SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠GBD=x+5/2°, the measure of ⌒ADC is 4*x+5, the measure of arc AGD is 50, BD is the tangent to ⊙A. Find the value of x.
问题：如图所示，∠GBD=x+5/2°，⌒ADC的角度为4*x+5，⌒AGD的角度为50，BD是⊙O的切线。求x的值。
Construction (CDL):
Shape(GB,BD,AGD)
Shape(ADC,CG,AGD)
Shape(ACG,GC)
Shape(CE)
Collinear(BGCE)
Cocircular(A,GDC)
Conditions (CDL):
Equal(MeasureOfAngle(GBD),x+5/2)
Equal(MeasureOfArc(ADC),4*x+5)
Equal(MeasureOfArc(AGD),50)
IsTangentOfCircle(BD,A)
Goal: Value(x)
Answer: 25
Theorem sequence:
circle_property_circular_power_tangent_and_segment_angle(2,BD,BGC,A)
Problem: As shown in the diagram, ∠FHE=15*x°, △EFG is an equilateral △, EH bisects ∠GEF. Find the measure of ∠GEH.
问题：如图所示，∠FHE=15*x°，三角形EFG为等边三角形，EH是∠GEF的角平分线。求∠GEH的大小。
Construction (CDL):
Shape(EF,FH,HE)
Shape(EH,HG,GE)
Collinear(FHG)
Conditions (CDL):
Equal(MeasureOfAngle(FHE),15*x)
EquilateralTriangle(EFG)
IsBisectorOfAngle(EH,GEF)
Goal: Value(MeasureOfAngle(GEH))
Answer: 30
Theorem sequence:
triangle_property_angle_sum(1,EHG)
isosceles_triangle_property_line_coincidence(3,EFG,H)
equilateral_triangle_property_angle(1,GEF)
Problem: As shown in the diagram, AB=16*sqrt(3), AC=y, BC=x, ∠BAC=60°, AC is perpendicular to BC. Find the value of y.
问题：如图所示，AB=16*sqrt(3)，AC=y，BC=x，∠BAC=60°，AC⊥BC。求y的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),16*sqrt(3))
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(BAC),60)
PerpendicularBetweenLine(AC,BC)
Goal: Value(y)
Answer: 8*sqrt(3)
Theorem sequence:
cosine_theorem(1,ACB)
cosine_theorem(1,CBA)
Problem: As shown in the diagram, HG=x, HJ=4, JK=6, JK is the tangent to ⊙A. Find the value of x.
问题：如图所示，HG=x，HJ=4，JK=6，⊙O的切线为JK。求x的值。
Construction (CDL):
Shape(KJ,JH,AKH)
Shape(AKH,HG,AGK)
Shape(AHG,GH)
Collinear(GHJ)
Cocircular(A,KHG)
Conditions (CDL):
Equal(LengthOfLine(HG),x)
Equal(LengthOfLine(HJ),4)
Equal(LengthOfLine(JK),6)
IsTangentOfCircle(JK,A)
Goal: Value(x)
Answer: 5
Theorem sequence:
line_addition(1,GH,HJ)
circle_property_circular_power_tangent_and_segment_line(1,JK,JHG,A)
Problem: As shown in the diagram, DF=e, ED=f, FE=d, ∠DFE=26°, ∠FED=108°, f=20. Find the value of d.
问题：如图所示，DF=e，ED=f，FE=d，∠DFE=26°，∠FED=108°，f=20。求d的值。
Construction (CDL):
Shape(ED,DF,FE)
Conditions (CDL):
Equal(LengthOfLine(DF),e)
Equal(LengthOfLine(ED),f)
Equal(LengthOfLine(FE),d)
Equal(MeasureOfAngle(DFE),26)
Equal(MeasureOfAngle(FED),108)
Equal(f,20)
Goal: Value(d)
Answer: 20*sin(23*pi/90)/sin(13*pi/90)
Theorem sequence:
triangle_property_angle_sum(1,EDF)
sine_theorem(1,EDF)
Problem: As shown in the diagram, AC=4, BA=16, CB=x, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AC=4，BA=16，CB=x，AC垂直于BC。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BA),16)
Equal(LengthOfLine(CB),x)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 4*sqrt(15)
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, FK=32, EH⊥GH, EHGF is a rectangle. Find the length of line EG.
