SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, TQ=3*x-8, TS=x+10, the tangent to ⊙A is TQ, the tangent to ⊙A is TR, TR is the tangent to circle B, TS is the tangent to circle B. Find the value of x.
问题：如图所示，TQ=3*x-8，TS=x+10，圆O的切线为TQ，⊙O的切线为TR，TR是⊙O的切线，圆O的切线为TS。求x的值。
Construction (CDL):
Shape(QT,TR,AQR)
Shape(AQR,ARQ)
Shape(RT,TS,BRS)
Shape(BRS,BSR)
Cocircular(A,QR)
Cocircular(B,RS)
Conditions (CDL):
Equal(LengthOfLine(TQ),3*x-8)
Equal(LengthOfLine(TS),x+10)
IsTangentOfCircle(TQ,A)
IsTangentOfCircle(TR,A)
IsTangentOfCircle(TR,B)
IsTangentOfCircle(TS,B)
Goal: Value(x)
Answer: 9
Theorem sequence:
tangent_of_circle_property_length_equal(1,TQ,TR,A)
tangent_of_circle_property_length_equal(1,TR,TS,B)
Problem: As shown in the diagram, PR=3, the radius of ⊙P is 5, P is the center of circle P, PR is perpendicular to QR. Find the length of line QS.
问题：如图所示，PR=3，⊙P的半径为5，P是⊙P的圆心，PR⊥QR。求直线QS的长度。
Construction (CDL):
Shape(PQ,PQS,SP)
Shape(PR,RQ,QP)
Shape(PS,SR,RP)
Shape(PSQ,QS)
Collinear(QRS)
Cocircular(P,SQ)
Conditions (CDL):
Equal(LengthOfLine(PR),3)
Equal(RadiusOfCircle(P),5)
IsCentreOfCircle(P,P)
PerpendicularBetweenLine(PR,QR)
Goal: Value(LengthOfLine(QS))
Answer: 8
Theorem sequence:
radius_of_circle_property_length_equal(1,PQ,P)
radius_of_circle_property_length_equal(1,PS,P)
adjacent_complementary_angle(1,SRP,PRQ)
right_triangle_judgment_angle(1,PRQ)
right_triangle_property_pythagorean(1,PRQ)
right_triangle_judgment_angle(1,SRP)
right_triangle_property_pythagorean(1,SRP)
line_addition(1,QR,RS)
Problem: As shown in the diagram, CB=18, CE=13, ED=24, the height of quadrilateral CEDB is CE, CE is perpendicular to DE, CB and ED are the parallel sides of trapezoid CEDB. Find the area of quadrilateral CEDB.
问题：如图所示，CB=18，CE=13，ED=24，CE是CEDB的高，CE垂直于DE，CE和DB是梯形CEDB的腰。求四边形CEDB的面积。
Construction (CDL):
Shape(CE,ED,DB,BC)
Conditions (CDL):
Equal(LengthOfLine(CB),18)
Equal(LengthOfLine(CE),13)
Equal(LengthOfLine(ED),24)
IsAltitudeOfQuadrilateral(CE,CEDB)
PerpendicularBetweenLine(CE,DE)
Trapezoid(CEDB)
Goal: Value(AreaOfQuadrilateral(CEDB))
Answer: 273
Theorem sequence:
right_trapezoid_judgment_right_angle(1,CEDB)
right_trapezoid_area_formular(1,CEDB)
Problem: As shown in the diagram, AB=96-y, AD=5*x-18, BC=2*x, DC=3*y, CBAD is a parallelogram. Find the value of y.
问题：如图所示，AB=96-y，AD=5*x-18，BC=2*x，DC=3*y，CBAD是平行四边形。求y的值。
Construction (CDL):
Shape(CB,BA,AD,DC)
Conditions (CDL):
Equal(LengthOfLine(AB),96-y)
Equal(LengthOfLine(AD),5*x-18)
Equal(LengthOfLine(BC),2*x)
Equal(LengthOfLine(DC),3*y)
Parallelogram(CBAD)
Goal: Value(y)
Answer: 24
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DCBA)
Problem: As shown in the diagram, AF=24, AH=25, BH=11, ∠ABH=30°, ∠GCH=28°, the incenter of triangle ABC is H, BG⊥HG, HD is perpendicular to BD, HF⊥AF. Find the length of line DH.
