Goal: Value(MeasureOfAngle(ADB))
Answer: 110
Theorem sequence:
flat_angle(1,CDB)
angle_addition(1,CDA,ADB)
Problem: As shown in the diagram, ∠DCE=64°, BA is parallel to CD. Find the measure of ∠ABC.
问题：如图所示，∠DCE=64°，BA∥CD。求∠ABC的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(DC,CE)
Collinear(BCE)
Conditions (CDL):
Equal(MeasureOfAngle(DCE),64)
ParallelBetweenLine(BA,CD)
Goal: Value(MeasureOfAngle(ABC))
Answer: 64
Theorem sequence:
parallel_property_corresponding_angle(2,BA,CD,E)
Problem: As shown in the diagram, RW=14, RZ=18, UZ=5, the centroid of △RST is Z. Find the length of line SR.
问题：如图所示，RW=14，RZ=18，UZ=5，Z是三角形RST的重心。求直线SR的长度。
Construction (CDL):
Shape(RW,WZ,ZR)
Shape(RZ,ZU,UR)
Shape(WS,SZ,ZW)
Shape(UZ,ZT,TU)
Shape(ZS,SV,VZ)
Shape(ZV,VT,TZ)
Collinear(RWS)
Collinear(RUT)
Collinear(SVT)
Collinear(RZU)
Collinear(WZT)
Collinear(SZU)
Conditions (CDL):
Equal(LengthOfLine(RW),14)
Equal(LengthOfLine(RZ),18)
Equal(LengthOfLine(UZ),5)
IsCentroidOfTriangle(Z,RST)
Goal: Value(LengthOfLine(SR))
Answer: 28
Theorem sequence:
centroid_of_triangle_property_intersection(1,Z,TRS,W)
line_addition(1,SW,WR)
Problem: As shown in the diagram, JH=10, KJ=x, ∠JHK=45°, ∠KJH=73°. Find the value of x.
问题：如图所示，JH=10，KJ=x，∠JHK=45°，∠KJH=73°。求x的值。
Construction (CDL):
Shape(HK,KJ,JH)
Conditions (CDL):
Equal(LengthOfLine(JH),10)
Equal(LengthOfLine(KJ),x)
Equal(MeasureOfAngle(JHK),45)
Equal(MeasureOfAngle(KJH),73)
Goal: Value(x)
Answer: 5*sqrt(2)/sin(31*pi/90)
Theorem sequence:
triangle_property_angle_sum(1,HKJ)
sine_theorem(1,JHK)
Problem: As shown in the diagram, CB=CD, CD=10, CD=2*x+4, DB=x+2. Find the value of x.
问题：如图所示，CB=CD，CD=10，CD=2*x+4，DB=x+2。求x的值。
Construction (CDL):
Shape(CB,BD,DC)
Conditions (CDL):
Equal(LengthOfLine(CB),LengthOfLine(CD))
Equal(LengthOfLine(CD),10)
Equal(LengthOfLine(CD),2*x+4)
Equal(LengthOfLine(DB),x+2)
Goal: Value(x)
Answer: 3
Theorem sequence:
Problem: As shown in the diagram, ∠JQR=131°, GT∥RQ, HR is parallel to TC. Find the measure of ∠BGT.
问题：如图所示，∠JQR=131°，GT∥RQ，HR平行于TC。求∠BGT的大小。
Construction (CDL):
Shape(JQ,QR)
Shape(QR,RH)
Shape(BG,GT)
Shape(GT,TC)
Shape(QT,TG,GR,RQ)
Collinear(JQTC)
Collinear(HRGB)
Conditions (CDL):
Equal(MeasureOfAngle(JQR),131)
ParallelBetweenLine(GT,RQ)
ParallelBetweenLine(HR,TC)
Goal: Value(MeasureOfAngle(BGT))
Answer: 131
Theorem sequence:
parallel_property_corresponding_angle(1,QR,TG,J)
parallel_property_collinear_extend(2,HR,TC,G)
parallel_property_collinear_extend(2,HG,TC,B)
adjacent_complementary_angle(1,QTG,GTC)
parallel_property_ipsilateral_internal_angle(1,GB,TC)
Problem: As shown in the diagram, ∠ADB=9*y-2°, ∠CEA=135°, ∠DAE=5*y+5°, ∠EDA=4*z+9°. Find the value of z.
