SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, WT=3, WX is perpendicular to YX, XY⊥ZY, YZ is perpendicular to WZ, ZW⊥XW, quadrilateral XYZW is a square. Find the length of line XY.
问题：如图所示，WT=3，WX⊥YX，XY垂直于ZY，YZ⊥WZ，ZW垂直于XW，XYZW是正方形。求直线XY的长度。
Construction (CDL):
Shape(XY,YT,TX)
Shape(TY,YZ,ZT)
Shape(TZ,ZW,WT)
Shape(XT,TW,WX)
Collinear(XTZ)
Collinear(WTY)
Conditions (CDL):
Equal(LengthOfLine(WT),3)
PerpendicularBetweenLine(WX,YX)
PerpendicularBetweenLine(XY,ZY)
PerpendicularBetweenLine(YZ,WZ)
PerpendicularBetweenLine(ZW,XW)
Square(XYZW)
Goal: Value(LengthOfLine(XY))
Answer: 3*sqrt(2)
Theorem sequence:
parallelogram_property_diagonal_bisection(1,YZWX,T)
line_addition(1,YT,TW)
right_triangle_judgment_angle(1,WXY)
right_triangle_property_pythagorean(1,WXY)
Problem: As shown in the diagram, ∠ABH=3*p-10°, ∠CDB=6*r+5°, ∠JAB=4*p+15°, AJ is parallel to BH, CI is parallel to AL. Find the value of r.
问题：如图所示，∠ABH=3*p-10°，∠CDB=6*r+5°，∠JAB=4*p+15°，AJ平行于BH，CI平行于AL。求r的值。
Construction (CDL):
Shape(LA,AJ)
Shape(JA,AB)
Shape(AB,BH)
Shape(HB,BE)
Shape(EB,BD)
Shape(BD,DK)
Shape(KD,DF)
Shape(FD,DC)
Shape(DC,CG)
Shape(GC,CI)
Shape(IC,CA)
Shape(CA,AL)
Shape(AC,CD,DB,BA)
Collinear(LABE)
Collinear(ICDK)
Collinear(JACG)
Collinear(HBDF)
Conditions (CDL):
Equal(MeasureOfAngle(ABH),3*p-10)
Equal(MeasureOfAngle(CDB),6*r+5)
Equal(MeasureOfAngle(JAB),4*p+15)
ParallelBetweenLine(AJ,BH)
ParallelBetweenLine(CI,AL)
Goal: Value(r)
Answer: 10
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AJ,BH)
adjacent_complementary_angle(1,DBA,ABH)
parallel_property_collinear_extend(2,LA,IC,B)
parallel_property_collinear_extend(1,CI,BA,D)
parallel_property_ipsilateral_internal_angle(1,DI,BA)
Problem: As shown in the diagram, DA=3, DC=CB, DC⊥AC. Find the length of line AB.
问题：如图所示，DA=3，DC=CB，DC⊥AC。求直线AB的长度。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AC,CB,BA)
Collinear(DCB)
Conditions (CDL):
Equal(LengthOfLine(DA),3)
Equal(LengthOfLine(DC),LengthOfLine(CB))
PerpendicularBetweenLine(DC,AC)
Goal: Value(LengthOfLine(AB))
Answer: 3
Theorem sequence:
adjacent_complementary_angle(1,DCA,ACB)
mirror_congruent_triangle_judgment_sas(1,CAD,CBA)
mirror_congruent_triangle_property_line_equal(1,CAD,CBA)
Problem: As shown in the diagram, ∠ADK=96°, ∠HGJ=42°, GA∥HD. Find the measure of ∠GHD.
问题：如图所示，∠ADK=96°，∠HGJ=42°，GA∥HD。求∠GHD的大小。
Construction (CDL):
Shape(KD,DL)
Shape(LD,DH)
Shape(DH,HI)
Shape(IH,HC)
Shape(CH,HG)
Shape(HG,GJ)
Shape(JG,GF)
Shape(FG,GA)
Shape(GA,AM)
Shape(MA,AF)
Shape(FA,AD)
Shape(AD,DK)
Shape(DA,AG,GH,HD)
Collinear(KDHC)
Collinear(EAGJ)
Collinear(LDAM)
Collinear(IHGF)
Conditions (CDL):
Equal(MeasureOfAngle(ADK),96)
Equal(MeasureOfAngle(HGJ),42)
ParallelBetweenLine(GA,HD)
Goal: Value(MeasureOfAngle(GHD))
Answer: 42
Theorem sequence:
parallel_property_collinear_extend(1,GA,HD,J)
parallel_property_alternate_interior_angle(2,DH,GJ)
Problem: As shown in the diagram, ∠GKM=62°. Find the measure of ∠BKG.
