SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BF=10, BG=8, CE=14, CF=x, ∠AEC=∠BGD. Find the value of x.
问题：如图所示，BF=10，BG=8，CE=14，CF=x，∠AEC=∠BGD。求x的值。
Construction (CDL):
Shape(CA,AE,EC)
Shape(CF,FA,AC)
Shape(DF,FB,BD)
Shape(GD,DB,BG)
Collinear(EAFB)
Collinear(CFDG)
Conditions (CDL):
Equal(LengthOfLine(BF),10)
Equal(LengthOfLine(BG),8)
Equal(LengthOfLine(CE),14)
Equal(LengthOfLine(CF),x)
Equal(MeasureOfAngle(AEC),MeasureOfAngle(BGD))
Goal: Value(x)
Answer: 35/2
Theorem sequence:
vertical_angle(1,CFE,GFB)
mirror_similar_triangle_judgment_aa(1,CFE,BGF)
mirror_similar_triangle_property_line_ratio(1,FEC,FBG)
mirror_similar_triangle_property_line_ratio(1,ECF,GFB)
Problem: As shown in the diagram, AB=y+2*x, AY=3*y+2*x, BC=5*y-2*x, YC=4, YA and CB are opposite sides of the ▱ AYCB. Find the value of x.
问题：如图所示，AB=y+2*x，AY=3*y+2*x，BC=5*y-2*x，YC=4，AB和YC是▱AYCB的一组对边。求x的值。
Construction (CDL):
Shape(AY,YC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),y+2*x)
Equal(LengthOfLine(AY),3*y+2*x)
Equal(LengthOfLine(BC),5*y-2*x)
Equal(LengthOfLine(YC),4)
Parallelogram(AYCB)
Goal: Value(x)
Answer: 1
Theorem sequence:
parallelogram_property_opposite_line_equal(1,AYCB)
parallelogram_property_opposite_line_equal(1,YCBA)
Problem: As shown in the diagram, AB=x, CA=7, CB=4, ∠BCA=62°. Find the value of x.
问题：如图所示，AB=x，CA=7，CB=4，∠BCA=62°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(CA),7)
Equal(LengthOfLine(CB),4)
Equal(MeasureOfAngle(BCA),62)
Goal: Value(x)
Answer: sqrt(65-56*cos(31*pi/90))
Theorem sequence:
cosine_theorem(1,CAB)
Problem: As shown in the diagram, AB=y+11, AC=x+7, BD=2*x+3, CD=3*y-5, quadrilateral DBAC is a parallelogram. Find the value of y.
问题：如图所示，AB=y+11，AC=x+7，BD=2*x+3，CD=3*y-5，DC和BA是▱DBAC的一组对边。求y的值。
Construction (CDL):
Shape(DB,BA,AC,CD)
Conditions (CDL):
Equal(LengthOfLine(AB),y+11)
Equal(LengthOfLine(AC),x+7)
Equal(LengthOfLine(BD),2*x+3)
Equal(LengthOfLine(CD),3*y-5)
Parallelogram(DBAC)
Goal: Value(y)
Answer: 8
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BACD)
Problem: As shown in the diagram, GH=7, GI=14, KJ=6, KH∥JI. Find the length of line GK.
问题：如图所示，GH=7，GI=14，KJ=6，KH平行于JI。求直线GK的长度。
Construction (CDL):
Shape(GK,KH,HG)
Shape(KJ,JI,IH,HK)
Collinear(GKJ)
Collinear(GHI)
Conditions (CDL):
Equal(LengthOfLine(GH),7)
Equal(LengthOfLine(GI),14)
Equal(LengthOfLine(KJ),6)
ParallelBetweenLine(KH,JI)
Goal: Value(LengthOfLine(GK))
Answer: 6
Theorem sequence:
parallel_property_corresponding_angle(1,KH,JI,G)
similar_triangle_judgment_aa(1,HGK,IGJ)
similar_triangle_property_line_ratio(1,HGK,IGJ)
similar_triangle_property_line_ratio(1,KHG,JIG)
line_addition(1,GK,KJ)
Problem: As shown in the diagram, the area of quadrilateral PSRQ is 250, QP=20, RS=30, PSRQ is a trapezoid. Find the height of PSRQ.