问题：如图所示，∠ADB=9*y-2°，∠CEA=135°，∠DAE=5*y+5°，∠EDA=4*z+9°。求z的值。
Construction (CDL):
Shape(AE,ED,DA)
Shape(CE,EA)
Shape(AD,DB)
Collinear(CEDB)
Conditions (CDL):
Equal(MeasureOfAngle(ADB),9*y-2)
Equal(MeasureOfAngle(CEA),135)
Equal(MeasureOfAngle(DAE),5*y+5)
Equal(MeasureOfAngle(EDA),4*z+9)
Goal: Value(z)
Answer: 14
Theorem sequence:
flat_angle(1,CED)
angle_addition(1,CEA,AED)
flat_angle(1,EDB)
angle_addition(1,EDA,ADB)
triangle_property_angle_sum(1,AED)
Problem: As shown in the diagram, CB=5, CD=4, DB=3, AD is perpendicular to CD, BC is perpendicular to AC. Find the perimeter of triangle ABC.
问题：如图所示，CB=5，CD=4，DB=3，AD⊥CD，BC垂直于AC。求三角形ABC的周长。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(CB),5)
Equal(LengthOfLine(CD),4)
Equal(LengthOfLine(DB),3)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(BC,AC)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 20
Theorem sequence:
adjacent_complementary_angle(1,ADC,CDB)
mirror_similar_triangle_judgment_aa(1,CDB,ABC)
triangle_perimeter_formula(1,ABC)
mirror_similar_triangle_property_line_ratio(1,CDB,ABC)
mirror_similar_triangle_property_line_ratio(1,DBC,CAB)
mirror_similar_triangle_property_line_ratio(1,BCD,BCA)
Problem: As shown in the diagram, AD=4, AE=2, BE=x, AD is the tangent to ⊙C. Find the value of x.
问题：如图所示，AD=4，AE=2，BE=x，圆O的切线为AD。求x的值。
Construction (CDL):
Shape(CBD,CDE,EB)
Shape(CDE,DA,AE)
Shape(CEB,BE)
Collinear(BEA)
Cocircular(C,DEB)
Conditions (CDL):
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(AE),2)
Equal(LengthOfLine(BE),x)
IsTangentOfCircle(AD,C)
Goal: Value(x)
Answer: 6
Theorem sequence:
circle_property_circular_power_tangent_and_segment_line(1,AD,AEB,C)
line_addition(1,BE,EA)
Problem: As shown in the diagram, ∠SVT=75°, ∠TVP=72°, V is the center of ⊙V, RV is perpendicular to SV. Find the measure of ⌒VPS.
问题：如图所示，∠SVT=75°，∠TVP=72°，V是圆V的圆心，RV垂直于SV。求弧VPS的角度。
Construction (CDL):
Shape(VR,VRQ,QV)
Shape(VQ,VQP,PV)
Shape(VP,VPT,TV)
Shape(VT,VTS,SV)
Shape(VS,VSR,RV)
Collinear(QVS)
Cocircular(V,RQPTS)
Conditions (CDL):
Equal(MeasureOfAngle(SVT),75)
Equal(MeasureOfAngle(TVP),72)
IsCentreOfCircle(V,V)
PerpendicularBetweenLine(RV,SV)
Goal: Value(MeasureOfArc(VPS))
Answer: 147
Theorem sequence:
angle_addition(1,SVT,TVP)
arc_property_center_angle(1,VPS,V)
Problem: As shown in the diagram, ∠ABF=6*x+12°, ∠FDA=4*x+8°, quadrilateral ABCD is a rectangle. Find the value of x.
