SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AE=4, DE=7, YE=11, quadrilateral DCYA is a kite. Find the area of ADCY.
问题：如图所示，AE=4，DE=7，YE=11，YC和YA是风筝形DCYA的一组临边。求四边形ADCY的面积。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DC,CE,ED)
Shape(AE,EY,YA)
Shape(EC,CY,YE)
Collinear(DEY)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AE),4)
Equal(LengthOfLine(DE),7)
Equal(LengthOfLine(YE),11)
Kite(DCYA)
Goal: Value(AreaOfQuadrilateral(ADCY))
Answer: 72
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,DCYA,E)
line_addition(1,DE,EY)
line_addition(1,AE,EC)
kite_area_formula_diagonal(1,DCYA)
Problem: As shown in the diagram, AB=3*y, AD=2*x, BC=5*x-18, DC=96-y, DA and CB are opposite sides of the parallelogram ADCB. Find the value of x.
问题：如图所示，AB=3*y，AD=2*x，BC=5*x-18，DC=96-y，ADCB是▱。求x的值。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),3*y)
Equal(LengthOfLine(AD),2*x)
Equal(LengthOfLine(BC),5*x-18)
Equal(LengthOfLine(DC),96-y)
Parallelogram(ADCB)
Goal: Value(x)
Answer: 6
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ADCB)
Problem: As shown in the diagram, PS=3, RY=5, WX=10, WY=8, XY=6, RP∥XW, BQ⊥SQ, SY is perpendicular to BY, YS⊥PS. Find the length of line SY.
问题：如图所示，PS=3，RY=5，WX=10，WY=8，XY=6，RP∥XW，BQ垂直于SQ，SY⊥BY，YS⊥PS。求直线SY的长度。
Construction (CDL):
Shape(YR,RB,BY)
Shape(YB,BQ,QS,SY)
Shape(PY,YS,SP)
Shape(BX,XQ,QB)
Shape(SQ,QW,WS)
Collinear(RYP)
Collinear(YSW)
Collinear(RBQ)
Collinear(XQW)
Collinear(YBX)
Collinear(PSQ)
Conditions (CDL):
Equal(LengthOfLine(PS),3)
Equal(LengthOfLine(RY),5)
Equal(LengthOfLine(WX),10)
Equal(LengthOfLine(WY),8)
Equal(LengthOfLine(XY),6)
ParallelBetweenLine(RP,XW)
PerpendicularBetweenLine(BQ,SQ)
PerpendicularBetweenLine(SY,BY)
PerpendicularBetweenLine(YS,PS)
Goal: Value(LengthOfLine(SY))
Answer: 4
Theorem sequence:
cosine_theorem(1,XWY)
parallel_property_collinear_extend(3,RP,XW,Y)
parallel_property_ipsilateral_internal_angle(1,YP,XW)
angle_addition(1,PYS,SYB)
triangle_property_angle_sum(1,PYS)
sine_theorem(1,SPY)
Problem: As shown in the diagram, the measure of arc FCB is 75, the measure of arc FDE is 45. Find the measure of ∠EAD.
问题：如图所示，弧FCB的角度为75，⌒FDE的角度为45。求∠EAD的大小。
Construction (CDL):
Shape(FDE,EA,AD)
Shape(FEC,CE)
Shape(FCB,BA,AC)
Shape(FBD,DA,AB)
Shape(AE,EC,CA)
Collinear(EAB)
Collinear(DAC)
Cocircular(F,DECB)
Conditions (CDL):
Equal(MeasureOfArc(FCB),75)
Equal(MeasureOfArc(FDE),45)
Goal: Value(MeasureOfAngle(EAD))
Answer: 60
Theorem sequence:
arc_property_circumference_angle_external(1,FDE,C)
arc_property_circumference_angle_external(1,FCB,E)
triangle_property_angle_sum(1,ECA)
adjacent_complementary_angle(1,CAE,EAD)
Problem: As shown in the diagram, JQ=2*y-1, PQ=8.1, ∠JMH=14*y-8°, ∠MHJ=13*y°, ∠PJQ=53°, △JMH is mirror congruent to △JQP. Find the value of y.
