SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠YZX=a/2°, ∠ZXY=2*a°, XY is perpendicular to ZY. Find the measure of ∠YZX.
问题：如图所示，∠YZX=a/2°，∠ZXY=2*a°，XY垂直于ZY。求∠YZX的大小。
Construction (CDL):
Shape(XY,YZ,ZX)
Conditions (CDL):
Equal(MeasureOfAngle(YZX),a/2)
Equal(MeasureOfAngle(ZXY),2*a)
PerpendicularBetweenLine(XY,ZY)
Goal: Value(MeasureOfAngle(YZX))
Answer: 18
Theorem sequence:
triangle_property_angle_sum(1,XYZ)
Problem: As shown in the diagram, BX=s-7, WB=8, YB=6, ZB=2*t-6, WYZX is a parallelogram. Find the value of s.
问题：如图所示，BX=s-7，WB=8，YB=6，ZB=2*t-6，四边形WYZX是▱。求s的值。
Construction (CDL):
Shape(WY,YB,BW)
Shape(BY,YZ,ZB)
Shape(BZ,ZX,XB)
Shape(BX,XW,WB)
Collinear(WBZ)
Collinear(YBX)
Conditions (CDL):
Equal(LengthOfLine(BX),s-7)
Equal(LengthOfLine(WB),8)
Equal(LengthOfLine(YB),6)
Equal(LengthOfLine(ZB),2*t-6)
Parallelogram(WYZX)
Goal: Value(s)
Answer: 13
Theorem sequence:
parallelogram_property_diagonal_bisection(1,YZXW,B)
Problem: As shown in the diagram, AD=2, BD=x, CB=y, CD=2*sqrt(3), AC is perpendicular to BC, BD⊥CD. Find the value of y.
问题：如图所示，AD=2，BD=x，CB=y，CD=2*sqrt(3)，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),2)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CB),y)
Equal(LengthOfLine(CD),2*sqrt(3))
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BD,CD)
Goal: Value(y)
Answer: 4*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,ACB)
right_triangle_judgment_angle(1,BDC)
line_addition(1,BD,DA)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,ACB)
right_triangle_property_pythagorean(1,BDC)
Problem: As shown in the diagram, AC=x, AD=12, BD=4, CD=11/2, ∠ACB=∠BCD. Find the value of x.
问题：如图所示，AC=x，AD=12，BD=4，CD=11/2，∠ACB=∠BCD。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Shape(BC,CD,DB)
Collinear(ABD)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(BD),4)
Equal(LengthOfLine(CD),11/2)
Equal(MeasureOfAngle(ACB),MeasureOfAngle(BCD))
Goal: Value(x)
Answer: 11
Theorem sequence:
line_addition(1,AB,BD)
bisector_of_angle_judgment_angle_equal(1,CB,ACD)
bisector_of_angle_property_line_ratio(1,CB,ACD)
Problem: As shown in the diagram, AB=26, AD=12, DG=4, EF=8, GF=14, ∠AGF=108°, quadrilateral AGFE is similar to quadrilateral ADCB. Find the perimeter of quadrilateral ADCB.
问题：如图所示，AB=26，AD=12，DG=4，EF=8，GF=14，∠AGF=108°，AGFE相似于ADCB。求四边形ADCB的周长。
Construction (CDL):
Shape(AG,GF,FE,EA)
Shape(GD,DC,CB,BE,EF,FG)
Collinear(AEB)
Collinear(AGD)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DG),4)
Equal(LengthOfLine(EF),8)
Equal(LengthOfLine(GF),14)
Equal(MeasureOfAngle(AGF),108)
SimilarBetweenQuadrilateral(AGFE,ADCB)
Goal: Value(PerimeterOfQuadrilateral(ADCB))
Answer: 71
Theorem sequence:
line_addition(1,AG,GD)
similar_quadrilateral_property_line_ratio(1,AGFE,ADCB)
similar_quadrilateral_property_line_ratio(1,GFEA,DCBA)
similar_quadrilateral_property_line_ratio(1,FEAG,CBAD)
quadrilateral_perimeter_formula(1,ADCB)
Problem: As shown in the diagram, CB=4*w-7, CD=11, ED=3*z+10, RS=2*w+13, RU=12, UT=z+16, ∠CBE=2*x+9°, ∠EDC=2*y-31°, ∠STU=y+11°, ∠URS=49°, BEDC is mirror congruent to RSTU. Find the value of y.
