SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AC=32, the center of ⊙O is O, the tangent to circle O is AE, AG is the tangent to ⊙O, BF is the tangent to ⊙O, BH is the tangent to circle O, the tangent to circle O is CE, CF is the tangent to ⊙O, the tangent to ⊙O is DG, DH is the tangent to circle O, EC is perpendicular to FC, FB⊥HB, GA is perpendicular to EA, HD is perpendicular to GD. Find the circumference of the ⊙O.
问题：如图所示，AC=32，O是⊙O的圆心，AE是⊙O的切线，AG是圆O的切线，圆O的切线为BF，圆O的切线为BH，圆O的切线为CE，圆O的切线为CF，DG是圆O的切线，圆O的切线为DH，EC垂直于FC，FB⊥HB，GA垂直于EA，HD⊥GD。求⊙O的周长。
Construction (CDL):
Shape(OHF,FB,BH)
Shape(HD,DG,OHG)
Shape(GA,AE,OGE)
Shape(OG,OGE,EO)
Shape(FO,OE,OEF)
Shape(EC,CF,OEF)
Shape(OF,OFH,OHG,GO)
Collinear(BHD)
Collinear(FOG)
Collinear(CEA)
Collinear(BFC)
Collinear(DGA)
Cocircular(O,FHGE)
Conditions (CDL):
Equal(LengthOfLine(AC),32)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AE,O)
IsTangentOfCircle(AG,O)
IsTangentOfCircle(BF,O)
IsTangentOfCircle(BH,O)
IsTangentOfCircle(CE,O)
IsTangentOfCircle(CF,O)
IsTangentOfCircle(DG,O)
IsTangentOfCircle(DH,O)
PerpendicularBetweenLine(EC,FC)
PerpendicularBetweenLine(FB,HB)
PerpendicularBetweenLine(GA,EA)
PerpendicularBetweenLine(HD,GD)
Goal: Value(PerimeterOfCircle(O))
Answer: 32*pi
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AE,O,O)
tangent_of_circle_property_perpendicular(1,AG,O,O)
tangent_of_circle_property_length_equal(1,AE,AG,O)
radius_of_circle_property_length_equal(1,OE,O)
radius_of_circle_property_length_equal(1,OG,O)
quadrilateral_property_angle_sum(1,OGAE)
parallelogram_judgment_angle_and_angle(1,OGAE)
kite_judgment_equal_and_equal(1,OGAE)
rhombus_judgment_parallelogram_and_kite(1,OGAE)
tangent_of_circle_property_perpendicular(2,CF,O,O)
tangent_of_circle_property_perpendicular(1,CE,O,O)
tangent_of_circle_property_length_equal(1,CF,CE,O)
radius_of_circle_property_length_equal(1,OF,O)
quadrilateral_property_angle_sum(1,OECF)
parallelogram_judgment_angle_and_angle(1,OECF)
kite_judgment_equal_and_equal(1,OECF)
rhombus_judgment_parallelogram_and_kite(1,OECF)
line_addition(1,CE,EA)
circle_perimeter_formula(1,O)
Problem: As shown in the diagram, AE=10, NB=7, NC=12, ABNC is a parallelogram, AE is perpendicular to NE. Find the perimeter of ABNC.
问题：如图所示，AE=10，NB=7，NC=12，四边形ABNC是平行四边形，AE⊥NE。求ABNC的周长。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,EN,NC,CA)
Collinear(BEN)
Conditions (CDL):
Equal(LengthOfLine(AE),10)
Equal(LengthOfLine(NB),7)
Equal(LengthOfLine(NC),12)
Parallelogram(ABNC)
PerpendicularBetweenLine(AE,NE)
Goal: Value(PerimeterOfQuadrilateral(ABNC))
Answer: 38
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABNC)
parallelogram_property_opposite_line_equal(1,BNCA)
quadrilateral_perimeter_formula(1,ABNC)
Problem: As shown in the diagram, ∠ADC=40°, the measure of ⌒BED is 98. Find the measure of arc BCF.