问题：如图所示，∠ABF=6*x+12°，∠FDA=4*x+8°，ABCD是长方形。求x的值。
Construction (CDL):
Shape(AB,BF,FA)
Shape(FB,BC,CF)
Shape(FC,CD,DF)
Shape(AF,FD,DA)
Collinear(AFC)
Collinear(BFD)
Conditions (CDL):
Equal(MeasureOfAngle(ABF),6*x+12)
Equal(MeasureOfAngle(FDA),4*x+8)
Rectangle(ABCD)
Goal: Value(x)
Answer: 7
Theorem sequence:
triangle_property_angle_sum(1,DAB)
Problem: As shown in the diagram, BC=x, BD=7, DE=13, DE is the tangent to ⊙A. Find the value of x.
问题：如图所示，BC=x，BD=7，DE=13，圆O的切线为DE。求x的值。
Construction (CDL):
Shape(BD,DE,ABE)
Shape(ABE,AEC,CB)
Shape(ACB,BC)
Collinear(DBC)
Cocircular(A,BEC)
Conditions (CDL):
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(BD),7)
Equal(LengthOfLine(DE),13)
IsTangentOfCircle(DE,A)
Goal: Value(x)
Answer: 120/7
Theorem sequence:
circle_property_circular_power_tangent_and_segment_line(1,DE,DBC,A)
line_addition(1,DB,BC)
Problem: As shown in the diagram, RQ=4*x, RS=2*x+1, SQ=6*x-1, triangle RQS is an equilateral triangle. Find the length of line RQ.
问题：如图所示，RQ=4*x，RS=2*x+1，SQ=6*x-1，三角形RQS为等边三角形。求直线RQ的长度。
Construction (CDL):
Shape(RQ,QS,SR)
Conditions (CDL):
Equal(LengthOfLine(RQ),4*x)
Equal(LengthOfLine(RS),2*x+1)
Equal(LengthOfLine(SQ),6*x-1)
EquilateralTriangle(RQS)
Goal: Value(LengthOfLine(RQ))
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, BN=EN, CB=BN, CN=2*sqrt(2), EN=CE, CB is perpendicular to NB, NE⊥CE. Find the area of CBNE.
问题：如图所示，BN=EN，CB=BN，CN=2*sqrt(2)，EN=CE，CB垂直于NB，NE垂直于CE。求CBNE的面积。
Construction (CDL):
Shape(CB,BN,NC)
Shape(CN,NE,EC)
Conditions (CDL):
Equal(LengthOfLine(BN),LengthOfLine(EN))
Equal(LengthOfLine(CB),LengthOfLine(BN))
Equal(LengthOfLine(CN),2*sqrt(2))
Equal(LengthOfLine(EN),LengthOfLine(CE))
PerpendicularBetweenLine(CB,NB)
PerpendicularBetweenLine(NE,CE)
Goal: Value(AreaOfQuadrilateral(CBNE))
Answer: 4
Theorem sequence:
right_triangle_judgment_angle(1,CBN)
right_triangle_property_pythagorean(1,CBN)
parallelogram_judgment_equal_and_equal(1,CBNE)
parallelogram_area_formula_sine(1,CBNE)
Problem: As shown in the diagram, AN=5, the center of ⊙N is N. Find the circumference of the ⊙N.
问题：如图所示，AN=5，N是圆N的圆心。求圆N的周长。
Construction (CDL):
Shape(NB,NBA,AN)
Shape(DN,NA,NAE,ED)
Shape(NED,DE)
Shape(ND,NDC,CN)
Shape(NC,CB,BN)
Shape(NCB,BC)
Collinear(AND)
Cocircular(N,AEDCB)
Conditions (CDL):
Equal(LengthOfLine(AN),5)
IsCentreOfCircle(N,N)
Goal: Value(PerimeterOfCircle(N))
Answer: 10*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,NA,N)
circle_perimeter_formula(1,N)
Problem: As shown in the diagram, RS=3*x+2, RS=RT, TR=2*x+4. Find the length of line RS.