问题：如图所示，CB=4*w-7，CD=11，ED=3*z+10，RS=2*w+13，RU=12，UT=z+16，∠CBE=2*x+9°，∠EDC=2*y-31°，∠STU=y+11°，∠URS=49°，四边形BEDC镜像全等于四边形RSTU。求y的值。
Construction (CDL):
Shape(BE,ED,DC,CB)
Shape(UR,RS,ST,TU)
Conditions (CDL):
Equal(LengthOfLine(CB),4*w-7)
Equal(LengthOfLine(CD),11)
Equal(LengthOfLine(ED),3*z+10)
Equal(LengthOfLine(RS),2*w+13)
Equal(LengthOfLine(RU),12)
Equal(LengthOfLine(UT),z+16)
Equal(MeasureOfAngle(CBE),2*x+9)
Equal(MeasureOfAngle(EDC),2*y-31)
Equal(MeasureOfAngle(STU),y+11)
Equal(MeasureOfAngle(URS),49)
MirrorCongruentBetweenQuadrilateral(BEDC,RSTU)
Goal: Value(y)
Answer: 42
Theorem sequence:
mirror_congruent_quadrilateral_property_angle_equal(1,DCBE,TURS)
Problem: As shown in the diagram, AC=14, CD=20, the center of ⊙B is B, ED is the diameter of circle B, quadrilateral ACDE is a rectangle. Find the sum of the area of the sector BDE and the area of ACDE.
问题：如图所示，AC=14，CD=20，圆B的圆心为B，圆B的直径为ED，四边形ACDE是矩形。求扇形BDE的面积与ACDE的面积之和。
Construction (CDL):
Shape(AC,CD,DE,EA)
Shape(BDE,EB,BD)
Collinear(EBD)
Cocircular(B,DE)
Conditions (CDL):
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(CD),20)
IsCentreOfCircle(B,B)
IsDiameterOfCircle(ED,B)
Rectangle(ACDE)
Goal: Value(Add(AreaOfSector(BDE),AreaOfQuadrilateral(ACDE)))
Answer: 49*pi/2+280
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ACDE)
diameter_of_circle_property_length_equal(1,ED,B)
circle_property_length_of_radius_and_diameter(1,B)
flat_angle(1,EBD)
sector_area_formula(1,BDE)
arc_property_center_angle(1,BDE,B)
parallelogram_area_formula_sine(1,ACDE)
Problem: As shown in the diagram, ∠KLM=57°, ∠LKJ=28°, ∠MKL=71°. Find the measure of ∠LMK.
问题：如图所示，∠KLM=57°，∠LKJ=28°，∠MKL=71°。求∠LMK的大小。
Construction (CDL):
Shape(JL,LK,KJ)
Shape(KL,LM,MK)
Collinear(JLM)
Conditions (CDL):
Equal(MeasureOfAngle(KLM),57)
Equal(MeasureOfAngle(LKJ),28)
Equal(MeasureOfAngle(MKL),71)
Goal: Value(MeasureOfAngle(LMK))
Answer: 52
Theorem sequence:
triangle_property_angle_sum(1,KLM)
Problem: As shown in the diagram, ∠BAD=x°, ∠CAB=x°, ∠DAC=x°. Find the value of x.
问题：如图所示，∠BAD=x°，∠CAB=x°，∠DAC=x°。求x的值。
Construction (CDL):
Shape(CA,CB)
Shape(BA,AD)
Shape(DA,AC)
Conditions (CDL):
Equal(MeasureOfAngle(BAD),x)
Equal(MeasureOfAngle(CAB),x)
Equal(MeasureOfAngle(DAC),x)
Goal: Value(x)
Answer: 120
Theorem sequence:
round_angle(1,CAD,DAC)
angle_addition(1,CAB,BAD)
Problem: As shown in the diagram, ∠UEN=64°, AK∥WJ, IL is parallel to BN. Find the measure of ∠DUA.
问题：如图所示，∠UEN=64°，AK∥WJ，IL∥BN。求∠DUA的大小。
Construction (CDL):
Shape(WD,DI)
Shape(BH,HJ)
Shape(KE,EN)
Shape(LU,UA)
Shape(DH,HE,EU,UD)
Collinear(IDUL)
Collinear(BHEN)
Collinear(WDHJ)
Collinear(AUEK)
Conditions (CDL):
Equal(MeasureOfAngle(UEN),64)
ParallelBetweenLine(AK,WJ)
ParallelBetweenLine(IL,BN)
Goal: Value(MeasureOfAngle(DUA))
Answer: 116
Theorem sequence:
parallel_property_collinear_extend(3,IL,BN,U)
parallel_property_collinear_extend(3,NB,LU,E)
parallel_property_corresponding_angle(1,UL,EN,A)
adjacent_complementary_angle(1,DUA,AUL)
Problem: As shown in the diagram, JK=3*f-6, ML=2*f+8, ∠JML=3*d-2°, ∠KJM=56°, quadrilateral JMLK is a ▱. Find the value of d.