问题：如图所示，∠ADC=40°，弧BED的角度为98。求⌒BCF的角度。
Construction (CDL):
Shape(BDC,CD)
Shape(AD,DC,CA)
Shape(BCF,FA,AC)
Shape(AF,BFE,EA)
Shape(AE,BED,DA)
Collinear(CAE)
Collinear(DAF)
Cocircular(B,DCFE)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),40)
Equal(MeasureOfArc(BED),98)
Goal: Value(MeasureOfArc(BCF))
Answer: 80
Theorem sequence:
arc_property_circumference_angle_external(1,BCF,D)
Problem: As shown in the diagram, AC=30-x, BD=4*x-60, quadrilateral DABC is a rectangle. Find the value of x.
问题：如图所示，AC=30-x，BD=4*x-60，DABC是长方形。求x的值。
Construction (CDL):
Shape(DA,AE,ED)
Shape(EA,AB,BE)
Shape(EB,BC,CE)
Shape(DE,EC,CD)
Collinear(DEB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AC),30-x)
Equal(LengthOfLine(BD),4*x-60)
Rectangle(DABC)
Goal: Value(x)
Answer: 18
Theorem sequence:
rectangle_property_diagonal_equal(1,DABC)
Problem: As shown in the diagram, AB=12, AD=10, BC=y, BD=x, CD=z, AB⊥CB, BD is perpendicular to AD. Find the value of y.
问题：如图所示，AB=12，AD=10，BC=y，BD=x，CD=z，AB⊥CB，BD⊥AD。求y的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CD),z)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(BD,AD)
Goal: Value(y)
Answer: 12*sqrt(11)/5
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,BDA,CAB)
right_triangle_judgment_angle(1,BDA)
right_triangle_property_pythagorean(1,BDA)
mirror_similar_triangle_property_line_ratio(1,BDA,CAB)
mirror_similar_triangle_property_line_ratio(1,ABD,ABC)
Problem: As shown in the diagram, AR=RC, AT=16, CT=TB, MC=7, RM=4, SA=SB, the centroid of △ACB is M. Find the length of line RB.
问题：如图所示，AR=RC，AT=16，CT=TB，MC=7，RM=4，SA=SB，三角形ACB的重心为O。求直线RB的长度。
Construction (CDL):
Shape(AR,RM,MA)
Shape(MR,RC,CM)
Shape(MC,CT,TM)
Shape(MT,TB,BM)
Shape(MB,BS,SM)
Shape(MS,SA,AM)
Collinear(ARC)
Collinear(CTB)
Collinear(BSA)
Collinear(AMT)
Collinear(BMR)
Collinear(CMS)
Conditions (CDL):
Equal(LengthOfLine(AR),LengthOfLine(RC))
Equal(LengthOfLine(AT),16)
Equal(LengthOfLine(CT),LengthOfLine(TB))
Equal(LengthOfLine(MC),7)
Equal(LengthOfLine(RM),4)
Equal(LengthOfLine(SA),LengthOfLine(SB))
IsCentroidOfTriangle(M,ACB)
Goal: Value(LengthOfLine(RB))
Answer: 12
Theorem sequence:
centroid_of_triangle_property_line_ratio(1,M,BAC,R)
line_addition(1,RM,MB)
Problem: As shown in the diagram, the diameter of ⊙O is 5, BY=5/2, the center of ⊙O is O, the diameter of ⊙O is DY, D bisects segment AC, E bisects segment CB, F is the midpoint of segment AB, the tangent to ⊙O is AD, the tangent to circle O is AF, BE is the tangent to circle O, the tangent to ⊙O is BF, CD is the tangent to ⊙O, the tangent to circle O is CE. Find the perimeter of triangle ACB.