问题：如图所示，RS=3*x+2，RS=RT，TR=2*x+4。求直线RS的长度。
Construction (CDL):
Shape(RT,TS,SR)
Conditions (CDL):
Equal(LengthOfLine(RS),3*x+2)
Equal(LengthOfLine(RS),LengthOfLine(RT))
Equal(LengthOfLine(TR),2*x+4)
Goal: Value(LengthOfLine(RS))
Answer: 8
Theorem sequence:
Problem: As shown in the diagram, AC=x, BC=4, ∠ACB=73°, ∠CBA=29°. Find the value of x.
问题：如图所示，AC=x，BC=4，∠ACB=73°，∠CBA=29°。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),4)
Equal(MeasureOfAngle(ACB),73)
Equal(MeasureOfAngle(CBA),29)
Goal: Value(x)
Answer: 32*sqrt(2)*sin(29*pi/180)/(-sqrt(2)+sqrt(10)+2*sqrt(3)*sqrt(sqrt(5)+5))
Theorem sequence:
triangle_property_angle_sum(1,BAC)
sine_theorem(1,CBA)
Problem: As shown in the diagram, ∠ZYX=120°, quadrilateral YXWZ is a kite, XW⊥ZW. Find the measure of ∠WZY.
问题：如图所示，∠ZYX=120°，四边形YXWZ是风筝形，XW垂直于ZW。求∠WZY的大小。
Construction (CDL):
Shape(YX,XW,WZ,ZY)
Conditions (CDL):
Equal(MeasureOfAngle(ZYX),120)
Kite(YXWZ)
PerpendicularBetweenLine(XW,ZW)
Goal: Value(MeasureOfAngle(WZY))
Answer: 75
Theorem sequence:
kite_property_opposite_angle_equal(1,YXWZ)
quadrilateral_property_angle_sum(1,YXWZ)
Problem: As shown in the diagram, AD=BA, CD=BC, ∠ADC=85°, ∠BAD=120°, quadrilateral ADCB is a kite. Find the measure of ∠DCB.
问题：如图所示，AD=BA，CD=BC，∠ADC=85°，∠BAD=120°，CD和CB是风筝形ADCB的一组临边。求∠DCB的大小。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(BA))
Equal(LengthOfLine(CD),LengthOfLine(BC))
Equal(MeasureOfAngle(ADC),85)
Equal(MeasureOfAngle(BAD),120)
Kite(ADCB)
Goal: Value(MeasureOfAngle(DCB))
Answer: 70
Theorem sequence:
kite_property_opposite_angle_equal(1,ADCB)
quadrilateral_property_angle_sum(1,ADCB)
Problem: As shown in the diagram, LJ=12, ML=MJ, MK is perpendicular to JK. Find the length of line KL.
问题：如图所示，LJ=12，ML=MJ，MK⊥JK。求直线KL的长度。
Construction (CDL):
Shape(JM,MK,KJ)
Shape(KM,ML,LK)
Collinear(JKL)
Conditions (CDL):
Equal(LengthOfLine(LJ),12)
Equal(LengthOfLine(ML),LengthOfLine(MJ))
PerpendicularBetweenLine(MK,JK)
Goal: Value(LengthOfLine(KL))
Answer: 6
Theorem sequence:
adjacent_complementary_angle(1,LKM,MKJ)
isosceles_triangle_judgment_line_equal(1,MLJ)
altitude_of_triangle_judgment(1,MK,MLJ)
isosceles_triangle_property_line_coincidence(1,MLJ,K)
line_addition(1,JK,KL)
Problem: As shown in the diagram, AB=4*y-8, AD=x+4, BC=2*x-4, DC=3*y-2, CD and BA are opposite sides of the ▱ CBAD. Find the value of y.
