SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠ABC=3*y-11°, ∠BAE=y+19°, ∠CBL=x°, ∠EBA=4*z+2°, CB is parallel to AE, AE is perpendicular to BE. Find the value of z.
问题：如图所示，∠ABC=3*y-11°，∠BAE=y+19°，∠CBL=x°，∠EBA=4*z+2°，CB∥AE，AE垂直于BE。求z的值。
Construction (CDL):
Shape(BA,AE,EB)
Shape(AB,BC)
Shape(DA,AB)
Shape(HA,AD)
Shape(EA,AH)
Shape(GE,EA)
Shape(MB,BE)
Shape(FB,BM)
Shape(LB,BF)
Shape(CB,BL)
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(CB,AE)
PerpendicularBetweenLine(AE,BE)
Goal: Value(z)
Answer: 27/2
Theorem sequence:
parallel_property_alternate_interior_angle(2,CB,AE)
triangle_property_angle_sum(1,BAE)
Problem: As shown in the diagram, AC=18, AD=x, ∠CAD=52°, AD⊥CD. Find the value of x.
问题：如图所示，AC=18，AD=x，∠CAD=52°，AD⊥CD。求x的值。
Construction (CDL):
Shape(AD,DC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),18)
Equal(LengthOfLine(AD),x)
Equal(MeasureOfAngle(CAD),52)
PerpendicularBetweenLine(AD,CD)
Goal: Value(x)
Answer: 18*cos(13*pi/45)
Theorem sequence:
cosine_theorem(1,ADC)
cosine_theorem(1,DCA)
Problem: As shown in the diagram, BX=x, DX=12*x+2, EX=x+1, XC=10*x. Find the value of x.
问题：如图所示，BX=x，DX=12*x+2，EX=x+1，XC=10*x。求x的值。
Construction (CDL):
Shape(DX,XC,ACD)
Shape(XE,AEB,BX)
Shape(XD,ADE,EX)
Shape(XB,ABC,CX)
Collinear(DXB)
Collinear(CXE)
Cocircular(A,DEBC)
Conditions (CDL):
Equal(LengthOfLine(BX),x)
Equal(LengthOfLine(DX),12*x+2)
Equal(LengthOfLine(EX),x+1)
Equal(LengthOfLine(XC),10*x)
Goal: Value(x)
Answer: 4
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,EXC,DXB,A)
Problem: As shown in the diagram, ∠DFH=4*x°, ∠HFA=2*x-6°. Find the measure of ∠HFA.
问题：如图所示，∠DFH=4*x°，∠HFA=2*x-6°。求∠HFA的大小。
Construction (CDL):
Shape(DF,FH)
Shape(BF,FD)
Shape(AF,FB)
Shape(HF,FA)
Collinear(DFA)
Collinear(BFH)
Conditions (CDL):
Equal(MeasureOfAngle(DFH),4*x)
Equal(MeasureOfAngle(HFA),2*x-6)
Goal: Value(MeasureOfAngle(HFA))
Answer: 56
Theorem sequence:
adjacent_complementary_angle(1,DFH,HFA)
Problem: As shown in the diagram, the area of MJKL is 138, the area of QRST is 5, JK=5, QT=x, MJKL is similar to STQR, MJKL is a trapezoid, quadrilateral QRST is a trapezoid. Find The ratio of similar quadrilaterals MJKL and STQR.
问题：如图所示，四边形MJKL的面积为138，QRST的面积为5，JK=5，QT=x，四边形MJKL相似于四边形STQR，MJ和KL是梯形MJKL的腰，QR和ST是梯形QRST的腰。求四边形MJKL和四边形STQR的相似比。
Construction (CDL):
Shape(MJ,JK,KL,LM)
Shape(QR,RS,ST,TQ)
Conditions (CDL):
Equal(AreaOfQuadrilateral(MJKL),138)
Equal(AreaOfQuadrilateral(QRST),5)
Equal(LengthOfLine(JK),5)
Equal(LengthOfLine(QT),x)
SimilarBetweenQuadrilateral(MJKL,STQR)
Trapezoid(MJKL)
Trapezoid(QRST)
Goal: Value(RatioOfSimilarQuadrilateral(MJKL,STQR))
Answer: sqrt(690)/5
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,MJKL,STQR)
Problem: As shown in the diagram, ST=11*x-2, TU=8*x+4, UV=15*x, △RST is mirror congruent to △VTU, RS is perpendicular to TS, TU⊥VU. Find the area of △RST.
