SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠HUN=123°, KA is parallel to UQ, UK is parallel to QA. Find the measure of ∠QAK.
问题：如图所示，∠HUN=123°，KA平行于UQ，UK∥QA。求∠QAK的大小。
Construction (CDL):
Shape(GK,KF)
Shape(HU,NU)
Shape(JA,AM)
Shape(EQ,QI)
Shape(KU,UQ,QA,AK)
Collinear(GKAM)
Collinear(NUQE)
Collinear(FKUH)
Collinear(JAQI)
Conditions (CDL):
Equal(MeasureOfAngle(HUN),123)
ParallelBetweenLine(KA,UQ)
ParallelBetweenLine(UK,QA)
Goal: Value(MeasureOfAngle(QAK))
Answer: 123
Theorem sequence:
vertical_angle(1,HUN,KUQ)
parallelogram_judgment_parallel_and_parallel(1,UQAK)
parallelogram_property_opposite_angle_equal(1,UQAK)
Problem: As shown in the diagram, PR=6*x-4, PS=5*y-12, QP=2*y-6, TP=4*x+2, quadrilateral QRST is a ▱. Find the value of x.
问题：如图所示，PR=6*x-4，PS=5*y-12，QP=2*y-6，TP=4*x+2，QRST是▱。求x的值。
Construction (CDL):
Shape(PQ,QR,RP)
Shape(PR,RS,SP)
Shape(PS,ST,TP)
Shape(PT,TQ,QP)
Collinear(QPS)
Collinear(TPR)
Conditions (CDL):
Equal(LengthOfLine(PR),6*x-4)
Equal(LengthOfLine(PS),5*y-12)
Equal(LengthOfLine(QP),2*y-6)
Equal(LengthOfLine(TP),4*x+2)
Parallelogram(QRST)
Goal: Value(x)
Answer: 3
Theorem sequence:
parallelogram_property_diagonal_bisection(1,TQRS,P)
Problem: As shown in the diagram, the diameter of ⊙X is 7, EF=2, the measure of ⌒XAE is 160, the center of circle X is B, the center of circle Y is B. Find the area of the sector XEA minus the area of the sector YFC.
问题：如图所示，圆X的直径为7，EF=2，⌒XAE的角度为160，⊙X的圆心为B，B是圆Y的圆心。求扇形XEA的面积减去扇形YFC的面积。
Construction (CDL):
Shape(XAE,EF,YCF,CA)
Shape(BC,YCF,FB)
Shape(FE,XEA,AC,YFC)
Shape(BF,YFC,CB)
Collinear(EFB)
Collinear(BCA)
Cocircular(Y,FC)
Cocircular(X,EA)
Conditions (CDL):
Equal(DiameterOfCircle(X),7)
Equal(LengthOfLine(EF),2)
Equal(MeasureOfArc(XAE),160)
IsCentreOfCircle(B,X)
IsCentreOfCircle(B,Y)
Goal: Value(Sub(AreaOfSector(XEA),AreaOfSector(YFC)))
Answer: 50*pi/9
Theorem sequence:
round_arc(1,XAE,XEA)
arc_property_center_angle(1,XEA,B)
arc_property_center_angle(1,YFC,B)
radius_of_circle_property_length_equal(1,BE,X)
circle_property_length_of_radius_and_diameter(1,X)
line_addition(1,EF,FB)
radius_of_circle_property_length_equal(1,BF,Y)
sector_area_formula(1,XEA)
sector_area_formula(1,YFC)
Problem: As shown in the diagram, PQ=2*x+1, PT=10, RQ=6, ST=4, QT is parallel to RS. Find the value of x.
问题：如图所示，PQ=2*x+1，PT=10，RQ=6，ST=4，QT∥RS。求x的值。
Construction (CDL):
Shape(RS,ST,TQ,QR)
Shape(QT,TP,PQ)
Collinear(RQP)
Collinear(STP)
Conditions (CDL):
Equal(LengthOfLine(PQ),2*x+1)
Equal(LengthOfLine(PT),10)
Equal(LengthOfLine(RQ),6)
Equal(LengthOfLine(ST),4)
ParallelBetweenLine(QT,RS)
Goal: Value(x)
Answer: 7
Theorem sequence:
parallel_property_corresponding_angle(1,QT,RS,P)
similar_triangle_judgment_aa(1,TPQ,SPR)
line_addition(1,PQ,QR)
line_addition(1,PT,TS)
similar_triangle_property_line_ratio(1,TPQ,SPR)
similar_triangle_property_line_ratio(1,QTP,RSP)
Problem: As shown in the diagram, AB=12, BD=6, BE=3, BX=x. Find the value of x.
