SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AC=4*y-8, AD=x+4, CB=2*x-4, DB=3*y-2, BD and CA are opposite sides of the ▱ BCAD. Find the value of x.
问题：如图所示，AC=4*y-8，AD=x+4，CB=2*x-4，DB=3*y-2，BCAD是平行四边形。求x的值。
Construction (CDL):
Shape(BC,CA,AD,DB)
Conditions (CDL):
Equal(LengthOfLine(AC),4*y-8)
Equal(LengthOfLine(AD),x+4)
Equal(LengthOfLine(CB),2*x-4)
Equal(LengthOfLine(DB),3*y-2)
Parallelogram(BCAD)
Goal: Value(x)
Answer: 8
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BCAD)
Problem: As shown in the diagram, ∠CFD=7*x-38°, ∠EHI=178-3*x°, BH∥DF. Find the value of x.
问题：如图所示，∠CFD=7*x-38°，∠EHI=178-3*x°，BH∥DF。求x的值。
Construction (CDL):
Shape(EH,HI)
Shape(BH,HE)
Shape(FH,HB)
Shape(IH,HF)
Shape(DF,FH)
Shape(HF,FG)
Shape(CF,FD)
Shape(GF,FC)
Collinear(BHI)
Collinear(DFG)
Collinear(EHFC)
Conditions (CDL):
Equal(MeasureOfAngle(CFD),7*x-38)
Equal(MeasureOfAngle(EHI),178-3*x)
ParallelBetweenLine(BH,DF)
Goal: Value(x)
Answer: 108/5
Theorem sequence:
parallel_property_corresponding_angle(1,FD,HB,C)
vertical_angle(1,EHI,FHB)
Problem: As shown in the diagram, ∠AEF=148°, ∠FAE=18°, ∠XYZ=3*x+y°, ∠XYZ=∠EFA, ∠YZX=5*x-y°, ∠ZXY=∠AEF. Find the value of x.
问题：如图所示，∠AEF=148°，∠FAE=18°，∠XYZ=3*x+y°，∠XYZ=∠EFA，∠YZX=5*x-y°，∠ZXY=∠AEF。求x的值。
Construction (CDL):
Shape(EF,FA,AE)
Shape(ZX,XY,YZ)
Conditions (CDL):
Equal(MeasureOfAngle(AEF),148)
Equal(MeasureOfAngle(FAE),18)
Equal(MeasureOfAngle(XYZ),3*x+y)
Equal(MeasureOfAngle(XYZ),MeasureOfAngle(EFA))
Equal(MeasureOfAngle(YZX),5*x-y)
Equal(MeasureOfAngle(ZXY),MeasureOfAngle(AEF))
Goal: Value(x)
Answer: 4
Theorem sequence:
triangle_property_angle_sum(1,ZXY)
Problem: As shown in the diagram, RS=3*x-5, RS=RT, TR=2*x+7, TS=22. Find the length of line RT.
问题：如图所示，RS=3*x-5，RS=RT，TR=2*x+7，TS=22。求直线RT的长度。
Construction (CDL):
Shape(RS,ST,TR)
Conditions (CDL):
Equal(LengthOfLine(RS),3*x-5)
Equal(LengthOfLine(RS),LengthOfLine(RT))
Equal(LengthOfLine(TR),2*x+7)
Equal(LengthOfLine(TS),22)
Goal: Value(LengthOfLine(RT))
Answer: 31
Theorem sequence:
Problem: As shown in the diagram, ∠AZB=104°, the measure of arc ZCB is 94, Z is the center of circle Z, the incenter of quadrilateral ADCB is Z, AB is parallel to DC. Find the measure of ⌒ZBA.
问题：如图所示，∠AZB=104°，⌒ZCB的角度为94，Z是⊙Z的圆心，Z是四边形ADCB的内心，AB∥DC。求弧ZBA的角度。
Construction (CDL):
Shape(ZCB,BC)
Shape(ZBA,AB)
Shape(ZAD,DA)
Shape(ZDC,CD)
Shape(EC,CB,BE)
Shape(DC,CE,ED)
Shape(ZB,BA,AZ)
Shape(EA,AD,DE)
Shape(EB,BZ,ZA,AE)
Collinear(BED)
Collinear(AEC)
Cocircular(Z,CBAD)
Conditions (CDL):
Equal(MeasureOfAngle(AZB),104)
Equal(MeasureOfArc(ZCB),94)
IsCentreOfCircle(Z,Z)
IsIncenterOfQuadrilateral(Z,ADCB)
ParallelBetweenLine(AB,DC)
Goal: Value(MeasureOfArc(ZBA))
Answer: 104
Theorem sequence:
arc_property_center_angle(1,ZBA,Z)
Problem: As shown in the diagram, MN=3*x-4, NQ=15.4, PN=2*y+5, PQ=11.1, RM=17.9, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, quadrilateral MRPN is a parallelogram. Find the measure of ∠MQN.