问题：如图所示，ST=11*x-2，TU=8*x+4，UV=15*x，△RST与△VTU是镜像全等三角形，RS垂直于TS，TU⊥VU。求三角形RST的面积。
Construction (CDL):
Shape(RS,ST,TR)
Shape(TU,UV,VT)
Collinear(STU)
Conditions (CDL):
Equal(LengthOfLine(ST),11*x-2)
Equal(LengthOfLine(TU),8*x+4)
Equal(LengthOfLine(UV),15*x)
MirrorCongruentBetweenTriangle(RST,VTU)
PerpendicularBetweenLine(RS,TS)
PerpendicularBetweenLine(TU,VU)
Goal: Value(AreaOfTriangle(RST))
Answer: 300
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,RST,VTU)
mirror_congruent_triangle_property_line_equal(1,TRS,TUV)
triangle_area_formula_sine(1,STR)
Problem: As shown in the diagram, AB=21, BD=17, DE=8, quadrilateral ACDB is a parallelogram, BE is perpendicular to DE. Find the perimeter of quadrilateral ACDB.
问题：如图所示，AB=21，BD=17，DE=8，AB和CD是▱ACDB的一组对边，BE⊥DE。求四边形ACDB的周长。
Construction (CDL):
Shape(AC,CE,EB,BA)
Shape(BE,ED,DB)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AB),21)
Equal(LengthOfLine(BD),17)
Equal(LengthOfLine(DE),8)
Parallelogram(ACDB)
PerpendicularBetweenLine(BE,DE)
Goal: Value(PerimeterOfQuadrilateral(ACDB))
Answer: 76
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ACDB)
parallelogram_property_opposite_line_equal(1,CDBA)
quadrilateral_perimeter_formula(1,ACDB)
Problem: As shown in the diagram, RS=x, RT=y+3, TS=49, UV=21, WU=29, WV=20, ∠RTS=∠WVU, ∠SRT=∠VUW, ∠TSR=∠VWU, triangle TRS is similar to triangle VUW.. Find the value of y.
问题：如图所示，RS=x，RT=y+3，TS=49，UV=21，WU=29，WV=20，∠RTS=∠WVU，∠SRT=∠VUW，∠TSR=∠VWU，三角形TRS相似于三角形VUW。求y的值。
Construction (CDL):
Shape(ST,TR,RS)
Shape(VU,UW,WV)
Conditions (CDL):
Equal(LengthOfLine(RS),x)
Equal(LengthOfLine(RT),y+3)
Equal(LengthOfLine(TS),49)
Equal(LengthOfLine(UV),21)
Equal(LengthOfLine(WU),29)
Equal(LengthOfLine(WV),20)
Equal(MeasureOfAngle(RTS),MeasureOfAngle(WVU))
Equal(MeasureOfAngle(SRT),MeasureOfAngle(VUW))
Equal(MeasureOfAngle(TSR),MeasureOfAngle(VWU))
SimilarBetweenTriangle(TRS,VUW)
Goal: Value(y)
Answer: 969/20
Theorem sequence:
similar_triangle_property_line_ratio(1,STR,WVU)
similar_triangle_property_line_ratio(1,RST,UWV)
Problem: As shown in the diagram, AC=AY, AD=x+2, CB=5/3*x+11, DB=3*y-9, DB=YD, YD=2*y+6. Find the value of y.
问题：如图所示，AC=AY，AD=x+2，CB=5/3*x+11，DB=3*y-9，DB=YD，YD=2*y+6。求y的值。
Construction (CDL):
Shape(CA,AD,DB,BC)
Shape(AY,YD,DA)
Collinear(CAY)
Collinear(BDY)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(AY))
Equal(LengthOfLine(AD),x+2)
Equal(LengthOfLine(CB),5/3*x+11)
Equal(LengthOfLine(DB),3*y-9)
Equal(LengthOfLine(DB),LengthOfLine(YD))
Equal(LengthOfLine(YD),2*y+6)
Goal: Value(y)
Answer: 15
Theorem sequence:
Problem: As shown in the diagram, ∠PTS=150°, RS is perpendicular to TS, ∠RST is the right angle of the right triangle RST. Find the measure of ∠TRS.