问题：如图所示，AB=12，BD=6，BE=3，BX=x。求x的值。
Construction (CDL):
Shape(CDE,EB,BD)
Shape(CEX,XB,BE)
Shape(CXA,AB,BX)
Shape(CAD,DB,BA)
Collinear(DBX)
Collinear(ABE)
Cocircular(C,DEXA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(BD),6)
Equal(LengthOfLine(BE),3)
Equal(LengthOfLine(BX),x)
Goal: Value(x)
Answer: 6
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,DBX,ABE,C)
Problem: As shown in the diagram, ST=a+15, SW=b+11, TV=3*b+5, VM=3a+11, quadrilateral TSWV is a ▱. Find the value of b.
问题：如图所示，ST=a+15，SW=b+11，TV=3*b+5，VM=3a+11，四边形TSWV是平行四边形。求b的值。
Construction (CDL):
Shape(TS,SW,WV,VT)
Conditions (CDL):
Equal(LengthOfLine(ST),a+15)
Equal(LengthOfLine(SW),b+11)
Equal(LengthOfLine(TV),3*b+5)
Equal(LengthOfLine(VM),3a+11)
Parallelogram(TSWV)
Goal: Value(b)
Answer: 3
Theorem sequence:
parallelogram_property_opposite_line_equal(1,VTSW)
Problem: As shown in the diagram, AD=3, BD=8, CA=y, CD=x, AC is perpendicular to BC, BD⊥CD. Find the value of y.
问题：如图所示，AD=3，BD=8，CA=y，CD=x，AC垂直于BC，BD垂直于CD。求y的值。
Construction (CDL):
Shape(CB,BD,DC)
Shape(CD,DA,AC)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(AD),3)
Equal(LengthOfLine(BD),8)
Equal(LengthOfLine(CA),y)
Equal(LengthOfLine(CD),x)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BD,CD)
Goal: Value(y)
Answer: sqrt(33)
Theorem sequence:
line_addition(1,BD,DA)
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, triangle RST is congruent to triangle XYZ, TR=x+21, ZX=2*x-14, ∠TRS=4*y-10°, ∠ZXY=3*y+5°. Find the value of x.
问题：如图所示，三角形RST与三角形XYZ是全等三角形，TR=x+21，ZX=2*x-14，∠TRS=4*y-10°，∠ZXY=3*y+5°。求x的值。
Construction (CDL):
Shape(RS,ST,TR)
Shape(ZX,XY,YZ)
Conditions (CDL):
CongruentBetweenTriangle(RST,XYZ)
Equal(LengthOfLine(TR),x+21)
Equal(LengthOfLine(ZX),2*x-14)
Equal(MeasureOfAngle(TRS),4*y-10)
Equal(MeasureOfAngle(ZXY),3*y+5)
Goal: Value(x)
Answer: 35
Theorem sequence:
congruent_triangle_property_line_equal(1,STR,YZX)
Problem: As shown in the diagram, AB=24, AC=20, CB=26, DB=a, ∠DAC=∠BAD. Find the value of a.
问题：如图所示，AB=24，AC=20，CB=26，DB=a，∠DAC=∠BAD。求a的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),24)
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(CB),26)
Equal(LengthOfLine(DB),a)
Equal(MeasureOfAngle(DAC),MeasureOfAngle(BAD))
Goal: Value(a)
Answer: 156/11
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, DC=6, IC=19, NB=19, NC=6, ND=6, Sub(AreaOfQuadrilateral(CNBI)=AreaOfTriangle(CND)), the center of circle E is E, IB is the diameter of circle E, CN⊥BN, IC is perpendicular to NC. Find the sum of x and the area of sector EBI
问题：如图所示，DC=6，IC=19，NB=19，NC=6，ND=6，Sub(AreaOfQuadrilateral(CNBI)=AreaOfTriangle(CND))，圆E的圆心为E，⊙E的直径为IB，CN⊥BN，IC⊥NC。求x与扇形EBI面积和。
Construction (CDL):
Shape(IC,CD,DN,NB,BI)
Shape(CN,ND,DC)
Shape(EBI,IB)
Collinear(IEB)
Cocircular(E,BI)
Conditions (CDL):
Equal(LengthOfLine(DC),6)
Equal(LengthOfLine(IC),19)
Equal(LengthOfLine(NB),19)
Equal(LengthOfLine(NC),6)
Equal(LengthOfLine(ND),6)
Equal(Sub(AreaOfQuadrilateral(CNBI),AreaOfTriangle(CND)),x)
IsCentreOfCircle(E,E)
IsDiameterOfCircle(IB,E)
PerpendicularBetweenLine(CN,BN)
PerpendicularBetweenLine(IC,NC)
Goal: Value(Add(x,AreaOfSector(EBI)))
Answer: -9*sqrt(3)+9*pi/2+114
Theorem sequence:
parallel_judgment_per_per(1,CI,NB)
parallelogram_judgment_parallel_and_equal(1,ICNB)
parallelogram_property_opposite_line_equal(1,CNBI)
parallelogram_area_formula_sine(1,CNBI)
isosceles_triangle_judgment_line_equal(1,CND)
isosceles_triangle_judgment_line_equal(1,NDC)
equilateral_triangle_judgment_isosceles_and_isosceles(1,CND)
triangle_area_formula_sine(1,DCN)
diameter_of_circle_property_length_equal(1,IB,E)
circle_property_length_of_radius_and_diameter(1,E)
sector_area_formula(1,EBI)
flat_angle(1,IEB)
arc_property_center_angle(1,EBI,E)
equilateral_triangle_property_angle(1,DCN)
Problem: As shown in the diagram, AE=10, BE=11, CE=y-7, DE=x+6, ABCD is a parallelogram. Find the value of y.