问题：如图所示，FK=32，EH垂直于GH，四边形EHGF是矩形。求直线EG的长度。
Construction (CDL):
Shape(EH,HK,KE)
Shape(KH,HG,GK)
Shape(KG,GF,FK)
Shape(EK,KF,FE)
Collinear(EKG)
Collinear(HKF)
Conditions (CDL):
Equal(LengthOfLine(FK),32)
PerpendicularBetweenLine(EH,GH)
Rectangle(EHGF)
Goal: Value(LengthOfLine(EG))
Answer: 64
Theorem sequence:
line_addition(1,FK,KH)
parallelogram_property_diagonal_bisection(1,HGFE,K)
rectangle_property_diagonal_equal(1,EHGF)
Problem: As shown in the diagram, AB=15, BC=y, CA=x, ∠BCA=60°, CA⊥BA. Find the value of x.
问题：如图所示，AB=15，BC=y，CA=x，∠BCA=60°，CA⊥BA。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CA),x)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 5*sqrt(3)
Theorem sequence:
cosine_theorem(1,ABC)
cosine_theorem(1,CAB)
Problem: As shown in the diagram, RT=3*x+3, RT=SR, SR=4*x-4, TS=12+x. Find the length of line ST.
问题：如图所示，RT=3*x+3，RT=SR，SR=4*x-4，TS=12+x。求直线ST的长度。
Construction (CDL):
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(RT),3*x+3)
Equal(LengthOfLine(RT),LengthOfLine(SR))
Equal(LengthOfLine(SR),4*x-4)
Equal(LengthOfLine(TS),12+x)
Goal: Value(LengthOfLine(ST))
Answer: 19
Theorem sequence:
Problem: As shown in the diagram, AB=x+4, CD=3*x+2, ST=9, S bisects segment AC, T is the midpoint of segment BD, AB and CD are the parallel sides of trapezoid ACDB. Find the length of line AB.
问题：如图所示，AB=x+4，CD=3*x+2，ST=9，S平分线段AC，T是线段BD的中点，AC和DB是梯形ACDB的腰。求直线AB的长度。
Construction (CDL):
Shape(AS,ST,TB,BA)
Shape(SC,CD,DT,TS)
Collinear(ASC)
Collinear(BTD)
Conditions (CDL):
Equal(LengthOfLine(AB),x+4)
Equal(LengthOfLine(CD),3*x+2)
Equal(LengthOfLine(ST),9)
IsMidpointOfLine(S,AC)
IsMidpointOfLine(T,BD)
Trapezoid(ACDB)
Goal: Value(LengthOfLine(AB))
Answer: 7
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,ST,ACDB)
midsegment_of_quadrilateral_property_length(1,ST,ACDB)
Problem: As shown in the diagram, ∠BCD=2*x+4°, ∠ECB=2*x-4°. Find the measure of ∠BCD.
问题：如图所示，∠BCD=2*x+4°，∠ECB=2*x-4°。求∠BCD的大小。
Construction (CDL):
Shape(BC,CD)
Shape(DC,CA)
Shape(EC,CB)
Shape(AC,CE)
Collinear(BCA)
Collinear(DCE)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),2*x+4)
Equal(MeasureOfAngle(ECB),2*x-4)
Goal: Value(MeasureOfAngle(BCD),2*x+4)
Answer: 94
Theorem sequence:
adjacent_complementary_angle(1,ECB,BCD)
Problem: As shown in the diagram, ∠DFE=2*x-3°, ∠GFD=x+12°, DG⊥FG, DGFE is a rectangle. Find the measure of ∠DFE.
问题：如图所示，∠DFE=2*x-3°，∠GFD=x+12°，DG垂直于FG，DGFE是长方形。求∠DFE的大小。
Construction (CDL):
Shape(AD,DG,GA)
Shape(AG,GF,FA)
Shape(AF,FE,EA)
Shape(AE,ED,DA)
Collinear(DAF)
Collinear(EAG)
Conditions (CDL):
Equal(MeasureOfAngle(DFE),2*x-3)
Equal(MeasureOfAngle(GFD),x+12)
PerpendicularBetweenLine(DG,FG)
Rectangle(DGFE)
Goal: Value(MeasureOfAngle(DFE))
Answer: 51
Theorem sequence:
angle_addition(1,GFD,DFE)
Problem: As shown in the diagram, BD=6, DA=10, quadrilateral BADC is a ▱, BD is perpendicular to CD, DB⊥AB. Find the area of quadrilateral BADC.
