SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, DG=9, DK=16, KF=12, the centroid of triangle CDF is K. Find the length of line CD.
问题：如图所示，DG=9，DK=16，KF=12，三角形CDF的重心为O。求直线CD的长度。
Construction (CDL):
Shape(CG,GK,KC)
Shape(GD,DK,KG)
Shape(KD,DJ,JK)
Shape(KJ,JF,FK)
Shape(KF,FH,HK)
Shape(CK,KH,HC)
Collinear(CGD)
Collinear(DJF)
Collinear(FHC)
Collinear(GKF)
Collinear(JKC)
Collinear(DKH)
Conditions (CDL):
Equal(LengthOfLine(DG),9)
Equal(LengthOfLine(DK),16)
Equal(LengthOfLine(KF),12)
IsCentroidOfTriangle(K,CDF)
Goal: Value(LengthOfLine(CD))
Answer: 18
Theorem sequence:
centroid_of_triangle_property_intersection(1,K,FCD,G)
line_addition(1,CG,GD)
Problem: As shown in the diagram, AB=z, AN=x, AY=5, YB=14, YN=y, BA is perpendicular to YA, BN is perpendicular to AN. Find the value of z.
问题：如图所示，AB=z，AN=x，AY=5，YB=14，YN=y，BA⊥YA，BN垂直于AN。求z的值。
Construction (CDL):
Shape(AY,YN,NA)
Shape(AN,NB,BA)
Collinear(YNB)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AN),x)
Equal(LengthOfLine(AY),5)
Equal(LengthOfLine(YB),14)
Equal(LengthOfLine(YN),y)
PerpendicularBetweenLine(BA,YA)
PerpendicularBetweenLine(BN,AN)
Goal: Value(z)
Answer: 3*sqrt(19)
Theorem sequence:
right_triangle_judgment_angle(1,BAY)
right_triangle_property_pythagorean(1,BAY)
Problem: As shown in the diagram, the area of the circle O is 18*pi, the center of ⊙O is O, quadrilateral CEDB is a square. Find the length of line CE.
问题：如图所示，⊙O的面积为18*pi，O是⊙O的圆心，CEDB是正方形。求直线CE的长度。
Construction (CDL):
Shape(CE,EO,OC)
Shape(OE,ED,DO)
Shape(OD,DB,BO)
Shape(CO,OB,BC)
Shape(OCE,EC)
Shape(OED,DE)
Shape(ODB,BD)
Shape(OBC,CB)
Collinear(COD)
Collinear(EOB)
Cocircular(O,CEDB)
Conditions (CDL):
Equal(AreaOfCircle(O),18*pi)
IsCentreOfCircle(O,O)
Square(CEDB)
Goal: Value(LengthOfLine(CE))
Answer: 6
Theorem sequence:
circle_area_formula(1,O)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OE,O)
kite_property_diagonal_perpendicular_bisection(1,CEDB,O)
right_triangle_judgment_angle(1,EOC)
right_triangle_property_pythagorean(1,EOC)
Problem: As shown in the diagram, GH=10, G is the center of circle G, J is the center of ⊙J, K is the center of circle K. Find the length of line JL.
问题：如图所示，GH=10，圆G的圆心为G，⊙J的圆心为J，⊙K的圆心为K。求直线JL的长度。
Construction (CDL):
Shape(GHB,BG,GH)
Shape(GBF,FG,GB)
Shape(GF,GFl,JGL)
Shape(JG,JGL,KJL)
Shape(KJ,KJL,LK)
Shape(KL,KLJ,JK)
Shape(JLG,GJ,KLJ)
Shape(GLH,HG,JLG)
Collinear(BGJKL)
Collinear(HGF)
Cocircular(G,HBFL)
Cocircular(J,GL)
Cocircular(K,DJICL)
Conditions (CDL):
Equal(LengthOfLine(GH),10)
IsCentreOfCircle(G,G)
IsCentreOfCircle(J,J)
IsCentreOfCircle(K,K)
Goal: Value(LengthOfLine(JL))
Answer: 5
Theorem sequence:
radius_of_circle_property_length_equal(1,GH,G)
radius_of_circle_property_length_equal(1,GL,G)
radius_of_circle_property_length_equal(1,JG,J)
radius_of_circle_property_length_equal(1,JL,J)
line_addition(1,BG,GJ)
line_addition(1,BG,GL)
line_addition(1,BJ,JL)
Problem: As shown in the diagram, AZ=2*b-6, WA=10, XA=a-7, YA=2, quadrilateral WYZX is a ▱. Find the value of a.