问题：如图所示，⊙O的直径为5，BY=5/2，⊙O的圆心为O，DY是⊙O的直径，D平分线段AC，E是线段CB的中点，F是线段AB的中点，AD是圆O的切线，圆O的切线为AF，⊙O的切线为BE，⊙O的切线为BF，⊙O的切线为CD，圆O的切线为CE。求三角形ACB的周长。
Construction (CDL):
Shape(AD,OXD,XA)
Shape(OX,OXD,DO)
Shape(AX,OFX,FA)
Shape(XO,OF,OFX)
Shape(OD,ODZ,ZO)
Shape(DC,CZ,ODZ)
Shape(OZ,OZE,EO)
Shape(ZC,CE,OZE)
Shape(OE,OEY,YO)
Shape(OEY,EB,BY)
Shape(OY,OYF,FO)
Shape(YB,BF,OYF)
Collinear(ADC)
Collinear(CEB)
Collinear(BFA)
Collinear(AXOE)
Collinear(CZOF)
Collinear(BYOD)
Cocircular(O,XDZEYF)
Conditions (CDL):
Equal(DiameterOfCircle(O),5)
Equal(LengthOfLine(BY),5/2)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(DY,O)
IsMidpointOfLine(D,AC)
IsMidpointOfLine(E,CB)
IsMidpointOfLine(F,AB)
IsTangentOfCircle(AD,O)
IsTangentOfCircle(AF,O)
IsTangentOfCircle(BE,O)
IsTangentOfCircle(BF,O)
IsTangentOfCircle(CD,O)
IsTangentOfCircle(CE,O)
Goal: Value(PerimeterOfTriangle(ACB))
Answer: 15*sqrt(3)
Theorem sequence:
diameter_of_circle_property_length_equal(1,DY,O)
line_addition(1,DY,YB)
line_addition(1,AD,DC)
line_addition(1,CE,EB)
line_addition(1,AF,FB)
tangent_of_circle_property_length_equal(1,AD,AF,O)
tangent_of_circle_property_length_equal(1,CD,CE,O)
isosceles_triangle_judgment_line_equal(1,ACB)
isosceles_triangle_judgment_line_equal(1,CBA)
equilateral_triangle_judgment_isosceles_and_isosceles(1,ACB)
equilateral_triangle_property_angle(1,CBA)
tangent_of_circle_property_perpendicular(1,CD,O,O)
sine_theorem(1,BDC)
triangle_perimeter_formula(1,ACB)
Problem: As shown in the diagram, ∠GMC=10*x°, ∠IMG=8*y+2°, ∠MGJ=25*y-20°, MI is parallel to GJ. Find the value of x.
问题：如图所示，∠GMC=10*x°，∠IMG=8*y+2°，∠MGJ=25*y-20°，MI∥GJ。求x的值。
Construction (CDL):
Shape(HM,MI)
Shape(IM,MG)
Shape(MG,GJ)
Shape(JG,GF)
Shape(FG,GD)
Shape(DG,GM)
Shape(GM,MC)
Shape(CM,MH)
Collinear(HMGF)
Collinear(IMC)
Collinear(JGD)
Conditions (CDL):
Equal(MeasureOfAngle(GMC),10*x)
Equal(MeasureOfAngle(IMG),8*y+2)
Equal(MeasureOfAngle(MGJ),25*y-20)
ParallelBetweenLine(MI,GJ)
Goal: Value(x)
Answer: 13
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,MI,GJ)
adjacent_complementary_angle(1,IMG,GMC)
Problem: As shown in the diagram, BZ=10, ∠ZXY=130°, the center of circle X is X, BZ is the diameter of circle X. Find the length of arc XYZ.
问题：如图所示，BZ=10，∠ZXY=130°，圆X的圆心为X，BZ是圆X的直径。求弧XYZ的长度。
Construction (CDL):
Shape(XB,XBY,YX)
Shape(XY,XYZ,ZX)
Shape(XZ,XZB,BX)
Collinear(BXZ)
Cocircular(X,BYZ)
Conditions (CDL):
Equal(LengthOfLine(BZ),10)
Equal(MeasureOfAngle(ZXY),130)
IsCentreOfCircle(X,X)
IsDiameterOfCircle(BZ,X)
Goal: Value(LengthOfArc(XYZ))
Answer: 65*pi/18
Theorem sequence:
diameter_of_circle_property_length_equal(1,BZ,X)
circle_property_length_of_radius_and_diameter(1,X)
arc_property_center_angle(1,XYZ,X)
arc_length_formula(1,XYZ)
Problem: As shown in the diagram, FL=LK, GF=x, GH=HJ, HL=15, JK=18, the midsegment of FKJG is LH, quadrilateral FKJG is a trapezoid. Find the value of x.
问题：如图所示，FL=LK，GF=x，GH=HJ，HL=15，JK=18，LH是四边形FKJG的中位线，四边形FKJG是梯形。求x的值。
Construction (CDL):
Shape(FL,LH,HG,GF)
Shape(LK,KJ,JH,HL)
Collinear(FLK)
Collinear(GHJ)
Conditions (CDL):
Equal(LengthOfLine(FL),LengthOfLine(LK))
Equal(LengthOfLine(GF),x)
Equal(LengthOfLine(GH),LengthOfLine(HJ))
Equal(LengthOfLine(HL),15)
Equal(LengthOfLine(JK),18)
IsMidsegmentOfQuadrilateral(LH,FKJG)
Trapezoid(FKJG)
Goal: Value(x)
Answer: 12
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,LH,FKJG)
Problem: As shown in the diagram, AC=7*sqrt(2), BC=x, ∠CBA=45°, BA⊥CA. Find the value of x.