问题：如图所示，JQ=2*y-1，PQ=8.1，∠JMH=14*y-8°，∠MHJ=13*y°，∠PJQ=53°，三角形JMH镜像全等于三角形JQP。求y的值。
Construction (CDL):
Shape(MH,HJ,JM)
Shape(JQ,QP,PJ)
Collinear(HJP)
Collinear(MJQ)
Conditions (CDL):
Equal(LengthOfLine(JQ),2*y-1)
Equal(LengthOfLine(PQ),8.1)
Equal(MeasureOfAngle(JMH),14*y-8)
Equal(MeasureOfAngle(MHJ),13*y)
Equal(MeasureOfAngle(PJQ),53)
MirrorCongruentBetweenTriangle(JMH,JQP)
Goal: Value(y)
Answer: 5
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,JMH,JQP)
triangle_property_angle_sum(1,JMH)
Problem: As shown in the diagram, TS=5, ∠ATS=135°, RS is perpendicular to TS. Find the length of line RT.
问题：如图所示，TS=5，∠ATS=135°，RS垂直于TS。求直线RT的长度。
Construction (CDL):
Shape(RS,ST,TR)
Shape(AT,TS)
Collinear(RTA)
Conditions (CDL):
Equal(LengthOfLine(TS),5)
Equal(MeasureOfAngle(ATS),135)
PerpendicularBetweenLine(RS,TS)
Goal: Value(LengthOfLine(RT))
Answer: 5*sqrt(2)
Theorem sequence:
adjacent_complementary_angle(1,ATS,STR)
triangle_property_angle_sum(1,RST)
sine_theorem(1,TRS)
Problem: As shown in the diagram, AB=21, SC=17, quadrilateral BSAC is a kite. Find the area of quadrilateral ACBS.
问题：如图所示，AB=21，SC=17，BSAC是风筝形。求四边形ACBS的面积。
Construction (CDL):
Shape(AC,CO,OA)
Shape(OC,CB,BO)
Shape(AO,OS,SA)
Shape(SO,OB,BS)
Collinear(AOB)
Collinear(COS)
Conditions (CDL):
Equal(LengthOfLine(AB),21)
Equal(LengthOfLine(SC),17)
Kite(BSAC)
Goal: Value(AreaOfQuadrilateral(ACBS))
Answer: 357/2
Theorem sequence:
kite_area_formula_diagonal(1,ACBS)
Problem: As shown in the diagram, ∠UEN=64°, DU is parallel to HE, HD is parallel to EU. Find the measure of ∠HDI.
问题：如图所示，∠UEN=64°，DU平行于HE，HD平行于EU。求∠HDI的大小。
Construction (CDL):
Shape(DH,HE,EU,UD)
Shape(ID,DW)
Shape(WD,DU)
Shape(DU,UA)
Shape(AU,UL)
Shape(HD,DI)
Shape(UD,DH)
Shape(EU,UD)
Shape(LU,UE)
Shape(BH,HD)
Shape(DH,HE)
Shape(HE,EU)
Shape(UE,EN)
Shape(JH,HB)
Shape(EH,HJ)
Shape(KE,EH)
Shape(NE,EK)
Collinear(IDUL)
Collinear(BHEN)
Collinear(WDHJ)
Collinear(AUEK)
Conditions (CDL):
Equal(MeasureOfAngle(UEN),64)
ParallelBetweenLine(DU,HE)
ParallelBetweenLine(HD,EU)
Goal: Value(MeasureOfAngle(HDI))
Answer: 64
Theorem sequence:
parallel_property_corresponding_angle(2,HD,EU,N)
parallel_property_corresponding_angle(1,DU,HE,W)
vertical_angle(1,WDU,HDI)
Problem: As shown in the diagram, AC=x, BC=17, ∠ABC=39°, ∠BCA=96°. Find the value of x.