问题：如图所示，AF=24，AH=25，BH=11，∠ABH=30°，∠GCH=28°，△ABC的内心为H，BG垂直于HG，HD垂直于BD，HF⊥AF。求直线DH的长度。
Construction (CDL):
Shape(AD,DH,HA)
Shape(DB,BH,HD)
Shape(BG,GH,HB)
Shape(GC,CH,HG)
Shape(CF,FH,HC)
Shape(HF,FA,AH)
Collinear(ADB)
Collinear(AFC)
Collinear(BGC)
Conditions (CDL):
Equal(LengthOfLine(AF),24)
Equal(LengthOfLine(AH),25)
Equal(LengthOfLine(BH),11)
Equal(MeasureOfAngle(ABH),30)
Equal(MeasureOfAngle(GCH),28)
IsIncenterOfTriangle(H,ABC)
PerpendicularBetweenLine(BG,HG)
PerpendicularBetweenLine(HD,BD)
PerpendicularBetweenLine(HF,AF)
Goal: Value(LengthOfLine(DH))
Answer: 11/2
Theorem sequence:
sine_theorem(1,HDB)
Problem: As shown in the diagram, AB=15, PB=12, ∠DBA=24°, quadrilateral ADCB is a rhombus. Find the length of line CP.
问题：如图所示，AB=15，PB=12，∠DBA=24°，四边形ADCB是菱形。求直线CP的长度。
Construction (CDL):
Shape(AD,DP,PA)
Shape(PD,DC,CP)
Shape(PC,CB,BP)
Shape(PB,BA,AP)
Collinear(APC)
Collinear(DPB)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(PB),12)
Equal(MeasureOfAngle(DBA),24)
Rhombus(ADCB)
Goal: Value(LengthOfLine(CP))
Answer: 9
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,BADC,P)
right_triangle_judgment_angle(1,APB)
right_triangle_property_pythagorean(1,APB)
Problem: As shown in the diagram, ∠DBA=17°, ∠DEA=29°, AD⊥BD. Find the measure of ∠AEB.
问题：如图所示，∠DBA=17°，∠DEA=29°，AD⊥BD。求∠AEB的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(AE,EB,BA)
Collinear(DEB)
Conditions (CDL):
Equal(MeasureOfAngle(DBA),17)
Equal(MeasureOfAngle(DEA),29)
PerpendicularBetweenLine(AD,BD)
Goal: Value(MeasureOfAngle(AEB))
Answer: 151
Theorem sequence:
angle_addition(1,BAE,EAD)
triangle_property_angle_sum(1,ADE)
triangle_property_angle_sum(1,AEB)
triangle_property_angle_sum(1,DBA)
Problem: As shown in the diagram, ∠FEI=2*x+70°, ∠HGL=2*y-20°, ∠RFN=3*x+40°, FN∥EI, GF is parallel to HE. Find the value of x.
问题：如图所示，∠FEI=2*x+70°，∠HGL=2*y-20°，∠RFN=3*x+40°，FN平行于EI，GF平行于HE。求x的值。
Construction (CDL):
Shape(LG,GQ)
Shape(HG,GL)
Shape(PH,HG)
Shape(MH,HP)
Shape(EH,HM)
Shape(QG,GF)
Shape(OE,EH)
Shape(GF,FR)
Shape(HE,EF,FG,GH)
Shape(RF,FN)
Shape(NF,FE)
Shape(FE,EI)
Shape(IE,EO)
Collinear(QGHM)
Collinear(RFEO)
Collinear(LGFN)
Collinear(PHEI)
Conditions (CDL):
Equal(MeasureOfAngle(FEI),2*x+70)
Equal(MeasureOfAngle(HGL),2*y-20)
Equal(MeasureOfAngle(RFN),3*x+40)
ParallelBetweenLine(FN,EI)
ParallelBetweenLine(GF,HE)
Goal: Value(x)
Answer: 30
Theorem sequence:
parallel_property_corresponding_angle(1,FN,EI,R)
Problem: As shown in the diagram, AB=5*x+5, AC=4*x, BC=8*x+9, the perimeter of triangle ABC is 65. Find the length of line AC.
问题：如图所示，AB=5*x+5，AC=4*x，BC=8*x+9，△ABC的周长为65。求直线AC的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),5*x+5)
Equal(LengthOfLine(AC),4*x)
Equal(LengthOfLine(BC),8*x+9)
Equal(PerimeterOfTriangle(ABC),65)
Goal: Value(LengthOfLine(AC))
Answer: 12
Theorem sequence:
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, ∠NMQ=10*x°, ∠PNM=20*x°, ∠PNM=∠MQP, ∠QPN=∠NMQ, quadrilateral MQPN is a ▱. Find the measure of ∠QPN.