问题：如图所示，PSRQ的面积为250，QP=20，RS=30，四边形PSRQ是梯形。求PSRQ的高。
Construction (CDL):
Shape(PS,SA,AP)
Shape(PA,AR,RQ,QP)
Collinear(SAR)
Conditions (CDL):
Equal(AreaOfQuadrilateral(PSRQ),250)
Equal(LengthOfLine(QP),20)
Equal(LengthOfLine(RS),30)
Trapezoid(PSRQ)
Goal: Value(HeightOfQuadrilateral(PSRQ))
Answer: 10
Theorem sequence:
trapezoid_area_formula(1,PSRQ)
Problem: As shown in the diagram, SR=x+3, ST=x, TV=9, the center of circle A is A, the tangent to ⊙A is TV. Find the value of x.
问题：如图所示，SR=x+3，ST=x，TV=9，⊙A的圆心为A，TV是⊙O的切线。求x的值。
Construction (CDL):
Shape(ASR,RA,AS)
Shape(AR,ARV,AVS,SA)
Shape(TS,AVS,VT)
Collinear(RAST)
Cocircular(A,RVS)
Conditions (CDL):
Equal(LengthOfLine(SR),x+3)
Equal(LengthOfLine(ST),x)
Equal(LengthOfLine(TV),9)
IsCentreOfCircle(A,A)
IsTangentOfCircle(TV,A)
Goal: Value(x)
Answer: -3/4+3*sqrt(73)/4
Theorem sequence:
line_addition(1,RS,ST)
circle_property_circular_power_tangent_and_segment_line(1,TV,TSR,A)
Problem: As shown in the diagram, ∠QTP=2*x**2-7°, ∠SRP=x**2+18°, quadrilateral QTSR is a rectangle. Find the measure of ∠QTP.
问题：如图所示，∠QTP=2*x**2-7°，∠SRP=x**2+18°，四边形QTSR是矩形。求∠QTP的大小。
Construction (CDL):
Shape(QT,TP,PQ)
Shape(PT,TS,SP)
Shape(QP,PR,RQ)
Shape(RP,PS,SR)
Collinear(QPS)
Collinear(TPR)
Conditions (CDL):
Equal(MeasureOfAngle(QTP),2*x**2-7)
Equal(MeasureOfAngle(SRP),x**2+18)
Rectangle(QTSR)
Goal: Value(MeasureOfAngle(QTP))
Answer: 43
Theorem sequence:
parallel_property_alternate_interior_angle(1,TQ,SR)
Problem: As shown in the diagram, XW=YZ, YX=XW, ∠XYZ=128°, ∠WXY and ∠XYW are the base angles of the isosceles triangle WXY, triangleXYZ is an isosceles triangle. Find the measure of ∠XWY.
问题：如图所示，XW=YZ，YX=XW，∠XYZ=128°，∠WXY和∠XYW是等腰△WXY的底角，XY和XZ是等腰△XYZ的腰。求∠XWY的大小。
Construction (CDL):
Shape(XW,WY,YX)
Shape(XY,YZ,ZX)
Collinear(WYZ)
Conditions (CDL):
Equal(LengthOfLine(XW),LengthOfLine(YZ))
Equal(LengthOfLine(YX),LengthOfLine(XW))
Equal(MeasureOfAngle(XYZ),128)
IsoscelesTriangle(WXY)
IsoscelesTriangle(XYZ)
Goal: Value(MeasureOfAngle(XWY))
Answer: 52
Theorem sequence:
adjacent_complementary_angle(1,WYX,XYZ)
isosceles_triangle_judgment_line_equal(1,XWY)
isosceles_triangle_property_angle_equal(1,XWY)
Problem: As shown in the diagram, XZ=14, ZY=10, XY is perpendicular to ZY. Find the measure of ∠YZX.
问题：如图所示，XZ=14，ZY=10，XY⊥ZY。求∠YZX的大小。
Construction (CDL):
Shape(XY,YZ,ZX)
Conditions (CDL):
Equal(LengthOfLine(XZ),14)
Equal(LengthOfLine(ZY),10)
PerpendicularBetweenLine(XY,ZY)
Goal: Value(MeasureOfAngle(YZX))
Answer: -180*asin(5/7)/pi+90
Theorem sequence:
sine_theorem(1,ZXY)
triangle_property_angle_sum(1,XYZ)
Problem: As shown in the diagram, the area of ABCD is 846, the area of quadrilateral EFGH is 376, CB=x, FG=24, quadrilateral ABCD is similar to quadrilateral EFGH. Find the value of x.