问题：如图所示，∠PTS=150°，RS⊥TS，RS和ST是直角△RST的两个直角边。求∠TRS的大小。
Construction (CDL):
Shape(RS,ST,TR)
Shape(PT,TS)
Collinear(RTP)
Conditions (CDL):
Equal(MeasureOfAngle(PTS),150)
PerpendicularBetweenLine(RS,TS)
RightTriangle(RST)
Goal: Value(MeasureOfAngle(TRS))
Answer: 60
Theorem sequence:
adjacent_complementary_angle(1,PTS,STR)
triangle_property_angle_sum(1,RST)
Problem: As shown in the diagram, ∠ADF=100°, ∠EFC=75°. Find the measure of ∠FDB.
问题：如图所示，∠ADF=100°，∠EFC=75°。求∠FDB的大小。
Construction (CDL):
Shape(DB,BF,FD)
Shape(GD,DA)
Shape(BD.DG)
Shape(AD,DF)
Shape(DF,FE)
Shape(CF,FB)
Shape(EF,FC)
Collinear(BDA)
Collinear(BFE)
Collinear(GDFC)
Conditions (CDL):
Equal(MeasureOfAngle(ADF),100)
Equal(MeasureOfAngle(EFC),75)
Goal: Value(MeasureOfAngle(FDB))
Answer: 80
Theorem sequence:
adjacent_complementary_angle(1,ADF,FDB)
Problem: As shown in the diagram, AB=4, CB=x, DC=x+2, BA is the tangent to circle E. Find the value of x.
问题：如图所示，AB=4，CB=x，DC=x+2，BA是圆O的切线。求x的值。
Construction (CDL):
Shape(CB,BA,ECA)
Shape(DE,EC,ECA,EAD)
Shape(CE,ED,EDC)
Collinear(DECB)
Cocircular(E,ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(CB),x)
Equal(LengthOfLine(DC),x+2)
IsTangentOfCircle(BA,E)
Goal: Value(x)
Answer: -1/2+sqrt(33)/2
Theorem sequence:
line_addition(1,DC,CB)
circle_property_circular_power_tangent_and_segment_line(1,BA,BCD,E)
Problem: As shown in the diagram, ∠ABC=x°, ∠BCD=2*x+18°, ∠CDA=x°, ∠DAB=4*x-26°. Find the measure of ∠BCD.
问题：如图所示，∠ABC=x°，∠BCD=2*x+18°，∠CDA=x°，∠DAB=4*x-26°。求∠BCD的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),x)
Equal(MeasureOfAngle(BCD),2*x+18)
Equal(MeasureOfAngle(CDA),x)
Equal(MeasureOfAngle(DAB),4*x-26)
Goal: Value(MeasureOfAngle(BCD))
Answer: 110
Theorem sequence:
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, DG=FG, FG=FH, FH=DG, ∠FDH=28°, ∠FGH and ∠GHF are the base angles of the isosceles △ FGH. Find the measure of ∠GFD.
问题：如图所示，DG=FG，FG=FH，FH=DG，∠FDH=28°，∠FGH和∠GHF是等腰三角形FGH的底角。求∠GFD的大小。
Construction (CDL):
Shape(DG,GF,FD)
Shape(GH,HF,FG)
Collinear(DGH)
Conditions (CDL):
Equal(LengthOfLine(DG),LengthOfLine(FG))
Equal(LengthOfLine(FG),LengthOfLine(FH))
Equal(LengthOfLine(FH),LengthOfLine(DG))
Equal(MeasureOfAngle(FDH),28)
IsoscelesTriangle(FGH))
Goal: Value(MeasureOfAngle(GFD))
Answer: 28
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,GFD)
isosceles_triangle_property_angle_equal(1,GFD)
Problem: As shown in the diagram, AC=x, BC=41, ∠ABC=111°, ∠CAB=19°. Find the value of x.
问题：如图所示，AC=x，BC=41，∠ABC=111°，∠CAB=19°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),41)
Equal(MeasureOfAngle(ABC),111)
Equal(MeasureOfAngle(CAB),19)
Goal: Value(x)
Answer: (-41*sqrt(5)*sqrt(sqrt(5)+5)-41*sqrt(3)*sqrt(sqrt(5)+5)+41*sqrt(2)+41*sqrt(6)+41*sqrt(sqrt(5)+5)+41*sqrt(10)+41*sqrt(30)+41*sqrt(15)*sqrt(sqrt(5)+5))/(16*sin(19*pi/180))
Theorem sequence:
sine_theorem(1,CAB)
Problem: As shown in the diagram, JZ=12, PZ=20, ∠JYP=27°, ∠YXP=31°, P is the incenter of triangle XZY, PJ⊥YJ, YK is perpendicular to PK. Find the length of line PK.