问题：如图所示，AE=10，BE=11，CE=y-7，DE=x+6，AD和BC是▱ABCD的一组对边。求y的值。
Construction (CDL):
Shape(EA,AB,BE)
Shape(EB,BC,CE)
Shape(EC,CD,DE)
Shape(ED,DA,AE)
Collinear(AEC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(AE),10)
Equal(LengthOfLine(BE),11)
Equal(LengthOfLine(CE),y-7)
Equal(LengthOfLine(DE),x+6)
Parallelogram(ABCD)
Goal: Value(y)
Answer: 17
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ABCD,E)
Problem: As shown in the diagram, ∠IFL=x°, the measure of ⌒BGH is 84, the measure of ⌒BHI is 52, the center of circle B is B. Find the value of x.
问题：如图所示，∠IFL=x°，弧BGH的角度为84，⌒BHI的角度为52，圆B的圆心为B。求x的值。
Construction (CDL):
Shape(BIL,IF,FL)
Shape(BHI,IH)
Shape(BLG,GL)
Shape(BGH,HB,BG)
Shape(IB,BH,HI)
Shape(BIL,LB,BI)
Collinear(FIH)
Collinear(FLBG)
Cocircular(B,HILG)
Conditions (CDL):
Equal(MeasureOfAngle(IFL),x)
Equal(MeasureOfArc(BGH),84)
Equal(MeasureOfArc(BHI),52)
IsCentreOfCircle(B,B)
Goal: Value(x)
Answer: 20
Theorem sequence:
arc_property_center_angle(1,BGH,B)
arc_property_center_angle(1,BHI,B)
angle_addition(1,IBH,HBG)
angle_addition(1,LBI,IBG)
flat_angle(1,LBG)
arc_property_center_angle(1,BIL,B)
circle_property_circular_power_segment_and_segment_angle(1,FIH,FLG,B)
Problem: As shown in the diagram, EH is perpendicular to GH, EHGF is a rectangle. Find the sum of the measure of ∠FEH and the measure of ∠GFE.
问题：如图所示，EH⊥GH，EHGF是长方形。求∠FEH的大小与∠GFE的大小之和。
Construction (CDL):
Shape(EH,HK,KE)
Shape(KH,HG,GK)
Shape(KG,GF,FK)
Shape(EK,KF,FE)
Collinear(EKG)
Collinear(HKF)
Conditions (CDL):
PerpendicularBetweenLine(EH,GH)
Rectangle(EHGF)
Goal: Value(Add(MeasureOfAngle(FEH),MeasureOfAngle(GFE)))
Answer: 180
Theorem sequence:
Problem: As shown in the diagram, ∠KLP=110°, quadrilateral LJMK is a ▱, quadrilateral PNML is a rectangle. Find the measure of ∠MKL.