问题：如图所示，∠NMQ=10*x°，∠PNM=20*x°，∠PNM=∠MQP，∠QPN=∠NMQ，四边形MQPN是平行四边形。求∠QPN的大小。
Construction (CDL):
Shape(MQ,QP,PN,NM)
Conditions (CDL):
Equal(MeasureOfAngle(NMQ),10*x)
Equal(MeasureOfAngle(PNM),20*x)
Equal(MeasureOfAngle(PNM),MeasureOfAngle(MQP))
Equal(MeasureOfAngle(QPN),MeasureOfAngle(NMQ))
Parallelogram(MQPN)
Goal: Value(MeasureOfAngle(QPN))
Answer: 60
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,NP,MQ)
Problem: As shown in the diagram, the area of quadrilateral BTAC is 153, AT=9, the height of quadrilateral BTAC is BD, BTAC is a parallelogram, BD⊥TD. Find the length of line BD.
问题：如图所示，BTAC的面积为153，AT=9，BD是四边形BTAC的高，BTAC是平行四边形，BD垂直于TD。求直线BD的长度。
Construction (CDL):
Shape(BD,DT,TB)
Shape(BT,TA,AC,CB)
Collinear(DTA)
Conditions (CDL):
Equal(AreaOfQuadrilateral(BTAC),153)
Equal(LengthOfLine(AT),9)
IsAltitudeOfQuadrilateral(BD,BTAC)
Parallelogram(BTAC)
PerpendicularBetweenLine(BD,TD)
Goal: Value(LengthOfLine(BD))
Answer: 17
Theorem sequence:
parallelogram_area_formula_common(1,BTAC)
Problem: As shown in the diagram, DC=3, ∠EDC=57°, the center of ⊙D is D. Find the area of the sector DCE.
问题：如图所示，DC=3，∠EDC=57°，D是圆D的圆心。求扇形DCE的面积。
Construction (CDL):
Shape(DC,DCE,ED)
Shape(DE,DEC,CD)
Cocircular(D,CE)
Conditions (CDL):
Equal(LengthOfLine(DC),3)
Equal(MeasureOfAngle(EDC),57)
IsCentreOfCircle(D,D)
Goal: Value(AreaOfSector(DCE))
Answer: 57*pi/40
Theorem sequence:
sector_area_formula(1,DCE)
arc_property_center_angle(1,DCE,D)
radius_of_circle_property_length_equal(1,DC,D)
Problem: As shown in the diagram, MO=6*x+14, PN=9*x+5, MNOP is a rectangle. Find the value of x.
问题：如图所示，MO=6*x+14，PN=9*x+5，MNOP是矩形。求x的值。
Construction (CDL):
Shape(AP,PM,MA)
Shape(AO,OP,PA)
Shape(AN,NO,OA)
Shape(AM,MN,NA)
Collinear(PAN)
Collinear(MAO)
Conditions (CDL):
Equal(LengthOfLine(MO),6*x+14)
Equal(LengthOfLine(PN),9*x+5)
Rectangle(MNOP)
Goal: Value(x)
Answer: 3
Theorem sequence:
rectangle_property_diagonal_equal(1,MNOP)
Problem: As shown in the diagram, ∠ADC=x°, ∠BAD=3*x°, CB is perpendicular to AB, DC⊥BC. Find the measure of ∠ADC.
问题：如图所示，∠ADC=x°，∠BAD=3*x°，CB⊥AB，DC垂直于BC。求∠ADC的大小。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),x)
Equal(MeasureOfAngle(BAD),3*x)
PerpendicularBetweenLine(CB,AB)
PerpendicularBetweenLine(DC,BC)
Goal: Value(MeasureOfAngle(ADC))
Answer: 45
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,CD,BA)
parallel_property_ipsilateral_internal_angle(1,AB,DC)
Problem: As shown in the diagram, ∠FJE=75°. Find the measure of ∠CJF.
问题：如图所示，∠FJE=75°。求∠CJF的大小。
Construction (CDL):
Shape(CJ,JF)
Shape(FJ,JE)
Collinear(CJE)
Conditions (CDL):
Equal(MeasureOfAngle(FJE),75)
Goal: Value(MeasureOfAngle(CJF))
Answer: 105
Theorem sequence:
adjacent_complementary_angle(1,CJF,FJE)
Problem: As shown in the diagram, ∠JKL=62°, A is the center of circle A. Find the measure of ⌒AJK.
问题：如图所示，∠JKL=62°，⊙A的圆心为A。求⌒AJK的角度。
Construction (CDL):
Shape(AL,ALJ,JA)
Shape(AJ,JK,KA)
Shape(AJK,KJ)
Shape(AK,AKL,LA)
Collinear(LAK)
Cocircular(A,LJK)
Conditions (CDL):
Equal(MeasureOfAngle(JKL),62)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfArc(AJK))
Answer: 56
Theorem sequence:
radius_of_circle_property_length_equal(1,AJ,A)
radius_of_circle_property_length_equal(1,AK,A)
isosceles_triangle_judgment_line_equal(1,AJK)
isosceles_triangle_property_angle_equal(1,AJK)
triangle_property_angle_sum(1,AJK)
arc_property_center_angle(1,AJK,A)
Problem: As shown in the diagram, AC=y, AD=8, BC=z, BD=25/2, CD=x, BC⊥AC, CD⊥BD. Find the value of z.