问题：如图所示，四边形ABCD的面积为846，四边形EFGH的面积为376，CB=x，FG=24，ABCD相似于EFGH。求x的值。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(EF,FG,GH,HE)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ABCD),846)
Equal(AreaOfQuadrilateral(EFGH),376)
Equal(LengthOfLine(CB),x)
Equal(LengthOfLine(FG),24)
SimilarBetweenQuadrilateral(ABCD,EFGH)
Goal: Value(x)
Answer: 36
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,BCDA,FGHE)
similar_quadrilateral_property_area_square_ratio(1,ABCD,EFGH)
Problem: As shown in the diagram, ∠BAF=12*x+4°, ∠FBA=16*x-12°, ADCB is a rectangle. Find the measure of ∠FBA.
问题：如图所示，∠BAF=12*x+4°，∠FBA=16*x-12°，ADCB是长方形。求∠FBA的大小。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(AF,FB,BA)
Shape(FC,CB,BF)
Collinear(AFC)
Collinear(DFB)
Conditions (CDL):
Equal(MeasureOfAngle(BAF),12*x+4)
Equal(MeasureOfAngle(FBA),16*x-12)
Rectangle(ADCB)
Goal: Value(MeasureOfAngle(FBA))
Answer: 52
Theorem sequence:
rectangle_property_diagonal_equal(1,ADCB)
parallelogram_property_diagonal_bisection(1,ADCB,F)
parallelogram_property_diagonal_bisection(1,DCBA,F)
line_addition(1,AF,FC)
line_addition(1,DF,FB)
isosceles_triangle_judgment_line_equal(1,FBA)
isosceles_triangle_property_angle_equal(1,FBA)
Problem: As shown in the diagram, AB=x, AC=y, BC=17, ∠ACB=60°, BA is perpendicular to CA. Find the value of y.
问题：如图所示，AB=x，AC=y，BC=17，∠ACB=60°，BA垂直于CA。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(BC),17)
Equal(MeasureOfAngle(ACB),60)
PerpendicularBetweenLine(BA,CA)
Goal: Value(y)
Answer: 17/2
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,CBA)
Problem: As shown in the diagram, AB=c, CA=b, CB=a, a=8, b=15, c=17, CB⊥CB. Find the value of sin(AB).
问题：如图所示，AB=c，CA=b，CB=a，a=8，b=15，c=17，CB⊥CB。求sin(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),c)
Equal(LengthOfLine(CA),b)
Equal(LengthOfLine(CB),a)
Equal(a,8)
Equal(b,15)
Equal(c,17)
PerpendicularBetweenLine(CB,CA)
Goal: Value(Sin(MeasureOfAngle(ABC)))
Answer: 15/17
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AB=x, BC=26, ∠CAB=43°, BC⊥AC. Find the value of x.
问题：如图所示，AB=x，BC=26，∠CAB=43°，BC⊥AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),26)
Equal(MeasureOfAngle(CAB),43)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 26/sin(43*pi/180)
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, AE=7, DC=32, FC=22, MF=3, CF⊥EF, FE⊥DE, quadrilateral CMAD is a trapezoid. Find the area of CMAD.
问题：如图所示，AE=7，DC=32，FC=22，MF=3，CF垂直于EF，FE垂直于DE，CM和AD是梯形CMAD的腰。求CMAD的面积。
Construction (CDL):
Shape(CM,MF,FC)
Shape(CF,FE,ED,DC)
Shape(DE,EA,AD)
Collinear(MFEA)
Conditions (CDL):
Equal(LengthOfLine(AE),7)
Equal(LengthOfLine(DC),32)
Equal(LengthOfLine(FC),22)
Equal(LengthOfLine(MF),3)
PerpendicularBetweenLine(CF,EF)
PerpendicularBetweenLine(FE,DE)
Trapezoid(CMAD)
Goal: Value(AreaOfQuadrilateral(CMAD))
Answer: 814
Theorem sequence:
parallel_property_collinear_extend(3,AM,DC,E)
parallel_property_collinear_extend(3,EM,DC,F)
parallel_judgment_per_per(1,FC,ED)
parallelogram_judgment_parallel_and_parallel(1,CFED)
parallelogram_property_opposite_line_equal(1,FEDC)
adjacent_complementary_angle(1,MFC,CFE)
altitude_of_quadrilateral_judgment_left_vertex(2,CF,CMAD)
line_addition(1,MF,FE)
line_addition(1,ME,EA)
trapezoid_area_formula(1,CMAD)
Problem: As shown in the diagram, AB=26, AE=24, BS=22, quadrilateral ABSC is a ▱, BE is perpendicular to AE. Find the perimeter of quadrilateral ABSC.