问题：如图所示，AZ=2*b-6，WA=10，XA=a-7，YA=2，WX和YZ是▱WYZX的一组对边。求a的值。
Construction (CDL):
Shape(WY,YA,AW)
Shape(AY,YZ,ZA)
Shape(AZ,ZX,XA)
Shape(WA,AX,XW)
Collinear(WAZ)
Collinear(YAX)
Conditions (CDL):
Equal(LengthOfLine(AZ),2*b-6)
Equal(LengthOfLine(WA),10)
Equal(LengthOfLine(XA),a-7)
Equal(LengthOfLine(YA),2)
Parallelogram(WYZX)
Goal: Value(a)
Answer: 9
Theorem sequence:
parallelogram_property_diagonal_bisection(1,YZXW,A)
Problem: As shown in the diagram, KN=9, LN=16, PM=Mul(LengthOfLine(KP),2), PR is parallel to KL, KN is perpendicular to QN, RM is perpendicular to PM. Find the length of line PR.
问题：如图所示，KN=9，LN=16，PM=Mul(LengthOfLine(KP),2)，PR平行于KL，KN垂直于QN，RM垂直于PM。求直线PR的长度。
Construction (CDL):
Shape(LR,RQ,QN,NL)
Shape(NQ,QP,PK,KN)
Shape(RM,MQ,QR)
Shape(QM,MP,PQ)
Collinear(LNK)
Collinear(RQP)
Collinear(LRM)
Collinear(KPM)
Conditions (CDL):
Equal(LengthOfLine(KN),9)
Equal(LengthOfLine(LN),16)
Equal(LengthOfLine(PM),Mul(LengthOfLine(KP),2))
ParallelBetweenLine(PR,KL)
PerpendicularBetweenLine(KN,QN)
PerpendicularBetweenLine(RM,PM)
Goal: Value(LengthOfLine(PR))
Answer: 50/3
Theorem sequence:
line_addition(1,LN,NK)
parallel_property_ipsilateral_internal_angle(1,LK,RP)
adjacent_complementary_angle(1,LRQ,QRM)
line_addition(1,KP,PM)
similar_triangle_judgment_aa(1,PRM,KLM)
similar_triangle_property_line_ratio(1,RMP,LMK)
similar_triangle_property_line_ratio(1,MPR,MKL)
Problem: As shown in the diagram, the area of the sector KLJ is 66, JL=x, ∠JKL=180°, the center of ⊙K is K. Find the value of x.
问题：如图所示，扇形KLJ的面积为66，JL=x，∠JKL=180°，圆K的圆心为K。求x的值。
Construction (CDL):
Shape(KJL,LK,KJ)
Shape(KLJ,JK,KL)
Collinear(JKL)
Cocircular(K,JL)
Conditions (CDL):
Equal(AreaOfSector(KLJ),66)
Equal(LengthOfLine(JL),x)
Equal(MeasureOfAngle(JKL),180)
IsCentreOfCircle(K,K)
Goal: Value(x)
Answer: 4*sqrt(33)/sqrt(pi)
Theorem sequence:
arc_property_center_angle(1,KLJ,K)
sector_area_formula(1,KLJ)
radius_of_circle_property_length_equal(1,KJ,K)
radius_of_circle_property_length_equal(1,KL,K)
line_addition(1,JK,KL)
Problem: As shown in the diagram, AD=41, BC=53, DC=48, ∠ADC and ∠DCB are the right angles of right trapezoid DCBA. Find the area of quadrilateral DCBA.
问题：如图所示，AD=41，BC=53，DC=48，四边形DCBA是直角梯形。求四边形DCBA的面积。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AD),41)
Equal(LengthOfLine(BC),53)
Equal(LengthOfLine(DC),48)
RightTrapezoid(DCBA)
Goal: Value(AreaOfQuadrilateral(DCBA))
Answer: 2256
Theorem sequence:
right_trapezoid_area_formular(1,DCBA)
Problem: As shown in the diagram, BA=15, BC=20, CA=25, CB⊥AB. Find the value of cos(BA).