问题：如图所示，AC=y，AD=8，BC=z，BD=25/2，CD=x，BC⊥AC，CD垂直于BD。求z的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BD),25/2)
Equal(LengthOfLine(CD),x)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(CD,BD)
Goal: Value(z)
Answer: 5*sqrt(41)/2
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,CDB,ABC)
line_addition(1,AD,DB)
mirror_similar_triangle_property_line_ratio(1,CDB,ABC)
mirror_similar_triangle_property_line_ratio(1,DBC,CAB)
Problem: As shown in the diagram, ∠RTS=47°, SU is perpendicular to TU, TS is perpendicular to RS, TV⊥UV. Find the measure of ∠VUT.
问题：如图所示，∠RTS=47°，SU⊥TU，TS⊥RS，TV垂直于UV。求∠VUT的大小。
Construction (CDL):
Shape(SR,RU,US)
Shape(SU,UV,VS)
Shape(VU,UT,TV)
Collinear(RUT)
Collinear(SVT)
Conditions (CDL):
Equal(MeasureOfAngle(RTS),47)
PerpendicularBetweenLine(SU,TU)
PerpendicularBetweenLine(TS,RS)
PerpendicularBetweenLine(TV,UV)
Goal: Value(MeasureOfAngle(VUT))
Answer: 43
Theorem sequence:
triangle_property_angle_sum(1,VUT)
Problem: As shown in the diagram, ∠AHD=14*x+9°, ∠FEG=5*x+90°, HD∥EB. Find the value of x.
问题：如图所示，∠AHD=14*x+9°，∠FEG=5*x+90°，HD∥EB。求x的值。
Construction (CDL):
Shape(CH,HA)
Shape(AH,HD)
Shape(EH,HC)
Shape(DH,HE)
Shape(GE,EH)
Shape(HE,EB)
Shape(FE,EG)
Shape(BE,EF)
Collinear(CHD)
Collinear(GEB)
Collinear(AHEF)
Conditions (CDL):
Equal(MeasureOfAngle(AHD),14*x+9)
Equal(MeasureOfAngle(FEG),5*x+90)
ParallelBetweenLine(HD,EB)
Goal: Value(x)
Answer: 9
Theorem sequence:
parallel_property_corresponding_angle(1,HD,EB,A)
vertical_angle(1,HEB,FEG)
Problem: As shown in the diagram, AD=24, the center of ⊙N is N, the diameter of ⊙N is AD. Find the length of line CN.
问题：如图所示，AD=24，⊙N的圆心为N，圆N的直径为AD。求直线CN的长度。
Construction (CDL):
Shape(AN,NB,NBA)
Shape(NC,CB,BN)
Shape(NCB,BC)
Shape(ND,NDC,CN)
Shape(NA,NAE,ED,DN)
Shape(NED,DE)
Collinear(AND)
Cocircular(N,AEDCB)
Conditions (CDL):
Equal(LengthOfLine(AD),24)
IsCentreOfCircle(N,N)
IsDiameterOfCircle(AD,N)
Goal: Value(LengthOfLine(CN))
Answer: 12
Theorem sequence:
diameter_of_circle_property_length_equal(1,AD,N)
circle_property_length_of_radius_and_diameter(1,N)
radius_of_circle_property_length_equal(1,NC,N)
Problem: As shown in the diagram, BF=2*y+4, DG=3/2*x+8, DG=FG, EB=3*y-6, FG=1/2*x+12. Find the value of x.
问题：如图所示，BF=2*y+4，DG=3/2*x+8，DG=FG，EB=3*y-6，FG=1/2*x+12。求x的值。
Construction (CDL):
Shape(FG,GB,BF)
Shape(BG,FD,DE,EB)
Collinear(FGD)
Collinear(FBE)
Conditions (CDL):
Equal(LengthOfLine(BF),2*y+4)
Equal(LengthOfLine(DG),3/2*x+8)
Equal(LengthOfLine(DG),LengthOfLine(FG))
Equal(LengthOfLine(EB),3*y-6)
Equal(LengthOfLine(FG),1/2*x+12)
Goal: Value(x)
Answer: 4
Theorem sequence:
Problem: As shown in the diagram, ∠TWV=3*x-4°, ∠UTW=x°, ∠VUT=3*x-4°, ∠WVU=x°. Find the measure of ∠VUT.