问题：如图所示，JZ=12，PZ=20，∠JYP=27°，∠YXP=31°，△XZY的内心为P，PJ⊥YJ，YK⊥PK。求直线PK的长度。
Construction (CDL):
Shape(XL,LP,PX)
Shape(LZ,ZP,PL)
Shape(ZJ,JP,PZ)
Shape(JY,YP,PJ)
Shape(YK,KP,PY)
Shape(KX,XP,PK)
Collinear(XLZ)
Collinear(ZJY)
Collinear(XKY)
Collinear(XPJ)
Collinear(LPY)
Collinear(ZPK)
Conditions (CDL):
Equal(LengthOfLine(JZ),12)
Equal(LengthOfLine(PZ),20)
Equal(MeasureOfAngle(JYP),27)
Equal(MeasureOfAngle(YXP),31)
IsIncenterOfTriangle(P,XZY)
PerpendicularBetweenLine(PJ,YJ)
PerpendicularBetweenLine(YK,PK)
Goal: Value(LengthOfLine(PK))
Answer: 16
Theorem sequence:
adjacent_complementary_angle(1,ZJP,PJY)
right_triangle_judgment_angle(1,ZJP)
right_triangle_property_pythagorean(1,ZJP)
bisector_of_angle_judgment_angle_equal(1,YP,JYK)
bisector_of_angle_property_distance_equal(1,YP,JYK)
Problem: As shown in the diagram, ∠ADC=x°, ∠BAD=x°, ∠CBA=2*x°, ∠DCB=2*x°. Find the measure of ∠ADC.
问题：如图所示，∠ADC=x°，∠BAD=x°，∠CBA=2*x°，∠DCB=2*x°。求∠ADC的大小。
Construction (CDL):
Shape(BA,AD,DC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),x)
Equal(MeasureOfAngle(BAD),x)
Equal(MeasureOfAngle(CBA),2*x)
Equal(MeasureOfAngle(DCB),2*x)
Goal: Value(MeasureOfAngle(ADC))
Answer: 60
Theorem sequence:
quadrilateral_property_angle_sum(1,BADC)
Problem: As shown in the diagram, KP=3, XJ=8, P is the centroid of triangle XYZ. Find the length of line XK.
问题：如图所示，KP=3，XJ=8，三角形XYZ的重心为O。求直线XK的长度。
Construction (CDL):
Shape(XJ,JP,PX)
Shape(JY,YP,PJ)
Shape(YK,KP,PY)
Shape(PK,KZ,ZP)
Shape(PZ,ZL,LP)
Shape(PL,LX,XP)
Collinear(YJX)
Collinear(YKZ)
Collinear(YPL)
Collinear(XPK)
Collinear(XLZ)
Collinear(JPZ)
Conditions (CDL):
Equal(LengthOfLine(KP),3)
Equal(LengthOfLine(XJ),8)
IsCentroidOfTriangle(P,XYZ)
Goal: Value(LengthOfLine(XK))
Answer: 9
Theorem sequence:
line_addition(1,XP,PK)
centroid_of_triangle_property_line_ratio(1,P,XYZ,K)
Problem: As shown in the diagram, AB=59, AC=39, BC=36, ∠ABC=x°. Find the value of x.
问题：如图所示，AB=59，AC=39，BC=36，∠ABC=x°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),59)
Equal(LengthOfLine(AC),39)
Equal(LengthOfLine(BC),36)
Equal(MeasureOfAngle(ABC),x)
Goal: Value(x)
Answer: 180*acos(407/531)/pi
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AC=12, CE=4, ∠CBE=35°, BCAD is a parallelogram, BE is perpendicular to CE. Find the perimeter of quadrilateral BCAD.
问题：如图所示，AC=12，CE=4，∠CBE=35°，CB和AD是▱BCAD的一组对边，BE垂直于CE。求BCAD的周长。
Construction (CDL):
Shape(BE,EC,CB)
Shape(BC,CA,AD,DB)
Collinear(ECA)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(CE),4)
Equal(MeasureOfAngle(CBE),35)
Parallelogram(BCAD)
PerpendicularBetweenLine(BE,CE)
Goal: Value(PerimeterOfQuadrilateral(BCAD))
Answer: 8/sin(7*pi/36)+24
Theorem sequence:
sine_theorem(1,BEC)
sine_theorem(1,ECB)
parallelogram_property_opposite_line_equal(1,BCAD)
parallelogram_property_opposite_line_equal(1,CADB)
quadrilateral_perimeter_formula(1,BCAD)
Problem: As shown in the diagram, ∠DBE=25°, the measure of arc CAG is 5*x, the measure of ⌒CED is 90. Find the value of x.