问题：如图所示，BD=6，DA=10，BC和AD是平行四边形BADC的一组对边，BD⊥CD，DB⊥AB。求四边形BADC的面积。
Construction (CDL):
Shape(BA,AD,DB)
Shape(BD,DC,CB)
Conditions (CDL):
Equal(LengthOfLine(BD),6)
Equal(LengthOfLine(DA),10)
Parallelogram(BADC)
PerpendicularBetweenLine(BD,CD)
PerpendicularBetweenLine(DB,AB)
Goal: Value(AreaOfQuadrilateral(BADC))
Answer: 48
Theorem sequence:
right_triangle_judgment_angle(1,DBA)
right_triangle_property_pythagorean(1,DBA)
altitude_of_quadrilateral_judgment_diagonal(3,CBAD)
parallelogram_area_formula_common(1,CBAD)
Problem: As shown in the diagram, ∠RTS=2*x°, ∠SRT=x°, the center of ⊙B is B, TR is the diameter of circle B. Find the value of x.
问题：如图所示，∠RTS=2*x°，∠SRT=x°，B是圆B的圆心，TR是圆B的直径。求x的值。
Construction (CDL):
Shape(SR,RB,BT,TS)
Shape(BSR,RS)
Shape(BRT,TB,BR)
Shape(BTS,ST)
Collinear(RBT)
Cocircular(B,SRT)
Conditions (CDL):
Equal(MeasureOfAngle(RTS),2*x)
Equal(MeasureOfAngle(SRT),x)
IsCentreOfCircle(B,B)
IsDiameterOfCircle(TR,B)
Goal: Value(x)
Answer: 30
Theorem sequence:
diameter_of_circle_property_right_angle(1,TSR,B)
triangle_property_angle_sum(1,SRT)
Problem: As shown in the diagram, FL=24, HJ=48, the measure of ⌒MHP is 65, the center of ⊙M is M, JN is perpendicular to PN, KL⊥GL. Find the length of line NJ.
问题：如图所示，FL=24，HJ=48，弧MHP的角度为65，M是⊙M的圆心，JN垂直于PN，KL⊥GL。求直线NJ的长度。
Construction (CDL):
Shape(MKF,FL,LK)
Shape(MFH,HN,NM,ML,LF)
Shape(MHP,PN,NH)
Shape(MPJ,JN,NP)
Shape(MJG,GL,LM,MN,NJ)
Shape(MGK,KL,LG)
Collinear(KLM)
Collinear(PNM)
Collinear(FLG)
Collinear(HNJ)
Cocircular(M,FHPJGK)
Conditions (CDL):
Equal(LengthOfLine(FL),24)
Equal(LengthOfLine(HJ),48)
Equal(MeasureOfArc(MHP),65)
IsCentreOfCircle(M,M)
PerpendicularBetweenLine(JN,PN)
PerpendicularBetweenLine(KL,GL)
Goal: Value(LengthOfLine(NJ))
Answer: 24
Theorem sequence:
line_addition(1,HN,NJ)
vertical_angle(1,HNM,JNP)
circle_property_chord_perpendicular_bisect_chord(1,M,MN,HJ)
Problem: As shown in the diagram, ∠FJE=75°, GD is parallel to IM, IM is parallel to CE. Find the measure of ∠MFJ.
问题：如图所示，∠FJE=75°，GD∥IM，IM∥CE。求∠MFJ的大小。
Construction (CDL):
Shape(LB,BD)
Shape(DB,BF)
Shape(GB,BL)
Shape(FB,BG)
Shape(BF,FM)
Shape(MF,FJ)
Shape(JF,FI)
Shape(IF,FB)
Shape(FJ,JE)
Shape(EJ,JT)
Shape(TJ,JC)
Shape(CJ,JF)
Collinear(LBFJT)
Collinear(DBG)
Collinear(MFI)
Collinear(EJC)
Conditions (CDL):
Equal(MeasureOfAngle(FJE),75)
ParallelBetweenLine(GD,IM)
ParallelBetweenLine(IM,CE)
Goal: Value(MeasureOfAngle(MFJ))
Answer: 105
Theorem sequence:
parallel_property_collinear_extend(3,IM,CE,F)
parallel_property_collinear_extend(3,EC,MF,J)
parallel_property_ipsilateral_internal_angle(1,FM,JE)
Problem: As shown in the diagram, AB=6, BC=15/2, CA=9, DE=9, ∠ABF=∠FBC, BF is the angle bisector of ∠ABC, ED∥AC, ED is parallel to FC. Find the length of line BD.