问题：如图所示，∠GKM=62°。求∠BKG的大小。
Construction (CDL):
Shape(GK,KM)
Shape(BK,KG)
Collinear(MKB)
Conditions (CDL):
Equal(MeasureOfAngle(GKM),62)
Goal: Value(MeasureOfAngle(BKG))
Answer: 118
Theorem sequence:
adjacent_complementary_angle(1,BKG,GKM)
Problem: As shown in the diagram, SR=5, TR=3, TS=4, RT is perpendicular to ST. Find the value of cos(TS).
问题：如图所示，SR=5，TR=3，TS=4，RT⊥ST。求cos(TS)的值。
Construction (CDL):
Shape(TS,SR,RT)
Conditions (CDL):
Equal(LengthOfLine(SR),5)
Equal(LengthOfLine(TR),3)
Equal(LengthOfLine(TS),4)
PerpendicularBetweenLine(RT,ST)
Goal: Value(Cos(MeasureOfAngle(TSR)))
Answer: 4/5
Theorem sequence:
cosine_theorem(1,SRT)
Problem: As shown in the diagram, AB=5*x, CA=3*x+4, WX=22, XY=22, the center of circle A is A, AB⊥XB, XC is perpendicular to AC. Find the length of line AB.
问题：如图所示，AB=5*x，CA=3*x+4，WX=22，XY=22，⊙A的圆心为A，AB垂直于XB，XC垂直于AC。求直线AB的长度。
Construction (CDL):
Shape(AYW,WB,BA,AC,CY)
Shape(BW,AWX,XB)
Shape(AB,BX,XA)
Shape(AX,XC,CA)
Shape(CX,AXY,YC)
Collinear(WBX)
Collinear(XCY)
Cocircular(A,WXY)
Conditions (CDL):
Equal(LengthOfLine(AB),5*x)
Equal(LengthOfLine(CA),3*x+4)
Equal(LengthOfLine(WX),22)
Equal(LengthOfLine(XY),22)
IsCentreOfCircle(A,A)
PerpendicularBetweenLine(AB,XB)
PerpendicularBetweenLine(XC,AC)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,WBA,ABX)
circle_property_chord_perpendicular_bisect_chord(1,A,AB,WX)
circle_property_chord_perpendicular_bisect_chord(1,A,AC,XY)
line_addition(1,WB,BX)
line_addition(1,XC,CY)
mirror_congruent_triangle_judgment_hl(1,ABX,AXC)
mirror_congruent_triangle_property_line_equal(1,XAB,XCA)
Problem: As shown in the diagram, AG=1/5*x+3, CJ=2*y+1, CJ=JE, EG=4*x-35, JE=5*y-8, AC∥GJ. Find the value of y.
问题：如图所示，AG=1/5*x+3，CJ=2*y+1，CJ=JE，EG=4*x-35，JE=5*y-8，AC平行于GJ。求y的值。
Construction (CDL):
Shape(AG,GJ,JC,CA)
Shape(GE,EJ,JG)
Collinear(AGE)
Collinear(CJE)
Conditions (CDL):
Equal(LengthOfLine(AG),1/5*x+3)
Equal(LengthOfLine(CJ),2*y+1)
Equal(LengthOfLine(CJ),LengthOfLine(JE))
Equal(LengthOfLine(EG),4*x-35)
Equal(LengthOfLine(JE),5*y-8)
ParallelBetweenLine(AC,GJ)
Goal: Value(y)
Answer: 3
Theorem sequence:
Problem: As shown in the diagram, AJ=2*x+3, BJ=5*x, JC=8*y-36, JD=4*y, CA and BD are opposite sides of the parallelogram ACBD. Find the value of x.
问题：如图所示，AJ=2*x+3，BJ=5*x，JC=8*y-36，JD=4*y，ACBD是▱。求x的值。
Construction (CDL):
Shape(AC,CJ,JA)
Shape(JC,CB,BJ)
Shape(JB,BD,DJ)
Shape(AJ,JD,DA)
Collinear(AJB)
Collinear(CJD)
Conditions (CDL):
Equal(LengthOfLine(AJ),2*x+3)
Equal(LengthOfLine(BJ),5*x)
Equal(LengthOfLine(JC),8*y-36)
Equal(LengthOfLine(JD),4*y)
Parallelogram(ACBD)
Goal: Value(x)
Answer: 1
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ACBD,J)
Problem: As shown in the diagram, AB=36, AD=22, ∠BCE=30°, quadrilateral BCDA is a parallelogram, CE is perpendicular to BE. Find the perimeter of BCDA.