问题：如图所示，∠DBE=25°，弧CAG的角度为5*x，弧CED的角度为90。求x的值。
Construction (CDL):
Shape(AB,BG,CAG)
Shape(EG,CGE)
Shape(CED,DA,CAG,GE)
Shape(AD,CDA)
Collinear(EGB)
Collinear(DAB)
Cocircular(C,GEDA)
Conditions (CDL):
Equal(MeasureOfAngle(DBE),25)
Equal(MeasureOfArc(CAG),5*x)
Equal(MeasureOfArc(CED),90)
Goal: Value(x)
Answer: 8
Theorem sequence:
circle_property_circular_power_segment_and_segment_angle(1,BAD,BGE,C)
Problem: As shown in the diagram, AB=12, AC=x, BC=x, AC⊥BC. Find the value of x.
问题：如图所示，AB=12，AC=x，BC=x，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 6*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, ∠CBA=x°, the measure of arc FCE is 125, the measure of arc FDA is 117. Find the value of x.
问题：如图所示，∠CBA=x°，⌒FCE的角度为125，⌒FDA的角度为117。求x的值。
Construction (CDL):
Shape(DA,AB,BD)
Shape(BA,AC,CB)
Shape(BC,CE,EB)
Shape(DB,BE,FED)
Shape(AD,FDA)
Shape(CA,FAC)
Shape(EC,FCE)
Collinear(DBC)
Collinear(EBA)
Cocircular(F,EDAC)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),x)
Equal(MeasureOfArc(FCE),125)
Equal(MeasureOfArc(FDA),117)
Goal: Value(x)
Answer: 59
Theorem sequence:
arc_property_circumference_angle_external(1,FCE,A)
arc_property_circumference_angle_external(1,FDA,C)
triangle_property_angle_sum(1,BAC)
Problem: As shown in the diagram, ∠EAD=x°, ∠FBG=120°, ∠HEB=160°, LB∥EG, AD⊥GD. Find the value of x.
问题：如图所示，∠EAD=x°，∠FBG=120°，∠HEB=160°，LB∥EG，AD⊥GD。求x的值。
Construction (CDL):
Shape(BE,EG,GB)
Shape(EA,AD,DG,GE)
Shape(EB,BL)
Shape(HE,EB)
Shape(AE,EH)
Shape(CG,GD)
Shape(BG,GC)
Shape(FB,BG)
Shape(GB,BL)
Collinear(LBF)
Collinear(HEGC)
Collinear(AEB)
Collinear(BGD)
Conditions (CDL):
Equal(MeasureOfAngle(EAD),x)
Equal(MeasureOfAngle(FBG),120)
Equal(MeasureOfAngle(HEB),160)
ParallelBetweenLine(LB,EG)
PerpendicularBetweenLine(AD,GD)
Goal: Value(x)
Answer: 50
Theorem sequence:
adjacent_complementary_angle(1,HEB,BEG)
parallel_property_alternate_interior_angle(2,LB,EG)
angle_addition(1,GBE,EBL)
adjacent_complementary_angle(1,FBG,GBL)
triangle_property_angle_sum(1,DBA)
Problem: As shown in the diagram, AC=20, AD=AB, AD=BC, AD=CD, BD=24, BE is perpendicular to CE. Find the area of quadrilateral BADC.
问题：如图所示，AC=20，AD=AB，AD=BC，AD=CD，BD=24，BE⊥CE。求四边形BADC的面积。
Construction (CDL):
Shape(BA,AE,EB)
Shape(EA,AD,DE)
Shape(ED,DC,CE)
Shape(BE,EC,CB)
Collinear(AEC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(AD),LengthOfLine(AB))
Equal(LengthOfLine(AD),LengthOfLine(BC))
Equal(LengthOfLine(AD),LengthOfLine(CD))
Equal(LengthOfLine(BD),24)
PerpendicularBetweenLine(BE,CE)
Goal: Value(AreaOfQuadrilateral(BADC))
Answer: 240
Theorem sequence:
kite_judgment_equal_and_equal(1,BADC)
kite_area_formula_diagonal(1,BADC)
Problem: As shown in the diagram, AB=2, AE=3, BC=6, ED=2*x-3, EB∥DC. Find the length of line ED.