问题：如图所示，∠KLP=110°，JL和MK是▱LJMK的一组对边，PNML是矩形。求∠MKL的大小。
Construction (CDL):
Shape(PN,NJ,JP)
Shape(JN,NM,MJ)
Shape(JM,ML,LJ)
Shape(PJ,JL,LP)
Shape(LM,MK,KL)
Collinear(PJM)
Collinear(NJL)
Conditions (CDL):
Equal(MeasureOfAngle(KLP),110)
Parallelogram(LJMK)
Rectangle(PNML)
Goal: Value(MeasureOfAngle(MKL))
Answer: 140
Theorem sequence:
line_addition(1,PJ,JM)
line_addition(1,NJ,JL)
rectangle_property_diagonal_equal(1,PNML)
parallelogram_property_diagonal_bisection(1,PNML,J)
parallelogram_property_diagonal_bisection(1,NMLP,J)
parallelogram_property_opposite_line_equal(1,LJMK)
parallelogram_property_opposite_line_equal(1,JMKL)
isosceles_triangle_judgment_line_equal(1,KLM)
isosceles_triangle_property_angle_equal(1,KLM)
angle_addition(1,KLM,MLP)
triangle_property_angle_sum(1,LMK)
Problem: As shown in the diagram, PT=4, RP=4, ∠RPT=60°. Find the length of line TR.
问题：如图所示，PT=4，RP=4，∠RPT=60°。求直线TR的长度。
Construction (CDL):
Shape(PT,TR,RP)
Conditions (CDL):
Equal(LengthOfLine(PT),4)
Equal(LengthOfLine(RP),4)
Equal(MeasureOfAngle(RPT),60)
Goal: Value(LengthOfLine(TR))
Answer: 4
Theorem sequence:
cosine_theorem(1,PTR)
Problem: As shown in the diagram, AW=4, FW=12, RW=4, WY=8. Find the length of line FY.
问题：如图所示，AW=4，FW=12，RW=4，WY=8。求直线FY的长度。
Construction (CDL):
Shape(AR,RL,LA)
Shape(AL,LW,WA)
Collinear(FLA)
Collinear(FYW)
Conditions (CDL):
Equal(LengthOfLine(AW),4)
Equal(LengthOfLine(FW),12)
Equal(LengthOfLine(RW),4)
Equal(LengthOfLine(WY),8)
Goal: Value(LengthOfLine(FY))
Answer: 4
Theorem sequence:
line_addition(1,FY,YW)
Problem: As shown in the diagram, DB=2*x-15, DC=x+7, ∠ADC=4*x-6°, AD is the altitude of △ ABC, BD⊥AD. Find the value of x.
问题：如图所示，DB=2*x-15，DC=x+7，∠ADC=4*x-6°，AD是三角形ABC的高，BD垂直于AD。求x的值。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AB,BD,DA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(DB),2*x-15)
Equal(LengthOfLine(DC),x+7)
Equal(MeasureOfAngle(ADC),4*x-6)
IsAltitudeOfTriangle(AD,ABC)
PerpendicularBetweenLine(BD,AD)
Goal: Value(x)
Answer: 24
Theorem sequence:
Problem: As shown in the diagram, CD=22, the radius of circle A is 14, the center of ⊙A is A, AE is perpendicular to DE. Find the length of line EB.
问题：如图所示，CD=22，⊙A的半径为14，A是⊙A的圆心，AE垂直于DE。求直线EB的长度。
Construction (CDL):
Shape(EC,ACB,BE)
Shape(AE,ED,DA)
Shape(EB,ABD,DE)
Shape(CE,EA,AD,ADC)
Collinear(CED)
Collinear(AEB)
Cocircular(A,CBD)
Conditions (CDL):
Equal(LengthOfLine(CD),22)
Equal(RadiusOfCircle(A),14)
IsCentreOfCircle(A,A)
PerpendicularBetweenLine(AE,DE)
Goal: Value(LengthOfLine(EB))
Answer: 14-5*sqrt(3)
Theorem sequence:
radius_of_circle_property_length_equal(1,AB,A)
radius_of_circle_property_length_equal(1,AD,A)
adjacent_complementary_angle(1,CEA,AED)
circle_property_chord_perpendicular_bisect_chord(1,A,AE,CD)
line_addition(1,CE,ED)
line_addition(1,AE,EB)
right_triangle_judgment_angle(1,AED)
right_triangle_property_pythagorean(1,AED)
Problem: As shown in the diagram, ∠ABD=3*x-60°, ∠ECF=2*x+15°, BD is parallel to CE. Find the measure of ∠ABD.
问题：如图所示，∠ABD=3*x-60°，∠ECF=2*x+15°，BD平行于CE。求∠ABD的大小。
Construction (CDL):
Shape(AB,BD)
Shape(DB,BC)
Shape(BC,CE)
Shape(EC,CF)
Collinear(ABCF)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),3*x-60)
Equal(MeasureOfAngle(ECF),2*x+15)
ParallelBetweenLine(BD,CE)
Goal: Value(MeasureOfAngle(ABD))
Answer: 75
Theorem sequence:
parallel_property_corresponding_angle(2,BD,CE,F)
adjacent_complementary_angle(1,ABD,DBC)
Problem: As shown in the diagram, the area of NQPO is 250, ON=20, QP=30, AH is the height of NQPO, NQ and PO are the non-parallel sides (legs) of trapezoid NQPO. Find the height of NQPO.