问题：如图所示，AB=36，AD=22，∠BCE=30°，BCDA是平行四边形，CE⊥BE。求BCDA的周长。
Construction (CDL):
Shape(BC,CE,EB)
Shape(BE,ED,DA,AB)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AB),36)
Equal(LengthOfLine(AD),22)
Equal(MeasureOfAngle(BCE),30)
Parallelogram(BCDA)
PerpendicularBetweenLine(CE,BE)
Goal: Value(PerimeterOfQuadrilateral(BCDA))
Answer: 116
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BCDA)
parallelogram_property_opposite_line_equal(1,CDAB)
quadrilateral_perimeter_formula(1,BCDA)
Problem: As shown in the diagram, CB=44, EA=19, ∠ACE=30°, BC and DA are opposite sides of the ▱ BDAC, DE⊥CE. Find the area of quadrilateral BDAC.
问题：如图所示，CB=44，EA=19，∠ACE=30°，BDAC是平行四边形，DE垂直于CE。求BDAC的面积。
Construction (CDL):
Shape(BD,DE,EC,CB)
Shape(CE,EA,AC)
Collinear(DEA)
Conditions (CDL):
Equal(LengthOfLine(CB),44)
Equal(LengthOfLine(EA),19)
Equal(MeasureOfAngle(ACE),30)
Parallelogram(BDAC)
PerpendicularBetweenLine(DE,CE)
Goal: Value(AreaOfQuadrilateral(BDAC))
Answer: 836*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,CEA)
sine_theorem(1,EAC)
altitude_of_quadrilateral_judgment_right_vertex(1,CE,BDAC)
parallelogram_property_opposite_line_equal(1,DACB)
parallelogram_area_formula_common(1,BDAC)
Problem: As shown in the diagram, AC=25, AD=21, EB=20, quadrilateral CADB is a parallelogram, DE is perpendicular to BE. Find the area of CADB.
问题：如图所示，AC=25，AD=21，EB=20，CB和AD是平行四边形CADB的一组对边，DE⊥BE。求CADB的面积。
Construction (CDL):
Shape(CA,AD,DB,BC)
Shape(BD,DE,EB)
Collinear(ADE)
Conditions (CDL):
Equal(LengthOfLine(AC),25)
Equal(LengthOfLine(AD),21)
Equal(LengthOfLine(EB),20)
Parallelogram(CADB)
PerpendicularBetweenLine(DE,BE)
Goal: Value(AreaOfQuadrilateral(CADB))
Answer: 420
Theorem sequence:
altitude_of_quadrilateral_judgment_right_vertex(5,BE,CADB)
parallelogram_area_formula_common(1,CADB)
Problem: As shown in the diagram, AC=b, BA=c, BC=a, ∠ABC=60°, ∠CAB=30°, b=3, BC is perpendicular to AC. Find the value of c.
问题：如图所示，AC=b，BA=c，BC=a，∠ABC=60°，∠CAB=30°，b=3，BC垂直于AC。求c的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(BA),c)
Equal(LengthOfLine(BC),a)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),30)
Equal(b,3)
PerpendicularBetweenLine(BC,AC)
Goal: Value(c)
Answer: 2*sqrt(3)
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, BC=3, ∠CBA=10°, B is the center of ⊙B. Find the area of the sector BAC.
问题：如图所示，BC=3，∠CBA=10°，B是⊙B的圆心。求扇形BAC的面积。
Construction (CDL):
Shape(BAC,CB,BA)
Shape(BCA,AB,BC)
Cocircular(B,AC)
Conditions (CDL):
Equal(LengthOfLine(BC),3)
Equal(MeasureOfAngle(CBA),10)
IsCentreOfCircle(B,B)
Goal: Value(AreaOfSector(BAC))
Answer: pi/4
Theorem sequence:
arc_property_center_angle(1,BAC,B)
radius_of_circle_property_length_equal(1,BC,B)
sector_area_formula(1,BAC)
Problem: As shown in the diagram, ∠BCH=23°, ∠HBC=32°, ∠HID=22°. Find the measure of ∠BHD.
问题：如图所示，∠BCH=23°，∠HBC=32°，∠HID=22°。求∠BHD的大小。
Construction (CDL):
Shape(BC,CH,HB)
Shape(HI,ID,DH)
Shape(BH,HD)
Shape(FD,DI)
Shape(IH,HC)
Collinear(CHDE)
Conditions (CDL):
Equal(MeasureOfAngle(BCH),23)
Equal(MeasureOfAngle(HBC),32)
Equal(MeasureOfAngle(HID),22)
Goal: Value(MeasureOfAngle(BHD))
Answer: 55
Theorem sequence:
triangle_property_angle_sum(1,BCH)
adjacent_complementary_angle(1,CHB,BHD)
Problem: As shown in the diagram, JK=33, LK=x, ∠KLJ=45°, JK is perpendicular to LK. Find the value of x.