问题：如图所示，AB=2，AE=3，BC=6，ED=2*x-3，EB平行于DC。求直线ED的长度。
Construction (CDL):
Shape(DC,CB,BE,ED)
Shape(EB,BA,AE)
Collinear(DEA)
Collinear(CBA)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(AE),3)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(ED),2*x-3)
ParallelBetweenLine(EB,DC)
Goal: Value(LengthOfLine(ED))
Answer: 9
Theorem sequence:
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
line_addition(1,AE,ED)
line_addition(1,AB,BC)
similar_triangle_property_line_ratio(1,BAE,CAD)
similar_triangle_property_line_ratio(1,EBA,DCA)
Problem: As shown in the diagram, ∠RWZ=45°, ∠TRW=4*x°, ∠WZT=2*x+30°. Find the measure of ∠WZT.
问题：如图所示，∠RWZ=45°，∠TRW=4*x°，∠WZT=2*x+30°。求∠WZT的大小。
Construction (CDL):
Shape(RW,WT,TR)
Shape(WZ,ZT,TW)
Shape(WR,ORW)
Shape(ZW,OWZ)
Shape(TZ,OZT)
Shape(RT,OTR)
Cocircular(O,RWZT)
Conditions (CDL):
Equal(MeasureOfAngle(RWZ),45)
Equal(MeasureOfAngle(TRW),4*x)
Equal(MeasureOfAngle(WZT),2*x+30)
Goal: Value(MeasureOfAngle(WZT))
Answer: 80
Theorem sequence:
arc_property_circumference_angle_external(1,OTW,Z)
arc_property_circumference_angle_internal(1,OTW,R)
Problem: As shown in the diagram, AF=2, BF=7, FC=x, FD=3. Find the value of x.
问题：如图所示，AF=2，BF=7，FC=x，FD=3。求x的值。
Construction (CDL):
Shape(EDB,BF,FD)
Shape(EBC,CF,FB)
Shape(ECA,AF,FC)
Shape(EAD,DF,FA)
Collinear(BFA)
Collinear(DFC)
Cocircular(E,DBCA)
Conditions (CDL):
Equal(LengthOfLine(AF),2)
Equal(LengthOfLine(BF),7)
Equal(LengthOfLine(FC),x)
Equal(LengthOfLine(FD),3)
Goal: Value(x)
Answer: 14/3
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,BFA,DFC,E)
Problem: As shown in the diagram, AH=4, BC=6, HC=BH, AH⊥BH. Find the perimeter of triangle ACB.
问题：如图所示，AH=4，BC=6，HC=BH，AH垂直于BH。求三角形ACB的周长。
Construction (CDL):
Shape(AC,CH,HA)
Shape(AH,HB,BA)
Collinear(CHB)
Conditions (CDL):
Equal(LengthOfLine(AH),4)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(HC),LengthOfLine(BH))
PerpendicularBetweenLine(AH,BH)
Goal: Value(PerimeterOfTriangle(ACB))
Answer: 16
Theorem sequence:
adjacent_complementary_angle(1,CHA,AHB)
line_addition(1,CH,HB)
right_triangle_judgment_angle(1,CHA)
right_triangle_judgment_angle(1,AHB)
right_triangle_property_pythagorean(1,CHA)
right_triangle_property_pythagorean(1,AHB)
triangle_perimeter_formula(1,ACB)
Problem: As shown in the diagram, ∠CGK=2*x-54°, ∠JHC=x°. Find the measure of ∠JHC.
问题：如图所示，∠CGK=2*x-54°，∠JHC=x°。求∠JHC的大小。
Construction (CDL):
Shape(AJH,HJ)
Shape(AHG,GC,CH)
Shape(AGK,KG)
Shape(AKJ,JC,CK)
Shape(CJ,JH,HC)
Shape(CG,GK,KC)
Collinear(HCK)
Collinear(GCJ)
Cocircular(A,HGKJ)
Conditions (CDL):
Equal(MeasureOfAngle(CGK),2*x-54)
Equal(MeasureOfAngle(JHC),x)
Goal: Value(MeasureOfAngle(JHC))
Answer: 54
Theorem sequence:
arc_property_circumference_angle_external(1,AKJ,H)
arc_property_circumference_angle_external(1,AKJ,G)
Problem: As shown in the diagram, ∠ACB=x°, ∠BAC=32°, ∠CBA=57°. Find the value of x.
问题：如图所示，∠ACB=x°，∠BAC=32°，∠CBA=57°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),x)
Equal(MeasureOfAngle(BAC),32)
Equal(MeasureOfAngle(CBA),57)
Goal: Value(x)
Answer: 91
Theorem sequence:
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, AB=41, AD=y, CH=5*x-8, DB=39, DC=72-3*x, the center of circle A is A, the tangent to circle A is BD, the tangent to circle A is CH. Find the value of y.