问题：如图所示，AB=26，AE=24，BS=22，四边形ABSC是▱，BE⊥AE。求四边形ABSC的周长。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,ES,SC,CA)
Collinear(BES)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AE),24)
Equal(LengthOfLine(BS),22)
Parallelogram(ABSC)
PerpendicularBetweenLine(BE,AE)
Goal: Value(PerimeterOfQuadrilateral(ABSC))
Answer: 96
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABSC)
parallelogram_property_opposite_line_equal(1,BSCA)
quadrilateral_perimeter_formula(1,ABSC)
Problem: As shown in the diagram, AC=16, AD=8, BC=y, BD=x, CD=z, BA is perpendicular to CA, DC is perpendicular to BC. Find the value of z.
问题：如图所示，AC=16，AD=8，BC=y，BD=x，CD=z，BA垂直于CA，DC垂直于BC。求z的值。
Construction (CDL):
Shape(DC,CA,AD)
Shape(AC,CB,BA)
Collinear(DAB)
Conditions (CDL):
Equal(LengthOfLine(AC),16)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CD),z)
PerpendicularBetweenLine(BA,CA)
PerpendicularBetweenLine(DC,BC)
Goal: Value(z)
Answer: 8*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,BAC,CAD)
right_triangle_judgment_angle(1,CAD)
right_triangle_property_pythagorean(1,CAD)
Problem: As shown in the diagram, AB=25, AC=z, AD=8, BC=y, CD=x, BA is perpendicular to CA, DC is perpendicular to BC. Find the value of z.
问题：如图所示，AB=25，AC=z，AD=8，BC=y，CD=x，BA⊥CA，DC⊥BC。求z的值。
Construction (CDL):
Shape(DC,CA,AD)
Shape(AC,CB,BA)
Collinear(DAB)
Conditions (CDL):
Equal(LengthOfLine(AB),25)
Equal(LengthOfLine(AC),z)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CD),x)
PerpendicularBetweenLine(BA,CA)
PerpendicularBetweenLine(DC,BC)
Goal: Value(z)
Answer: 10*sqrt(2)
Theorem sequence:
adjacent_complementary_angle(1,BAC,CAD)
right_triangle_judgment_angle(1,BAC)
right_triangle_judgment_angle(1,CAD)
right_triangle_judgment_angle(1,DCB)
right_triangle_property_pythagorean(1,BAC)
right_triangle_property_pythagorean(1,CAD)
right_triangle_property_pythagorean(1,DCB)
line_addition(1,DA,AB)
Problem: As shown in the diagram, AB=10, AC=x, BC=24, AB is perpendicular to CB. Find the value of x.
问题：如图所示，AB=10，AC=x，BC=24，AB垂直于CB。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),24)
PerpendicularBetweenLine(AB,CB)
Goal: Value(x)
Answer: 26
Theorem sequence:
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,ABC)
Problem: As shown in the diagram, ∠ABD=140°, ∠ACD=x°, ∠BAC=80°. Find the value of x.
问题：如图所示，∠ABD=140°，∠ACD=x°，∠BAC=80°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Shape(AB,BD)
Collinear(CBD)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),140)
Equal(MeasureOfAngle(ACD),x)
Equal(MeasureOfAngle(BAC),80)
Goal: Value(x)
Answer: 60
Theorem sequence:
adjacent_complementary_angle(1,CBA,ABD)
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, AB=5*y, CD=y+8, ∠CDB=4*x°, ∠DBA=2*x-6°, quadrilateral DBAC is a ▱. Find the value of x.
问题：如图所示，AB=5*y，CD=y+8，∠CDB=4*x°，∠DBA=2*x-6°，四边形DBAC是平行四边形。求x的值。
Construction (CDL):
Shape(DB,BA,AC,CD)
Conditions (CDL):
Equal(LengthOfLine(AB),5*y)
Equal(LengthOfLine(CD),y+8)
Equal(MeasureOfAngle(CDB),4*x)
Equal(MeasureOfAngle(DBA),2*x-6)
Parallelogram(DBAC)
Goal: Value(x)
Answer: 31
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,DC,BA)
Problem: As shown in the diagram, the area of triangle ABC is 165, AC=x, AD=11, DB=41, DA⊥CA. Find the value of x.