问题：如图所示，MN=3*x-4，NQ=15.4，PN=2*y+5，PQ=11.1，RM=17.9，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，RM和PN是平行四边形MRPN的一组对边。求∠MQN的大小。
Construction (CDL):
Shape(MR,RQ,QM)
Shape(QR,RP,PQ)
Shape(MQ,QN,NM)
Shape(NQ,QP,PN)
Collinear(MQP)
Collinear(RQN)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(NQ),15.4)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),11.1)
Equal(LengthOfLine(RM),17.9)
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(MeasureOfAngle(MQN))
Answer: 97
Theorem sequence:
flat_angle(1,MQP)
angle_addition(1,MQN,NQP)
Problem: As shown in the diagram, ∠DHK=120°, ∠IKH=54°, ∠JKI=36°, JI is parallel to KH. Find the measure of ∠KIJ.
问题：如图所示，∠DHK=120°，∠IKH=54°，∠JKI=36°，JI平行于KH。求∠KIJ的大小。
Construction (CDL):
Shape(KH,HI,IK)
Shape(KI,IJ,JK)
Shape(DH,HK)
Collinear(DHI)
Conditions (CDL):
Equal(MeasureOfAngle(DHK),120)
Equal(MeasureOfAngle(IKH),54)
Equal(MeasureOfAngle(JKI),36)
ParallelBetweenLine(JI,KH)
Goal: Value(MeasureOfAngle(KIJ))
Answer: 54
Theorem sequence:
parallel_property_alternate_interior_angle(2,JI,KH)
Problem: As shown in the diagram, RQ=21, SQ=33, SR=53, ∠QRS=x°. Find the value of x.
问题：如图所示，RQ=21，SQ=33，SR=53，∠QRS=x°。求x的值。
Construction (CDL):
Shape(QR,RS,SQ)
Conditions (CDL):
Equal(LengthOfLine(RQ),21)
Equal(LengthOfLine(SQ),33)
Equal(LengthOfLine(SR),53)
Equal(MeasureOfAngle(QRS),x)
Goal: Value(x)
Answer: 180*acos(2161/2226)/pi
Theorem sequence:
cosine_theorem(1,RSQ)
Problem: As shown in the diagram, ∠FHM=94°, DH∥EC. Find the measure of ∠ICE.
问题：如图所示，∠FHM=94°，DH平行于EC。求∠ICE的大小。
Construction (CDL):
Shape(FH,HM)
Shape(MH,HC)
Shape(HC,CA)
Shape(AC,CI)
Shape(IC,CE)
Shape(EC,CH)
Shape(CH,HD)
Shape(DH,HF)
Collinear(FHCI)
Collinear(MHD)
Collinear(ACE)
Conditions (CDL):
Equal(MeasureOfAngle(FHM),94)
ParallelBetweenLine(DH,EC)
Goal: Value(MeasureOfAngle(ICE))
Answer: 94
Theorem sequence:
vertical_angle(1,FHM,CHD)
parallel_property_corresponding_angle(1,CE,HD,I)
Problem: As shown in the diagram, BD=4*x+9, CD=7*x-6, ∠CDA=2*y-6°, AD is the median of triangle ACB. Find the value of x.