问题：如图所示，JK=33，LK=x，∠KLJ=45°，JK⊥LK。求x的值。
Construction (CDL):
Shape(JK,KL,LJ)
Conditions (CDL):
Equal(LengthOfLine(JK),33)
Equal(LengthOfLine(LK),x)
Equal(MeasureOfAngle(KLJ),45)
PerpendicularBetweenLine(JK,LK)
Goal: Value(x)
Answer: 33
Theorem sequence:
triangle_property_angle_sum(1,JKL)
sine_theorem(1,KLJ)
Problem: As shown in the diagram, JK=JL, JK=KL, JK=x+7, JL=4*x-8. Find the length of line JK.
问题：如图所示，JK=JL，JK=KL，JK=x+7，JL=4*x-8。求直线JK的长度。
Construction (CDL):
Shape(KJ,JL,LK)
Conditions (CDL):
Equal(LengthOfLine(JK),LengthOfLine(JL))
Equal(LengthOfLine(JK),LengthOfLine(KL))
Equal(LengthOfLine(JK),x+7)
Equal(LengthOfLine(JL),4*x-8)
Goal: Value(LengthOfLine(JK))
Answer: 12
Theorem sequence:
Problem: As shown in the diagram, AB=1/4*x+5, BC=1/2*x-7, ED=66-2/3*y, FE=1/3*y-6, FE=ED, AB is perpendicular to EB, BC⊥DC. Find the value of y.
问题：如图所示，AB=1/4*x+5，BC=1/2*x-7，ED=66-2/3*y，FE=1/3*y-6，FE=ED，AB⊥EB，BC垂直于DC。求y的值。
Construction (CDL):
Shape(AB,BE,EF,FA)
Shape(BC,CD,DE,EB)
Collinear(ABC)
Collinear(FED)
Conditions (CDL):
Equal(LengthOfLine(AB),1/4*x+5)
Equal(LengthOfLine(BC),1/2*x-7)
Equal(LengthOfLine(ED),66-2/3*y)
Equal(LengthOfLine(FE),1/3*y-6)
Equal(LengthOfLine(FE),LengthOfLine(ED))
PerpendicularBetweenLine(AB,EB)
PerpendicularBetweenLine(BC,DC)
Goal: Value(y)
Answer: 72
Theorem sequence:
Problem: As shown in the diagram, RS=3*x-5, RT=2*x+7, RT=RS, ST=22. Find the value of x.
问题：如图所示，RS=3*x-5，RT=2*x+7，RT=RS，ST=22。求x的值。
Construction (CDL):
Shape(RS,ST,TR)
Conditions (CDL):
Equal(LengthOfLine(RS),3*x-5)
Equal(LengthOfLine(RT),2*x+7)
Equal(LengthOfLine(RT),LengthOfLine(RS))
Equal(LengthOfLine(ST),22)
Goal: Value(x)
Answer: 12
Theorem sequence:
Problem: As shown in the diagram, CB=6, ∠DAB=Mul(MeasureOfAngle(CDA),2)°, ABCD is a rhombus. Find the length of line DA.
问题：如图所示，CB=6，∠DAB=Mul(MeasureOfAngle(CDA),2)°，ABCD是菱形。求直线DA的长度。
Construction (CDL):
Shape(DA,AE,ED)
Shape(EA,AB,BE)
Shape(DE,EC,CD)
Shape(CE,EB,BC)
Collinear(DEB)
Collinear(CEA)
Conditions (CDL):
Equal(LengthOfLine(CB),6)
Equal(MeasureOfAngle(DAB),Mul(MeasureOfAngle(CDA),2))
Rhombus(ABCD)
Goal: Value(LengthOfLine(DA))
Answer: 6
Theorem sequence:
Problem: As shown in the diagram, AB=x, AC=x, BC=5*sqrt(2), CA is perpendicular to BA. Find the value of x.
问题：如图所示，AB=x，AC=x，BC=5*sqrt(2)，CA⊥BA。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),5*sqrt(2))
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 5
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, AC=b, BA=c, BC=a, a=14, b=48, c=50, BC is perpendicular to AC. Find the value of tan(AB).
问题：如图所示，AC=b，BA=c，BC=a，a=14，b=48，c=50，BC垂直于AC。求tan(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(BA),c)
Equal(LengthOfLine(BC),a)
Equal(a,14)
Equal(b,48)
Equal(c,50)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Tan(MeasureOfAngle(ABC)))
Answer: 24/7
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, BA=5, ∠ABC=46°, B is the center of circle B. Find the area of the sector BCA.