问题：如图所示，三角形ABC的面积为165，AC=x，AD=11，DB=41，DA⊥CA。求x的值。
Construction (CDL):
Shape(DA,AC,CD)
Shape(CA,AB,BC)
Collinear(DAB)
Conditions (CDL):
Equal(AreaOfTriangle(ABC),165)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),11)
Equal(LengthOfLine(DB),41)
PerpendicularBetweenLine(DA,CA)
Goal: Value(x)
Answer: 11
Theorem sequence:
line_addition(1,DA,AB)
adjacent_complementary_angle(1,DAC,CAB)
triangle_area_formula_sine(1,ABC)
Problem: As shown in the diagram, AB=32, FC=11, the center of circle C is C, the diameter of ⊙C is AB, CF⊥EF. Find the length of line FE.
问题：如图所示，AB=32，FC=11，⊙C的圆心为C，⊙C的直径为AB，CF垂直于EF。求直线FE的长度。
Construction (CDL):
Shape(FA,CAE,EF)
Shape(AF,FD,CDA)
Shape(FE,EC,CF)
Shape(DF,FC,CD)
Shape(CE,CEB,BC)
Shape(DC,CB,CBD)
Collinear(AFCB)
Collinear(EFD)
Cocircular(C,AEBD)
Conditions (CDL):
Equal(LengthOfLine(AB),32)
Equal(LengthOfLine(FC),11)
IsCentreOfCircle(C,C)
IsDiameterOfCircle(AB,C)
PerpendicularBetweenLine(CF,EF)
Goal: Value(LengthOfLine(FE))
Answer: 3*sqrt(15)
Theorem sequence:
diameter_of_circle_property_length_equal(1,AB,C)
circle_property_length_of_radius_and_diameter(1,C)
radius_of_circle_property_length_equal(1,CE,C)
right_triangle_judgment_angle(1,CFE)
right_triangle_property_pythagorean(1,CFE)
Problem: As shown in the diagram, MN=18, ON=18, F is the center of circle F, the diameter of ⊙F is HJ, NH is the tangent to circle F, BO⊥HO, GM is perpendicular to JM, HN is perpendicular to GN, JP⊥BP. Find the area of quadrilateral ONMP minus the area of the circle F.
问题：如图所示，MN=18，ON=18，F是⊙F的圆心，⊙F的直径为HJ，圆O的切线为NH，BO垂直于HO，GM垂直于JM，HN垂直于GN，JP垂直于BP。求ONMP的面积减去圆F的面积。
Construction (CDL):
Shape(OH,FBH,BO)
Shape(PB,FJB,JP)
Shape(FJB,FBH,HF,FJ)
Shape(FH,FHG,FGJ,JF)
Shape(HN,NG,FHG)
Shape(GM,MJ,FGJ)
Collinear(OBP)
Collinear(OHN)
Collinear(NGM)
Collinear(MJP)
Collinear(HFJ)
Cocircular(F,BHGJ)
Conditions (CDL):
Equal(LengthOfLine(MN),18)
Equal(LengthOfLine(ON),18)
IsCentreOfCircle(F,F)
IsDiameterOfCircle(HJ,F)
IsTangentOfCircle(NH,F)
PerpendicularBetweenLine(BO,HO)
PerpendicularBetweenLine(GM,JM)
PerpendicularBetweenLine(HN,GN)
PerpendicularBetweenLine(JP,BP)
Goal: Value(Sub(AreaOfQuadrilateral(ONMP),AreaOfCircle(F)))
Answer: 324-81*pi
Theorem sequence:
parallelogram_judgment_angle_and_angle(1,ONMP)
tangent_of_circle_property_perpendicular(1,NH,F,F)
quadrilateral_property_angle_sum(1,HNMJ)
parallelogram_judgment_angle_and_angle(1,HNMJ)
parallelogram_property_opposite_line_equal(1,NMJH)
diameter_of_circle_property_length_equal(1,HJ,F)
circle_property_length_of_radius_and_diameter(1,F)
circle_area_formula(1,F)
parallelogram_area_formula_sine(1,ONMP)
Problem: As shown in the diagram, ∠AFD=50°, F is the center of ⊙F, CF is perpendicular to BF. Find the measure of ⌒FEB.