问题：如图所示，AB=4*y-8，AD=x+4，BC=2*x-4，DC=3*y-2，四边形CBAD是▱。求y的值。
Construction (CDL):
Shape(CB,BA,AD,DC)
Conditions (CDL):
Equal(LengthOfLine(AB),4*y-8)
Equal(LengthOfLine(AD),x+4)
Equal(LengthOfLine(BC),2*x-4)
Equal(LengthOfLine(DC),3*y-2)
Parallelogram(CBAD)
Goal: Value(y)
Answer: 6
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DCBA)
Problem: As shown in the diagram, DC=3*y+13, EO=2*y+19, ∠COE=2*x+41°, ∠EDC=115°, DE and CO are opposite sides of the parallelogram EDCO. Find the value of y.
问题：如图所示，DC=3*y+13，EO=2*y+19，∠COE=2*x+41°，∠EDC=115°，EDCO是▱。求y的值。
Construction (CDL):
Shape(ED,DC,CO,OE)
Conditions (CDL):
Equal(LengthOfLine(DC),3*y+13)
Equal(LengthOfLine(EO),2*y+19)
Equal(MeasureOfAngle(COE),2*x+41)
Equal(MeasureOfAngle(EDC),115)
Parallelogram(EDCO)
Goal: Value(y)
Answer: 6
Theorem sequence:
parallelogram_property_opposite_line_equal(1,OEDC)
Problem: As shown in the diagram, AB=10, the center of ⊙A is A. Find the circumference of the ⊙A.
问题：如图所示，AB=10，A是⊙A的圆心。求圆A的周长。
Construction (CDL):
Shape(AB)
Cocircular(A,B)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
IsCentreOfCircle(A,A)
Goal: Value(PerimeterOfCircle(A))
Answer: 20*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,AB,A)
circle_perimeter_formula(1,A)
Problem: As shown in the diagram, AB=x+1, CD=x-1, FE=8, HG=5, ∠ADC=∠EHG, ∠BAD=∠FEH, ∠CBA=∠GFE, ∠DCB=∠HGF, BADC is similar to FEHG. Find the length of line AB.
问题：如图所示，AB=x+1，CD=x-1，FE=8，HG=5，∠ADC=∠EHG，∠BAD=∠FEH，∠CBA=∠GFE，∠DCB=∠HGF，BADC相似于FEHG。求直线AB的长度。
Construction (CDL):
Shape(BA,AD,DC,CB)
Shape(EH,HG,GF,FE)
Conditions (CDL):
Equal(LengthOfLine(AB),x+1)
Equal(LengthOfLine(CD),x-1)
Equal(LengthOfLine(FE),8)
Equal(LengthOfLine(HG),5)
Equal(MeasureOfAngle(ADC),MeasureOfAngle(EHG))
Equal(MeasureOfAngle(BAD),MeasureOfAngle(FEH))
Equal(MeasureOfAngle(CBA),MeasureOfAngle(GFE))
Equal(MeasureOfAngle(DCB),MeasureOfAngle(HGF))
SimilarBetweenQuadrilateral(BADC,FEHG)
Goal: Value(LengthOfLine(AB))
Answer: 16/3
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,BADC,FEHG)
similar_quadrilateral_property_line_ratio(1,DCBA,HGFE)
Problem: As shown in the diagram, AC=3*x-6, AD=21, BD=BC, EG=x+4, EH=14, HF=GF, ∠ACB=∠EGF, ∠BDA=∠FHE. Find the value of x.