问题：如图所示，∠TWV=3*x-4°，∠UTW=x°，∠VUT=3*x-4°，∠WVU=x°。求∠VUT的大小。
Construction (CDL):
Shape(UT,TW,WV,VU)
Conditions (CDL):
Equal(MeasureOfAngle(TWV),3*x-4)
Equal(MeasureOfAngle(UTW),x)
Equal(MeasureOfAngle(VUT),3*x-4)
Equal(MeasureOfAngle(WVU),x)
Goal: Value(MeasureOfAngle(VUT))
Answer: 134
Theorem sequence:
parallelogram_judgment_angle_and_angle(1,UTWV)
parallel_property_ipsilateral_internal_angle(1,UV,TW)
Problem: As shown in the diagram, LJ=KL, ∠JKL=70°. Find the measure of ∠KLJ.
问题：如图所示，LJ=KL，∠JKL=70°。求∠KLJ的大小。
Construction (CDL):
Shape(JK,KL,LJ)
Conditions (CDL):
Equal(LengthOfLine(LJ),LengthOfLine(KL))
Equal(MeasureOfAngle(JKL),70)
Goal: Value(MeasureOfAngle(KLJ))
Answer: 40
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,LJK)
isosceles_triangle_property_angle_equal(1,LJK)
triangle_property_angle_sum(1,JKL)
Problem: As shown in the diagram, BA=BC, BC=1, BC=AC, ED is perpendicular to CD, quadrilateral EDCB is a square. Find the ratio of the area of triangle ABC to the area of quadrilateral EDCB.
问题：如图所示，BA=BC，BC=1，BC=AC，ED垂直于CD，EDCB是正方形。求三角形ABC的面积与EDCB的面积之比。
Construction (CDL):
Shape(ED,DC,CB,BE)
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(BA),LengthOfLine(BC))
Equal(LengthOfLine(BC),1)
Equal(LengthOfLine(BC),LengthOfLine(AC))
PerpendicularBetweenLine(ED,CD)
Square(EDCB)
Goal: Value(Div(AreaOfTriangle(ABC),AreaOfQuadrilateral(EDCB)))
Answer: sqrt(3)/4
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_judgment_line_equal(1,BCA)
equilateral_triangle_judgment_isosceles_and_isosceles(1,ABC)
equilateral_triangle_property_angle(1,ABC)
triangle_area_formula_sine(1,ABC)
parallelogram_area_formula_sine(1,EDCB)
Problem: As shown in the diagram, the diameter of circle A is 10, the diameter of circle B is 30, the diameter of ⊙C is 10, AD=CH, CH=2, the center of ⊙A is A, the center of ⊙B is B, C is the center of ⊙C. Find the length of line BX.
问题：如图所示，⊙A的直径为10，⊙B的直径为30，圆C的直径为10，AD=CH，CH=2，圆A的圆心为A，圆B的圆心为B，C是⊙C的圆心。求直线BX的长度。
Construction (CDL):
Shape(ED,BJD,AJE)
Shape(DE,AEK,BDK)
Shape(DX,AXJ,BJD)
Shape(XD,BDK,AKX)
Shape(XY,CMY,BMJ,AXJ)
Shape(YX,AKX,BKN,CYN)
Shape(CMY,YH,BHM)
Shape(HY,CYN,BNH)
Shape(HF,CFM,BHM)
Shape(FH,BNH,CNF)
Collinear(EADXBYHCF)
Cocircular(A,EKXJ)
Cocircular(B,JDKNHM)
Cocircular(C,MYNF)
Conditions (CDL):
Equal(DiameterOfCircle(A),10)
Equal(DiameterOfCircle(B),30)
Equal(DiameterOfCircle(C),10)
Equal(LengthOfLine(AD),LengthOfLine(CH))
Equal(LengthOfLine(CH),2)
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
Goal: Value(LengthOfLine(BX))
Answer: 12
Theorem sequence:
line_addition(1,AD,DX)
line_addition(1,DX,XB)
radius_of_circle_property_length_equal(1,AX,A)
radius_of_circle_property_length_equal(1,BD,B)
circle_property_length_of_radius_and_diameter(1,A)
circle_property_length_of_radius_and_diameter(1,B)
Problem: As shown in the diagram, AC=x+7, CB=4, CE=12-x, DC=6, AB⊥CB, ED⊥CD, △ABC is similar to △EDC.. Find the length of line CE.