问题：如图所示，BA=15，BC=20，CA=25，CB垂直于AB。求cos(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(BA),15)
Equal(LengthOfLine(BC),20)
Equal(LengthOfLine(CA),25)
PerpendicularBetweenLine(CB,AB)
Goal: Value(Cos(MeasureOfAngle(BAC)))
Answer: 3/5
Theorem sequence:
cosine_theorem(1,ACB)
Problem: As shown in the diagram, AB=3*y-8, AD=13, CB=2*x+7, CD=10, ∠ADF=59°, ∠BFC=49°, ∠FBA=20°, DA and CB are opposite sides of the parallelogram ADCB. Find the measure of ∠CAD.
问题：如图所示，AB=3*y-8，AD=13，CB=2*x+7，CD=10，∠ADF=59°，∠BFC=49°，∠FBA=20°，DA和CB是平行四边形ADCB的一组对边。求∠CAD的大小。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(FC,CB,BF)
Shape(AF,FB,BA)
Collinear(AFC)
Collinear(BFD)
Conditions (CDL):
Equal(LengthOfLine(AB),3*y-8)
Equal(LengthOfLine(AD),13)
Equal(LengthOfLine(CB),2*x+7)
Equal(LengthOfLine(CD),10)
Equal(MeasureOfAngle(ADF),59)
Equal(MeasureOfAngle(BFC),49)
Equal(MeasureOfAngle(FBA),20)
Parallelogram(ADCB)
Goal: Value(MeasureOfAngle(CAD))
Answer: 72
Theorem sequence:
vertical_angle(1,DFA,BFC)
triangle_property_angle_sum(1,ADF)
Problem: As shown in the diagram, LN=3*x+5, QS=2*x+10, ∠LNM=y+6°, ∠RSQ=2*y-12°, triangle NLM is mirror congruent to triangle SRQ. Find the value of x.
问题：如图所示，LN=3*x+5，QS=2*x+10，∠LNM=y+6°，∠RSQ=2*y-12°，△NLM镜像全等于△SRQ。求x的值。
Construction (CDL):
Shape(NL,LM,MN)
Shape(SR,RQ,QS)
Conditions (CDL):
Equal(LengthOfLine(LN),3*x+5)
Equal(LengthOfLine(QS),2*x+10)
Equal(MeasureOfAngle(LNM),y+6)
Equal(MeasureOfAngle(RSQ),2*y-12)
MirrorCongruentBetweenTriangle(NLM,SRQ)
Goal: Value(x)
Answer: 5
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,MNL,RQS)
Problem: As shown in the diagram, AB=28, BC=9, ∠CBA=45°, CBAD is a ▱, CE⊥AE. Find the area of CBAD.
问题：如图所示，AB=28，BC=9，∠CBA=45°，BC和AD是▱CBAD的一组对边，CE⊥AE。求四边形CBAD的面积。
Construction (CDL):
Shape(CB,BE,EC)
Shape(CE,EA,AD,DC)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(AB),28)
Equal(LengthOfLine(BC),9)
Equal(MeasureOfAngle(CBA),45)
Parallelogram(CBAD)
PerpendicularBetweenLine(CE,AE)
Goal: Value(AreaOfQuadrilateral(CBAD))
Answer: 126*sqrt(2)
Theorem sequence:
parallelogram_area_formula_sine(1,CBAD)
Problem: As shown in the diagram, ∠BAD=58°, ∠DBE=26°, ∠EBA=47°. Find the measure of ∠ADB.
问题：如图所示，∠BAD=58°，∠DBE=26°，∠EBA=47°。求∠ADB的大小。
Construction (CDL):
Shape(BA,AE,EB)
Shape(BE,ED,DB)
Collinear(AED)
Conditions (CDL):
Equal(MeasureOfAngle(BAD),58)
Equal(MeasureOfAngle(DBE),26)
Equal(MeasureOfAngle(EBA),47)
Goal: Value(MeasureOfAngle(ADB))
Answer: 49
Theorem sequence:
angle_addition(1,DBE,EBA)
triangle_property_angle_sum(1,BAD)
Problem: As shown in the diagram, AB=y, AC=24, BC=x, ∠ABC=30°, BC⊥AC. Find the value of y.
问题：如图所示，AB=y，AC=24，BC=x，∠ABC=30°，BC⊥AC。求y的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),24)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 48
Theorem sequence:
sine_theorem(1,CAB)
sine_theorem(1,BCA)
Problem: As shown in the diagram, FN=1/4*x+6, IP=DP, NA=2*x-29, PD=16-5*y, PI=12-3*y, FI∥NP, NP is parallel to AD. Find the value of y.