问题：如图所示，AC=3*x-6，AD=21，BD=BC，EG=x+4，EH=14，HF=GF，∠ACB=∠EGF，∠BDA=∠FHE。求x的值。
Construction (CDL):
Shape(DA,AB,BD)
Shape(BA,AC,CB)
Shape(HE,EF,FH)
Shape(FE,EG,GF)
Collinear(DBC)
Collinear(HFG)
Conditions (CDL):
Equal(LengthOfLine(AC),3*x-6)
Equal(LengthOfLine(AD),21)
Equal(LengthOfLine(BD),LengthOfLine(BC))
Equal(LengthOfLine(EG),x+4)
Equal(LengthOfLine(EH),14)
Equal(LengthOfLine(HF),LengthOfLine(GF))
Equal(MeasureOfAngle(ACB),MeasureOfAngle(EGF))
Equal(MeasureOfAngle(BDA),MeasureOfAngle(FHE))
Goal: Value(x)
Answer: 8
Theorem sequence:
similar_triangle_judgment_aa(1,ACD,EGH)
similar_triangle_property_line_ratio(1,DAC,HEG)
similar_triangle_property_line_ratio(1,CDA,GHE)
Problem: As shown in the diagram, ∠HUN=123°, UT∥QA. Find the measure of ∠AQE.
问题：如图所示，∠HUN=123°，UT平行于QA。求∠AQE的大小。
Construction (CDL):
Shape(NU,UT)
Shape(TU,UQ)
Shape(UQ,QA)
Shape(AQ,QE)
Shape(EQ,QB)
Shape(BQ,QU)
Shape(QU,UH)
Shape(HU,UN)
Collinear(NUQE)
Collinear(TUH)
Collinear(AQB)
Conditions (CDL):
Equal(MeasureOfAngle(HUN),123)
ParallelBetweenLine(UT,QA)
Goal: Value(MeasureOfAngle(AQE))
Answer: 123
Theorem sequence:
vertical_angle(1,HUN,TUQ)
parallel_property_corresponding_angle(2,UT,QA,E)
Problem: As shown in the diagram, ∠PXR=x°, ∠QPX=4*x°, ∠RQP=2*x°, ∠XRQ=5*x°. Find the measure of ∠XRQ.
问题：如图所示，∠PXR=x°，∠QPX=4*x°，∠RQP=2*x°，∠XRQ=5*x°。求∠XRQ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(XRQ))
Answer: 150
Theorem sequence:
quadrilateral_property_angle_sum(1,XRQP)
Problem: As shown in the diagram, ∠JKL=x+10°, ∠KLM=x°, ∠LMJ=2*x-8°, ∠MJK=3*x-6°. Find the measure of ∠JKL.
问题：如图所示，∠JKL=x+10°，∠KLM=x°，∠LMJ=2*x-8°，∠MJK=3*x-6°。求∠JKL的大小。
Construction (CDL):
Shape(JK,KL,LM,MJ)
Conditions (CDL):
Equal(MeasureOfAngle(JKL),x+10)
Equal(MeasureOfAngle(KLM),x)
Equal(MeasureOfAngle(LMJ),2*x-8)
Equal(MeasureOfAngle(MJK),3*x-6)
Goal: Value(MeasureOfAngle(JKL))
Answer: 62
Theorem sequence:
quadrilateral_property_angle_sum(1,JKLM)
Problem: As shown in the diagram, AE=2*x+6, BE=y+3, CE=4*y-9, DE=2*y+5, quadrilateral BCAD is a ▱. Find the value of y.
问题：如图所示，AE=2*x+6，BE=y+3，CE=4*y-9，DE=2*y+5，四边形BCAD是▱。求y的值。
Construction (CDL):
Shape(BC,CE,EB)
Shape(EC,CA,AE)
Shape(EA,AD,DE)
Shape(BE,ED,DB)
Collinear(BEA)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AE),2*x+6)
Equal(LengthOfLine(BE),y+3)
Equal(LengthOfLine(CE),4*y-9)
Equal(LengthOfLine(DE),2*y+5)
Parallelogram(BCAD)
Goal: Value(y)
Answer: 7
Theorem sequence:
parallelogram_property_diagonal_bisection(1,CADB,E)
Problem: As shown in the diagram, AB=25, AC=12, AC is perpendicular to BC. Find the measure of ∠CBA.