问题：如图所示，AB=6，BC=15/2，CA=9，DE=9，∠ABF=∠FBC，BF是∠ABC的角平分线，ED∥AC，ED∥FC。求直线BD的长度。
Construction (CDL):
Shape(AB,BF,FA)
Shape(FB,BC,CF)
Shape(FC,CD,DE,EF)
Collinear(AFC)
Collinear(BFE)
Collinear(BCD)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(BC),15/2)
Equal(LengthOfLine(CA),9)
Equal(LengthOfLine(DE),9)
Equal(MeasureOfAngle(ABF),MeasureOfAngle(FBC))
IsBisectorOfAngle(BF,ABC)
ParallelBetweenLine(ED,AC)
ParallelBetweenLine(ED,FC)
Goal: Value(LengthOfLine(BD))
Answer: 27/2
Theorem sequence:
line_addition(1,CF,FA)
bisector_of_angle_property_line_ratio(1,BF,ABC)
parallel_property_corresponding_angle(1,CA,DE,B)
parallel_property_corresponding_angle(2,ED,FC,B)
similar_triangle_judgment_aa(1,BDE,BCF)
similar_triangle_property_line_ratio(1,BDE,BCF)
similar_triangle_property_line_ratio(1,EBD,FBC)
Problem: As shown in the diagram, ∠ABF=4*x+11°, ∠CAB=8*x+1°, EC is parallel to DF. Find the value of x.
问题：如图所示，∠ABF=4*x+11°，∠CAB=8*x+1°，EC∥DF。求x的值。
Construction (CDL):
Shape(GB,BD)
Shape(DB,BA)
Shape(AB,BF)
Shape(FB,BG)
Shape(BA,AE)
Shape(EA,AH)
Shape(HA,AC)
Shape(CA,AB)
Collinear(GBAH)
Collinear(DBF)
Collinear(EAC)
Conditions (CDL):
Equal(MeasureOfAngle(ABF),4*x+11)
Equal(MeasureOfAngle(CAB),8*x+1)
ParallelBetweenLine(EC,DF)
Goal: Value(x)
Answer: 14
Theorem sequence:
parallel_property_collinear_extend(3,EC,DF,A)
parallel_property_collinear_extend(3,FD,CA,B)
parallel_property_ipsilateral_internal_angle(1,AC,BF)
Problem: As shown in the diagram, BG=15, CB=30, FE=15, BG is the median of △ BAC, EH is the median of △ EDF, triangle BAC is similar to triangle EDF.. Find the length of line EH.
问题：如图所示，BG=15，CB=30，FE=15，BG是△BAC的中线，EH是△EDF的中线，△BAC与△EDF是相似三角形。求直线EH的长度。
Construction (CDL):
Shape(BA,AG,GB)
Shape(BG,GC,CB)
Shape(ED,DH,HE)
Shape(EH,HF,FE)
Collinear(AGC)
Collinear(DHF)
Conditions (CDL):
Equal(LengthOfLine(BG),15)
Equal(LengthOfLine(CB),30)
Equal(LengthOfLine(FE),15)
IsMedianOfTriangle(BG,BAC)
IsMedianOfTriangle(EH,EDF)
SimilarBetweenTriangle(BAC,EDF)
Goal: Value(LengthOfLine(EH))
Answer: 15/2
Theorem sequence:
line_addition(1,AG,GC)
line_addition(1,DH,HF)
similar_triangle_property_line_ratio(1,BAC,EDF)
similar_triangle_property_line_ratio(1,ACB,DFE)
similar_triangle_property_angle_equal(1,CBA,FED)
similar_triangle_judgment_sas(1,CBG,FEH)
similar_triangle_property_line_ratio(1,CBG,FEH)
similar_triangle_property_line_ratio(1,GCB,HFE)
Problem: As shown in the diagram, ∠GJH=x°, ∠HGB=2*x°, the center of circle A is A, GB is perpendicular to HB, JH is perpendicular to GH. Find the measure of ∠GJH.
问题：如图所示，∠GJH=x°，∠HGB=2*x°，圆A的圆心为A，GB垂直于HB，JH⊥GH。求∠GJH的大小。
Construction (CDL):
Shape(AGF,FB,BG)
Shape(AFJ,JA,AB,BF)
Shape(AJH,HJ)
Shape(HB,BA,AJ,JH)
Shape(GB,BH,HG)
Shape(AHG,GH)
Collinear(FBH)
Collinear(GBAJ)
Cocircular(A,FJHG)
Conditions (CDL):
Equal(MeasureOfAngle(GJH),x)
Equal(MeasureOfAngle(HGB),2*x)
IsCentreOfCircle(A,A)
PerpendicularBetweenLine(GB,HB)
PerpendicularBetweenLine(JH,GH)
Goal: Value(MeasureOfAngle(GJH))
Answer: 30
Theorem sequence:
triangle_property_angle_sum(1,GJH)
Problem: As shown in the diagram, RT=13, XY=7, ∠ZXY=124°, XZ is the midsegment of △ SRT, YX is the midsegment of triangle RTS. Find the length of line TS.