问题：如图所示，NQPO的面积为250，ON=20，QP=30，四边形NQPO的高为AH，NQPO是梯形。求四边形NQPO的高。
Construction (CDL):
Shape(NQ,QH,HA,AN)
Shape(AH,HP,PO,OA)
Collinear(NAO)
Collinear(QHP)
Conditions (CDL):
Equal(AreaOfQuadrilateral(NQPO),250)
Equal(LengthOfLine(ON),20)
Equal(LengthOfLine(QP),30)
IsAltitudeOfQuadrilateral(AH,NQPO)
Trapezoid(NQPO)
Goal: Value(HeightOfQuadrilateral(NQPO))
Answer: 10
Theorem sequence:
trapezoid_area_formula(1,NQPO)
Problem: As shown in the diagram, ∠ABD=3*x-60°, ∠DBC=y°, ∠ECF=2*x+15°, BD∥CE. Find the value of x.
问题：如图所示，∠ABD=3*x-60°，∠DBC=y°，∠ECF=2*x+15°，BD平行于CE。求x的值。
Construction (CDL):
Shape(AB,BD)
Shape(DB,BC)
Shape(BC,CE)
Shape(EC,CF)
Collinear(ABCF)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),3*x-60)
Equal(MeasureOfAngle(DBC),y)
Equal(MeasureOfAngle(ECF),2*x+15)
ParallelBetweenLine(BD,CE)
Goal: Value(x)
Answer: 45
Theorem sequence:
parallel_property_corresponding_angle(2,BD,CE,F)
adjacent_complementary_angle(1,ABD,DBC)
Problem: As shown in the diagram, ∠AMD=56°, ∠KDJ=4*x°, DJ is parallel to MB. Find the value of x.
问题：如图所示，∠AMD=56°，∠KDJ=4*x°，DJ∥MB。求x的值。
Construction (CDL):
Shape(PD,DK)
Shape(KD,DJ)
Shape(MD,DP)
Shape(JD,DM)
Shape(AM,MD)
Shape(DM,MB)
Collinear(KDM)
Collinear(PDJ)
Collinear(AMB)
Conditions (CDL):
Equal(MeasureOfAngle(AMD),56)
Equal(MeasureOfAngle(KDJ),4*x)
ParallelBetweenLine(DJ,MB)
Goal: Value(x)
Answer: 31
Theorem sequence:
parallel_property_collinear_extend(1,DJ,MB,P)
parallel_property_collinear_extend(2,BM,DP,A)
parallel_property_corresponding_angle(2,MA,DP,K)
adjacent_complementary_angle(1,PDK,KDJ)
Problem: As shown in the diagram, JL=2*x+5, KM=4*x-11, quadrilateral JKLM is a rectangle. Find the value of x.
问题：如图所示，JL=2*x+5，KM=4*x-11，JKLM是矩形。求x的值。
Construction (CDL):
Shape(MJ,JN,NM)
Shape(NJ,JK,KN)
Shape(NK,KL,LN)
Shape(MN,NL,LM)
Collinear(MNK)
Collinear(JNL)
Conditions (CDL):
Equal(LengthOfLine(JL),2*x+5)
Equal(LengthOfLine(KM),4*x-11)
Rectangle(JKLM)
Goal: Value(x)
Answer: 8
Theorem sequence:
rectangle_property_diagonal_equal(1,JKLM)
Problem: As shown in the diagram, AD=3*x+5, CD=5*x-1, BD is the median of △ BAC. Find the length of line AC.
问题：如图所示，AD=3*x+5，CD=5*x-1，BD是△BAC的中线。求直线AC的长度。
Construction (CDL):
Shape(BA,AD,DB)
Shape(BD,DC,CB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AD),3*x+5)
Equal(LengthOfLine(CD),5*x-1)
IsMedianOfTriangle(BD,BAC)
Goal: Value(LengthOfLine(AC))
Answer: 28
Theorem sequence:
line_addition(1,AD,DC)
Problem: As shown in the diagram, ∠EGB=x+55°, ∠GEA=2*x-15°, AH∥CB. Find the value of x.