问题：如图所示，FN=1/4*x+6，IP=DP，NA=2*x-29，PD=16-5*y，PI=12-3*y，FI∥NP，NP∥AD。求y的值。
Construction (CDL):
Shape(IP,PN,NF,FI)
Shape(PD,DA,AN,NP)
Collinear(BIFC)
Collinear(OPNH)
Collinear(GDAL)
Collinear(EIPDM)
Collinear(KFNAJ)
Conditions (CDL):
Equal(LengthOfLine(FN),1/4*x+6)
Equal(LengthOfLine(IP),LengthOfLine(DP))
Equal(LengthOfLine(NA),2*x-29)
Equal(LengthOfLine(PD),16-5*y)
Equal(LengthOfLine(PI),12-3*y)
ParallelBetweenLine(FI,NP)
ParallelBetweenLine(NP,AD)
Goal: Value(y)
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, ∠GHA=2*y+32°, ∠HBD=3*y-3°, AC is parallel to ED. Find the value of y.
问题：如图所示，∠GHA=2*y+32°，∠HBD=3*y-3°，AC平行于ED。求y的值。
Construction (CDL):
Shape(BH,HA)
Shape(HB,BD)
Collinear(CHA)
Collinear(DBE)
Collinear(FHBG)
Conditions (CDL):
Equal(MeasureOfAngle(GHA),2*y+32)
Equal(MeasureOfAngle(HBD),3*y-3)
ParallelBetweenLine(AC,ED)
Goal: Value(y)
Answer: 35
Theorem sequence:
parallel_property_collinear_extend(3,AC,ED,H)
parallel_property_collinear_extend(3,DE,HA,B)
parallel_property_alternate_interior_angle(2,DB,HA)
Problem: As shown in the diagram, EF=4*x-6, HG=x+3, EH⊥GH, quadrilateral EHGF is a rectangle. Find the length of line EF.
问题：如图所示，EF=4*x-6，HG=x+3，EH⊥GH，四边形EHGF是长方形。求直线EF的长度。
Construction (CDL):
Shape(EH,HG,GF,FE)
Conditions (CDL):
Equal(LengthOfLine(EF),4*x-6)
Equal(LengthOfLine(HG),x+3)
PerpendicularBetweenLine(EH,GH)
Rectangle(EHGF)
Goal: Value(LengthOfLine(EF))
Answer: 6
Theorem sequence:
parallelogram_property_opposite_line_equal(1,HGFE)
Problem: As shown in the diagram, ∠ACB=52°, ∠BAC=x°, ∠CBA=35°. Find the value of x.
问题：如图所示，∠ACB=52°，∠BAC=x°，∠CBA=35°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),52)
Equal(MeasureOfAngle(BAC),x)
Equal(MeasureOfAngle(CBA),35)
Goal: Value(x)
Answer: 93
Theorem sequence:
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, AD=4, CD=CB, AC is perpendicular to DC. Find the length of line AB.
问题：如图所示，AD=4，CD=CB，AC⊥DC。求直线AB的长度。
Construction (CDL):
Shape(BC,CA,AB)
Shape(AC,CD,DA)
Collinear(FCAE)
Collinear(BCD)
Conditions (CDL):
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(CD),LengthOfLine(CB))
PerpendicularBetweenLine(AC,DC)
Goal: Value(LengthOfLine(AB))
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,BCA,ACD)
mirror_congruent_triangle_judgment_sas(1,CAB,CDA)
mirror_congruent_triangle_property_line_equal(1,CAB,CDA)
Problem: As shown in the diagram, AB=48, CA=50, CB=14, BC is perpendicular to AC. Find the value of cos(CA).
问题：如图所示，AB=48，CA=50，CB=14，BC⊥AC。求cos(CA)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),48)
Equal(LengthOfLine(CA),50)
Equal(LengthOfLine(CB),14)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Cos(MeasureOfAngle(CAB)))
Answer: 24/25
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, AB=h, BC=4, ∠ACB=90°, ∠CBA=60°, AC is perpendicular to BC. Find the value of h.
问题：如图所示，AB=h，BC=4，∠ACB=90°，∠CBA=60°，AC垂直于BC。求h的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),h)
Equal(LengthOfLine(BC),4)
Equal(MeasureOfAngle(ACB),90)
Equal(MeasureOfAngle(CBA),60)
PerpendicularBetweenLine(AC,BC)
Goal: Value(h)
Answer: 8
Theorem sequence:
sine_theorem(1,BAC)
triangle_property_angle_sum(1,BAC)
Problem: As shown in the diagram, AB=y, AC=5, CD=x, ∠ADB=45°, AB is perpendicular to CB, CA is perpendicular to DA. Find the value of y.