问题：如图所示，RT=13，XY=7，∠ZXY=124°，XZ是△SRT的中位线，YX是三角形RTS的中位线。求直线TS的长度。
Construction (CDL):
Shape(RY,YX,XR)
Shape(XY,YT,TZ,ZX)
Shape(SX,XZ,ZS)
Collinear(RXS)
Collinear(RYT)
Collinear(TZS)
Conditions (CDL):
Equal(LengthOfLine(RT),13)
Equal(LengthOfLine(XY),7)
Equal(MeasureOfAngle(ZXY),124)
IsMidsegmentOfTriangle(XZ,SRT)
IsMidsegmentOfTriangle(YX,RTS)
Goal: Value(LengthOfLine(TS))
Answer: 14
Theorem sequence:
midsegment_of_triangle_property_length(1,YX,RTS)
Problem: As shown in the diagram, DE=7, EX=24, the center of circle D is D, DA⊥XA, XE is perpendicular to DE. Find the length of line TX.
问题：如图所示，DE=7，EX=24，D是⊙D的圆心，DA⊥XA，XE⊥DE。求直线TX的长度。
Construction (CDL):
Shape(AX,XQ,DAQ)
Shape(DAQ,QD,DA)
Shape(QX,XE,DQE)
Shape(DQE,ED,DQ)
Shape(DTA,AD,DT)
Shape(DET,TD,DE)
Collinear(XQDT)
Cocircular(D,AQET)
Conditions (CDL):
Equal(LengthOfLine(DE),7)
Equal(LengthOfLine(EX),24)
IsCentreOfCircle(D,D)
PerpendicularBetweenLine(DA,XA)
PerpendicularBetweenLine(XE,DE)
Goal: Value(LengthOfLine(TX))
Answer: 32
Theorem sequence:
radius_of_circle_property_length_equal(1,DE,D)
radius_of_circle_property_length_equal(1,DT,D)
right_triangle_judgment_angle(1,XED)
right_triangle_property_pythagorean(1,XED)
line_addition(1,XD,DT)
Problem: As shown in the diagram, ∠AGE=55°, ∠BGA=x°, ∠GBH=y°, HD is parallel to CA. Find the value of x.
问题：如图所示，∠AGE=55°，∠BGA=x°，∠GBH=y°，HD∥CA。求x的值。
Construction (CDL):
Shape(HB,BF)
Shape(FB,BD)
Shape(DB,BG)
Shape(GB,BH)
Shape(CG,GB)
Shape(BG,GA)
Shape(AG,GE)
Shape(EG,GC)
Collinear(HBD)
Collinear(CGA)
Collinear(FBGE)
Conditions (CDL):
Equal(MeasureOfAngle(AGE),55)
Equal(MeasureOfAngle(BGA),x)
Equal(MeasureOfAngle(GBH),y)
ParallelBetweenLine(HD,CA)
Goal: Value(x)
Answer: 125
Theorem sequence:
adjacent_complementary_angle(1,BGA,AGE)
Problem: As shown in the diagram, AC=6, BA=x, CB=12, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AC=6，BA=x，CB=12，BA⊥CA。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(BA),x)
Equal(LengthOfLine(CB),12)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 6*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, KL=NL, NM=ML, ∠JLK=25°, ∠KLN=18°, ∠NKJ=130°, ∠NLM=20°, LN and LK are the legs of the isosceles triangle LNK, MN and ML are the legs of the isosceles △ MNL. Find the measure of ∠KJL.
问题：如图所示，KL=NL，NM=ML，∠JLK=25°，∠KLN=18°，∠NKJ=130°，∠NLM=20°，∠LNK和∠NKL是等腰三角形LNK的底角，MN和ML是等腰三角形MNL的腰。求∠KJL的大小。
Construction (CDL):
Shape(JL,LK,KJ)
Shape(KL,LN,NK)
Shape(NL,LM,MN)
Conditions (CDL):
Equal(LengthOfLine(KL),LengthOfLine(NL))
Equal(LengthOfLine(NM),LengthOfLine(ML))
Equal(MeasureOfAngle(JLK),25)
Equal(MeasureOfAngle(KLN),18)
Equal(MeasureOfAngle(NKJ),130)
Equal(MeasureOfAngle(NLM),20)
IsoscelesTriangle(LNK)
IsoscelesTriangle(MNL)
Goal: Value(MeasureOfAngle(KJL))
Answer: 106
Theorem sequence:
isosceles_triangle_property_angle_equal(1,LNK)
triangle_property_angle_sum(1,JLK)
triangle_property_angle_sum(1,KLN)
angle_addition(1,NKL,LKJ)
Problem: As shown in the diagram, PQ=4*x-8, QM=2*x+2, ∠MNQ=∠QNP, NP is perpendicular to QP, QM is perpendicular to NM. Find the length of line QM.