问题：如图所示，BD=4*x+9，CD=7*x-6，∠CDA=2*y-6°，AD是三角形ACB的中线。求x的值。
Construction (CDL):
Shape(BA,AD,DB)
Shape(DA,AC,CD)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(BD),4*x+9)
Equal(LengthOfLine(CD),7*x-6)
Equal(MeasureOfAngle(CDA),2*y-6)
IsMedianOfTriangle(AD,ACB)
Goal: Value(x)
Answer: 5
Theorem sequence:
Problem: As shown in the diagram, AB=x, CA=z, CB=12, PA=y, PB=8, AP⊥CP, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AB=x，CA=z，CB=12，PA=y，PB=8，AP垂直于CP，CA垂直于BA。求x的值。
Construction (CDL):
Shape(CA,AP,PC)
Shape(PA,AB,BP)
Collinear(CPB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(CA),z)
Equal(LengthOfLine(CB),12)
Equal(LengthOfLine(PA),y)
Equal(LengthOfLine(PB),8)
PerpendicularBetweenLine(AP,CP)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 4*sqrt(6)
Theorem sequence:
adjacent_complementary_angle(1,BPA,APC)
right_triangle_judgment_angle(1,APC)
right_triangle_judgment_angle(1,CAB)
right_triangle_judgment_angle(1,BPA)
line_addition(1,CP,PB)
right_triangle_property_pythagorean(1,APC)
right_triangle_property_pythagorean(1,CAB)
right_triangle_property_pythagorean(1,BPA)
Problem: As shown in the diagram, ∠EYQ=3*y+1°, ∠MAQ=3*x+11°, ∠YQF=4*x-5°, EF is parallel to MA, EF is parallel to YQ, YM∥QA, YQ∥MA. Find the value of x.
问题：如图所示，∠EYQ=3*y+1°，∠MAQ=3*x+11°，∠YQF=4*x-5°，EF平行于MA，EF∥YQ，YM平行于QA，YQ平行于MA。求x的值。
Construction (CDL):
Shape(EY,YQ,QF,FE)
Shape(YM,MA,AQ,QY)
Collinear(EYM)
Collinear(FQA)
Conditions (CDL):
Equal(MeasureOfAngle(EYQ),3*y+1)
Equal(MeasureOfAngle(MAQ),3*x+11)
Equal(MeasureOfAngle(YQF),4*x-5)
ParallelBetweenLine(EF,MA)
ParallelBetweenLine(EF,YQ)
ParallelBetweenLine(YM,QA)
ParallelBetweenLine(YQ,MA)
Goal: Value(x)
Answer: 16
Theorem sequence:
parallel_property_corresponding_angle(2,AM,QY,F)
Problem: As shown in the diagram, QP=17, RP=8, RQ=15, QR is perpendicular to PR. Find the value of sin(PQ).
问题：如图所示，QP=17，RP=8，RQ=15，QR垂直于PR。求sin(PQ)的值。
Construction (CDL):
Shape(QR,RP,PQ)
Conditions (CDL):
Equal(LengthOfLine(QP),17)
Equal(LengthOfLine(RP),8)
Equal(LengthOfLine(RQ),15)
PerpendicularBetweenLine(QR,PR)
Goal: Value(Sin(MeasureOfAngle(PQR)))
Answer: 8/17
Theorem sequence:
sine_theorem(1,PQR)
Problem: As shown in the diagram, ∠AFB=70°, ∠ECF=35°, AD⊥ED, EC is perpendicular to BC. Find the measure of ∠FCB.
问题：如图所示，∠AFB=70°，∠ECF=35°，AD⊥ED，EC垂直于BC。求∠FCB的大小。
Construction (CDL):
Shape(AD,DE,EF,FA)
Shape(FE,EC,CF)
Shape(BF,FC,CB)
Shape(AF,FB)
Collinear(DEC)
Collinear(AFC)
Collinear(EFB)
Conditions (CDL):
Equal(MeasureOfAngle(AFB),70)
Equal(MeasureOfAngle(ECF),35)
PerpendicularBetweenLine(AD,ED)
PerpendicularBetweenLine(EC,BC)
Goal: Value(MeasureOfAngle(FCB))
Answer: 55
Theorem sequence:
angle_addition(1,ECF,FCB)
Problem: As shown in the diagram, AC=y, AD=9, BC=z, CD=x, DB=16, AD is perpendicular to CD, BC is perpendicular to AC. Find the value of y.
问题：如图所示，AC=y，AD=9，BC=z，CD=x，DB=16，AD⊥CD，BC⊥AC。求y的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AD),9)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(CD),x)
Equal(LengthOfLine(DB),16)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 15
Theorem sequence:
adjacent_complementary_angle(1,ADC,CDB)
right_triangle_judgment_angle(1,ADC)
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,BCA)
line_addition(1,AD,DB)
right_triangle_property_pythagorean(1,ADC)
right_triangle_property_pythagorean(1,CDB)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠CAB=50°, ∠DAC=x°. Find the value of x.
问题：如图所示，∠CAB=50°，∠DAC=x°。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Shape(DA,AC)
Collinear(DAB)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),50)
Equal(MeasureOfAngle(DAC),x)
Goal: Value(x)
Answer: 130
Theorem sequence:
adjacent_complementary_angle(1,DAC,CAB)
Problem: As shown in the diagram, AC=15, ∠AFB=y**2+26°, ∠BAF=2*x+20°, ∠FAD=5*x-4°, ADCB is a rhombus. Find the length of line AF.