问题：如图所示，AC=7*sqrt(2)，BC=x，∠CBA=45°，BA⊥CA。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AC),7*sqrt(2))
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 14
Theorem sequence:
sine_theorem(1,CBA)
Problem: As shown in the diagram, AB=10, CE=9, AD is the median of △ ACB, CE is the median of triangle CBA, AF⊥EF. Find the length of line CA.
问题：如图所示，AB=10，CE=9，AD是三角形ACB的中线，CE是三角形CBA的中线，AF⊥EF。求直线CA的长度。
Construction (CDL):
Shape(AC,CF,FA)
Shape(FC,CD,DF)
Shape(AF,FE,EA)
Shape(EF,FD,DB,BE)
Collinear(AEB)
Collinear(CDB)
Collinear(AFD)
Collinear(CFE)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(CE),9)
IsMedianOfTriangle(AD,ACB)
IsMedianOfTriangle(CE,CBA)
PerpendicularBetweenLine(AF,EF)
Goal: Value(LengthOfLine(CA))
Answer: 2*sqrt(13)
Theorem sequence:
line_addition(1,AE,EB)
centroid_of_triangle_judgment_intersection(1,F,BAC,E,D)
centroid_of_triangle_property_line_ratio(1,F,CBA,E)
line_addition(1,CF,FE)
right_triangle_judgment_angle(1,AFE)
right_triangle_property_pythagorean(1,AFE)
adjacent_complementary_angle(1,CFA,AFE)
right_triangle_judgment_angle(1,CFA)
right_triangle_property_pythagorean(1,CFA)
Problem: As shown in the diagram, ∠BCA=40°, ∠CAB=x°, ∠CBD=115°. Find the value of x.
问题：如图所示，∠BCA=40°，∠CAB=x°，∠CBD=115°。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Shape(CB,BD)
Collinear(ABD)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),40)
Equal(MeasureOfAngle(CAB),x)
Equal(MeasureOfAngle(CBD),115)
Goal: Value(x)
Answer: 75
Theorem sequence:
adjacent_complementary_angle(1,ABC,CBD)
triangle_property_angle_sum(1,CAB)
Problem: As shown in the diagram, FG=12, ∠FGH=133°, the center of circle G is G. Find the area of the sector GHF.
问题：如图所示，FG=12，∠FGH=133°，G是⊙G的圆心。求扇形GHF的面积。
Construction (CDL):
Shape(GF,GFH,HG)
Shape(GH,GHF,FG)
Cocircular(G,HF)
Conditions (CDL):
Equal(LengthOfLine(FG),12)
Equal(MeasureOfAngle(FGH),133)
IsCentreOfCircle(G,G)
Goal: Value(AreaOfSector(GHF))
Answer: 266*pi/5
Theorem sequence:
arc_property_center_angle(1,GHF,G)
radius_of_circle_property_length_equal(1,GF,G)
sector_area_formula(1,GHF)
Problem: As shown in the diagram, BA=5*x, BC=3*y-4, DA=29, DC=25, ∠ADF=34°, ∠DCF=54°, ∠FAD=49°, AB and DC are opposite sides of the parallelogram BADC. Find the measure of ∠FCB.
问题：如图所示，BA=5*x，BC=3*y-4，DA=29，DC=25，∠ADF=34°，∠DCF=54°，∠FAD=49°，BADC是平行四边形。求∠FCB的大小。
Construction (CDL):
Shape(BA,AF,FB)
Shape(FA,AD,DF)
Shape(FD,DC,CF)
Shape(FC,CB,BF)
Collinear(BFD)
Collinear(AFC)
Conditions (CDL):
Equal(LengthOfLine(BA),5*x)
Equal(LengthOfLine(BC),3*y-4)
Equal(LengthOfLine(DA),29)
Equal(LengthOfLine(DC),25)
Equal(MeasureOfAngle(ADF),34)
Equal(MeasureOfAngle(DCF),54)
Equal(MeasureOfAngle(FAD),49)
Parallelogram(BADC)
Goal: Value(MeasureOfAngle(FCB))
Answer: 49
Theorem sequence:
parallel_property_alternate_interior_angle(2,BC,AD)
Problem: As shown in the diagram, AD=DB, AF=FC, CA=15, CE=EB, EF=9.2, ∠BDE=82°. Find the measure of ∠FED.