问题：如图所示，AC=x+7，CB=4，CE=12-x，DC=6，AB垂直于CB，ED⊥CD，三角形ABC与三角形EDC是相似三角形。求直线CE的长度。
Construction (CDL):
Shape(AB,BC,CA)
Shape(ED,DC,CE)
Collinear(DCB)
Collinear(ACE)
Conditions (CDL):
Equal(LengthOfLine(AC),x+7)
Equal(LengthOfLine(CB),4)
Equal(LengthOfLine(CE),12-x)
Equal(LengthOfLine(DC),6)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(ED,CD)
SimilarBetweenTriangle(ABC,EDC)
Goal: Value(LengthOfLine(CE))
Answer: 57/5
Theorem sequence:
similar_triangle_property_line_ratio(1,BCA,DCE)
similar_triangle_property_line_ratio(1,ABC,EDC)
Problem: As shown in the diagram, AB=26, AC=24, BC=10, BC is perpendicular to ，C. Find the value of tan(CA).
问题：如图所示，AB=26，AC=24，BC=10，BC⊥，C。求tan(CA)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AC),24)
Equal(LengthOfLine(BC),10)
PerpendicularBetweenLine(BC，AC)
Goal: Value(Tan(MeasureOfAngle(CAB)))
Answer: 5/12
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, PQ=21, PT=8, TR=15, PQ∥TR. Find the length of line TS.
问题：如图所示，PQ=21，PT=8，TR=15，PQ平行于TR。求直线TS的长度。
Construction (CDL):
Shape(QP,PT,TR,RQ)
Shape(RT,TS,SR)
Collinear(QRS)
Collinear(PTS)
Conditions (CDL):
Equal(LengthOfLine(PQ),21)
Equal(LengthOfLine(PT),8)
Equal(LengthOfLine(TR),15)
ParallelBetweenLine(PQ,TR)
Goal: Value(LengthOfLine(TS))
Answer: 20
Theorem sequence:
line_addition(1,PT,TS)
parallel_property_corresponding_angle(1,RT,QP,S)
similar_triangle_judgment_aa(1,TSR,PSQ)
similar_triangle_property_line_ratio(1,SRT,SQP)
similar_triangle_property_line_ratio(1,RTS,QPS)
Problem: As shown in the diagram, AB=15, AE=18, BD=9, DF=16, RE=r, RF=2, BA is the tangent to circle O. Find the value of r.
问题：如图所示，AB=15，AE=18，BD=9，DF=16，RE=r，RF=2，圆O的切线为BA。求r的值。
Construction (CDL):
Shape(DB,BA,CDA)
Shape(DF,CFD)
Shape(CEF,FD,CDA,AE)
Shape(CAE,EA)
Shape(CEF,ER,RF)
Collinear(BDFR)
Collinear(AER)
Cocircular(C,DAEF)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AE),18)
Equal(LengthOfLine(BD),9)
Equal(LengthOfLine(DF),16)
Equal(LengthOfLine(RE),r)
Equal(LengthOfLine(RF),2)
IsTangentOfCircle(BA,O)
Goal: Value(r)
Answer: -9+3*sqrt(13)
Theorem sequence:
circle_property_circular_power_segment_and_segment_line(1,REA,RFD,C)
line_addition(1,RF,FD)
line_addition(1,RE,EA)
Problem: As shown in the diagram, BN=9, MO=16, MNOB is a kite. Find the area of quadrilateral MNOB.
问题：如图所示，BN=9，MO=16，MNOB是风筝形。求MNOB的面积。
Construction (CDL):
Shape(MN,NA,AM)
Shape(AN,NO,OA)
Shape(AO,OB,BA)
Shape(AB,BM,MA)
Collinear(MAO)
Collinear(NAB)
Conditions (CDL):
Equal(LengthOfLine(BN),9)
Equal(LengthOfLine(MO),16)
Kite(MNOB)
Goal: Value(AreaOfQuadrilateral(MNOB))
Answer: 72
Theorem sequence:
kite_area_formula_diagonal(1,MNOB)
Problem: As shown in the diagram, ∠AGC=40°, ∠DGF=53°, CB⊥GB, FG⊥CG, GF⊥DF. Find the measure of ∠GCB.
问题：如图所示，∠AGC=40°，∠DGF=53°，CB垂直于GB，FG垂直于CG，GF垂直于DF。求∠GCB的大小。
Construction (CDL):
Shape(DG,GF,FD)
Shape(GA,AF,FG)
Shape(GC,CB,BG)
Shape(CA,AB,BC)
Collinear(DFA)
Collinear(GBA)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),40)
Equal(MeasureOfAngle(DGF),53)
PerpendicularBetweenLine(CB,GB)
PerpendicularBetweenLine(FG,CG)
PerpendicularBetweenLine(GF,DF)
Goal: Value(MeasureOfAngle(GCB))
Answer: 50
Theorem sequence:
triangle_property_angle_sum(1,GCB)
Problem: As shown in the diagram, AC=2/3*x-4, AD=1/3*x+2, AD=AC, DE=5*y, EB=7/3*y+8, ED=EB. Find the value of y.