问题：如图所示，AC=15，∠AFB=y**2+26°，∠BAF=2*x+20°，∠FAD=5*x-4°，ADCB是菱形。求直线AF的长度。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(BA,AF,FB)
Shape(BF,FC,CB)
Collinear(AFC)
Collinear(BFD)
Conditions (CDL):
Equal(LengthOfLine(AC),15)
Equal(MeasureOfAngle(AFB),y**2+26)
Equal(MeasureOfAngle(BAF),2*x+20)
Equal(MeasureOfAngle(FAD),5*x-4)
Rhombus(ADCB)
Goal: Value(LengthOfLine(AF))
Answer: 15/2
Theorem sequence:
line_addition(1,AF,FC)
parallelogram_property_diagonal_bisection(1,ADCB,F)
Problem: As shown in the diagram, AB=x, AC=14, BC=48, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=x，AC=14，BC=48，BC垂直于AC。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(BC),48)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 50
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠MRP=5*x+7°, ∠NPR=7*x-21°, LR∥NP. Find the measure of ∠MRP.
问题：如图所示，∠MRP=5*x+7°，∠NPR=7*x-21°，LR∥NP。求∠MRP的大小。
Construction (CDL):
Shape(CR,RM)
Shape(MR,RP)
Shape(RP,PJ)
Shape(JP,PQ)
Shape(LR,RC)
Shape(PR,RL)
Shape(NP,PR)
Shape(QP,PN)
Collinear(CRPQ)
Collinear(MRL)
Collinear(JPN)
Conditions (CDL):
Equal(MeasureOfAngle(MRP),5*x+7)
Equal(MeasureOfAngle(NPR),7*x-21)
ParallelBetweenLine(LR,NP)
Goal: Value(MeasureOfAngle(MRP))
Answer: 77
Theorem sequence:
vertical_angle(1,MRP,LRC)
parallel_property_corresponding_angle(2,PN,RL,C)
Problem: As shown in the diagram, AB=15, AE=18.5, BD=q, DF=16, RE=r, RF=2, BA is the tangent to ⊙C. Find the value of q.
问题：如图所示，AB=15，AE=18.5，BD=q，DF=16，RE=r，RF=2，圆O的切线为BA。求q的值。
Construction (CDL):
Shape(CEF,ER,RF)
Shape(CFD,DF)
Shape(CAE,EA)
Shape(CEF,FD,CDA,AE)
Shape(CDA,DB,DA)
Collinear(RFDB)
Collinear(REA)
Cocircular(C,EFDA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AE),18.5)
Equal(LengthOfLine(BD),q)
Equal(LengthOfLine(DF),16)
Equal(LengthOfLine(RE),r)
Equal(LengthOfLine(RF),2)
IsTangentOfCircle(BA,C)
Goal: Value(q)
Answer: 9
Theorem sequence:
circle_property_circular_power_tangent_and_segment_line(1,BA,BDF,C)
line_addition(1,BD,DF)
Problem: As shown in the diagram, XB=12, XC=13, YE=x, YF=13/2, ∠YFE=∠XCB, CB is perpendicular to XB, FE⊥YE. Find the value of x.
问题：如图所示，XB=12，XC=13，YE=x，YF=13/2，∠YFE=∠XCB，CB⊥XB，FE垂直于YE。求x的值。
Construction (CDL):
Shape(AY,YE,EA)
Shape(EY,YF,FE)
Collinear(AEF)
Shape(DX,XB,BD)
Shape(BX,XC,CB)
Collinear(DBC)
Conditions (CDL):
Equal(LengthOfLine(XB),12)
Equal(LengthOfLine(XC),13)
Equal(LengthOfLine(YE),x)
Equal(LengthOfLine(YF),13/2)
Equal(MeasureOfAngle(YFE),MeasureOfAngle(XCB))
PerpendicularBetweenLine(CB,XB)
PerpendicularBetweenLine(FE,YE)
Goal: Value(x)
Answer: 6
Theorem sequence:
similar_triangle_judgment_aa(1,YFE,XCB)
similar_triangle_property_line_ratio(1,EYF,BXC)
similar_triangle_property_line_ratio(1,FEY,CBX)
Problem: As shown in the diagram, ∠ALJ=133°, ∠EKI=126°, BL is parallel to KJ. Find the measure of ∠KEJ.