问题：如图所示，∠AFD=50°，⊙F的圆心为F，CF垂直于BF。求弧FEB的角度。
Construction (CDL):
Shape(FC,FCD,DF)
Shape(FD,FDA,AF)
Shape(FA,FAE,EF)
Shape(FE,FEB,BF)
Shape(FB,FBC,CF)
Collinear(AFB)
Collinear(DFE)
Cocircular(F,AEBCD)
Conditions (CDL):
Equal(MeasureOfAngle(AFD),50)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(CF,BF)
Goal: Value(MeasureOfArc(FEB))
Answer: 50
Theorem sequence:
vertical_angle(1,AFD,BFE)
arc_property_center_angle(1,FEB,F)
Problem: As shown in the diagram, RT=15, ∠RPS=130°, ∠TPQ=112°, the center of circle P is P, the diameter of circle P is TR. Find the length of arc PSR.
问题：如图所示，RT=15，∠RPS=130°，∠TPQ=112°，⊙P的圆心为P，圆P的直径为TR。求弧PSR的长度。
Construction (CDL):
Shape(PR,PRQ,QP)
Shape(PQ,PQT,TP)
Shape(PT,PTS,SP)
Shape(PS,PSR,RP)
Collinear(RPT)
Cocircular(P,QTSR)
Conditions (CDL):
Equal(LengthOfLine(RT),15)
Equal(MeasureOfAngle(RPS),130)
Equal(MeasureOfAngle(TPQ),112)
IsCentreOfCircle(P,P)
IsDiameterOfCircle(TR,P)
Goal: Value(LengthOfArc(PSR))
Answer: 65*pi/12
Theorem sequence:
diameter_of_circle_property_length_equal(1,TR,P)
circle_property_length_of_radius_and_diameter(1,P)
arc_property_center_angle(1,PSR,P)
arc_length_formula(1,PSR)
Problem: As shown in the diagram, the measure of arc HBC is 99, the center of circle H is H, the tangent to circle H is AB, AC is the tangent to circle H. Find the measure of ∠BAC.
问题：如图所示，弧HBC的角度为99，H是⊙H的圆心，AB是⊙O的切线，AC是⊙O的切线。求∠BAC的大小。
Construction (CDL):
Shape(HCB,BH,HC)
Shape(HB,HBC,CH)
Shape(HBC,BA,AC)
Cocircular(H,BC)
Conditions (CDL):
Equal(MeasureOfArc(HBC),99)
IsCentreOfCircle(H,H)
IsTangentOfCircle(AB,H)
IsTangentOfCircle(AC,H)
Goal: Value(MeasureOfAngle(BAC))
Answer: 81
Theorem sequence:
arc_property_center_angle(1,HBC,H)
tangent_of_circle_property_perpendicular(1,AB,H,H)
tangent_of_circle_property_perpendicular(2,AC,H,H)
quadrilateral_property_angle_sum(1,CHBA)
Problem: As shown in the diagram, BA=5*x-1, BC=4*x+3, the measure of arc OBA is 85, the measure of arc OCB is 85. Find the value of x.
问题：如图所示，BA=5*x-1，BC=4*x+3，⌒OBA的角度为85，弧OCB的角度为85。求x的值。
Construction (CDL):
Shape(O)
Shape(AB,OBA)
Shape(BC,OCB)
Shape(CB,BA,OAC)
Cocircular(O,CBA)
Conditions (CDL):
Equal(LengthOfLine(BA),5*x-1)
Equal(LengthOfLine(BC),4*x+3)
Equal(MeasureOfArc(OBA),85)
Equal(MeasureOfArc(OCB),85)
Goal: Value(x)
Answer: 4
Theorem sequence:
congruent_arc_judgment_measure_equal(1,OCB,OBA)
congruent_arc_property_chord_equal(1,OCB,OBA)
Problem: As shown in the diagram, ∠ABC=43°, ∠DAB=22°. Find the measure of ∠DCB.
问题：如图所示，∠ABC=43°，∠DAB=22°。求∠DCB的大小。
Construction (CDL):
Shape(CA,AB,BC)
Shape(DC,CB)
Collinear(DCA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),43)
Equal(MeasureOfAngle(DAB),22)
Goal: Value(MeasureOfAngle(DCB))
Answer: 65
Theorem sequence:
triangle_property_angle_sum(1,ABC)
adjacent_complementary_angle(1,DCB,BCA)
Problem: As shown in the diagram, BC=31, BD=20, DA=x, the center of ⊙X is X, the tangent to circle X is BC. Find the value of x.