问题：如图所示，AD=DB，AF=FC，CA=15，CE=EB，EF=9.2，∠BDE=82°。求∠FED的大小。
Construction (CDL):
Shape(AF,FE,ED,DA)
Shape(FC,CE,EF)
Shape(DE,EB,BD)
Collinear(AFC)
Collinear(CEB)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(DB))
Equal(LengthOfLine(AF),LengthOfLine(FC))
Equal(LengthOfLine(CA),15)
Equal(LengthOfLine(CE),LengthOfLine(EB))
Equal(LengthOfLine(EF),9.2)
Equal(MeasureOfAngle(BDE),82)
Goal: Value(MeasureOfAngle(FED))
Answer: 82
Theorem sequence:
line_addition(1,CF,FA)
line_addition(1,CE,EB)
similar_triangle_judgment_sas(1,CEF,CBA)
similar_triangle_property_angle_equal(1,FCE,ACB)
parallel_judgment_corresponding_angle(2,AD,FE,C)
adjacent_complementary_angle(1,BDE,EDA)
parallel_property_ipsilateral_internal_angle(1,EF,DA)
Problem: As shown in the diagram, AB=x, AD=32, BD=40, AD is perpendicular to BD, BC⊥AC. Find the value of x.
问题：如图所示，AB=x，AD=32，BD=40，AD⊥BD，BC⊥AC。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Shape(BA,AD,DB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),32)
Equal(LengthOfLine(BD),40)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 8*sqrt(41)
Theorem sequence:
right_triangle_judgment_angle(1,ADB)
right_triangle_property_pythagorean(1,ADB)
Problem: As shown in the diagram, AB=10, AP=13, EP=15, ∠DCP=28°, ∠PAF=33°, the incenter of △AEC is P, ED is perpendicular to PD, PB is perpendicular to AB, PF⊥EF. Find the length of line DE.
问题：如图所示，AB=10，AP=13，EP=15，∠DCP=28°，∠PAF=33°，三角形AEC的内切圆圆心为P，ED⊥PD，PB垂直于AB，PF⊥EF。求直线DE的长度。
Construction (CDL):
Shape(AF,FP,PA)
Shape(AP,PB,BA)
Shape(FE,EP,PF)
Shape(PE,ED,DP)
Shape(PD,DC,CP)
Shape(PC,CB,BP)
Collinear(AFE)
Collinear(ABC)
Collinear(EDC)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AP),13)
Equal(LengthOfLine(EP),15)
Equal(MeasureOfAngle(DCP),28)
Equal(MeasureOfAngle(PAF),33)
IsIncenterOfTriangle(P,AEC)
PerpendicularBetweenLine(ED,PD)
PerpendicularBetweenLine(PB,AB)
PerpendicularBetweenLine(PF,EF)
Goal: Value(LengthOfLine(DE))
Answer: 15*sin(61*pi/180)
Theorem sequence:
angle_addition(1,BAP,PAF)
angle_addition(1,FEP,PED)
angle_addition(1,DCP,PCB)
triangle_property_angle_sum(1,AEC)
triangle_property_angle_sum(1,PED)
sine_theorem(1,EDP)
Problem: As shown in the diagram, AE=6, DB=7, FC=16, EA⊥BA, FB is perpendicular to DB. Find the sum of the area of △DFC and the area of triangle ECF.
问题：如图所示，AE=6，DB=7，FC=16，EA⊥BA，FB⊥DB。求三角形DFC的面积与△ECF的面积之和。
Construction (CDL):
Shape(DF,FB,BD)
Shape(DB,BA,AC,CD)
Shape(FE,EA,AB,BF)
Shape(AE,EC,CA)
Collinear(FBAC)
Conditions (CDL):
Equal(LengthOfLine(AE),6)
Equal(LengthOfLine(DB),7)
Equal(LengthOfLine(FC),16)
PerpendicularBetweenLine(EA,BA)
PerpendicularBetweenLine(FB,DB)
Goal: Value(Add(AreaOfTriangle(DFC),AreaOfTriangle(ECF)))
Answer: 104
Theorem sequence:
adjacent_complementary_angle(1,CAE,EAF)
altitude_of_triangle_judgment(1,DB,DFC)
altitude_of_triangle_judgment(1,EA,ECF)
triangle_area_formula_common(1,DFC)
triangle_area_formula_common(1,ECF)
Problem: As shown in the diagram, AB=y, AD=z, BD=9, CA=x, CD=4, BD⊥AD, CA is perpendicular to BA. Find the value of y.