问题：如图所示，BA=5，∠ABC=46°，B是⊙B的圆心。求扇形BCA的面积。
Construction (CDL):
Shape(BCA,AB,BC)
Shape(BA,BAC,CB)
Cocircular(B,CA)
Conditions (CDL):
Equal(LengthOfLine(BA),5)
Equal(MeasureOfAngle(ABC),46)
IsCentreOfCircle(B,B)
Goal: Value(AreaOfSector(BCA))
Answer: 115*pi/36
Theorem sequence:
radius_of_circle_property_length_equal(1,BA,B)
arc_property_center_angle(1,BCA,B)
sector_area_formula(1,BCA)
Problem: As shown in the diagram, ∠MLN=31°, ∠QPN=22°, NM⊥LM. Find the measure of ∠LNM.
问题：如图所示，∠MLN=31°，∠QPN=22°，NM垂直于LM。求∠LNM的大小。
Construction (CDL):
Shape(LN,NM,ML)
Shape(NQ,QP,PN)
Collinear(MNQ)
Collinear(LNP)
Conditions (CDL):
Equal(MeasureOfAngle(MLN),31)
Equal(MeasureOfAngle(QPN),22)
PerpendicularBetweenLine(NM,LM)
Goal: Value(MeasureOfAngle(LNM))
Answer: 59
Theorem sequence:
triangle_property_angle_sum(1,LNM)
Problem: As shown in the diagram, SR=5, TR=3, TS=4, RT⊥ST. Find the value of sin(SR).
问题：如图所示，SR=5，TR=3，TS=4，RT垂直于ST。求sin(SR)的值。
Construction (CDL):
Shape(TS,SR,RT)
Conditions (CDL):
Equal(LengthOfLine(SR),5)
Equal(LengthOfLine(TR),3)
Equal(LengthOfLine(TS),4)
PerpendicularBetweenLine(RT,ST)
Goal: Value(Sin(MeasureOfAngle(SRT)))
Answer: 4/5
Theorem sequence:
cosine_theorem(1,RTS)
Problem: As shown in the diagram, BA=3, BA=AC, BC=4*y-5, ∠BAC=2*x°, ∠CBA=2*x°. Find the value of y.
问题：如图所示，BA=3，BA=AC，BC=4*y-5，∠BAC=2*x°，∠CBA=2*x°。求y的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(BA),3)
Equal(LengthOfLine(BA),LengthOfLine(AC))
Equal(LengthOfLine(BC),4*y-5)
Equal(MeasureOfAngle(BAC),2*x)
Equal(MeasureOfAngle(CBA),2*x)
Goal: Value(y)
Answer: 2
Theorem sequence:
isosceles_triangle_judgment_angle_equal(1,CBA)
Problem: As shown in the diagram, AF=1/2*y+20, CF=3*y, CG=20-3*x, FC=AF, HG=2*x-5, FG∥AH. Find the value of y.
问题：如图所示，AF=1/2*y+20，CF=3*y，CG=20-3*x，FC=AF，HG=2*x-5，FG∥AH。求y的值。
Construction (CDL):
Shape(CF,FG,GC)
Shape(FA,AH,HG,GF)
Collinear(CFA)
Collinear(CGH)
Conditions (CDL):
Equal(LengthOfLine(AF),1/2*y+20)
Equal(LengthOfLine(CF),3*y)
Equal(LengthOfLine(CG),20-3*x)
Equal(LengthOfLine(FC),LengthOfLine(AF))
Equal(LengthOfLine(HG),2*x-5)
ParallelBetweenLine(FG,AH)
Goal: Value(y)
Answer: 8
Theorem sequence:
Problem: As shown in the diagram, ∠BCA=5*x°, ∠DCB=3*x°. Find the value of x.
问题：如图所示，∠BCA=5*x°，∠DCB=3*x°。求x的值。
Construction (CDL):
Shape(DC,CB)
Shape(BC,CA)
Collinear(DCA)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),5*x)
Equal(MeasureOfAngle(DCB),3*x)
Goal: Value(x)
Answer: 45/2
Theorem sequence:
adjacent_complementary_angle(1,DCB,BCA)
Problem: As shown in the diagram, ∠ACF=140°, ∠IGE=9*x°, GD is parallel to CA. Find the value of x.
问题：如图所示，∠ACF=140°，∠IGE=9*x°，GD平行于CA。求x的值。
Construction (CDL):
Shape(EG,GD)
Shape(DG,GC)
Shape(GC,CA)
Shape(AC,CF)
Shape(FC,CB)
Shape(BC,CG)
Shape(CG,GI)
Shape(IG,GE)
Collinear(IGD)
Collinear(BCA)
Collinear(EGCF)
Conditions (CDL):
Equal(MeasureOfAngle(ACF),140)
Equal(MeasureOfAngle(IGE),9*x)
ParallelBetweenLine(GD,CA)
Goal: Value(x)
Answer: 140/9
Theorem sequence:
parallel_property_corresponding_angle(2,GD,CA,F)
vertical_angle(1,IGE,DGC)
Problem: As shown in the diagram, ∠DAF=4*x+5°, ∠FAB=9*x+20°, ∠FBC=4*y+4°, ∠FDA=y**2-1°, quadrilateral ABCD is a rectangle. Find the value of y.