问题：如图所示，BC=31，BD=20，DA=x，X是⊙X的圆心，圆O的切线为BC。求x的值。
Construction (CDL):
Shape(X)
Shape(DA,XAD)
Shape(AD,XDC,XCA)
Shape(DB,BC,XDC)
Collinear(ADB)
Cocircular(X,DCA)
Conditions (CDL):
Equal(LengthOfLine(BC),31)
Equal(LengthOfLine(BD),20)
Equal(LengthOfLine(DA),x)
IsCentreOfCircle(X,X)
IsTangentOfCircle(BC,X)
Goal: Value(x)
Answer: 561/20
Theorem sequence:
line_addition(1,AD,DB)
circle_property_circular_power_tangent_and_segment_line(1,BC,BDA,X)
Problem: As shown in the diagram, AC=8, AD=8, CB=14, AC⊥BC, AD and CB are the parallel sides of trapezoid ACBD. Find the area of ACBD.
问题：如图所示，AC=8，AD=8，CB=14，AC垂直于BC，AD和CB是梯形ACBD的平行边。求四边形ACBD的面积。
Construction (CDL):
Shape(AC,CB,BD,DA)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(CB),14)
PerpendicularBetweenLine(AC,BC)
Trapezoid(ACBD)
Goal: Value(AreaOfQuadrilateral(ACBD))
Answer: 88
Theorem sequence:
right_trapezoid_judgment_right_angle(1,ACBD)
right_trapezoid_area_formular(1,ACBD)
Problem: As shown in the diagram, AB=z, AC=5, AT=q, BC=15, BT=x, CT=y, AT⊥BT, BA⊥CA. Find the value of x.
问题：如图所示，AB=z，AC=5，AT=q，BC=15，BT=x，CT=y，AT垂直于BT，BA⊥CA。求x的值。
Construction (CDL):
Shape(BA,AT,TB)
Shape(AC,CT,TA)
Collinear(BTC)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(AT),q)
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(BT),x)
Equal(LengthOfLine(CT),y)
PerpendicularBetweenLine(AT,BT)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 40/3
Theorem sequence:
adjacent_complementary_angle(1,CTA,ATB)
right_triangle_judgment_angle(1,ATB)
right_triangle_judgment_angle(1,CTA)
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,ATB)
right_triangle_property_pythagorean(1,CTA)
right_triangle_property_pythagorean(1,BAC)
line_addition(1,BT,TC)
Problem: As shown in the diagram, the diameter of circle B is 16, Sub(AreaOfSector(BHI)=AreaOfTriangle(IBH)), △AHI is an equilateral △, the center of circle B is B, B is the circumcenter of △IAH, BC⊥HC. Find the value of x.
问题：如图所示，圆B的直径为16，Sub(AreaOfSector(BHI)=AreaOfTriangle(IBH))，三角形AHI为等边三角形，B是圆B的圆心，△IAH的外接圆圆心为B，BC垂直于HC。求x的值。
Construction (CDL):
Shape(AB,BI,IA)
Shape(BC,CH,HB)
Shape(AC,CB,BA)
Shape(IB,BH,HI)
Shape(AI,BIA)
Shape(HC,CA,BAH)
Shape(IH,BHI)
Collinear(IBC)
Collinear(ACH)
Cocircular(B,AHI)
Conditions (CDL):
Equal(DiameterOfCircle(B),16)
Equal(Sub(AreaOfSector(BHI),AreaOfTriangle(IBH)),x)
EquilateralTriangle(AHI)
IsCentreOfCircle(B,B)
IsCircumcenterOfTriangle(B,IAH)
PerpendicularBetweenLine(BC,HC)
Goal: Value(x)
Answer: -16*sqrt(3)+64*pi/3
Theorem sequence:
circle_property_length_of_radius_and_diameter(1,B)
radius_of_circle_property_length_equal(1,BH,B)
radius_of_circle_property_length_equal(1,BI,B)
equilateral_triangle_property_angle(1,HIA)
isosceles_triangle_judgment_line_equal(1,BHI)
isosceles_triangle_property_angle_equal(1,BHI)
triangle_property_angle_sum(1,IBH)
triangle_property_angle_sum(1,CHI)
sine_theorem(1,IBH)
triangle_area_formula_sine(1,IBH)
arc_property_center_angle(1,BHI,B)
sector_area_formula(1,BHI)
Problem: As shown in the diagram, CX=15, QC=5, XQ=x. Find the value of x.