问题：如图所示，AB=y，AC=5，CD=x，∠ADB=45°，AB⊥CB，CA垂直于DA。求y的值。
Construction (CDL):
Shape(AD,DB,BA)
Shape(BC,CA,AB)
Collinear(DBC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(CD),x)
Equal(MeasureOfAngle(ADB),45)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(CA,DA)
Goal: Value(y)
Answer: 5*sqrt(2)/2
Theorem sequence:
triangle_property_angle_sum(1,ADC)
triangle_property_angle_sum(1,ABC)
isosceles_triangle_judgment_angle_equal(1,BCA)
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,ABC)
Problem: As shown in the diagram, BA=x, BC=30, CD=18, the center of ⊙A is A, CB is the tangent to ⊙A. Find the value of x.
问题：如图所示，BA=x，BC=30，CD=18，圆A的圆心为A，CB是圆O的切线。求x的值。
Construction (CDL):
Shape(AB,ABD,DA)
Shape(ADB,BA,AD)
Shape(CB,ADB,DC)
Collinear(ADC)
Cocircular(A,DB)
Conditions (CDL):
Equal(LengthOfLine(BA),x)
Equal(LengthOfLine(BC),30)
Equal(LengthOfLine(CD),18)
IsCentreOfCircle(A,A)
IsTangentOfCircle(CB,A)
Goal: Value(x)
Answer: 16
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CB,A,A)
radius_of_circle_property_length_equal(1,AD,A)
radius_of_circle_property_length_equal(1,AB,A)
right_triangle_judgment_angle(1,CBA)
line_addition(1,AD,DC)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, AG=12, HA=15, HG and HJ are one pair of adjacent sides of the kite FGHJ. Find the length of line GH.
问题：如图所示，AG=12，HA=15，四边形FGHJ是风筝形。求直线GH的长度。
Construction (CDL):
Shape(FG,GA,AF)
Shape(FA,AJ,JF)
Shape(AG,GH,HA)
Shape(AH,HJ,JA)
Collinear(GAJ)
Collinear(FAH)
Conditions (CDL):
Equal(LengthOfLine(AG),12)
Equal(LengthOfLine(HA),15)
Kite(FGHJ)
Goal: Value(LengthOfLine(GH))
Answer: 3*sqrt(41)
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,HJFG,A)
right_triangle_judgment_angle(1,HAG)
right_triangle_property_pythagorean(1,HAG)
Problem: As shown in the diagram, ∠BDC=68°, ∠CBD=3*x-15°, ∠DCB=y*y°, ∠EAC=2*x°, DB is parallel to AE. Find the value of x.
问题：如图所示，∠BDC=68°，∠CBD=3*x-15°，∠DCB=y*y°，∠EAC=2*x°，DB平行于AE。求x的值。
Construction (CDL):
Shape(BD,DA,AE,EB)
Shape(EA,AC,CE)
Collinear(BEC)
Collinear(DAC)
Conditions (CDL):
Equal(MeasureOfAngle(BDC),68)
Equal(MeasureOfAngle(CBD),3*x-15)
Equal(MeasureOfAngle(DCB),y*y)
Equal(MeasureOfAngle(EAC),2*x)
ParallelBetweenLine(DB,AE)
Goal: Value(x)
Answer: 34
Theorem sequence:
parallel_property_corresponding_angle(2,DB,AE,C)
Problem: As shown in the diagram, AD=12, DC=10, ∠DCB=60°, DA and CB are opposite sides of the parallelogram DCBA. Find the perimeter of DCBA.
问题：如图所示，AD=12，DC=10，∠DCB=60°，CD和BA是平行四边形DCBA的一组对边。求DCBA的周长。
Construction (CDL):
Shape(DC,CB,BE,ED)
Shape(AE,EB,BA)
Collinear(DEA)
Conditions (CDL):
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DC),10)
Equal(MeasureOfAngle(DCB),60)
Parallelogram(DCBA)
Goal: Value(PerimeterOfQuadrilateral(DCBA))
Answer: 44
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DCBA)
parallelogram_property_opposite_line_equal(1,CBAD)
quadrilateral_perimeter_formula(1,DCBA)
Problem: As shown in the diagram, AD=12, BD=24, CD=t+1, BD⊥AD, CA⊥BA. Find the value of t.