问题：如图所示，AB=41，AD=y，CH=5*x-8，DB=39，DC=72-3*x，A是⊙A的圆心，BD是⊙O的切线，CH是⊙O的切线。求y的值。
Construction (CDL):
Shape(HC,CD,AHD)
Shape(AFD,DA,AF)
Shape(ADF,FA,AD)
Shape(ADF,DB,BF)
Collinear(CDB)
Collinear(AFB)
Cocircular(A,HDF)
Conditions (CDL):
Equal(LengthOfLine(AB),41)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(CH),5*x-8)
Equal(LengthOfLine(DB),39)
Equal(LengthOfLine(DC),72-3*x)
IsCentreOfCircle(A,A)
IsTangentOfCircle(BD,A)
IsTangentOfCircle(CH,A)
Goal: Value(y)
Answer: 4*sqrt(10)
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BD,A,A)
right_triangle_judgment_angle(1,ADB)
right_triangle_property_pythagorean(1,ADB)
Problem: As shown in the diagram, SU=20, YW=20, the measure of ⌒RYX is 45, the center of circle R is R, SV is perpendicular to RV, YZ⊥RZ. Find the measure of ⌒RYW.
问题：如图所示，SU=20，YW=20，⌒RYX的角度为45，⊙R的圆心为R，SV⊥RV，YZ垂直于RZ。求⌒RYW的角度。
Construction (CDL):
Shape(RSY,YR,RV,VS)
Shape(RY,YZ,ZR)
Shape(RZ,ZW,WR)
Shape(RYX,XZ,ZY)
Shape(RXW,WZ,ZX)
Shape(RWU,UV,VR,RW)
Shape(RUT,TV,VU)
Shape(RTS,SV,VT)
Collinear(SVU)
Collinear(YZW)
Collinear(XZR)
Collinear(TVR)
Cocircular(R,YXWUTS)
Conditions (CDL):
Equal(LengthOfLine(SU),20)
Equal(LengthOfLine(YW),20)
Equal(MeasureOfArc(RYX),45)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(SV,RV)
PerpendicularBetweenLine(YZ,RZ)
Goal: Value(MeasureOfArc(RYW))
Answer: 90
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,R,RZ,YW)
mirror_congruent_triangle_judgment_sas(1,ZRY,ZWR)
mirror_congruent_triangle_property_angle_equal(1,RYZ,RZW)
angle_addition(1,WRZ,ZRY)
arc_property_center_angle(1,RYX,R)
arc_property_center_angle(1,RYW,R)
Problem: As shown in the diagram, BF=6, the measure of ⌒BAE is 57, the measure of ⌒BCF is 50, the measure of ⌒BDG is 80, the measure of ⌒BED is 57, the measure of ⌒BFA is 46, the measure of arc BGC is 70, the center of ⊙B is B. Find the area of the sector BFA.
问题：如图所示，BF=6，⌒BAE的角度为57，弧BCF的角度为50，⌒BDG的角度为80，弧BED的角度为57，⌒BFA的角度为46，⌒BGC的角度为70，⊙B的圆心为B。求扇形BFA的面积。
Construction (CDL):
Shape(BFA,AB,BF)
Shape(BAE,EB,BA)
Shape(BED,DB,BE)
Shape(BDG,GB,BD)
Shape(BGC,CB,BG)
Shape(BCF,FB,BC)
Cocircular(B,AEDGCF)
Conditions (CDL):
Equal(LengthOfLine(BF),6)
Equal(MeasureOfArc(BAE),57)
Equal(MeasureOfArc(BCF),50)
Equal(MeasureOfArc(BDG),80)
Equal(MeasureOfArc(BED),57)
Equal(MeasureOfArc(BFA),46)
Equal(MeasureOfArc(BGC),70)
IsCentreOfCircle(B,B)
Goal: Value(AreaOfSector(BFA))
Answer: 23*pi/5
Theorem sequence:
radius_of_circle_property_length_equal(1,BF,B)
sector_area_formula(1,BFA)
Problem: As shown in the diagram, ∠ABC=6*x+14°, ∠BCD=7*y+2°, ∠CDA=8*x-8°, ∠DAB=6*y+16°, AD and BC are opposite sides of the ▱ DABC. Find the value of x.