问题：如图所示，AB=y，AD=z，BD=9，CA=x，CD=4，BD垂直于AD，CA垂直于BA。求y的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(DA,AB,BD)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AD),z)
Equal(LengthOfLine(BD),9)
Equal(LengthOfLine(CA),x)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 3*sqrt(13)
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADC)
line_addition(1,CD,DB)
angle_addition(1,CAD,DAB)
triangle_property_angle_sum(1,DAB)
mirror_similar_triangle_judgment_aa(1,ADC,BCA)
similar_triangle_judgment_aa(1,CAD,ABD)
mirror_similar_triangle_property_line_ratio(1,ADC,BCA)
mirror_similar_triangle_property_line_ratio(1,DCA,ABC)
mirror_similar_triangle_property_line_ratio(1,CAD,CAB)
similar_triangle_property_line_ratio(1,CAD,ABD)
similar_triangle_property_line_ratio(1,ADC,BDA)
Problem: As shown in the diagram, AB=z, AC=x, AY=17, BC=y, YC=6, AY is perpendicular to BY, BA is perpendicular to CA. Find the value of z.
问题：如图所示，AB=z，AC=x，AY=17，BC=y，YC=6，AY垂直于BY，BA垂直于CA。求z的值。
Construction (CDL):
Shape(BA,AY,YB)
Shape(AC,CY,YA)
Collinear(BYC)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AY),17)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(YC),6)
PerpendicularBetweenLine(AY,BY)
PerpendicularBetweenLine(BA,CA)
Goal: Value(z)
Answer: 85*sqrt(13)/6
Theorem sequence:
adjacent_complementary_angle(1,CYA,AYB)
mirror_similar_triangle_judgment_aa(1,ACY,BAC)
right_triangle_judgment_angle(1,CYA)
right_triangle_property_pythagorean(1,CYA)
mirror_similar_triangle_property_line_ratio(1,ACY,BAC)
mirror_similar_triangle_property_line_ratio(1,CYA,CBA)
Problem: As shown in the diagram, AB=17, AC=12, ∠ABC=x°, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=17，AC=12，∠ABC=x°，BC⊥AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),17)
Equal(LengthOfLine(AC),12)
Equal(MeasureOfAngle(ABC),x)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 180*asin(12/17)/pi
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, AB=25, BC=15, CA=x, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=25，BC=15，CA=x，BC⊥AC。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Shape(BA,AD,DB)
Conditions (CDL):
Equal(LengthOfLine(AB),25)
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(CA),x)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 20
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠BIG=24°, ∠DIB=84°, ∠GIE=x°, EI⊥DI. Find the value of x.
问题：如图所示，∠BIG=24°，∠DIB=84°，∠GIE=x°，EI垂直于DI。求x的值。
Construction (CDL):
Shape(ID,IDE,EI)
Shape(IE,IEG,GI)
Shape(IG,IGB,BI)
Shape(IB,IBD,DI)
Cocircular(I,EGBD)
Conditions (CDL):
Equal(MeasureOfAngle(BIG),24)
Equal(MeasureOfAngle(DIB),84)
Equal(MeasureOfAngle(GIE),x)
PerpendicularBetweenLine(EI,DI)
Goal: Value(x)
Answer: 162
Theorem sequence:
angle_addition(1,EID,DIB)
angle_addition(1,EIB,BIG)
round_angle(1,GIE,EIG)
Problem: As shown in the diagram, ∠BGC=40°, ∠DGF=53°, CB⊥GB, FG⊥CG, GF is perpendicular to DF. Find the measure of ∠GCB.