问题：如图所示，PQ=4*x-8，QM=2*x+2，∠MNQ=∠QNP，NP垂直于QP，QM垂直于NM。求直线QM的长度。
Construction (CDL):
Shape(MN,NQ,QM)
Shape(QN,NP,PQ)
Collinear(NMB)
Collinear(NQC)
Collinear(NPA)
Conditions (CDL):
Equal(LengthOfLine(PQ),4*x-8)
Equal(LengthOfLine(QM),2*x+2)
Equal(MeasureOfAngle(MNQ),MeasureOfAngle(QNP))
PerpendicularBetweenLine(NP,QP)
PerpendicularBetweenLine(QM,NM)
Goal: Value(LengthOfLine(QM))
Answer: 12
Theorem sequence:
mirror_congruent_triangle_judgment_aas(1,QNP,QMN)
mirror_congruent_triangle_property_line_equal(1,NPQ,NQM)
Problem: As shown in the diagram, ∠EDA=3*x°, the measure of ⌒FCB is 4*x+50, the measure of arc FEA is 30, F is the center of ⊙F. Find the value of x.
问题：如图所示，∠EDA=3*x°，弧FCB的角度为4*x+50，弧FEA的角度为30，⊙F的圆心为F。求x的值。
Construction (CDL):
Shape(FAC,CA)
Shape(FCB,BE,FEA,AC)
Shape(FBE,EB)
Shape(FEA,ED,DA)
Collinear(CAD)
Collinear(BED)
Cocircular(F,CBEA)
Conditions (CDL):
Equal(MeasureOfAngle(EDA),3*x)
Equal(MeasureOfArc(FCB),4*x+50)
Equal(MeasureOfArc(FEA),30)
IsCentreOfCircle(F,F)
Goal: Value(x)
Answer: 10
Theorem sequence:
circle_property_circular_power_segment_and_segment_angle(2,DAC,DEB,F)
Problem: As shown in the diagram, ∠BAC=2*x-15°, ∠DBA=x-5°, ∠DCA=148°. Find the measure of ∠DBA.
问题：如图所示，∠BAC=2*x-15°，∠DBA=x-5°，∠DCA=148°。求∠DBA的大小。
Construction (CDL):
Shape(AC,CB,BA)
Shape(DC,CA)
Collinear(DCB)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),2*x-15)
Equal(MeasureOfAngle(DBA),x-5)
Equal(MeasureOfAngle(DCA),148)
Goal: Value(MeasureOfAngle(DBA))
Answer: 51
Theorem sequence:
adjacent_complementary_angle(1,DCA,ACB)
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, BF=x, FC=x+8, FD=5, FE=4, A is the center of ⊙A. Find the value of x.
问题：如图所示，BF=x，FC=x+8，FD=5，FE=4，⊙A的圆心为A。求x的值。
Construction (CDL):
Shape(ADC,CF,FD)
Shape(ACE,EF,FC)
Shape(AEB,BF,FE)
Shape(ABD,DF,FB)
Collinear(DFE)
Collinear(CFB)
Cocircular(A,DCEB)
Conditions (CDL):
Equal(LengthOfLine(BF),x)
Equal(LengthOfLine(FC),x+8)
Equal(LengthOfLine(FD),5)
Equal(LengthOfLine(FE),4)
IsCentreOfCircle(A,A)
Goal: Value(x)
Answer: 2
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,CFB,DFE,A)
Problem: As shown in the diagram, ∠ADB=11*x-1°, ∠BDF=7*x+9°, ∠DFH=7*y-4°, ∠GAD=2*y+5°, ∠HFC=z°, AG∥DB, DB∥FH. Find the value of x.