问题：如图所示，∠ABC=6*x+14°，∠BCD=7*y+2°，∠CDA=8*x-8°，∠DAB=6*y+16°，四边形DABC是平行四边形。求x的值。
Construction (CDL):
Shape(DA,AB,BC,CD)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),6*x+14)
Equal(MeasureOfAngle(BCD),7*y+2)
Equal(MeasureOfAngle(CDA),8*x-8)
Equal(MeasureOfAngle(DAB),6*y+16)
Parallelogram(DABC)
Goal: Value(x)
Answer: 11
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BCDA)
Problem: As shown in the diagram, BD=6, DA=2, AC is perpendicular to BC, BD⊥CD. Find the length of line CD.
问题：如图所示，BD=6，DA=2，AC⊥BC，BD垂直于CD。求直线CD的长度。
Construction (CDL):
Shape(AC,CD,DA)
Shape(CB,BD,DC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(BD),6)
Equal(LengthOfLine(DA),2)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BD,CD)
Goal: Value(LengthOfLine(CD))
Answer: 2*sqrt(3)
Theorem sequence:
line_addition(1,AD,DB)
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, AB=13, AL=5, BN=x, CN=7, DL=6, the tangent to circle Q is AL, the tangent to ⊙Q is AM, BM is the tangent to circle Q, BN is the tangent to circle Q, the tangent to ⊙Q is CN, CP is the tangent to ⊙Q, the tangent to circle Q is DL, DP is the tangent to circle Q. Find the value of x.
问题：如图所示，AB=13，AL=5，BN=x，CN=7，DL=6，圆O的切线为AL，圆O的切线为AM，BM是⊙O的切线，圆O的切线为BN，⊙O的切线为CN，⊙O的切线为CP，DL是⊙O的切线，⊙O的切线为DP。求x的值。
Construction (CDL):
Shape(QML,MA,AL)
Shape(QLP,LD,DP)
Shape(QPN,PC,CN)
Shape(QNM,NB,BM)
Collinear(AMB)
Collinear(ALD)
Collinear(DPC)
Collinear(CNB)
Cocircular(Q,MLPN)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AL),5)
Equal(LengthOfLine(BN),x)
Equal(LengthOfLine(CN),7)
Equal(LengthOfLine(DL),6)
IsTangentOfCircle(AL,Q)
IsTangentOfCircle(AM,Q)
IsTangentOfCircle(BM,Q)
IsTangentOfCircle(BN,Q)
IsTangentOfCircle(CN,Q)
IsTangentOfCircle(CP,Q)
IsTangentOfCircle(DL,Q)
IsTangentOfCircle(DP,Q)
Goal: Value(x)
Answer: 8
Theorem sequence:
tangent_of_circle_property_length_equal(1,AL,AM,Q)
tangent_of_circle_property_length_equal(1,BM,BN,Q)
line_addition(1,AM,MB)
Problem: As shown in the diagram, KJ=4, the measure of ⌒KJL is 53, K is the center of circle K. Find the area of the sector KJL.
问题：如图所示，KJ=4，弧KJL的角度为53，K是圆K的圆心。求扇形KJL的面积。
Construction (CDL):
Shape(KJL,LK,KJ)
Shape(KLJ,JK,KL)
Cocircular(K,JL)
Conditions (CDL):
Equal(LengthOfLine(KJ),4)
Equal(MeasureOfArc(KJL),53)
IsCentreOfCircle(K,K)
Goal: Value(AreaOfSector(KJL))
Answer: 106*pi/45
Theorem sequence:
radius_of_circle_property_length_equal(1,KJ,K)
sector_area_formula(1,KJL)
Problem: As shown in the diagram, AB=35, BC=22, triangle ACB is a right triangle. Find the measure of ∠BAC.
问题：如图所示，AB=35，BC=22，AC和CB是直角三角形ACB的两个直角边。求∠BAC的大小。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),35)
Equal(LengthOfLine(BC),22)
RightTriangle(ACB)
Goal: Value(MeasureOfAngle(BAC))
Answer: 180*acos(sqrt(741)/35)/pi
Theorem sequence:
cosine_theorem(1,ACB)
cosine_theorem(1,CBA)
Problem: As shown in the diagram, AB=x, BC=24, ∠ACB=47°, ∠BAC=30°. Find the value of x.
问题：如图所示，AB=x，BC=24，∠ACB=47°，∠BAC=30°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),24)
Equal(MeasureOfAngle(ACB),47)
Equal(MeasureOfAngle(BAC),30)
Goal: Value(x)
Answer: 48*sin(47*pi/180)
Theorem sequence:
sine_theorem(1,BAC)