问题：如图所示，JK=3*f-6，ML=2*f+8，∠JML=3*d-2°，∠KJM=56°，MJ和LK是▱JMLK的一组对边。求d的值。
Construction (CDL):
Shape(JM,ML,LK,KJ)
Conditions (CDL):
Equal(LengthOfLine(JK),3*f-6)
Equal(LengthOfLine(ML),2*f+8)
Equal(MeasureOfAngle(JML),3*d-2)
Equal(MeasureOfAngle(KJM),56)
Parallelogram(JMLK)
Goal: Value(d)
Answer: 42
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,JK,ML)
Problem: As shown in the diagram, ∠ABC=3*y-11°, ∠BAE=y+19°, ∠CBL=x°, ∠EBA=4*z+2°, ED is parallel to MC, AE is perpendicular to BE. Find the value of x.
问题：如图所示，∠ABC=3*y-11°，∠BAE=y+19°，∠CBL=x°，∠EBA=4*z+2°，ED平行于MC，AE⊥BE。求x的值。
Construction (CDL):
Shape(FB,BM)
Shape(LB,BM)
Shape(CB,BL)
Shape(AB,BC)
Shape(HA,AD)
Shape(EA,AH)
Shape(GE,EA)
Shape(BA,AE,EB)
Collinear(LBEG)
Collinear(FBAH)
Collinear(MBC)
Collinear(EAD)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),3*y-11)
Equal(MeasureOfAngle(BAE),y+19)
Equal(MeasureOfAngle(CBL),x)
Equal(MeasureOfAngle(EBA),4*z+2)
ParallelBetweenLine(ED,MC)
PerpendicularBetweenLine(AE,BE)
Goal: Value(x)
Answer: 90
Theorem sequence:
parallel_property_collinear_extend(3,CM,DE,B)
parallel_property_corresponding_angle(2,ED,BC,L)
Problem: As shown in the diagram, CD=46, HE=38, the midsegment of quadrilateral CGFD is HE, CG and FD are the non-parallel sides (legs) of trapezoid CGFD. Find the length of line GF.
问题：如图所示，CD=46，HE=38，四边形CGFD的中位线为HE，CGFD是梯形。求直线GF的长度。
Construction (CDL):
Shape(CH,HE,ED,DC)
Shape(HG,GF,FE,EH)
Collinear(CHG)
Collinear(DEF)
Conditions (CDL):
Equal(LengthOfLine(CD),46)
Equal(LengthOfLine(HE),38)
IsMidsegmentOfQuadrilateral(HE,CGFD)
Trapezoid(CGFD)
Goal: Value(LengthOfLine(GF))
Answer: 30
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,HE,CGFD)
Problem: As shown in the diagram, ED=6.5, GD=6.5, GF=5, DF is perpendicular to EF. Find the length of line EG.
问题：如图所示，ED=6.5，GD=6.5，GF=5，DF⊥EF。求直线EG的长度。
Construction (CDL):
Shape(DG,GF,FD)
Shape(DF,FE,ED)
Collinear(GFE)
Conditions (CDL):
Equal(LengthOfLine(ED),6.5)
Equal(LengthOfLine(GD),6.5)
Equal(LengthOfLine(GF),5)
PerpendicularBetweenLine(DF,EF)
Goal: Value(LengthOfLine(EG))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,GFD,DFE)
mirror_congruent_triangle_judgment_hl(1,GFD,EDF)
mirror_congruent_triangle_property_line_equal(1,DGF,DFE)
line_addition(1,GF,FE)
Problem: As shown in the diagram, NM=11, ∠NMP=∠MPN, ∠PNM=120°. Find the measure of ∠NMP.
问题：如图所示，NM=11，∠NMP=∠MPN，∠PNM=120°。求∠NMP的大小。
Construction (CDL):
Shape(PN,NM,MP)
Conditions (CDL):
Equal(LengthOfLine(NM),11)
Equal(MeasureOfAngle(NMP),MeasureOfAngle(MPN))
Equal(MeasureOfAngle(PNM),120)
Goal: Value(MeasureOfAngle(NMP))
Answer: 30
Theorem sequence:
triangle_property_angle_sum(1,PNM)
Problem: As shown in the diagram, EN=13, N is the center of circle N. Find the diameter of circle N.