问题：如图所示，∠ALJ=133°，∠EKI=126°，BL平行于KJ。求∠KEJ的大小。
Construction (CDL):
Shape(JK,KE,EJ)
Shape(DL,LA)
Shape(BL,LD)
Shape(BL,LJ)
Shape(JL,LA)
Shape(LJ,JF)
Shape(KJ,JL)
Shape(EJ,JK)
Shape(JE,EG)
Shape(GE,EC)
Shape(CE,EK)
Shape(HK,KJ)
Shape(IK,KH)
Shape(EK,KI)
Collinear(DLJEC)
Collinear(ALB)
Collinear(FJKI)
Collinear(HKEG)
Conditions (CDL):
Equal(MeasureOfAngle(ALJ),133)
Equal(MeasureOfAngle(EKI),126)
ParallelBetweenLine(BL,KJ)
Goal: Value(MeasureOfAngle(KEJ))
Answer: 79
Theorem sequence:
parallel_property_collinear_extend(1,JK,LB,F)
parallel_property_collinear_extend(2,BL,JF,A)
parallel_property_corresponding_angle(2,LA,JF,E)
flat_angle(1,FJK)
angle_addition(1,FJE,EJK)
flat_angle(1,JKI)
angle_addition(1,JKE,EKI)
triangle_property_angle_sum(1,JKE)
Problem: As shown in the diagram, AD=9, BD=x, DC=6, DE=3. Find the value of x.
问题：如图所示，AD=9，BD=x，DC=6，DE=3。求x的值。
Construction (CDL):
Shape(FBE,ED,DB)
Shape(FEA,AD,DE)
Shape(FAC,CD,DA)
Shape(FCB,BD,DC)
Collinear(BDA)
Collinear(EDC)
Cocircular(F,CBEA)
Conditions (CDL):
Equal(LengthOfLine(AD),9)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(DC),6)
Equal(LengthOfLine(DE),3)
Goal: Value(x)
Answer: 2
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,CDE,BDA,F)
Problem: As shown in the diagram, KJ=7*x, KJ=LJ, KL=11*x-8, KL=KJ, LJ=x+12. Find the length of line JK.
问题：如图所示，KJ=7*x，KJ=LJ，KL=11*x-8，KL=KJ，LJ=x+12。求直线JK的长度。
Construction (CDL):
Shape(KJ,JL,LK)
Conditions (CDL):
Equal(LengthOfLine(KJ),7*x)
Equal(LengthOfLine(KJ),LengthOfLine(LJ))
Equal(LengthOfLine(KL),11*x-8)
Equal(LengthOfLine(KL),LengthOfLine(KJ))
Equal(LengthOfLine(LJ),x+12)
Goal: Value(LengthOfLine(JK))
Answer: 14
Theorem sequence:
Problem: As shown in the diagram, the length of ⌒ONM is equal to the length of ⌒OPQ, NM=26, PQ=3*x+5. Find the value of x.
问题：如图所示，弧ONM的长度与⌒OPQ的长度相等，NM=26，PQ=3*x+5。求x的值。
Construction (CDL):
Shape(ONM,MN)
Shape(OPQ,QP)
Shape(OQN,NM,OMP,PQ)
Cocircular(O,NMPQ)
Conditions (CDL):
Equal(LengthOfArc(ONM),LengthOfArc(OPQ))
Equal(LengthOfLine(NM),26)
Equal(LengthOfLine(PQ),3*x+5)
Goal: Value(x)
Answer: 7
Theorem sequence:
congruent_arc_judgment_length_equal(1,ONM,OPQ)
congruent_arc_property_chord_equal(1,ONM,OPQ)
Problem: As shown in the diagram, ∠ALC=35°, ∠GCL=28°, ∠LDE=25°, ∠LEG=51°, CA is perpendicular to LA, LC⊥FC, LG⊥CG. Find the measure of ∠LCA.