问题：如图所示，∠EGB=x+55°，∠GEA=2*x-15°，AH∥CB。求x的值。
Construction (CDL):
Shape(AE,EI)
Shape(IE,EH)
Shape(GE,EA)
Shape(HE,EG)
Shape(CG,GE)
Shape(EG,GB)
Collinear(IEG)
Collinear(AEH)
Collinear(CGB)
Conditions (CDL):
Equal(MeasureOfAngle(EGB),x+55)
Equal(MeasureOfAngle(GEA),2*x-15)
ParallelBetweenLine(AH,CB)
Goal: Value(x)
Answer: 70
Theorem sequence:
parallel_property_collinear_extend(3,BC,HA,G)
parallel_property_collinear_extend(3,AH,GB,E)
parallel_property_alternate_interior_angle(2,AE,GB)
Problem: As shown in the diagram, LN=NM, ML=17, NL=3*x-4, NM=2*x+7. Find the length of line LN.
问题：如图所示，LN=NM，ML=17，NL=3*x-4，NM=2*x+7。求直线LN的长度。
Construction (CDL):
Shape(LN,NM,ML)
Conditions (CDL):
Equal(LengthOfLine(LN),LengthOfLine(NM))
Equal(LengthOfLine(ML),17)
Equal(LengthOfLine(NL),3*x-4)
Equal(LengthOfLine(NM),2*x+7)
Goal: Value(LengthOfLine(LN))
Answer: 29
Theorem sequence:
Problem: As shown in the diagram, GR=7, PG=z+4, PT=3*y-1, TQ=2*x, TR=8, TF perpendicular bisects RQ, TG perpendicular bisects PR, TH is the perpendicular bisector of QP. Find the value of y.
问题：如图所示，GR=7，PG=z+4，PT=3*y-1，TQ=2*x，TR=8，TF垂直平分AB，TG是PR的垂直平分线，TH垂直平分AB。求y的值。
Construction (CDL):
Shape(QH,HT,TQ)
Shape(HP,PT,TH)
Shape(TP,PG,GT)
Shape(TG,GR,RT)
Shape(TR,RF,FT)
Shape(TF,FQ,QT)
Collinear(QHP)
Collinear(PGR)
Collinear(RFQ)
Conditions (CDL):
Equal(LengthOfLine(GR),7)
Equal(LengthOfLine(PG),z+4)
Equal(LengthOfLine(PT),3*y-1)
Equal(LengthOfLine(TQ),2*x)
Equal(LengthOfLine(TR),8)
IsPerpendicularBisectorOfLine(TF,RQ)
IsPerpendicularBisectorOfLine(TG,PR)
IsPerpendicularBisectorOfLine(TH,QP)
Goal: Value(y)
Answer: 3
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,TG,PR)
Problem: As shown in the diagram, ∠EAB=40°, AB is perpendicular to DB, quadrilateral ABDC is a rectangle. Find the measure of ∠EBD.
问题：如图所示，∠EAB=40°，AB垂直于DB，ABDC是矩形。求∠EBD的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BD,DE)
Shape(ED,DC,CE)
Shape(AE,EC,CA)
Collinear(BEC)
Collinear(AED)
Conditions (CDL):
Equal(MeasureOfAngle(EAB),40)
PerpendicularBetweenLine(AB,DB)
Rectangle(ABDC)
Goal: Value(MeasureOfAngle(EBD))
Answer: 50
Theorem sequence:
rectangle_property_diagonal_equal(1,ABDC)
parallelogram_property_diagonal_bisection(1,ABDC,E)
parallelogram_property_diagonal_bisection(1,BDCA,E)
line_addition(1,BE,EC)
line_addition(1,AE,ED)
isosceles_triangle_judgment_line_equal(1,EAB)
isosceles_triangle_property_angle_equal(1,EAB)
angle_addition(1,ABE,EBD)
Problem: As shown in the diagram, ∠RTS=47°, RU is perpendicular to SU, TV⊥UV, VS⊥RS, TS and SR are the legs of the right triangle TSR. Find the measure of ∠SUV.
问题：如图所示，∠RTS=47°，RU垂直于SU，TV⊥UV，VS⊥RS，∠TSR是直角△TSR的直角。求∠SUV的大小。
Construction (CDL):
Shape(SR,RU,US)
Shape(SU,UV,VS)
Shape(UT,TV,VU)
Collinear(RUT)
Collinear(SVT)
Conditions (CDL):
Equal(MeasureOfAngle(RTS),47)
PerpendicularBetweenLine(RU,SU)
PerpendicularBetweenLine(TV,UV)
PerpendicularBetweenLine(VS,RS)
RightTriangle(TSR)
Goal: Value(MeasureOfAngle(SUV))
Answer: 47
Theorem sequence:
adjacent_complementary_angle(1,RUS,SUT)
triangle_property_angle_sum(1,UTV)
angle_addition(1,SUV,VUT)
Problem: As shown in the diagram, CB=2*x+7, YA=x+9, ∠BYA=2*y+21°, ∠CBY=2*y-5°, BC and YA are opposite sides of the parallelogram BYAC. Find the value of x.