问题：如图所示，EN=13，N是⊙N的圆心。求⊙N的直径。
Construction (CDL):
Shape(NED,DN,NE)
Shape(NDC,CN,ND)
Shape(NCF,FN,NC)
Shape(NFE,EF)
Shape(NF,FE,EN)
Cocircular(N,DCFE)
Conditions (CDL):
Equal(LengthOfLine(EN),13)
IsCentreOfCircle(N,N)
Goal: Value(DiameterOfCircle(N))
Answer: 26
Theorem sequence:
radius_of_circle_property_length_equal(1,NE,N)
circle_property_length_of_radius_and_diameter(1,N)
Problem: As shown in the diagram, ∠AGC=60°, DG is perpendicular to AG. Find the measure of ∠DGC.
问题：如图所示，∠AGC=60°，DG垂直于AG。求∠DGC的大小。
Construction (CDL):
Shape(GCA,AG,GC)
Shape(GAD,DG,GA)
Shape(GDE,EG,GD)
Shape(GEB,BG,GE)
Shape(GBC,CG,GB)
Cocircular(G,ADEBC)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),60)
PerpendicularBetweenLine(DG,AG)
Goal: Value(MeasureOfAngle(DGC))
Answer: 150
Theorem sequence:
angle_addition(1,DGA,AGC)
Problem: As shown in the diagram, AB=5, AC=15, OB=21.5, OC=8, OC is the altitude of △ OAB, OC is perpendicular to AC. Find the area of triangle OAB.
问题：如图所示，AB=5，AC=15，OB=21.5，OC=8，OC是三角形OAB的高，OC⊥AC。求△OAB的面积。
Construction (CDL):
Shape(OC,CA,AO)
Shape(OA,AB,BO)
Collinear(CAB)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),15)
Equal(LengthOfLine(OB),21.5)
Equal(LengthOfLine(OC),8)
IsAltitudeOfTriangle(OC,OAB)
PerpendicularBetweenLine(OC,AC)
Goal: Value(AreaOfTriangle(OAB))
Answer: 20
Theorem sequence:
triangle_area_formula_common(1,OAB)
Problem: As shown in the diagram, MP=3, MQ=6, QP=8, ∠PMQ=x°. Find the value of x.
问题：如图所示，MP=3，MQ=6，QP=8，∠PMQ=x°。求x的值。
Construction (CDL):
Shape(MQ,QP,PM)
Conditions (CDL):
Equal(LengthOfLine(MP),3)
Equal(LengthOfLine(MQ),6)
Equal(LengthOfLine(QP),8)
Equal(MeasureOfAngle(PMQ),x)
Goal: Value(x)
Answer: 180*acos(-19/36)/pi
Theorem sequence:
cosine_theorem(1,MQP)
Problem: As shown in the diagram, AC=10, AC=AI, AC=CE, AC=EI, the center of circle N is N, IC is the diameter of circle N, IA⊥CA. Find the circumference of the circle N.
问题：如图所示，AC=10，AC=AI，AC=CE，AC=EI，圆N的圆心为N，圆N的直径为IC，IA垂直于CA。求⊙N的周长。
Construction (CDL):
Shape(NIA,AI)
Shape(NAC,CA)
Shape(NCE,EC)
Shape(BEJ,JE)
Shape(IA,AC,CN,NI)
Shape(CE,EI,IN,NC)
Collinear(INC)
Cocircular(N,IACE)
Conditions (CDL):
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(AC),LengthOfLine(AI))
Equal(LengthOfLine(AC),LengthOfLine(CE))
Equal(LengthOfLine(AC),LengthOfLine(EI))
IsCentreOfCircle(N,N)
IsDiameterOfCircle(IC,N)
PerpendicularBetweenLine(IA,CA)
Goal: Value(PerimeterOfCircle(N))
Answer: 10*sqrt(2)*pi
Theorem sequence:
right_triangle_judgment_angle(1,IAC)
right_triangle_property_pythagorean(1,IAC)
diameter_of_circle_property_length_equal(1,IC,N)
circle_property_length_of_radius_and_diameter(1,N)
circle_perimeter_formula(1,N)
Problem: As shown in the diagram, FG=9*x-6, FG=HG, HF=17, HG=7*x+4. Find the length of line GH.
问题：如图所示，FG=9*x-6，FG=HG，HF=17，HG=7*x+4。求直线GH的长度。
Construction (CDL):
Shape(FG,GH,HF)
Conditions (CDL):
Equal(LengthOfLine(FG),9*x-6)
Equal(LengthOfLine(FG),LengthOfLine(HG))
Equal(LengthOfLine(HF),17)
Equal(LengthOfLine(HG),7*x+4)
Goal: Value(LengthOfLine(GH))
Answer: 39
Theorem sequence:
Problem: As shown in the diagram, the area of quadrilateral EFGH is 60, AD=12, EM=6, TA and BC are opposite sides of the parallelogram ATBC, EFGH is a ▱, FM is perpendicular to EM, TD is perpendicular to AD, EFGH is similar to ATBC. Find the area of quadrilateral ATBC.