问题：如图所示，CX=15，QC=5，XQ=x。求x的值。
Construction (CDL):
Shape(PB,BX,XD,DP)
Shape(AQ,QE,EF,FA)
Shape(QC,CE)
Collinear(XQC)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(CX),15)
Equal(LengthOfLine(QC),5)
Equal(LengthOfLine(XQ),x)
Goal: Value(x)
Answer: 10
Theorem sequence:
line_addition(1,XQ,QC)
Problem: As shown in the diagram, PD=10, PQ=10, QE=20, P is the center of ⊙P, CQ is perpendicular to PQ, PD is perpendicular to BD. Find the length of line AB.
问题：如图所示，PD=10，PQ=10，QE=20，P是圆P的圆心，CQ垂直于PQ，PD⊥BD。求直线AB的长度。
Construction (CDL):
Shape(BD,DG,PGB)
Shape(GD,DA,PAG)
Shape(AD,DP,PA)
Shape(CQ,QP,PD,DB,PBC)
Shape(FQ,QC,PCF)
Shape(EQ,QF,PFE)
Shape(PQ,QE,EP)
Shape(AP,PE,PEA)
Collinear(ADB)
Collinear(GDP)
Collinear(CQE)
Collinear(PQF)
Cocircular(P,AGBCFE)
Conditions (CDL):
Equal(LengthOfLine(PD),10)
Equal(LengthOfLine(PQ),10)
Equal(LengthOfLine(QE),20)
IsCentreOfCircle(P,P)
PerpendicularBetweenLine(CQ,PQ)
PerpendicularBetweenLine(PD,BD)
Goal: Value(LengthOfLine(AB))
Answer: 40
Theorem sequence:
line_addition(1,AD,DB)
adjacent_complementary_angle(1,ADP,PDB)
adjacent_complementary_angle(1,CQP,PQE)
circle_property_chord_perpendicular_bisect_chord(1,P,PD,AB)
radius_of_circle_property_length_equal(1,PA,P)
radius_of_circle_property_length_equal(1,PE,P)
mirror_congruent_triangle_judgment_hl(1,ADP,EPQ)
mirror_congruent_triangle_property_line_equal(1,PAD,PQE)
Problem: As shown in the diagram, RQ=9/2, TR=TS, RQ is perpendicular to TQ. Find the length of line RS.
问题：如图所示，RQ=9/2，TR=TS，RQ⊥TQ。求直线RS的长度。
Construction (CDL):
Shape(RQ,QT,TR)
Shape(QS,ST,TQ)
Collinear(RQS)
Conditions (CDL):
Equal(LengthOfLine(RQ),9/2)
Equal(LengthOfLine(TR),LengthOfLine(TS))
PerpendicularBetweenLine(RQ,TQ)
Goal: Value(LengthOfLine(RS))
Answer: 9
Theorem sequence:
altitude_of_triangle_judgment(1,TQ,TRS)
isosceles_triangle_judgment_line_equal(1,TRS)
isosceles_triangle_property_line_coincidence(1,TRS,Q)
line_addition(1,RQ,QS)
Problem: As shown in the diagram, the measure of ⌒CAE is 70, the measure of ⌒CDB is 80, the center of ⊙C is C. Find the measure of ∠ECA.
问题：如图所示，⌒CAE的角度为70，⌒CDB的角度为80，圆C的圆心为C。求∠ECA的大小。
Construction (CDL):
Shape(AC,CB,CBA)
Shape(BC,CD,CDB)
Shape(DC,CE,CED)
Shape(EC,CA,CAE)
Collinear(ACD)
Cocircular(C,AEDB)
Conditions (CDL):
Equal(MeasureOfArc(CAE),70)
Equal(MeasureOfArc(CDB),80)
IsCentreOfCircle(C,C)
Goal: Value(MeasureOfAngle(ECA))
Answer: 70
Theorem sequence:
arc_property_center_angle(1,CAE,C)
Problem: As shown in the diagram, AB=y, BC=18, CA=x, ∠BCA=60°, CA⊥BA. Find the value of y.
问题：如图所示，AB=y，BC=18，CA=x，∠BCA=60°，CA⊥BA。求y的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(BC),18)
Equal(LengthOfLine(CA),x)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 9*sqrt(3)
Theorem sequence:
sine_theorem(1,BCA)