问题：如图所示，AD=12，BD=24，CD=t+1，BD⊥AD，CA垂直于BA。求t的值。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AD,DC,CA)
Shape(AB,BD,DA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(BD),24)
Equal(LengthOfLine(CD),t+1)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(t)
Answer: 5
Theorem sequence:
line_addition(1,BD,DC)
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,CAB)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,ADC)
right_triangle_property_pythagorean(1,CAB)
right_triangle_property_pythagorean(1,ADC)
right_triangle_property_pythagorean(1,BDA)
Problem: As shown in the diagram, ∠WZY=4*x°, ∠XWZ=3*x°, ∠YXW=x°, ∠ZYX=2*x°. Find the measure of ∠XWZ.
问题：如图所示，∠WZY=4*x°，∠XWZ=3*x°，∠YXW=x°，∠ZYX=2*x°。求∠XWZ的大小。
Construction (CDL):
Shape(XW,WZ,ZY,YX)
Conditions (CDL):
Equal(MeasureOfAngle(WZY),4*x)
Equal(MeasureOfAngle(XWZ),3*x)
Equal(MeasureOfAngle(YXW),x)
Equal(MeasureOfAngle(ZYX),2*x)
Goal: Value(MeasureOfAngle(XWZ))
Answer: 108
Theorem sequence:
quadrilateral_property_angle_sum(1,XWZY)
Problem: As shown in the diagram, ∠EAG=136°, ∠EBG=47°. Find the measure of ∠AEB.
问题：如图所示，∠EAG=136°，∠EBG=47°。求∠AEB的大小。
Construction (CDL):
Shape(EB,BA,AE)
Shape(EA,AG)
Collinear(BAG)
Conditions (CDL):
Equal(MeasureOfAngle(EAG),136)
Equal(MeasureOfAngle(EBG),47)
Goal: Value(MeasureOfAngle(AEB))
Answer: 89
Theorem sequence:
flat_angle(1,BAG)
angle_addition(1,BAE,EAG)
triangle_property_angle_sum(1,EBA)
Problem: As shown in the diagram, MN=39, MP=36, NP=15, MP is perpendicular to NP. Find the value of cos(PN).
问题：如图所示，MN=39，MP=36，NP=15，MP垂直于NP。求cos(PN)的值。
Construction (CDL):
Shape(MP,PN,NM)
Conditions (CDL):
Equal(LengthOfLine(MN),39)
Equal(LengthOfLine(MP),36)
Equal(LengthOfLine(NP),15)
PerpendicularBetweenLine(MP,NP)
Goal: Value(Cos(MeasureOfAngle(PNM)))
Answer: 5/13
Theorem sequence:
cosine_theorem(1,NMP)
Problem: As shown in the diagram, BC=18, DC=12, ∠BAD=115°, quadrilateral ADCB is a parallelogram. Find the length of line AD.
问题：如图所示，BC=18，DC=12，∠BAD=115°，四边形ADCB是平行四边形。求直线AD的长度。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(BC),18)
Equal(LengthOfLine(DC),12)
Equal(MeasureOfAngle(BAD),115)
Parallelogram(ADCB)
Goal: Value(LengthOfLine(AD))
Answer: 18
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ADCB)
Problem: As shown in the diagram, ∠FEL=55°, CK∥FG. Find the measure of ∠KAJ.
问题：如图所示，∠FEL=55°，CK平行于FG。求∠KAJ的大小。
Construction (CDL):
Shape(IA,AE,EH,HI)
Shape(FE,EL)
Shape(CA,AL)
Shape(KA,AJ)
Collinear(CIAK)
Collinear(FHEG)
Collinear(DIHB)
Collinear(JAEL)
Conditions (CDL):
Equal(MeasureOfAngle(FEL),55)
ParallelBetweenLine(CK,FG)
Goal: Value(MeasureOfAngle(KAJ))
Answer: 55
Theorem sequence:
flat_angle(1,CAK)
flat_angle(1,LEJ)
angle_addition(1,CAK,KAJ)
angle_addition(1,FEL,LEJ)
parallel_property_collinear_extend(3,CK,FG,A)
parallel_property_collinear_extend(3,GF,AC,E)
parallel_property_corresponding_angle(2,EF,AC,J)
Problem: As shown in the diagram, ∠CBD=55°, B is the center of ⊙B. Find the measure of arc BAG.