问题：如图所示，四边形EFGH的面积为60，AD=12，EM=6，AC和TB是▱ATBC的一组对边，EFGH是平行四边形，FM垂直于EM，TD垂直于AD，四边形EFGH相似于四边形ATBC。求ATBC的面积。
Construction (CDL):
Shape(EF,FM,ME)
Shape(EM,MG,GH,HE)
Shape(AT,TD,DA)
Shape(AD,DB,BC,CA)
Collinear(FMG)
Collinear(TDB)
Conditions (CDL):
Equal(AreaOfQuadrilateral(EFGH),60)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(EM),6)
Parallelogram(ATBC)
Parallelogram(EFGH)
PerpendicularBetweenLine(FM,EM)
PerpendicularBetweenLine(TD,AD)
SimilarBetweenQuadrilateral(EFGH,ATBC)
Goal: Value(AreaOfQuadrilateral(ATBC))
Answer: 240
Theorem sequence:
altitude_of_quadrilateral_judgment_left_vertex(1,EM,EFGH)
altitude_of_quadrilateral_judgment_left_vertex(1,AD,ATBC)
parallelogram_area_formula_common(1,EFGH)
parallelogram_area_formula_common(1,ATBC)
similar_quadrilateral_property_line_ratio(1,FGHE,TBCA)
similar_quadrilateral_property_area_square_ratio(1,EFGH,ATBC)
Problem: As shown in the diagram, ∠KOM=43°, BF is parallel to MQ, KC∥GJ. Find the measure of ∠ANQ.
问题：如图所示，∠KOM=43°，BF平行于MQ，KC∥GJ。求∠ANQ的大小。
Construction (CDL):
Shape(KO,OM)
Shape(GN,NM)
Shape(AN,NQ)
Collinear(BPHF)
Collinear(MONQ)
Collinear(KPOC)
Collinear(GHRNAIJ)
Conditions (CDL):
Equal(MeasureOfAngle(KOM),43)
ParallelBetweenLine(BF,MQ)
ParallelBetweenLine(KC,GJ)
Goal: Value(MeasureOfAngle(ANQ))
Answer: 43
Theorem sequence:
parallel_property_collinear_extend(3,KC,GJ,O)
parallel_property_collinear_extend(3,JG,OK,N)
parallel_property_corresponding_angle(2,NG,OK,M)
vertical_angle(1,GNM,ANQ)
Problem: As shown in the diagram, ∠AZB=104°, the measure of arc ZCB is 94, the center of circle Z is Z, AB∥DC. Find the measure of ∠ADB.
问题：如图所示，∠AZB=104°，弧ZCB的角度为94，Z是⊙Z的圆心，AB平行于DC。求∠ADB的大小。
Construction (CDL):
Shape(ZBA,AB)
Shape(ZB,BA,AZ)
Shape(EB,BZ,ZA,AE)
Shape(EA,AD,DE)
Shape(ZAD,DA)
Shape(ED,DC,CE)
Shape(ZDC,CD)
Shape(BE,EC,CB)
Shape(ECB,BC)
Collinear(BED)
Collinear(CEA)
Cocircular(Z,ADCB)
Conditions (CDL):
Equal(MeasureOfAngle(AZB),104)
Equal(MeasureOfArc(ZCB),94)
IsCentreOfCircle(Z,Z)
ParallelBetweenLine(AB,DC)
Goal: Value(MeasureOfAngle(ADB))
Answer: 52
Theorem sequence:
arc_property_center_angle(1,ZBA,Z)
arc_property_circumference_angle_external(1,ZBA,D)
Problem: As shown in the diagram, AD and BC are opposite sides of the parallelogram ABCD. Prove that ∠DAB+∠ABC=180°.
问题：如图所示，BA和CD是▱ABCD的一组对边。求证∠DAB+∠ABC=180°。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Parallelogram(ABCD)
Goal: Equal(Add(MeasureOfAngle(DAB),MeasureOfAngle(ABC)),180)
Answer: 0
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AD,BC)
Problem: As shown in the diagram, ∠JKL=x+10°, ∠KLM=x°, ∠LMJ=2*x-8°, ∠MJK=3*x-6°. Find the measure of ∠LMJ.
问题：如图所示，∠JKL=x+10°，∠KLM=x°，∠LMJ=2*x-8°，∠MJK=3*x-6°。求∠LMJ的大小。
Construction (CDL):
Shape(JK,KL,LM,MJ)
Conditions (CDL):
Equal(MeasureOfAngle(JKL),x+10)
Equal(MeasureOfAngle(KLM),x)
Equal(MeasureOfAngle(LMJ),2*x-8)
Equal(MeasureOfAngle(MJK),3*x-6)
Goal: Value(MeasureOfAngle(LMJ))
Answer: 96
Theorem sequence:
quadrilateral_property_angle_sum(1,MJKL)
Problem: As shown in the diagram, JP=20, JQ=25, ∠QKM=29°, ∠QLP=26°, Q is the incenter of △LJK, LP is perpendicular to QP, QM⊥LM, QN⊥KN. Find the measure of ∠QJK.