问题：如图所示，AC=x，BC=17，∠ABC=39°，∠BCA=96°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),17)
Equal(MeasureOfAngle(ABC),39)
Equal(MeasureOfAngle(BCA),96)
Goal: Value(x)
Answer: -17*sqrt(30*sqrt(5)+150)/16-17*sqrt(2*sqrt(5)+10)/16+17/8+17*sqrt(3)/8+17*sqrt(5)/8+17*sqrt(6*sqrt(5)+30)/16+17*sqrt(15)/8+17*sqrt(10*sqrt(5)+50)/16
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,CAB)
Problem: As shown in the diagram, MP=MR, MR=RP. Find the measure of ∠MRP.
问题：如图所示，MP=MR，MR=RP。求∠MRP的大小。
Construction (CDL):
Shape(MR,RP,PM)
Conditions (CDL):
Equal(LengthOfLine(MP),LengthOfLine(MR))
Equal(LengthOfLine(MR),LengthOfLine(RP))
Goal: Value(MeasureOfAngle(MRP))
Answer: 60
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,MRP)
isosceles_triangle_judgment_line_equal(1,RPM)
equilateral_triangle_judgment_isosceles_and_isosceles(1,MRP)
equilateral_triangle_property_angle(1,RPM)
Problem: As shown in the diagram, BC=11, BO=35, CA=23, OA=25, OA is perpendicular to CA. Find the area of triangle OCB.
问题：如图所示，BC=11，BO=35，CA=23，OA=25，OA⊥CA。求△OCB的面积。
Construction (CDL):
Shape(OA,AC,CO)
Shape(OC,CB,BO)
Collinear(ACB)
Conditions (CDL):
Equal(LengthOfLine(BC),11)
Equal(LengthOfLine(BO),35)
Equal(LengthOfLine(CA),23)
Equal(LengthOfLine(OA),25)
PerpendicularBetweenLine(OA,CA)
Goal: Value(AreaOfTriangle(OCB))
Answer: 275/2
Theorem sequence:
altitude_of_triangle_judgment(2,OA,OCB)
triangle_area_formula_common(1,OCB)
Problem: As shown in the diagram, ∠UVS=27°, ∠VWS=72°, TU⊥VU, UV is perpendicular to WV. Find the measure of ∠WSV.
问题：如图所示，∠UVS=27°，∠VWS=72°，TU⊥VU，UV垂直于WV。求∠WSV的大小。
Construction (CDL):
Shape(TU,US,ST)
Shape(UV,VS,SU)
Shape(SV,VW,WS)
Collinear(USW)
Collinear(TSV)
Conditions (CDL):
Equal(MeasureOfAngle(UVS),27)
Equal(MeasureOfAngle(VWS),72)
PerpendicularBetweenLine(TU,VU)
PerpendicularBetweenLine(UV,WV)
Goal: Value(MeasureOfAngle(WSV))
Answer: 45
Theorem sequence:
angle_addition(1,UVS,SVW)
triangle_property_angle_sum(1,SVW)
Problem: As shown in the diagram, ∠AHN=37°, ∠EKL=110°, AQ is parallel to HN, KL is parallel to AQ, KL is parallel to HN. Find the measure of ∠HAK.