问题：如图所示，∠BGC=40°，∠DGF=53°，CB垂直于GB，FG⊥CG，GF⊥DF。求∠GCB的大小。
Construction (CDL):
Shape(DG,GF,FD)
Shape(FG,GB,BA,AF)
Shape(BG,GC,CB)
Shape(AB,BC,CA)
Collinear(DFA)
Collinear(GBA)
Conditions (CDL):
Equal(MeasureOfAngle(BGC),40)
Equal(MeasureOfAngle(DGF),53)
PerpendicularBetweenLine(CB,GB)
PerpendicularBetweenLine(FG,CG)
PerpendicularBetweenLine(GF,DF)
Goal: Value(MeasureOfAngle(GCB))
Answer: 50
Theorem sequence:
triangle_property_angle_sum(1,BGC)
Problem: As shown in the diagram, ∠ACB=2*x°, ∠DCA=x°. Find the value of x.
问题：如图所示，∠ACB=2*x°，∠DCA=x°。求x的值。
Construction (CDL):
Shape(DC,CA)
Shape(AC,CB)
Collinear(DCB)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),2*x)
Equal(MeasureOfAngle(DCA),x)
Goal: Value(x)
Answer: 60
Theorem sequence:
adjacent_complementary_angle(1,DCA,ACB)
Problem: As shown in the diagram, PQ=25, RS=10, RT=x, SQ=5, PQ is parallel to TS, TR⊥SR. Find the value of x.
问题：如图所示，PQ=25，RS=10，RT=x，SQ=5，PQ平行于TS，TR⊥SR。求x的值。
Construction (CDL):
Shape(TR,RS,ST)
Shape(PT,TS,SQ,QP)
Collinear(PTR)
Collinear(RSQ)
Conditions (CDL):
Equal(LengthOfLine(PQ),25)
Equal(LengthOfLine(RS),10)
Equal(LengthOfLine(RT),x)
Equal(LengthOfLine(SQ),5)
ParallelBetweenLine(PQ,TS)
PerpendicularBetweenLine(TR,SR)
Goal: Value(x)
Answer: 40/3
Theorem sequence:
parallel_property_corresponding_angle(1,ST,QP,R)
similar_triangle_judgment_aa(1,TRS,PRQ)
line_addition(1,RS,SQ)
right_triangle_judgment_angle(1,TRS)
right_triangle_property_pythagorean(1,TRS)
similar_triangle_property_line_ratio(1,RST,RQP)
similar_triangle_property_line_ratio(1,TRS,PRQ)
Problem: As shown in the diagram, the measure of ⌒BJL is 58, the measure of arc BMJ is 164, KJ is the tangent to circle B. Find the measure of ∠JKL.
问题：如图所示，⌒BJL的角度为58，弧BMJ的角度为164，圆O的切线为KJ。求∠JKL的大小。
Construction (CDL):
Shape(BLM,ML)
Shape(BMJ,BJL,LM)
Shape(KL,BJL,JK)
Collinear(KLM)
Cocircular(B,MJL)
Conditions (CDL):
Equal(MeasureOfArc(BJL),58)
Equal(MeasureOfArc(BMJ),164)
IsTangentOfCircle(KJ,B)
Goal: Value(MeasureOfAngle(JKL))
Answer: 53
Theorem sequence:
circle_property_circular_power_tangent_and_segment_angle(1,KJ,KLM,B)
Problem: As shown in the diagram, AB=y, CD=9, CE=x, DE=7, XA=20, XB=14, triangle CED is similar to triangle AXB.. Find the value of x.
问题：如图所示，AB=y，CD=9，CE=x，DE=7，XA=20，XB=14，△CED与△AXB是相似三角形。求x的值。
Construction (CDL):
Shape(CE,ED,DC)
Shape(AX,XB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(CD),9)
Equal(LengthOfLine(CE),x)
Equal(LengthOfLine(DE),7)
Equal(LengthOfLine(XA),20)
Equal(LengthOfLine(XB),14)
SimilarBetweenTriangle(CED,AXB)
Goal: Value(x)
Answer: 10
Theorem sequence:
similar_triangle_property_line_ratio(1,CED,AXB)
similar_triangle_property_line_ratio(1,DCE,BAX)
Problem: As shown in the diagram, QR=3*y-1, QR=RS, QS=4*y-9, RS=y+11. Find the value of y.
问题：如图所示，QR=3*y-1，QR=RS，QS=4*y-9，RS=y+11。求y的值。
Construction (CDL):
Shape(RQ,QS,SR)
Conditions (CDL):
Equal(LengthOfLine(QR),3*y-1)
Equal(LengthOfLine(QR),LengthOfLine(RS))
Equal(LengthOfLine(QS),4*y-9)
Equal(LengthOfLine(RS),y+11)
Goal: Value(y)
Answer: 6
Theorem sequence:
Problem: As shown in the diagram, HW=20, XZ=39, YG=21, WH is perpendicular to XH, YG⊥ZG. Find the sum of the area of triangle YXZ and the area of △WZX.