问题：如图所示，JP=20，JQ=25，∠QKM=29°，∠QLP=26°，△LJK的内心为Q，LP垂直于QP，QM垂直于LM，QN⊥KN。求∠QJK的大小。
Construction (CDL):
Shape(PQ,QL,LP)
Shape(PJ,JQ,QP)
Shape(QJ,JN,NQ)
Shape(QN,NK,KQ)
Shape(QK,KM,MQ)
Shape(QM,ML,LQ)
Collinear(JPL)
Collinear(JNK)
Collinear(LMK)
Conditions (CDL):
Equal(LengthOfLine(JP),20)
Equal(LengthOfLine(JQ),25)
Equal(MeasureOfAngle(QKM),29)
Equal(MeasureOfAngle(QLP),26)
IsIncenterOfTriangle(Q,LJK)
PerpendicularBetweenLine(LP,QP)
PerpendicularBetweenLine(QM,LM)
PerpendicularBetweenLine(QN,KN)
Goal: Value(MeasureOfAngle(QJK))
Answer: 35
Theorem sequence:
angle_addition(1,MLQ,QLP)
angle_addition(1,PJQ,QJN)
angle_addition(1,NKQ,QKM)
triangle_property_angle_sum(1,LJK)
Problem: As shown in the diagram, AB=9, AC=12, BC=9, BD=x, AD⊥BD. Find the value of x.
问题：如图所示，AB=9，AC=12，BC=9，BD=x，AD⊥BD。求x的值。
Construction (CDL):
Shape(BA,AD,DB)
Shape(BD,DC,CB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),9)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(BC),9)
Equal(LengthOfLine(BD),x)
PerpendicularBetweenLine(AD,BD)
Goal: Value(x)
Answer: 3*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,ADB,BDC)
right_triangle_judgment_angle(1,BDC)
mirror_congruent_triangle_judgment_hl(1,ADB,CBD)
mirror_congruent_triangle_property_line_equal(1,BAD,BDC)
line_addition(1,AD,DC)
right_triangle_property_pythagorean(1,BDC)
Problem: As shown in the diagram, BD=14, FB=26, ∠ADC=30°, quadrilateral CFBD is a parallelogram, CA is perpendicular to DA. Find the area of DCFB.
问题：如图所示，BD=14，FB=26，∠ADC=30°，CD和FB是平行四边形CFBD的一组对边，CA⊥DA。求四边形DCFB的面积。
Construction (CDL):
Shape(DC,CA,AD)
Shape(DA,AF,FB,BD)
Collinear(CAF)
Conditions (CDL):
Equal(LengthOfLine(BD),14)
Equal(LengthOfLine(FB),26)
Equal(MeasureOfAngle(ADC),30)
Parallelogram(CFBD)
PerpendicularBetweenLine(CA,DA)
Goal: Value(AreaOfQuadrilateral(DCFB))
Answer: 182*sqrt(3)
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DCFB)
parallelogram_property_opposite_line_equal(1,BDCF)
triangle_property_angle_sum(1,DCA)
parallelogram_area_formula_sine(1,DCFB)
Problem: As shown in the diagram, JM=3*a, KJ=2*b+3, LK=21, ML=45, ∠MLJ=30°, ∠RLK=70°, MJ and LK are opposite sides of the parallelogram JMLK. Find the value of b.
问题：如图所示，JM=3*a，KJ=2*b+3，LK=21，ML=45，∠MLJ=30°，∠RLK=70°，JK和ML是平行四边形JMLK的一组对边。求b的值。
Construction (CDL):
Shape(JM,MR,RJ)
Shape(RM,ML,LR)
Shape(RL,LK,KR)
Shape(RK,KJ,JR)
Collinear(JRL)
Collinear(MRK)
Conditions (CDL):
Equal(LengthOfLine(JM),3*a)
Equal(LengthOfLine(KJ),2*b+3)
Equal(LengthOfLine(LK),21)
Equal(LengthOfLine(ML),45)
Equal(MeasureOfAngle(MLJ),30)
Equal(MeasureOfAngle(RLK),70)
Parallelogram(JMLK)
Goal: Value(b)
Answer: 21
Theorem sequence:
parallelogram_property_opposite_line_equal(1,MLKJ)
Problem: As shown in the diagram, ∠AMD=56°, ∠ELO=3*y-11°, ∠KDJ=4*x°, AB is parallel to OG, PJ∥AB, PJ is parallel to OG. Find the value of y.