问题：如图所示，AC=2/3*x-4，AD=1/3*x+2，AD=AC，DE=5*y，EB=7/3*y+8，ED=EB。求y的值。
Construction (CDL):
Shape(DA,AE,ED)
Shape(EA,AC,CB,BE)
Collinear(DAC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(AC),2/3*x-4)
Equal(LengthOfLine(AD),1/3*x+2)
Equal(LengthOfLine(AD),LengthOfLine(AC))
Equal(LengthOfLine(DE),5*y)
Equal(LengthOfLine(EB),7/3*y+8)
Equal(LengthOfLine(ED),LengthOfLine(EB))
Goal: Value(y)
Answer: 3
Theorem sequence:
Problem: As shown in the diagram, AB=y, AD=x, BC=z, BD=8, ∠BAC=45°, ∠BDA=60°, AC⊥BC, DA⊥BA. Find the value of z.
问题：如图所示，AB=y，AD=x，BC=z，BD=8，∠BAC=45°，∠BDA=60°，AC垂直于BC，DA⊥BA。求z的值。
Construction (CDL):
Shape(DA,AB,BD)
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BD),8)
Equal(MeasureOfAngle(BAC),45)
Equal(MeasureOfAngle(BDA),60)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(DA,BA)
Goal: Value(z)
Answer: 2*sqrt(6)
Theorem sequence:
sine_theorem(1,BDA)
sine_theorem(1,BAC)
Problem: As shown in the diagram, ∠ABC=x°, ∠BCD=2*x+18°, ∠CDA=x°, ∠DAB=4*x-26°. Find the measure of ∠ABC.
问题：如图所示，∠ABC=x°，∠BCD=2*x+18°，∠CDA=x°，∠DAB=4*x-26°。求∠ABC的大小。
Construction (CDL):
Shape(BC,CD,DA,AB)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),x)
Equal(MeasureOfAngle(BCD),2*x+18)
Equal(MeasureOfAngle(CDA),x)
Equal(MeasureOfAngle(DAB),4*x-26)
Goal: Value(MeasureOfAngle(ABC))
Answer: 46
Theorem sequence:
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, △QPS is congruent to △RQS, QP=2*x, QR=3*y+8, SP=x, SR=2*y, QS⊥RS. Find the value of x.
问题：如图所示，三角形QPS与三角形RQS是全等三角形，QP=2*x，QR=3*y+8，SP=x，SR=2*y，QS⊥RS。求x的值。
Construction (CDL):
Shape(QP,PS,SQ)
Shape(QS,SR,RQ)
Collinear(PSR)
Conditions (CDL):
CongruentBetweenTriangle(QPS,RQS)
Equal(LengthOfLine(QP),2*x)
Equal(LengthOfLine(QR),3*y+8)
Equal(LengthOfLine(SP),x)
Equal(LengthOfLine(SR),2*y)
PerpendicularBetweenLine(QS,RS)
Goal: Value(x)
Answer: 16
Theorem sequence:
congruent_triangle_property_line_equal(1,QPS,RQS)
congruent_triangle_property_line_equal(1,PSQ,QSR)
congruent_triangle_property_line_equal(1,SQP,SRQ)
Problem: As shown in the diagram, NB=17, ND=13, ∠NAD=40°, ND is perpendicular to BD. Find the area of triangle ABN.
问题：如图所示，NB=17，ND=13，∠NAD=40°，ND垂直于BD。求三角形ABN的面积。
Construction (CDL):
Shape(AD,DN,NA)
Shape(DB,BN,ND)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(NB),17)
Equal(LengthOfLine(ND),13)
Equal(MeasureOfAngle(NAD),40)
PerpendicularBetweenLine(ND,BD)
Goal: Value(AreaOfTriangle(ABN))
Answer: 13*sqrt(30)+169/(2*tan(2*pi/9))
Theorem sequence:
right_triangle_judgment_angle(1,NDB)
right_triangle_property_pythagorean(1,NDB)
adjacent_complementary_angle(1,ADN,NDB)
triangle_property_angle_sum(1,NAD)
sine_theorem(1,NAD)
sine_theorem(1,DNA)
line_addition(1,AD,DB)
triangle_area_formula_sine(1,ABN)
Problem: As shown in the diagram, GH=9, JR=5, KL=4*x+1, SJ=5, the center of circle J is J, HR is perpendicular to JR, JS⊥LS. Find the value of x.