问题：如图所示，∠ALC=35°，∠GCL=28°，∠LDE=25°，∠LEG=51°，CA⊥LA，LC⊥FC，LG垂直于CG。求∠LCA的大小。
Construction (CDL):
Shape(LD,DE,EL)
Shape(LE,EG,GL)
Shape(LG,GC,CL)
Shape(LC,CA,AL)
Shape(AC,CF,FA)
Collinear(DEGC)
Collinear(LAF)
Conditions (CDL):
Equal(MeasureOfAngle(ALC),35)
Equal(MeasureOfAngle(GCL),28)
Equal(MeasureOfAngle(LDE),25)
Equal(MeasureOfAngle(LEG),51)
PerpendicularBetweenLine(CA,LA)
PerpendicularBetweenLine(LC,FC)
PerpendicularBetweenLine(LG,CG)
Goal: Value(MeasureOfAngle(LCA))
Answer: 55
Theorem sequence:
triangle_property_angle_sum(1,ALC)
Problem: As shown in the diagram, JO=12, KM=7, NL=x+3, OL=4*x-9, R is the center of the inscribed circle of △KJL, the tangent to ⊙R is JM, JO is the tangent to circle R, the tangent to circle R is KM, KN is the tangent to ⊙R, the tangent to ⊙R is LN, the tangent to circle R is LO. Find the perimeter of triangle JLK.
问题：如图所示，JO=12，KM=7，NL=x+3，OL=4*x-9，△KJL的内心为R，圆O的切线为JM，JO是⊙O的切线，KM是⊙O的切线，KN是⊙O的切线，⊙O的切线为LN，LO是⊙O的切线。求三角形JLK的周长。
Construction (CDL):
Shape(RNM,RMO,RON)
Shape(RNM,NK,KM)
Shape(RMO,MJ,JO)
Shape(RON,OL,LN)
Collinear(JMK)
Collinear(JOL)
Collinear(KNL)
Cocircular(R,MON)
Conditions (CDL):
Equal(LengthOfLine(JO),12)
Equal(LengthOfLine(KM),7)
Equal(LengthOfLine(NL),x+3)
Equal(LengthOfLine(OL),4*x-9)
IsIncenterOfTriangle(R,KJL)
IsTangentOfCircle(JM,R)
IsTangentOfCircle(JO,R)
IsTangentOfCircle(KM,R)
IsTangentOfCircle(KN,R)
IsTangentOfCircle(LN,R)
IsTangentOfCircle(LO,R)
Goal: Value(PerimeterOfTriangle(JLK))
Answer: 52
Theorem sequence:
tangent_of_circle_property_length_equal(1,JM,JO,R)
tangent_of_circle_property_length_equal(1,KM,KN,R)
tangent_of_circle_property_length_equal(1,LO,LN,R)
triangle_perimeter_formula(1,JLK)
line_addition(1,JM,MK)
line_addition(1,JO,OL)
line_addition(1,LN,NK)
Problem: As shown in the diagram, BC=BA, BC=y, BD=12, CA=x, ∠BCD=60°, BD⊥AD. Find the value of x.
问题：如图所示，BC=BA，BC=y，BD=12，CA=x，∠BCD=60°，BD⊥AD。求x的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(BC),LengthOfLine(BA))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),12)
Equal(LengthOfLine(CA),x)
Equal(MeasureOfAngle(BCD),60)
PerpendicularBetweenLine(BD,AD)
Goal: Value(x)
Answer: 8*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,CDB,BDA)
triangle_property_angle_sum(1,CDB)
sine_theorem(1,CDB)
sine_theorem(1,BCD)
perpendicular_bisector_judgment_distance_equal(1,BD,CA)
line_addition(1,CD,DA)
Problem: As shown in the diagram, ∠HPM=4*y°, ∠MPR=68°, ∠PRC=x°, ∠SCR=5*z+2°, CR∥MP, MC∥PR. Find the value of x.
问题：如图所示，∠HPM=4*y°，∠MPR=68°，∠PRC=x°，∠SCR=5*z+2°，CR平行于MP，MC∥PR。求x的值。
Construction (CDL):
Shape(CM,MP,PR,RC)
Shape(SC,CR)
Shape(NC,CS)
Shape(MC,CN)
Shape(CR,RG)
Shape(GR,RI)
Shape(IR,RP)
Shape(EM,MD)
Shape(DM,MC)
Shape(PM,ME)
Shape(HP,PM)
Shape(LP,PH)
Shape(RP,PL)
Collinear(NCRI)
Collinear(DMPL)
Collinear(EMCS)
Collinear(HPRG)
Conditions (CDL):
Equal(MeasureOfAngle(HPM),4*y)
Equal(MeasureOfAngle(MPR),68)
Equal(MeasureOfAngle(PRC),x)
Equal(MeasureOfAngle(SCR),5*z+2)
ParallelBetweenLine(CR,MP)
ParallelBetweenLine(MC,PR)
Goal: Value(x)
Answer: 112
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,PM,RC)
Problem: As shown in the diagram, PQ=10, ∠QPM=81°, the center of circle P is P. Find the area of the sector PQM.