问题：如图所示，∠DAF=4*x+5°，∠FAB=9*x+20°，∠FBC=4*y+4°，∠FDA=y**2-1°，ABCD是矩形。求y的值。
Construction (CDL):
Shape(AB,BF,FA)
Shape(FB,BC,CF)
Shape(FC,CD,DF)
Shape(AF,FD,DA)
Collinear(AFC)
Collinear(BFD)
Conditions (CDL):
Equal(MeasureOfAngle(DAF),4*x+5)
Equal(MeasureOfAngle(FAB),9*x+20)
Equal(MeasureOfAngle(FBC),4*y+4)
Equal(MeasureOfAngle(FDA),y**2-1)
Rectangle(ABCD)
Goal: Value(y)
Answer: 5
Theorem sequence:
parallel_property_alternate_interior_angle(2,AD,BC)
Problem: As shown in the diagram, AH=RH, HA=6-x, IJ=4/3*y+1, JE=2*y, RH=2*x+3, HJ is parallel to AE, RI is parallel to HJ. Find the value of y.
问题：如图所示，AH=RH，HA=6-x，IJ=4/3*y+1，JE=2*y，RH=2*x+3，HJ平行于AE，RI∥HJ。求y的值。
Construction (CDL):
Shape(RH,HJ,JI,IR)
Shape(HA,AE,EJ,JH)
Collinear(RHA)
Collinear(IJE)
Conditions (CDL):
Equal(LengthOfLine(AH),LengthOfLine(RH))
Equal(LengthOfLine(HA),6-x)
Equal(LengthOfLine(IJ),4/3*y+1)
Equal(LengthOfLine(JE),2*y)
Equal(LengthOfLine(RH),2*x+3)
ParallelBetweenLine(HJ,AE)
ParallelBetweenLine(RI,HJ)
Goal: Value(y)
Answer: 3/2
Theorem sequence:
parallel_judgment_par_par(1,RI,HJ,AE)
trapezoid_judgment_parallel(1,RAEI)
midsegment_of_quadrilateral_judgment_parallel(1,HJ,RAEI)
Problem: As shown in the diagram, AD=27, BA=CD, CH=7, ∠ABC=135°, IH is parallel to BC, CH is perpendicular to IH, HI⊥BI. Find the perimeter of ABCD.
问题：如图所示，AD=27，BA=CD，CH=7，∠ABC=135°，IH∥BC，CH⊥IH，HI垂直于BI。求四边形ABCD的周长。
Construction (CDL):
Shape(AB,BI,IA)
Shape(IB,BC,CH,HI)
Shape(HC,CD,DH)
Collinear(AIHD)
Conditions (CDL):
Equal(LengthOfLine(AD),27)
Equal(LengthOfLine(BA),LengthOfLine(CD))
Equal(LengthOfLine(CH),7)
Equal(MeasureOfAngle(ABC),135)
ParallelBetweenLine(IH,BC)
PerpendicularBetweenLine(CH,IH)
PerpendicularBetweenLine(HI,BI)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 14*sqrt(2)+40
Theorem sequence:
adjacent_complementary_angle(1,HIB,BIA)
adjacent_complementary_angle(1,DHC,CHA)
parallel_judgment_ipsilateral_internal_angle(1,HC,IB)
parallelogram_judgment_parallel_and_parallel(1,IBCH)
parallelogram_property_opposite_line_equal(1,IBCH)
parallel_property_ipsilateral_internal_angle(1,IH,BC)
angle_addition(1,ABI,IBC)
triangle_property_angle_sum(1,ABI)
sine_theorem(1,IAB)
sine_theorem(1,BIA)
mirror_congruent_triangle_judgment_hl(2,BIA,CDH)
mirror_congruent_triangle_property_line_equal(1,BIA,CDH)
line_addition(1,AI,IH)
line_addition(1,AH,HD)
parallelogram_property_opposite_line_equal(1,BCHI)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, ∠WZY=4*x°, ∠XWZ=3*x°, ∠YXW=x°, ∠ZYX=2*x°. Find the measure of ∠ZYX.