问题：如图所示，GH=9，JR=5，KL=4*x+1，SJ=5，圆J的圆心为J，HR垂直于JR，JS垂直于LS。求x的值。
Construction (CDL):
Shape(JH,HR,RJ)
Shape(JR,RG,GJ)
Shape(JG,JGK,KJ)
Shape(JK,KS,SJ)
Shape(JS,SL,LJ)
Shape(JL,JLH,HJ)
Shape(JHG,GH)
Shape(JKL,LK)
Collinear(HRG)
Collinear(KSL)
Cocircular(J,HGKL)
Conditions (CDL):
Equal(LengthOfLine(GH),9)
Equal(LengthOfLine(JR),5)
Equal(LengthOfLine(KL),4*x+1)
Equal(LengthOfLine(SJ),5)
IsCentreOfCircle(J,J)
PerpendicularBetweenLine(HR,JR)
PerpendicularBetweenLine(JS,LS)
Goal: Value(x)
Answer: 2
Theorem sequence:
adjacent_complementary_angle(1,HRJ,JRG)
adjacent_complementary_angle(1,KSJ,JSL)
radius_of_circle_property_length_equal(1,JH,J)
radius_of_circle_property_length_equal(1,JG,J)
radius_of_circle_property_length_equal(1,JK,J)
radius_of_circle_property_length_equal(1,JL,J)
right_triangle_judgment_angle(1,HRJ)
right_triangle_judgment_angle(1,JRG)
right_triangle_judgment_angle(1,KSJ)
right_triangle_judgment_angle(1,JSL)
right_triangle_property_pythagorean(1,HRJ)
right_triangle_property_pythagorean(1,KSJ)
right_triangle_property_pythagorean(1,JRG)
right_triangle_property_pythagorean(1,JSL)
line_addition(1,HR,RG)
line_addition(1,KS,SL)
Problem: As shown in the diagram, BC=4*x-17, DF=2*x-1, ∠CBF=3*y+5°, ∠DFB=5*y-13°, BC and DF are opposite sides of the ▱ BDFC. Find the value of y.
问题：如图所示，BC=4*x-17，DF=2*x-1，∠CBF=3*y+5°，∠DFB=5*y-13°，四边形BDFC是平行四边形。求y的值。
Construction (CDL):
Shape(BD,DF,FB)
Shape(BF,FC,CB)
Conditions (CDL):
Equal(LengthOfLine(BC),4*x-17)
Equal(LengthOfLine(DF),2*x-1)
Equal(MeasureOfAngle(CBF),3*y+5)
Equal(MeasureOfAngle(DFB),5*y-13)
Parallelogram(BDFC)
Goal: Value(y)
Answer: 9
Theorem sequence:
parallel_property_alternate_interior_angle(1,BC,DF)
Problem: As shown in the diagram, AB=x, AC=x-5, BC=54, BD=28, ∠DAC=∠BAD. Find the value of x.
问题：如图所示，AB=x，AC=x-5，BC=54，BD=28，∠DAC=∠BAD。求x的值。
Construction (CDL):
Shape(CD,DA,AC)
Shape(DB,BA,AD)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),x-5)
Equal(LengthOfLine(BC),54)
Equal(LengthOfLine(BD),28)
Equal(MeasureOfAngle(DAC),MeasureOfAngle(BAD))
Goal: Value(x)
Answer: 70
Theorem sequence:
line_addition(1,CD,DB)
bisector_of_angle_judgment_angle_equal(1,AD,BAC)
bisector_of_angle_property_line_ratio(1,AD,BAC)
Problem: As shown in the diagram, ∠AFG=2*x°, ∠BID=x+5°, ∠CGB=2*x+1°, ∠HDF=3*x+2°. Find the value of x.
问题：如图所示，∠AFG=2*x°，∠BID=x+5°，∠CGB=2*x+1°，∠HDF=3*x+2°。求x的值。
Construction (CDL):
Shape(HD,DF)
Shape(BI,ID)
Shape(CG,GI)
Shape(AF,FG)
Shape(DI,IG,GF,FD)
Collinear(HDI)
Collinear(BIG)
Collinear(CGF)
Collinear(DFA)
Conditions (CDL):
Equal(MeasureOfAngle(AFG),2*x)
Equal(MeasureOfAngle(BID),x+5)
Equal(MeasureOfAngle(CGB),2*x+1)
Equal(MeasureOfAngle(HDF),3*x+2)
Goal: Value(x)
Answer: 44
Theorem sequence:
adjacent_complementary_angle(1,HDF,FDI)
adjacent_complementary_angle(1,BID,DIG)
adjacent_complementary_angle(1,CGI,IGF)
adjacent_complementary_angle(1,AFG,GFD)
quadrilateral_property_angle_sum(1,DIGF)