问题：如图所示，PQ=10，∠QPM=81°，P是⊙P的圆心。求扇形PQM的面积。
Construction (CDL):
Shape(PQM,MP,PQ)
Shape(PMQ,QP,PM)
Cocircular(P,QM)
Conditions (CDL):
Equal(LengthOfLine(PQ),10)
Equal(MeasureOfAngle(QPM),81)
IsCentreOfCircle(P,P)
Goal: Value(AreaOfSector(PQM))
Answer: 155*pi/2
Theorem sequence:
radius_of_circle_property_length_equal(1,PQ,P)
round_angle(1,MPQ,QPM)
arc_property_center_angle(1,PQM,P)
sector_area_formula(1,PQM)
Problem: As shown in the diagram, AN=19, BF=10, CE=2, FN=20, AE is perpendicular to FE, EF⊥NF, AC and BN are the non-parallel sides (legs) of trapezoid ACBN. Find the area of quadrilateral ACBN.
问题：如图所示，AN=19，BF=10，CE=2，FN=20，AE垂直于FE，EF垂直于NF，AN和CB是梯形ACBN的平行边。求ACBN的面积。
Construction (CDL):
Shape(AC,CE,EA)
Shape(AE,EF,FN,NA)
Shape(NF,FB,BN)
Collinear(CEFB)
Conditions (CDL):
Equal(LengthOfLine(AN),19)
Equal(LengthOfLine(BF),10)
Equal(LengthOfLine(CE),2)
Equal(LengthOfLine(FN),20)
PerpendicularBetweenLine(AE,FE)
PerpendicularBetweenLine(EF,NF)
Trapezoid(ACBN)
Goal: Value(AreaOfQuadrilateral(ACBN))
Answer: 500
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,EA,FN)
parallel_property_collinear_extend(3,BC,NA,E)
parallel_property_collinear_extend(1,EC,NA,F)
parallelogram_judgment_parallel_and_parallel(1,AEFN)
parallelogram_property_opposite_line_equal(1,EFNA)
line_addition(1,CE,EF)
line_addition(1,CF,FB)
altitude_of_quadrilateral_judgment_right_vertex(2,NF,ACBN)
trapezoid_area_formula(1,ACBN)
Problem: As shown in the diagram, MPON is a rhombus. Find the measure of ∠MRN.
问题：如图所示，四边形MPON是菱形。求∠MRN的大小。
Construction (CDL):
Shape(MP,PR,RM)
Shape(RP,PO,OR)
Shape(RO,ON,NR)
Shape(RN,NM,MR)
Collinear(MRO)
Collinear(PRN)
Conditions (CDL):
Rhombus(MPON)
Goal: Value(MeasureOfAngle(MRN))
Answer: 90
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,MPON,R)
Problem: As shown in the diagram, QT=18, RT=34, QS⊥RS, RT⊥ST. Find the height of triangle SRQ.
问题：如图所示，QT=18，RT=34，QS垂直于RS，RT垂直于ST。求△SRQ的高。
Construction (CDL):
Shape(SR,RT,TS)
Shape(ST,TQ,QS)
Collinear(RTQ)
Conditions (CDL):
Equal(LengthOfLine(QT),18)
Equal(LengthOfLine(RT),34)
PerpendicularBetweenLine(QS,RS)
PerpendicularBetweenLine(RT,ST)
Goal: Value(HeightOfTriangle(SRQ))
Answer: 6*sqrt(17)
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,SRT,QSR)
altitude_of_triangle_judgment(1,ST,SRQ)
line_addition(1,QT,TR)
right_triangle_judgment_angle(1,RTS)
right_triangle_property_pythagorean(1,RTS)
mirror_similar_triangle_property_line_ratio(1,SRT,QSR)
mirror_similar_triangle_property_line_ratio(1,TSR,SRQ)
Problem: As shown in the diagram, VW=3, XV=2*a-2, ZV=Div(5*a+1,4), ∠ZYV=53°, XWZY is a rhombus. Find the measure of ∠VZY.