问题：如图所示，∠WZY=4*x°，∠XWZ=3*x°，∠YXW=x°，∠ZYX=2*x°。求∠ZYX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(ZYX))
Answer: 72
Theorem sequence:
quadrilateral_property_angle_sum(1,XWZY)
Problem: As shown in the diagram, BC=CE, CE=4*sqrt(2), the center of ⊙D is D, BE is the diameter of circle D, EC is perpendicular to BC. Find the circumference of the ⊙D.
问题：如图所示，BC=CE，CE=4*sqrt(2)，⊙D的圆心为D，BE是圆D的直径，EC垂直于BC。求圆D的周长。
Construction (CDL):
Shape(DCB,BC)
Shape(DEC,CE)
Shape(CB,BD,DE,EC)
Shape(DB,DBE,ED)
Collinear(BDE)
Cocircular(D,CBE)
Conditions (CDL):
Equal(LengthOfLine(BC),LengthOfLine(CE))
Equal(LengthOfLine(CE),4*sqrt(2))
IsCentreOfCircle(D,D)
IsDiameterOfCircle(BE,D)
PerpendicularBetweenLine(EC,BC)
Goal: Value(PerimeterOfCircle(D))
Answer: 8*pi
Theorem sequence:
right_triangle_judgment_angle(1,ECB)
right_triangle_property_pythagorean(1,ECB)
diameter_of_circle_property_length_equal(1,BE,D)
circle_property_length_of_radius_and_diameter(1,D)
circle_perimeter_formula(1,D)
Problem: As shown in the diagram, ∠HPM=4*y°, ∠MPR=68°, ∠PRC=x°, ∠SCR=5*z+2°, MC is parallel to PR, PM is parallel to RC. Find the value of z.
问题：如图所示，∠HPM=4*y°，∠MPR=68°，∠PRC=x°，∠SCR=5*z+2°，MC平行于PR，PM平行于RC。求z的值。
Construction (CDL):
Shape(NC,CS)
Shape(SC,CR)
Shape(CR,RG)
Shape(GR,RI)
Shape(IR,RP)
Shape(RP,PL)
Shape(LP,PH)
Shape(HP,PM)
Shape(PM,ME)
Shape(EM,MD)
Shape(DM,MC)
Shape(MC,CN)
Shape(CM,MP,PR,RC)
Collinear(NCRI)
Collinear(DMPL)
Collinear(SCME)
Collinear(GRPH)
Conditions (CDL):
Equal(MeasureOfAngle(HPM),4*y)
Equal(MeasureOfAngle(MPR),68)
Equal(MeasureOfAngle(PRC),x)
Equal(MeasureOfAngle(SCR),5*z+2)
ParallelBetweenLine(MC,PR)
ParallelBetweenLine(PM,RC)
Goal: Value(z)
Answer: 22
Theorem sequence:
parallelogram_judgment_parallel_and_parallel(1,CMPR)
parallelogram_property_opposite_angle_equal(1,CMPR)
adjacent_complementary_angle(1,SCR,RCM)
Problem: As shown in the diagram, ∠DCG=3*x°, ∠GBA=x+24°. Find the measure of ∠GBA.
问题：如图所示，∠DCG=3*x°，∠GBA=x+24°。求∠GBA的大小。
Construction (CDL):
Shape(OBA,AB)
Shape(GB,BA,AG)
Shape(OAD,DG,GA)
Shape(GD,DC,CG)
Shape(ODC,CD)
Shape(OCB.BG,GC)
Collinear(BGD)
Collinear(CGA)
Cocircular(O,BADC)
Conditions (CDL):
Equal(MeasureOfAngle(DCG),3*x)
Equal(MeasureOfAngle(GBA),x+24)
Goal: Value(MeasureOfAngle(GBA))
Answer: 36
Theorem sequence:
arc_property_circumference_angle_external(1,OAD,B)
arc_property_circumference_angle_external(1,OAD,C)
Problem: As shown in the diagram, ∠CBD=55°, ∠FBG=35°, B is the center of ⊙B. Find the measure of ⌒BFG.
问题：如图所示，∠CBD=55°，∠FBG=35°，⊙B的圆心为B。求⌒BFG的角度。
Construction (CDL):
Shape(BC,BCA,AB)
Shape(BA,BAG,GB)
Shape(BG,BGF,FB)
Shape(BF,BFD,DB)
Shape(BD,BDC,DB)
Collinear(CBG)
Collinear(ABD)
Cocircular(B,AGFDC)
Conditions (CDL):
Equal(MeasureOfAngle(CBD),55)
Equal(MeasureOfAngle(FBG),35)
IsCentreOfCircle(B,B)
Goal: Value(MeasureOfArc(BFG))
Answer: 325
Theorem sequence:
round_angle(1,FBG,GBF)
arc_property_center_angle(1,BFG,B)
Problem: As shown in the diagram, AE=6, DH=6, EF=6, AB is perpendicular to FB, DH is perpendicular to BH, EA is perpendicular to BA, FE is perpendicular to AE. Find the sum of the area of triangle DFB and the area of FEAB.