问题：如图所示，AB=25，AC=12，AC垂直于BC。求∠CBA的大小。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),25)
Equal(LengthOfLine(AC),12)
PerpendicularBetweenLine(AC,BC)
Goal: Value(MeasureOfAngle(CBA))
Answer: 180*asin(12/25)/pi
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, ∠JQR=131°, QJ∥RH, QR is parallel to TS. Find the measure of ∠SRQ.
问题：如图所示，∠JQR=131°，QJ∥RH，QR平行于TS。求∠SRQ的大小。
Construction (CDL):
Shape(JQ,QR)
Shape(QR,RH)
Shape(ST,TC)
Shape(BS,ST)
Shape(QT,TS,SR,RQ)
Collinear(JQTC)
Collinear(HRSB)
Conditions (CDL):
Equal(MeasureOfAngle(JQR),131)
ParallelBetweenLine(QJ,RH)
ParallelBetweenLine(QR,TS)
Goal: Value(MeasureOfAngle(SRQ))
Answer: 131
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,QJ,RH)
adjacent_complementary_angle(1,SRQ,QRH)
Problem: As shown in the diagram, BC=CD, CB=2*x+4, CD=10, DB=x+2. Find the length of line DB.
问题：如图所示，BC=CD，CB=2*x+4，CD=10，DB=x+2。求直线DB的长度。
Construction (CDL):
Shape(CB,BD,DC)
Conditions (CDL):
Equal(LengthOfLine(BC),LengthOfLine(CD))
Equal(LengthOfLine(CB),2*x+4)
Equal(LengthOfLine(CD),10)
Equal(LengthOfLine(DB),x+2)
Goal: Value(LengthOfLine(DB))
Answer: 5
Theorem sequence:
Problem: As shown in the diagram, ∠FEQ=100°, ∠QEC=160°, the center of ⊙E is E, the tangent to circle E is DC, GC is the tangent to ⊙E. Find the measure of ∠GCF.
问题：如图所示，∠FEQ=100°，∠QEC=160°，⊙E的圆心为E，DC是圆O的切线，⊙O的切线为GC。求∠GCF的大小。
Construction (CDL):
Shape(CF,EFC)
Shape(EF,FC,CE)
Shape(EC,ECQ,QE)
Shape(EQ,EQF,FE)
Shape(GC,CF)
Shape(FC,CE)
Shape(EC,CD)
Collinear(GCD)
Cocircular(E,FCQ)
Conditions (CDL):
Equal(MeasureOfAngle(FEQ),100)
Equal(MeasureOfAngle(QEC),160)
IsCentreOfCircle(E,E)
IsTangentOfCircle(DC,E)
IsTangentOfCircle(GC,E)
Goal: Value(MeasureOfAngle(GCF))
Answer: 50
Theorem sequence:
radius_of_circle_property_length_equal(1,EF,E)
radius_of_circle_property_length_equal(1,EC,E)
angle_addition(1,FEQ,QEC)
round_angle(1,FEC,CEF)
isosceles_triangle_judgment_line_equal(1,EFC)
isosceles_triangle_property_angle_equal(1,EFC)
triangle_property_angle_sum(1,EFC)
tangent_of_circle_property_perpendicular(2,GC,E,E)
angle_addition(1,GCF,FCE)
Problem: As shown in the diagram, the height of DCBA is 5, DA=10, DC=4, DA and CB are opposite sides of the parallelogram ADCB. Find the area of DCBA.