问题：如图所示，VW=3，XV=2*a-2，ZV=Div(5*a+1,4)，∠ZYV=53°，XWZY是菱形。求∠VZY的大小。
Construction (CDL):
Shape(XV,VY,YX)
Shape(XW,WV,VX)
Shape(YV,VZ,ZY)
Shape(VW,WZ,ZV)
Collinear(XVZ)
Collinear(YVW)
Conditions (CDL):
Equal(LengthOfLine(VW),3)
Equal(LengthOfLine(XV),2*a-2)
Equal(LengthOfLine(ZV),Div(5*a+1,4))
Equal(MeasureOfAngle(ZYV),53)
Rhombus(XWZY)
Goal: Value(MeasureOfAngle(VZY))
Answer: 37
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,ZYXW,V)
triangle_property_angle_sum(1,YVZ)
Problem: As shown in the diagram, GJ=40, the radius of circle L is 32, L is the center of circle L, JK⊥LK. Find the length of line LK.
问题：如图所示，GJ=40，圆L的半径为32，圆L的圆心为L，JK垂直于LK。求直线LK的长度。
Construction (CDL):
Shape(LHG,GK,KH)
Shape(LJH,HK,KJ)
Shape(KL,LJ,JK)
Shape(LGJ,JL,LK,KG)
Collinear(GKJ)
Collinear(LKH)
Cocircular(L,JHG)
Conditions (CDL):
Equal(LengthOfLine(GJ),40)
Equal(RadiusOfCircle(L),32)
IsCentreOfCircle(L,L)
PerpendicularBetweenLine(JK,LK)
Goal: Value(LengthOfLine(LK))
Answer: 4*sqrt(39)
Theorem sequence:
line_addition(1,GK,KJ)
circle_property_chord_perpendicular_bisect_chord(1,L,LK,JG)
radius_of_circle_property_length_equal(1,LJ,L)
right_triangle_judgment_angle(1,JKL)
right_triangle_property_pythagorean(1,JKL)
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 x.
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x的值。
Construction (CDL):
Shape(DA,AE,ED)
Shape(AC,CB,BE,EA)
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(x)
Answer: 18
Theorem sequence:
Problem: As shown in the diagram, ∠AEJ=3*y+44°, ∠CAE=94°, ∠EJC=2*x°, ∠JCA=96°, AC∥EJ. Find the value of y.
问题：如图所示，∠AEJ=3*y+44°，∠CAE=94°，∠EJC=2*x°，∠JCA=96°，AC平行于EJ。求y的值。
Construction (CDL):
Shape(CA,AE,EJ,JC)
Conditions (CDL):
Equal(MeasureOfAngle(AEJ),3*y+44)
Equal(MeasureOfAngle(CAE),94)
Equal(MeasureOfAngle(EJC),2*x)
Equal(MeasureOfAngle(JCA),96)
ParallelBetweenLine(AC,EJ)
Goal: Value(y)
Answer: 14
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AC,EJ)
Problem: As shown in the diagram, BD=2*x, CD=3*x-7, ED=14, BD⊥ED, CE is perpendicular to BE. Find the value of x.
问题：如图所示，BD=2*x，CD=3*x-7，ED=14，BD垂直于ED，CE垂直于BE。求x的值。
Construction (CDL):
Shape(CE,ED,DC)
Shape(DE,EB,BD)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(BD),2*x)
Equal(LengthOfLine(CD),3*x-7)
Equal(LengthOfLine(ED),14)
PerpendicularBetweenLine(BD,ED)
PerpendicularBetweenLine(CE,BE)
Goal: Value(x)
Answer: 7
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,EBD,CEB)
line_addition(1,CD,DB)
right_triangle_judgment_angle(1,BDE)
right_triangle_property_pythagorean(1,BDE)
mirror_similar_triangle_property_line_ratio(1,EBD,CEB)
mirror_similar_triangle_property_line_ratio(1,DEB,EBC)
Problem: As shown in the diagram, AC=5, BA=15, BC=z, YA=y, BC is perpendicular to AC, CY is perpendicular to BY. Find the value of y.
问题：如图所示，AC=5，BA=15，BC=z，YA=y，BC⊥AC，CY垂直于BY。求y的值。
Construction (CDL):
Shape(CA,AY,YC)
Shape(CY,YB,BC)
Collinear(AYB)
Conditions (CDL):
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BA),15)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(YA),y)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(CY,BY)
Goal: Value(y)
Answer: 5/3
Theorem sequence:
adjacent_complementary_angle(1,AYC,CYB)
right_triangle_judgment_angle(1,AYC)
right_triangle_judgment_angle(1,CYB)
right_triangle_judgment_angle(1,BCA)
line_addition(1,AY,YB)
right_triangle_property_pythagorean(1,AYC)
right_triangle_property_pythagorean(1,CYB)
right_triangle_property_pythagorean(1,BCA)