问题：如图所示，∠AMD=56°，∠ELO=3*y-11°，∠KDJ=4*x°，AB平行于OG，PJ平行于AB，PJ平行于OG。求y的值。
Construction (CDL):
Shape(AM,MK)
Shape(OL,LM)
Shape(EL,LO)
Collinear(KDMLE)
Collinear(JDP)
Collinear(BMA)
Collinear(GLO)
Conditions (CDL):
Equal(MeasureOfAngle(AMD),56)
Equal(MeasureOfAngle(ELO),3*y-11)
Equal(MeasureOfAngle(KDJ),4*x)
ParallelBetweenLine(AB,OG)
ParallelBetweenLine(PJ,AB)
ParallelBetweenLine(PJ,OG)
Goal: Value(y)
Answer: 45
Theorem sequence:
parallel_property_collinear_extend(3,AB,OG,M)
parallel_property_collinear_extend(3,GO,MA,L)
parallel_property_corresponding_angle(2,LO,MA,K)
adjacent_complementary_angle(1,ELO,OLK)
Problem: As shown in the diagram, BA=AD, CB=DC, ∠BAD=38°, ∠DCB=50°. Find the measure of ∠ADC.
问题：如图所示，BA=AD，CB=DC，∠BAD=38°，∠DCB=50°。求∠ADC的大小。
Construction (CDL):
Shape(BA,AT,TB)
Shape(AD,DT,TA)
Shape(TD,DC,CT)
Shape(BT,TC,CB)
Collinear(BTD)
Collinear(ATC)
Conditions (CDL):
Equal(LengthOfLine(BA),LengthOfLine(AD))
Equal(LengthOfLine(CB),LengthOfLine(DC))
Equal(MeasureOfAngle(BAD),38)
Equal(MeasureOfAngle(DCB),50)
Goal: Value(MeasureOfAngle(ADC))
Answer: 136
Theorem sequence:
mirror_congruent_triangle_judgment_sss(1,BAC,DCA)
mirror_congruent_triangle_property_angle_equal(1,CBA,CAD)
mirror_congruent_triangle_property_angle_equal(1,ACB,ADC)
angle_addition(1,BAT,TAD)
angle_addition(1,DCT,TCB)
triangle_property_angle_sum(1,ADC)
Problem: As shown in the diagram, ∠CBA=2*x-7°, ∠DCB=2*x+3°, DP is perpendicular to AP, quadrilateral ADCB is a rhombus. Find the measure of ∠BAD.
问题：如图所示，∠CBA=2*x-7°，∠DCB=2*x+3°，DP⊥AP，ADCB是菱形。求∠BAD的大小。
Construction (CDL):
Shape(AD,DP,PA)
Shape(DC,CP,PD)
Shape(PC,CB,BP)
Shape(AP,PB,BA)
Collinear(APC)
Collinear(BPD)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),2*x-7)
Equal(MeasureOfAngle(DCB),2*x+3)
PerpendicularBetweenLine(DP,AP)
Rhombus(ADCB)
Goal: Value(MeasureOfAngle(BAD))
Answer: 95
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,ADCB)
parallelogram_property_opposite_angle_equal(1,DCBA)
quadrilateral_property_angle_sum(1,ADCB)
Problem: As shown in the diagram, RS=6*x+2, RT=4*x+3, RW=12, WS=16, WT=9, RW is perpendicular to TW. Find the length of line RT.
问题：如图所示，RS=6*x+2，RT=4*x+3，RW=12，WS=16，WT=9，RW⊥TW。求直线RT的长度。
Construction (CDL):
Shape(SR,RW,WS)
Shape(RW,WT,TR)
Collinear(SWT)
Conditions (CDL):
Equal(LengthOfLine(RS),6*x+2)
Equal(LengthOfLine(RT),4*x+3)
Equal(LengthOfLine(RW),12)
Equal(LengthOfLine(WS),16)
Equal(LengthOfLine(WT),9)
PerpendicularBetweenLine(RW,TW)
Goal: Value(LengthOfLine(RT))
Answer: 15
Theorem sequence:
right_triangle_judgment_angle(1,RWT)
right_triangle_property_pythagorean(1,RWT)
Problem: As shown in the diagram, ∠ACB=5*y-3°, ∠ADB=4*x+7°, ∠CAD=5*x°, ∠CBD=7*x-8°. Find the measure of ∠CAD.