问题：如图所示，CB=2*x+7，YA=x+9，∠BYA=2*y+21°，∠CBY=2*y-5°，YB和AC是平行四边形BYAC的一组对边。求x的值。
Construction (CDL):
Shape(BY,YA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(CB),2*x+7)
Equal(LengthOfLine(YA),x+9)
Equal(MeasureOfAngle(BYA),2*y+21)
Equal(MeasureOfAngle(CBY),2*y-5)
Parallelogram(BYAC)
Goal: Value(x)
Answer: 2
Theorem sequence:
parallelogram_property_opposite_line_equal(1,YACB)
Problem: As shown in the diagram, PS=x, QP=2*x, RQ=3*y+8, RS=2*y, △QPS is mirror congruent to △QSR, QS⊥RS. Find the value of y.
问题：如图所示，PS=x，QP=2*x，RQ=3*y+8，RS=2*y，△QPS镜像全等于△QSR，QS垂直于RS。求y的值。
Construction (CDL):
Shape(QP,PS,SQ)
Shape(QS,SR,RQ)
Collinear(PSR)
Conditions (CDL):
Equal(LengthOfLine(PS),x)
Equal(LengthOfLine(QP),2*x)
Equal(LengthOfLine(RQ),3*y+8)
Equal(LengthOfLine(RS),2*y)
MirrorCongruentBetweenTriangle(QPS,QSR)
PerpendicularBetweenLine(QS,RS)
Goal: Value(y)
Answer: 8
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,QPS,QSR)
mirror_congruent_triangle_property_line_equal(1,SQP,SRQ)
Problem: As shown in the diagram, AB=5, AC=x, BC=6, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=5，AC=x，BC=6，BA⊥CA。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),6)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: sqrt(11)
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, ∠FGC=6*x-144°, ∠NEH=2*x°, FT is parallel to IN. Find the value of x.
问题：如图所示，∠FGC=6*x-144°，∠NEH=2*x°，FT平行于IN。求x的值。
Construction (CDL):
Shape(CG,GT)
Shape(TG,GE)
Shape(GE,EN)
Shape(NE,EH)
Shape(FG,GC)
Shape(EG,GF)
Shape(IE,EG)
Shape(HE,EI)
Collinear(CGEH)
Collinear(FGT)
Collinear(IEN)
Conditions (CDL):
Equal(MeasureOfAngle(FGC),6*x-144)
Equal(MeasureOfAngle(NEH),2*x)
ParallelBetweenLine(FT,IN)
Goal: Value(x)
Answer: 36
Theorem sequence:
parallel_property_collinear_extend(3,NI,TF,E)
vertical_angle(1,FGC,TGE)
parallel_property_collinear_extend(3,FT,EN,G)
parallel_property_corresponding_angle(2,GT,EN,H)
Problem: As shown in the diagram, RS=4*x-4, RS=RT, RT=3*x+3, TS=12+x. Find the length of line SR.
问题：如图所示，RS=4*x-4，RS=RT，RT=3*x+3，TS=12+x。求直线SR的长度。
Construction (CDL):
Shape(RT,TS,SR)
Conditions (CDL):
Equal(LengthOfLine(RS),4*x-4)
Equal(LengthOfLine(RS),LengthOfLine(RT))
Equal(LengthOfLine(RT),3*x+3)
Equal(LengthOfLine(TS),12+x)
Goal: Value(LengthOfLine(SR))
Answer: 24
Theorem sequence:
Problem: As shown in the diagram, the area of △WYX is 248, AX=16, XA is perpendicular to YA. Find the length of line WY.
问题：如图所示，△WYX的面积为248，AX=16，XA垂直于YA。求直线WY的长度。
Construction (CDL):
Shape(XW,WA,AX)
Shape(XA,AY,YX)
Collinear(WAY)
Conditions (CDL):
Equal(AreaOfTriangle(WYX),248)
Equal(LengthOfLine(AX),16)
PerpendicularBetweenLine(XA,YA)
Goal: Value(LengthOfLine(WY))
Answer: 31
Theorem sequence:
adjacent_complementary_angle(1,WAX,XAY)
altitude_of_triangle_judgment(1,XA,XWY)
triangle_area_formula_common(1,XWY)
Problem: As shown in the diagram, AC=60, BC=x, ∠ACB=62°, CB⊥AB. Find the value of x.