问题：如图所示，AE=6，DH=6，EF=6，AB⊥FB，DH⊥BH，EA垂直于BA，FE⊥AE。求三角形DFB的面积与FEAB的面积之和。
Construction (CDL):
Shape(DF,FH,HD)
Shape(DH,HB,BD)
Shape(FE,EA,AB,BH,HF)
Collinear(FHB)
Conditions (CDL):
Equal(LengthOfLine(AE),6)
Equal(LengthOfLine(DH),6)
Equal(LengthOfLine(EF),6)
PerpendicularBetweenLine(AB,FB)
PerpendicularBetweenLine(DH,BH)
PerpendicularBetweenLine(EA,BA)
PerpendicularBetweenLine(FE,AE)
Goal: Value(Add(AreaOfTriangle(DFB),AreaOfQuadrilateral(FEAB)))
Answer: 54
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,EF,AB)
parallel_judgment_ipsilateral_internal_angle(1,AE,BF)
parallelogram_judgment_parallel_and_parallel(1,FEAB)
parallelogram_property_opposite_line_equal(1,EABF)
parallelogram_area_formula_sine(1,FEAB)
adjacent_complementary_angle(1,FHD,DHB)
altitude_of_triangle_judgment(1,DH,DFB)
triangle_area_formula_common(1,DFB)
Problem: As shown in the diagram, ∠ANE=30°, ∠CEB=110°, ∠ECA=∠AFN, ∠ENJ=130°. Find the measure of ∠CAE.
问题：如图所示，∠ANE=30°，∠CEB=110°，∠ECA=∠AFN，∠ENJ=130°。求∠CAE的大小。
Construction (CDL):
Shape(CA,AE,EC)
Shape(EA,AN,NE)
Shape(AF,FN,NA)
Shape(CE,EB)
Shape(BE,EN)
Shape(EN,NJ)
Collinear(CAF)
Collinear(CEN)
Collinear(AEB)
Collinear(FNJ)
Conditions (CDL):
Equal(MeasureOfAngle(ANE),30)
Equal(MeasureOfAngle(CEB),110)
Equal(MeasureOfAngle(ECA),MeasureOfAngle(AFN))
Equal(MeasureOfAngle(ENJ),130)
Goal: Value(MeasureOfAngle(CAE))
Answer: 45
Theorem sequence:
adjacent_complementary_angle(1,AEC,CEB)
adjacent_complementary_angle(1,FNE,ENJ)
triangle_property_angle_sum(1,CFN)
triangle_property_angle_sum(1,CAE)
Problem: As shown in the diagram, KJ=11, KL=11, ML=5.5, ∠KJM=60°, KM is perpendicular to LM. Find the length of line JM.
问题：如图所示，KJ=11，KL=11，ML=5.5，∠KJM=60°，KM垂直于LM。求直线JM的长度。
Construction (CDL):
Shape(KJ,JM,MK)
Shape(KM,ML,LK)
Collinear(JML)
Conditions (CDL):
Equal(LengthOfLine(KJ),11)
Equal(LengthOfLine(KL),11)
Equal(LengthOfLine(ML),5.5)
Equal(MeasureOfAngle(KJM),60)
PerpendicularBetweenLine(KM,LM)
Goal: Value(LengthOfLine(JM))
Answer: 11/2
Theorem sequence:
adjacent_complementary_angle(1,JMK,KML)
triangle_property_angle_sum(1,JMK)
sine_theorem(1,JMK)
Problem: As shown in the diagram, ∠CAD=42°, ∠CFG=77°, △CDE is an equilateral △, △CAB is an isosceles △, triangleCFG is an isosceles triangle. Find the measure of ∠GCF.
问题：如图所示，∠CAD=42°，∠CFG=77°，三角形CDE为等边三角形，CA和CB是等腰三角形CAB的腰，△CFG为等腰△。求∠GCF的大小。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DF,FC)
Shape(CF,FG,GC)
Shape(CG,GE,EC)
Shape(CE,EB,BC)
Collinear(ADFGEB)
Conditions (CDL):
Equal(MeasureOfAngle(CAD),42)
Equal(MeasureOfAngle(CFG),77)
EquilateralTriangle(CDE)
IsoscelesTriangle(CAB)
IsoscelesTriangle(CFG)
Goal: Value(MeasureOfAngle(GCF))
Answer: 26
Theorem sequence:
triangle_property_angle_sum(1,CFG)
isosceles_triangle_property_angle_equal(1,CFG)