问题：如图所示，∠AHN=37°，∠EKL=110°，AQ平行于HN，KL∥AQ，KL平行于HN。求∠HAK的大小。
Construction (CDL):
Shape(JK,KE)
Shape(EK,KL)
Shape(AK,KJ)
Shape(LK,KA)
Shape(CH,HA)
Shape(AH,HN)
Shape(FH,HC)
Shape(NH,HF)
Shape(PA,AK)
Shape(KA,AQ)
Shape(HA,AP)
Shape(QA,AH)
Collinear(JKL)
Collinear(CHN)
Collinear(EKA)
Collinear(AHF)
Collinear(PAQ)
Conditions (CDL):
Equal(MeasureOfAngle(AHN),37)
Equal(MeasureOfAngle(EKL),110)
ParallelBetweenLine(AQ,HN)
ParallelBetweenLine(KL,AQ)
ParallelBetweenLine(KL,HN)
Goal: Value(MeasureOfAngle(HAK))
Answer: 107
Theorem sequence:
parallel_property_corresponding_angle(1,KL,AQ,E)
flat_angle(1,PAQ)
angle_addition(1,PAK,KAQ)
parallel_property_collinear_extend(1,AQ,HN,P)
parallel_property_alternate_interior_angle(2,PA,HN)
angle_addition(1,HAP,PAK)
Problem: As shown in the diagram, CD=30, ED=10, ∠DEF=60°, ED and BC are opposite sides of the ▱ DEBC, EF is perpendicular to DF. Find the area of quadrilateral DEBC.
问题：如图所示，CD=30，ED=10，∠DEF=60°，四边形DEBC是平行四边形，EF垂直于DF。求DEBC的面积。
Construction (CDL):
Shape(DE,EF,FD)
Shape(DF,FB,BC,CD)
Collinear(EFB)
Conditions (CDL):
Equal(LengthOfLine(CD),30)
Equal(LengthOfLine(ED),10)
Equal(MeasureOfAngle(DEF),60)
Parallelogram(DEBC)
PerpendicularBetweenLine(EF,DF)
Goal: Value(AreaOfQuadrilateral(DEBC))
Answer: 150*sqrt(3)
Theorem sequence:
sine_theorem(1,DEF)
altitude_of_quadrilateral_judgment_left_vertex(1,DF,DEBC)
parallelogram_property_opposite_line_equal(1,EBCD)
parallelogram_area_formula_common(1,DEBC)
Problem: As shown in the diagram, ∠BAD=2*x+33°, ∠CBA=4*x+11°, quadrilateral CBAD is a isosceles trapezoid. Find the value of x.
问题：如图所示，∠BAD=2*x+33°，∠CBA=4*x+11°，CD和BA是等腰梯形CBAD的平行边。求x的值。
Construction (CDL):
Shape(BA,AP,PB)
Shape(AD,DP,PA)
Shape(BP,PC,CB)
Shape(PD,DC,CP)
Collinear(APC)
Collinear(DPB)
Conditions (CDL):
Equal(MeasureOfAngle(BAD),2*x+33)
Equal(MeasureOfAngle(CBA),4*x+11)
IsoscelesTrapezoid(CBAD)
Goal: Value(x)
Answer: 11
Theorem sequence:
isosceles_trapezoid_property_angle_equal(1,CBAD)
Problem: As shown in the diagram, AC=5, AD=9, ∠DAC=60°, AD and CB are opposite sides of the ▱ DACB, AC⊥DC. Find the area of DACB.
问题：如图所示，AC=5，AD=9，∠DAC=60°，DACB是▱，AC⊥DC。求四边形DACB的面积。
Construction (CDL):
Shape(DA,AC,CD)
Shape(DC,CB,BD)
Conditions (CDL):
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(AD),9)
Equal(MeasureOfAngle(DAC),60)
Parallelogram(DACB)
PerpendicularBetweenLine(AC,DC)
Goal: Value(AreaOfQuadrilateral(DACB))
Answer: 45*sqrt(3)/2
Theorem sequence:
sine_theorem(1,DAC)
altitude_of_quadrilateral_judgment_diagonal(1,DACB)
parallelogram_area_formula_common(1,DACB)
Problem: As shown in the diagram, AB=12, DA=5, P is the center of ⊙P, the diameter of ⊙P is BD, the diameter of ⊙P is DB. Find the circumference of the circle P.