问题：如图所示，AC=60，BC=x，∠ACB=62°，CB垂直于AB。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),60)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ACB),62)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 60*sin(7*pi/45)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,CBA)
Problem: As shown in the diagram, DG=6*y-42, ED=6*x-12, FE=4*y, GF=2*x+36, EDGF is a parallelogram. Find the value of x.
问题：如图所示，DG=6*y-42，ED=6*x-12，FE=4*y，GF=2*x+36，EF和DG是▱EDGF的一组对边。求x的值。
Construction (CDL):
Shape(ED,DG,GF,FE)
Conditions (CDL):
Equal(LengthOfLine(DG),6*y-42)
Equal(LengthOfLine(ED),6*x-12)
Equal(LengthOfLine(FE),4*y)
Equal(LengthOfLine(GF),2*x+36)
Parallelogram(EDGF)
Goal: Value(x)
Answer: 12
Theorem sequence:
parallelogram_property_opposite_line_equal(1,EDGF)
Problem: As shown in the diagram, ∠ADC=2*y+16°, ∠BAD=58°, ∠CBA=3*y+4°, ∠DCB=2*x°. Find the measure of ∠DCB.
问题：如图所示，∠ADC=2*y+16°，∠BAD=58°，∠CBA=3*y+4°，∠DCB=2*x°。求∠DCB的大小。
Construction (CDL):
Shape(OBA,AB)
Shape(OAD,DA)
Shape(ODC,CD)
Shape(OCB,BC)
Shape(BA,AD,DC,CB)
Cocircular(O,ADCB)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),2*y+16)
Equal(MeasureOfAngle(BAD),58)
Equal(MeasureOfAngle(CBA),3*y+4)
Equal(MeasureOfAngle(DCB),2*x)
Goal: Value(MeasureOfAngle(DCB))
Answer: 122
Theorem sequence:
arc_property_circumference_angle_external(1,ODB,A)
arc_property_circumference_angle_internal(1,ODB,C)
Problem: As shown in the diagram, AB=x, AC=21, ∠ABC=37°, ∠BCA=87°. Find the value of x.
问题：如图所示，AB=x，AC=21，∠ABC=37°，∠BCA=87°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),21)
Equal(MeasureOfAngle(ABC),37)
Equal(MeasureOfAngle(BCA),87)
Goal: Value(x)
Answer: (-21*sqrt(10)-21*sqrt(6)+21*sqrt(2)+21*sqrt(5-sqrt(5))+21*sqrt(3)*sqrt(5-sqrt(5))+21*sqrt(5)*sqrt(5-sqrt(5))+21*sqrt(30)+21*sqrt(15)*sqrt(5-sqrt(5)))/(16*sin(37*pi/180))
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, ∠LBQ=4*x+61°, ∠MBL=3*x°. Find the measure of ∠QBP.
问题：如图所示，∠LBQ=4*x+61°，∠MBL=3*x°。求∠QBP的大小。
Construction (CDL):
Shape(MB,BL,BLM)
Shape(BM,BMP,PB)
Shape(BP,BPQ,QB)
Shape(BQ,BQK,KB)
Shape(LB,BK,BKL)
Collinear(MBQ)
Collinear(LBP)
Cocircular(B,LMPQK)
Conditions (CDL):
Equal(MeasureOfAngle(LBQ),4*x+61)
Equal(MeasureOfAngle(MBL),3*x)
Goal: Value(MeasureOfAngle(QBP))
Answer: 51
Theorem sequence:
adjacent_complementary_angle(1,MBL,LBQ)
adjacent_complementary_angle(1,PBM,MBL)
adjacent_complementary_angle(1,QBP,PBM)
Problem: As shown in the diagram, JM=1, KJ=2, KM=2, ∠KJM=x°. Find the value of x.
问题：如图所示，JM=1，KJ=2，KM=2，∠KJM=x°。求x的值。
Construction (CDL):
Shape(JM,MK,KJ)
Conditions (CDL):
Equal(LengthOfLine(JM),1)
Equal(LengthOfLine(KJ),2)
Equal(LengthOfLine(KM),2)
Equal(MeasureOfAngle(KJM),x)
Goal: Value(x)
Answer: 180*acos(1/4)/pi
Theorem sequence:
cosine_theorem(1,JMK)