问题：如图所示，∠ADB=11*x-1°，∠BDF=7*x+9°，∠DFH=7*y-4°，∠GAD=2*y+5°，∠HFC=z°，AG∥DB，DB平行于FH。求x的值。
Construction (CDL):
Shape(HF,FC)
Shape(DF,FH)
Shape(BD,DF)
Shape(AD,DB)
Shape(GA,AD)
Shape(EA,AG)
Collinear(EAD)
Collinear(CFD)
Conditions (CDL):
Equal(MeasureOfAngle(ADB),11*x-1)
Equal(MeasureOfAngle(BDF),7*x+9)
Equal(MeasureOfAngle(DFH),7*y-4)
Equal(MeasureOfAngle(GAD),2*y+5)
Equal(MeasureOfAngle(HFC),z)
ParallelBetweenLine(AG,DB)
ParallelBetweenLine(DB,FH)
Goal: Value(x)
Answer: 14
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,DB,FH)
parallel_property_ipsilateral_internal_angle(1,AG,DB)
Problem: As shown in the diagram, BA=DA, CB=CD, CE=12, ED=5, AE⊥DE. Find the length of line BC.
问题：如图所示，BA=DA，CB=CD，CE=12，ED=5，AE垂直于DE。求直线BC的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,ED,DA)
Shape(BC,CE,EB)
Shape(EC,CD,DE)
Collinear(BED)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(BA),LengthOfLine(DA))
Equal(LengthOfLine(CB),LengthOfLine(CD))
Equal(LengthOfLine(CE),12)
Equal(LengthOfLine(ED),5)
PerpendicularBetweenLine(AE,DE)
Goal: Value(LengthOfLine(BC))
Answer: 13
Theorem sequence:
adjacent_complementary_angle(1,AED,DEC)
right_triangle_judgment_angle(1,DEC)
right_triangle_property_pythagorean(1,DEC)
Problem: As shown in the diagram, BC=22, CA=h, ∠ABC=30°, CA is perpendicular to BA. Find the value of h.
问题：如图所示，BC=22，CA=h，∠ABC=30°，CA⊥BA。求h的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(BC),22)
Equal(LengthOfLine(CA),h)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(CA,BA)
Goal: Value(h)
Answer: 11
Theorem sequence:
sine_theorem(1,CAB)
Problem: As shown in the diagram, AB=27, BC=18, BC⊥AC. Find the measure of ∠CAB.
问题：如图所示，AB=27，BC=18，BC垂直于AC。求∠CAB的大小。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),27)
Equal(LengthOfLine(BC),18)
PerpendicularBetweenLine(BC,AC)
Goal: Value(MeasureOfAngle(CAB))
Answer: 180*asin(2/3)/pi
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, ∠ABD=4*x-23°, ∠UEC=2*x+17°, AC∥FU. Find the value of x.
问题：如图所示，∠ABD=4*x-23°，∠UEC=2*x+17°，AC平行于FU。求x的值。
Construction (CDL):
Shape(AB,BD)
Shape(DB,BC)
Shape(CB,BE)
Shape(EB,BA)
Shape(FE,EB)
Shape(BE,EU)
Shape(UE,EC)
Shape(CE,EF)
Collinear(ABC)
Collinear(FEU)
Collinear(DBEC)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),4*x-23)
Equal(MeasureOfAngle(UEC),2*x+17)
ParallelBetweenLine(AC,FU)
Goal: Value(x)
Answer: 20
Theorem sequence:
parallel_property_collinear_extend(3,AC,FU,B)
parallel_property_collinear_extend(3,UF,BA,E)
vertical_angle(1,FED,UEC)
parallel_property_corresponding_angle(2,EF,BA,D)
Problem: As shown in the diagram, ∠BAF=x°, ∠DAC=67°, A is the center of circle A, FA is perpendicular to DA. Find the value of x.
问题：如图所示，∠BAF=x°，∠DAC=67°，圆A的圆心为A，FA垂直于DA。求x的值。
Construction (CDL):
Shape(ACD,DA,AC)
Shape(ADF,FA,AD)
Shape(AFB,BA,AF)
Shape(ABE,EA,AB)
Shape(AEC,CA,AE)
Collinear(BAC)
Collinear(DAE)
Cocircular(A,DFBEC)
Conditions (CDL):
Equal(MeasureOfAngle(BAF),x)
Equal(MeasureOfAngle(DAC),67)
IsCentreOfCircle(A,A)
PerpendicularBetweenLine(FA,DA)
Goal: Value(x)
Answer: 23
Theorem sequence:
vertical_angle(1,DAC,EAB)
adjacent_complementary_angle(1,EAF,FAD)
angle_addition(1,EAB,BAF)
Problem: As shown in the diagram, AB=y, BC=x, CA=24, ∠ABC=30°, BC⊥AC. Find the value of x.