问题：如图所示，四边形DCBA的高为5，DA=10，DC=4，ADCB是平行四边形。求DCBA的面积。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(HeightOfQuadrilateral(DCBA),5)
Equal(LengthOfLine(DA),10)
Equal(LengthOfLine(DC),4)
Parallelogram(ADCB)
Goal: Value(AreaOfQuadrilateral(DCBA))
Answer: 50
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ADCB)
parallelogram_area_formula_common(1,DCBA)
Problem: As shown in the diagram, ∠GKM=62°, ∠HBF=38°, NO is parallel to MH. Find the measure of ∠CBH.
问题：如图所示，∠GKM=62°，∠HBF=38°，NO平行于MH。求∠CBH的大小。
Construction (CDL):
Shape(NC,CJ)
Shape(JC,CI)
Shape(IC,CO)
Shape(KC,CN)
Shape(CK,KB,BC)
Shape(OC,CB)
Shape(MK,KC)
Shape(CB,BH)
Shape(GK,KM)
Shape(BK,KG)
Shape(FB,BK)
Shape(HB,BF)
Collinear(NCO)
Collinear(MKBH)
Collinear(JCBF)
Collinear(ICKG)
Conditions (CDL):
Equal(MeasureOfAngle(GKM),62)
Equal(MeasureOfAngle(HBF),38)
ParallelBetweenLine(NO,MH)
Goal: Value(MeasureOfAngle(CBH))
Answer: 142
Theorem sequence:
adjacent_complementary_angle(1,CBH,HBF)
Problem: As shown in the diagram, ∠CDG=78°, ∠GDE=x°. Find the value of x.
问题：如图所示，∠CDG=78°，∠GDE=x°。求x的值。
Construction (CDL):
Shape(CD,DG)
Shape(GD,DE)
Collinear(CDE)
Conditions (CDL):
Equal(MeasureOfAngle(CDG),78)
Equal(MeasureOfAngle(GDE),x)
Goal: Value(x)
Answer: 102
Theorem sequence:
adjacent_complementary_angle(1,CDG,GDE)
Problem: As shown in the diagram, ∠BAE=58°, ∠CBE=26°, ∠EBA=47°, AB is parallel to EC. Find the measure of ∠CED.
问题：如图所示，∠BAE=58°，∠CBE=26°，∠EBA=47°，AB∥EC。求∠CED的大小。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BE,EC,CB)
Shape(CE,ED,DC)
Collinear(AED)
Collinear(BCD)
Conditions (CDL):
Equal(MeasureOfAngle(BAE),58)
Equal(MeasureOfAngle(CBE),26)
Equal(MeasureOfAngle(EBA),47)
ParallelBetweenLine(AB,EC)
Goal: Value(MeasureOfAngle(CED))
Answer: 58
Theorem sequence:
parallel_property_corresponding_angle(2,AB,EC,D)
Problem: As shown in the diagram, QR=RS, QT=TS, ∠QTS=68°, ∠SRQ=136°, QR and QT are one pair of adjacent sides of the kite QRST. Find the measure of ∠TSR.
问题：如图所示，QR=RS，QT=TS，∠QTS=68°，∠SRQ=136°，QRST是风筝形。求∠TSR的大小。
Construction (CDL):
Shape(RQ,QS,SR)
Shape(QT,TS,SQ)
Conditions (CDL):
Equal(LengthOfLine(QR),LengthOfLine(RS))
Equal(LengthOfLine(QT),LengthOfLine(TS))
Equal(MeasureOfAngle(QTS),68)
Equal(MeasureOfAngle(SRQ),136)
Kite(QRST)
Goal: Value(MeasureOfAngle(TSR))
Answer: 78
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,RQS)
isosceles_triangle_judgment_line_equal(1,TSQ)
isosceles_triangle_property_angle_equal(1,RQS)
isosceles_triangle_property_angle_equal(1,TSQ)
triangle_property_angle_sum(1,RQS)
triangle_property_angle_sum(1,TSQ)
angle_addition(1,TSQ,QSR)