问题：如图所示，HW=20，XZ=39，YG=21，WH垂直于XH，YG垂直于ZG。求三角形YXZ的面积与三角形WZX的面积之和。
Construction (CDL):
Shape(YX,XH,HG,GY)
Shape(YG,GZ,ZY)
Shape(XW,WH,HX)
Shape(HW,WZ,ZG,GH)
Collinear(XHGZ)
Conditions (CDL):
Equal(LengthOfLine(HW),20)
Equal(LengthOfLine(XZ),39)
Equal(LengthOfLine(YG),21)
PerpendicularBetweenLine(WH,XH)
PerpendicularBetweenLine(YG,ZG)
Goal: Value(Add(AreaOfTriangle(YXZ),AreaOfTriangle(WZX)))
Answer: 1599/2
Theorem sequence:
adjacent_complementary_angle(1,GHW,WHX)
adjacent_complementary_angle(1,XGY,YGZ)
altitude_of_triangle_judgment(1,YG,YXZ)
altitude_of_triangle_judgment(1,WH,WZX)
triangle_area_formula_common(1,YXZ)
triangle_area_formula_common(1,WZX)
Problem: As shown in the diagram, AB=30, CD=30, the measure of arc XCZ is 40, X is the center of ⊙X, AM⊥YM, DN is perpendicular to ZN. Find the measure of arc XBY.
问题：如图所示，AB=30，CD=30，弧XCZ的角度为40，圆X的圆心为X，AM⊥YM，DN⊥ZN。求⌒XBY的角度。
Construction (CDL):
Shape(XYA,AM,MY)
Shape(XM,MA,XAC,CN,NX)
Shape(NC,XCZ,ZN)
Shape(NZ,XZD,DN)
Shape(XN,ND,XDB,BM,MX)
Shape(YM,MB,XBY)
Collinear(AMB)
Collinear(YMX)
Collinear(XNZ)
Collinear(CND)
Cocircular(X,ACZDBY)
Conditions (CDL):
Equal(LengthOfLine(AB),30)
Equal(LengthOfLine(CD),30)
Equal(MeasureOfArc(XCZ),40)
IsCentreOfCircle(X,X)
PerpendicularBetweenLine(AM,YM)
PerpendicularBetweenLine(DN,ZN)
Goal: Value(MeasureOfArc(XBY))
Answer: 40
Theorem sequence:
vertical_angle(1,CNX,DNZ)
circle_property_chord_perpendicular_bisect_arc(1,XCD,XNZ)
congruent_arc_judgment_length_equal(1,XCZ,XZD)
congruent_arc_property_measure_equal(1,XCZ,XZD)
arc_addition_measure(1,XCZ,XZD)
congruent_arc_judgment_chord_equal(1,XBA,XCD)
congruent_arc_property_measure_equal(1,XBA,XCD)
vertical_angle(1,AMY,BMX)
circle_property_chord_perpendicular_bisect_arc(1,XBA,XMY)
congruent_arc_judgment_length_equal(1,XBY,XYA)
congruent_arc_property_measure_equal(1,XBY,XYA)
arc_addition_measure(1,XBY,XYA)
Problem: As shown in the diagram, ∠AXW=57°, the center of ⊙A is A, XY is the diameter of circle A. Find the measure of ⌒AXW.
问题：如图所示，∠AXW=57°，A是圆A的圆心，XY是圆A的直径。求弧AXW的角度。
Construction (CDL):
Shape(AXW,WX)
Shape(AX,XW,WY,YA)
Shape(AWY,YW)
Shape(XA,AY,AYX)
Collinear(XAY)
Cocircular(A,XWY)
Conditions (CDL):
Equal(MeasureOfAngle(AXW),57)
IsCentreOfCircle(A,A)
IsDiameterOfCircle(XY,A)
Goal: Value(MeasureOfArc(AXW))
Answer: 66
Theorem sequence:
diameter_of_circle_property_right_angle(1,XWY,A)
triangle_property_angle_sum(1,XWY)
arc_property_circumference_angle_external(1,AXW,Y)