问题：如图所示，AB=y，BC=x，CA=24，∠ABC=30°，BC⊥AC。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CA),24)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 24*sqrt(3)
Theorem sequence:
sine_theorem(1,ABC)
cosine_theorem(1,CAB)
Problem: As shown in the diagram, AD=9, BA=z, BD=y, CB=8, DC=x, AD⊥BD, CB is perpendicular to AB. Find the value of x.
问题：如图所示，AD=9，BA=z，BD=y，CB=8，DC=x，AD垂直于BD，CB⊥AB。求x的值。
Construction (CDL):
Shape(CB,BD,DC)
Shape(BA,AD,DB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AD),9)
Equal(LengthOfLine(BA),z)
Equal(LengthOfLine(BD),y)
Equal(LengthOfLine(CB),8)
Equal(LengthOfLine(DC),x)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: -9/2+sqrt(337)/2
Theorem sequence:
adjacent_complementary_angle(1,ADB,BDC)
line_addition(1,CD,DA)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,ADB)
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,ADB)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, ∠AZB=104°, the measure of ⌒ZCB is 94, Z is the center of ⊙Z, AB is parallel to DC. Find the measure of ∠ZAC.
问题：如图所示，∠AZB=104°，弧ZCB的角度为94，Z是⊙Z的圆心，AB∥DC。求∠ZAC的大小。
Construction (CDL):
Shape(ZB,BA,AZ)
Shape(EB,BZ,ZA,AE)
Shape(ZBA,AB)
Shape(ZAD,DA)
Shape(ZDC,CD)
Shape(ZCB,BC)
Shape(EC,CB,BE)
Shape(CE,ED,DC)
Shape(EA,AD,DE)
Collinear(BED)
Collinear(CEA)
Cocircular(Z,BADC)
Conditions (CDL):
Equal(MeasureOfAngle(AZB),104)
Equal(MeasureOfArc(ZCB),94)
IsCentreOfCircle(Z,Z)
ParallelBetweenLine(AB,DC)
Goal: Value(MeasureOfAngle(ZAC))
Answer: 9
Theorem sequence:
radius_of_circle_property_length_equal(1,ZB,Z)
radius_of_circle_property_length_equal(1,ZA,Z)
isosceles_triangle_judgment_line_equal(1,ZBA)
isosceles_triangle_property_angle_equal(1,ZBA)
triangle_property_angle_sum(1,ZBA)
arc_property_circumference_angle_external(1,ZCB,A)
angle_addition(1,BAZ,ZAE)
Problem: As shown in the diagram, AB=50, BC=14, CA=48, BC⊥AC. Find the value of sin(CA).
问题：如图所示，AB=50，BC=14，CA=48，BC垂直于AC。求sin(CA)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),50)
Equal(LengthOfLine(BC),14)
Equal(LengthOfLine(CA),48)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Sin(MeasureOfAngle(CAB)))
Answer: 7/25
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, ∠AXW=30°, WY is perpendicular to ZY, quadrilateral WYZX is a rectangle. Find the measure of ∠WYA.
问题：如图所示，∠AXW=30°，WY⊥ZY，四边形WYZX是矩形。求∠WYA的大小。
Construction (CDL):
Shape(AW,WY,YA)
Shape(AY,YZ,ZA)
Shape(AZ,ZX,XA)
Shape(WA,AX,XW)
Collinear(WAZ)
Collinear(YAX)
Conditions (CDL):
Equal(MeasureOfAngle(AXW),30)
PerpendicularBetweenLine(WY,ZY)
Rectangle(WYZX)
Goal: Value(MeasureOfAngle(WYA))
Answer: 60
Theorem sequence:
triangle_property_angle_sum(1,WYX)
Problem: As shown in the diagram, AB=13, AC=BD, AD=AC, AD=x, BC=AD, ∠ABC=y°, AD⊥BD. Find the value of y.
问题：如图所示，AB=13，AC=BD，AD=AC，AD=x，BC=AD，∠ABC=y°，AD⊥BD。求y的值。
Construction (CDL):
Shape(DB,BA,AD)
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),LengthOfLine(BD))
Equal(LengthOfLine(AD),LengthOfLine(AC))
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BC),LengthOfLine(AD))
Equal(MeasureOfAngle(ABC),y)
PerpendicularBetweenLine(AD,BD)
Goal: Value(y)
Answer: 45
Theorem sequence:
parallelogram_judgment_equal_and_equal(1,DBCA)
rectangle_judgment_right_angle(1,ADBC)
angle_addition(1,DBA,ABC)
isosceles_triangle_judgment_line_equal(1,CAB)
isosceles_triangle_property_angle_equal(1,CAB)
parallel_property_alternate_interior_angle(1,BD,CA)