问题：如图所示，∠CBD=55°，B是圆B的圆心。求⌒BAG的角度。
Construction (CDL):
Shape(BCA,AB,BC)
Shape(BAG,GB,BA)
Shape(BGD,DB,BG)
Shape(BDC,CB,BD)
Collinear(ABD)
Collinear(CBG)
Cocircular(B,DCAG)
Conditions (CDL):
Equal(MeasureOfAngle(CBD),55)
IsCentreOfCircle(B,B)
Goal: Value(MeasureOfArc(BAG))
Answer: 55
Theorem sequence:
vertical_angle(1,CBD,GBA)
arc_property_center_angle(1,BAG,B)
Problem: As shown in the diagram, AB=x, AD=4, BE=12, EC=5, CD∥EA. Find the value of x.
问题：如图所示，AB=x，AD=4，BE=12，EC=5，CD平行于EA。求x的值。
Construction (CDL):
Shape(DC,CE,EA,AD)
Shape(AE,EB,BA)
Collinear(DAB)
Collinear(CEB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(BE),12)
Equal(LengthOfLine(EC),5)
ParallelBetweenLine(CD,EA)
Goal: Value(x)
Answer: 48/5
Theorem sequence:
parallel_property_corresponding_angle(2,CD,EA,B)
similar_triangle_judgment_aa(1,AEB,DCB)
line_addition(1,DA,AB)
line_addition(1,CE,EB)
similar_triangle_property_line_ratio(1,AEB,DCB)
similar_triangle_property_line_ratio(1,EBA,CBD)
Problem: As shown in the diagram, the area of quadrilateral DACB is 148, AC=8, BE=x, quadrilateral DACB is a parallelogram, CE is perpendicular to BE. Find the value of x.
问题：如图所示，四边形DACB的面积为148，AC=8，BE=x，DB和AC是▱DACB的一组对边，CE⊥BE。求x的值。
Construction (CDL):
Shape(DA,AC,CB,BD)
Shape(BC,CE,EB)
Collinear(ACE)
Conditions (CDL):
Equal(AreaOfQuadrilateral(DACB),148)
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BE),x)
Parallelogram(DACB)
PerpendicularBetweenLine(CE,BE)
Goal: Value(x)
Answer: 37/2
Theorem sequence:
altitude_of_quadrilateral_judgment_right_vertex(5,BE,DACB)
parallelogram_area_formula_common(1,DACB)
Problem: As shown in the diagram, ∠ABE=24°, ∠BCE=109°, ∠EAB=33°, ∠EDA=95°. Find the measure of ∠DAE.
问题：如图所示，∠ABE=24°，∠BCE=109°，∠EAB=33°，∠EDA=95°。求∠DAE的大小。
Construction (CDL):
Shape(DA,AE,ED)
Shape(EA,AB,BE)
Shape(CE,EB,BC)
Collinear(DEB)
Collinear(CEA)
Conditions (CDL):
Equal(MeasureOfAngle(ABE),24)
Equal(MeasureOfAngle(BCE),109)
Equal(MeasureOfAngle(EAB),33)
Equal(MeasureOfAngle(EDA),95)
Goal: Value(MeasureOfAngle(DAE))
Answer: 28
Theorem sequence:
angle_addition(1,DAE,EAB)
triangle_property_angle_sum(1,DAB)
Problem: As shown in the diagram, AB=32, DC=21, quadrilateral ADBC is a kite. Find the area of ADBC.
问题：如图所示，AB=32，DC=21，四边形ADBC是风筝形。求四边形ADBC的面积。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DB,BE)
Shape(CE,EB,BC)
Shape(AE,EC,CA)
Collinear(AEB)
Collinear(DEC)
Conditions (CDL):
Equal(LengthOfLine(AB),32)
Equal(LengthOfLine(DC),21)
Kite(ADBC)
Goal: Value(AreaOfQuadrilateral(ADBC))
Answer: 336
Theorem sequence:
kite_area_formula_diagonal(1,ADBC)
Problem: As shown in the diagram, ∠DAC=218°, the center of circle A is A. Find the measure of ∠DBC.
问题：如图所示，∠DAC=218°，A是圆A的圆心。求∠DBC的大小。
Construction (CDL):
Shape(AD,DB,BC,CA)
Shape(ACD,DA,AC)
Shape(ADB,BD)
Shape(ABC,CB)
Cocircular(A,DBC)
Conditions (CDL):
Equal(MeasureOfAngle(DAC),218)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfAngle(DBC))
Answer: 109
Theorem sequence:
arc_property_center_angle(1,ACD,A)
arc_property_circumference_angle_external(1,ACD,B)
Problem: As shown in the diagram, AB=y, AC=6, BC=x, ∠CBA=30°, BA⊥CA. Find the value of y.