问题：如图所示，∠ACB=5*y-3°，∠ADB=4*x+7°，∠CAD=5*x°，∠CBD=7*x-8°。求∠CAD的大小。
Construction (CDL):
Shape(JBA,AI,IB)
Shape(JAD,DA)
Shape(JDC,CI,ID)
Shape(JCB,BC)
Shape(IA,AD,DI)
Shape(IC,CB,BI)
Shape(CJ,JD)
Collinear(AIC)
Collinear(BID)
Cocircular(J,BADC)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),5*y-3)
Equal(MeasureOfAngle(ADB),4*x+7)
Equal(MeasureOfAngle(CAD),5*x)
Equal(MeasureOfAngle(CBD),7*x-8)
Goal: Value(MeasureOfAngle(CAD))
Answer: 20
Theorem sequence:
arc_property_circumference_angle_external(1,JDC,A)
arc_property_circumference_angle_external(1,JDC,B)
Problem: As shown in the diagram, ∠UZY=2*x+24°, ∠VZU=4*x°, ∠XZW=∠YZX, Z is the center of ⊙Z. Find the measure of arc ZWX.
问题：如图所示，∠UZY=2*x+24°，∠VZU=4*x°，∠XZW=∠YZX，Z是⊙Z的圆心。求⌒ZWX的角度。
Construction (CDL):
Shape(ZVW,WZ,ZV)
Shape(ZWX,XZ,ZW)
Shape(ZXY,YZ,ZX)
Shape(ZYU,UZ,ZY)
Shape(ZUV,VZ,ZU)
Collinear(VZY)
Collinear(WZU)
Cocircular(Z,VWXYU)
Conditions (CDL):
Equal(MeasureOfAngle(UZY),2*x+24)
Equal(MeasureOfAngle(VZU),4*x)
Equal(MeasureOfAngle(XZW),MeasureOfAngle(YZX))
IsCentreOfCircle(Z,Z)
Goal: Value(MeasureOfArc(ZWX))
Answer: 52
Theorem sequence:
adjacent_complementary_angle(1,VZU,UZY)
adjacent_complementary_angle(1,UZY,YZW)
angle_addition(1,YZX,XZW)
arc_property_center_angle(1,ZWX,Z)
Problem: As shown in the diagram, the measure of ⌒DCB is 96. Find the measure of ∠BAC.
问题：如图所示，弧DCB的角度为96。求∠BAC的大小。
Construction (CDL):
Shape(DBA,AB)
Shape(DCB,BA,AC)
Shape(DAC,CD,DA)
Cocircular(D,BAC)
Conditions (CDL):
Equal(MeasureOfArc(DCB),96)
Goal: Value(MeasureOfAngle(BAC))
Answer: 48
Theorem sequence:
arc_property_circumference_angle_external(1,DCB,A)
Problem: As shown in the diagram, BC=24, FB=14, IF=16, IH=12, FI is parallel to BC, BH is perpendicular to IH. Find the area of IFBC.
问题：如图所示，BC=24，FB=14，IF=16，IH=12，FI∥BC，BH垂直于IH。求IFBC的面积。
Construction (CDL):
Shape(IF,FB,BH,HI)
Shape(IH,HC,CI)
Collinear(BHC)
Conditions (CDL):
Equal(LengthOfLine(BC),24)
Equal(LengthOfLine(FB),14)
Equal(LengthOfLine(IF),16)
Equal(LengthOfLine(IH),12)
ParallelBetweenLine(FI,BC)
PerpendicularBetweenLine(BH,IH)
Goal: Value(AreaOfQuadrilateral(IFBC))
Answer: 240
Theorem sequence:
trapezoid_judgment_parallel(1,CIFB)
altitude_of_quadrilateral_judgment_right_vertex(2,IH,FBCI)
trapezoid_area_formula(1,FBCI)
Problem: As shown in the diagram, ∠FJE=75°, DG∥MI, EC is parallel to MI, GD is parallel to CE. Find the measure of ∠TBG.
问题：如图所示，∠FJE=75°，DG∥MI，EC平行于MI，GD∥CE。求∠TBG的大小。
Construction (CDL):
Shape(FJ,JE)
Shape(BF,FM)
Shape(LB,BD)
Shape(TB,BG)
Collinear(LBFJT)
Collinear(DBG)
Collinear(MFI)
Collinear(EJC)
Conditions (CDL):
Equal(MeasureOfAngle(FJE),75)
ParallelBetweenLine(DG,MI)
ParallelBetweenLine(EC,MI)
ParallelBetweenLine(GD,CE)
Goal: Value(MeasureOfAngle(TBG))
Answer: 75
Theorem sequence:
parallel_property_collinear_extend(3,GD,CE,B)
parallel_property_collinear_extend(3,EC,BG,J)
parallel_property_alternate_interior_angle(2,GB,JE)