问题：如图所示，AB=y，AC=6，BC=x，∠CBA=30°，BA垂直于CA。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(CBA),30)
PerpendicularBetweenLine(BA,CA)
Goal: Value(y)
Answer: 6*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,ACB)
Problem: As shown in the diagram, KN=9, LN=16, PM=Mul(LengthOfLine(KP),2), PR is parallel to KL, KN⊥MN, LM⊥KM. Find the length of line MN.
问题：如图所示，KN=9，LN=16，PM=Mul(LengthOfLine(KP),2)，PR∥KL，KN垂直于MN，LM垂直于KM。求直线MN的长度。
Construction (CDL):
Shape(LR,RQ,QN,NL)
Shape(NQ,QP,PK,KN)
Shape(RM,MQ,QR)
Shape(QM,MP,PQ)
Collinear(LNK)
Collinear(RQP)
Collinear(NQM)
Collinear(LRM)
Collinear(MPK)
Conditions (CDL):
Equal(LengthOfLine(KN),9)
Equal(LengthOfLine(LN),16)
Equal(LengthOfLine(PM),Mul(LengthOfLine(KP),2))
ParallelBetweenLine(PR,KL)
PerpendicularBetweenLine(KN,MN)
PerpendicularBetweenLine(LM,KM)
Goal: Value(LengthOfLine(MN))
Answer: 12
Theorem sequence:
line_addition(1,LN,NK)
adjacent_complementary_angle(1,KNQ,QNL)
right_triangle_judgment_angle(1,KNM)
right_triangle_judgment_angle(1,MNL)
right_triangle_judgment_angle(1,LMK)
right_triangle_property_pythagorean(1,KNM)
right_triangle_property_pythagorean(1,MNL)
right_triangle_property_pythagorean(1,LMK)
Problem: As shown in the diagram, PQ=9, QR=7, SR=7, PR⊥QR, SR⊥PR. Find the length of line PS.
问题：如图所示，PQ=9，QR=7，SR=7，PR垂直于QR，SR垂直于PR。求直线PS的长度。
Construction (CDL):
Shape(PS,SQ,QP)
Shape(PR,RQ,QP)
Shape(PS,SR,RP)
Collinear(SRQ)
Conditions (CDL):
Equal(LengthOfLine(PQ),9)
Equal(LengthOfLine(QR),7)
Equal(LengthOfLine(SR),7)
PerpendicularBetweenLine(PR,QR)
PerpendicularBetweenLine(SR,PR)
Goal: Value(LengthOfLine(PS))
Answer: 9
Theorem sequence:
right_triangle_judgment_angle(1,SRP)
right_triangle_judgment_angle(1,PRQ)
right_triangle_property_pythagorean(1,PRQ)
right_triangle_property_pythagorean(1,SRP)
Problem: As shown in the diagram, DF=6, FC=8, the perimeter of triangle BCF is 27, triangle EDF is similar to triangle BCF.. Find the perimeter of △EDF.
问题：如图所示，DF=6，FC=8，三角形BCF的周长为27，三角形EDF相似于三角形BCF。求△EDF的周长。
Construction (CDL):
Shape(ED,DF,FE)
Shape(BC,CF,FB)
Conditions (CDL):
Equal(LengthOfLine(DF),6)
Equal(LengthOfLine(FC),8)
Equal(PerimeterOfTriangle(BCF),27)
SimilarBetweenTriangle(EDF,BCF)
Goal: Value(PerimeterOfTriangle(EDF))
Answer: 81/4
Theorem sequence:
similar_triangle_property_line_ratio(1,EDF,BCF)
similar_triangle_property_perimeter_ratio(1,EDF,BCF)
Problem: As shown in the diagram, ∠BCD=67°. Find the measure of ∠ACB.
问题：如图所示，∠BCD=67°。求∠ACB的大小。
Construction (CDL):
Shape(AC,CB)
Shape(BC,CD)
Collinear(ACD)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),67)
Goal: Value(MeasureOfAngle(ACB))
Answer: 113
Theorem sequence:
flat_angle(1,ACD)
angle_addition(1,ACB,BCD)