问题：如图所示，AB=12，DA=5，P是圆P的圆心，⊙P的直径为BD，⊙P的直径为DB。求圆P的周长。
Construction (CDL):
Shape(PBC,CB)
Shape(PDA,AD)
Shape(PCD,DC)
Shape(PAB,BA)
Shape(CD,DP,PB,BC)
Shape(AB,BP,PD,DA)
Collinear(DPB)
Cocircular(P,ABCD)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(DA),5)
IsCentreOfCircle(P,P)
IsDiameterOfCircle(BD,P)
IsDiameterOfCircle(DB,P)
Goal: Value(PerimeterOfCircle(P))
Answer: 13*pi
Theorem sequence:
diameter_of_circle_property_right_angle(1,DAB,P)
right_triangle_judgment_angle(1,DAB)
right_triangle_property_pythagorean(1,DAB)
diameter_of_circle_property_length_equal(1,BD,P)
circle_property_length_of_radius_and_diameter(1,P)
circle_perimeter_formula(1,P)
Problem: As shown in the diagram, AB=25, CA=z, CB=y, DA=8, DC=x, BA is perpendicular to CA, DC⊥BC. Find the value of y.
问题：如图所示，AB=25，CA=z，CB=y，DA=8，DC=x，BA⊥CA，DC垂直于BC。求y的值。
Construction (CDL):
Shape(DC,CA,AD)
Shape(AC,CB,BA)
Collinear(DAB)
Conditions (CDL):
Equal(LengthOfLine(AB),25)
Equal(LengthOfLine(CA),z)
Equal(LengthOfLine(CB),y)
Equal(LengthOfLine(DA),8)
Equal(LengthOfLine(DC),x)
PerpendicularBetweenLine(BA,CA)
PerpendicularBetweenLine(DC,BC)
Goal: Value(y)
Answer: 5*sqrt(33)
Theorem sequence:
adjacent_complementary_angle(1,BAC,CAD)
right_triangle_judgment_angle(1,DCB)
right_triangle_judgment_angle(1,BAC)
right_triangle_judgment_angle(1,CAD)
line_addition(1,DA,AB)
right_triangle_property_pythagorean(1,DCB)
right_triangle_property_pythagorean(1,BAC)
right_triangle_property_pythagorean(1,CAD)
Problem: As shown in the diagram, AB=BC, AC=12, BD=h, ∠CAB=60°, CD is perpendicular to BD. Find the value of h.
问题：如图所示，AB=BC，AC=12，BD=h，∠CAB=60°，CD垂直于BD。求h的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(BC))
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(BD),h)
Equal(MeasureOfAngle(CAB),60)
PerpendicularBetweenLine(CD,BD)
Goal: Value(h)
Answer: 6*sqrt(3)
Theorem sequence:
perpendicular_bisector_judgment_distance_equal(1,BD,CA)
line_addition(1,CD,DA)
triangle_property_angle_sum(1,BDA)
sine_theorem(1,BDA)
sine_theorem(1,ABD)
Problem: As shown in the diagram, BE=7, AFDE is a square. Find the area of AFDE.
问题：如图所示，BE=7，四边形AFDE是正方形。求AFDE的面积。
Construction (CDL):
Shape(FB,BA,AF)
Shape(FD,DB,BF)
Shape(BD,DE,EB)
Shape(AB,BE,EA)
Collinear(FBE)
Collinear(DBA)
Conditions (CDL):
Equal(LengthOfLine(BE),7)
Square(AFDE)
Goal: Value(AreaOfQuadrilateral(AFDE))
Answer: 98
Theorem sequence:
parallelogram_property_diagonal_bisection(1,FDEA,B)
rectangle_property_diagonal_equal(1,AFDE)
line_addition(1,FB,BE)
kite_area_formula_diagonal(1,AFDE)
Problem: As shown in the diagram, AB=18, AD=y, BC=x, BD=z, ∠ABC=45°, ∠ADB=60°, BA is perpendicular to DA, BC⊥AC. Find the value of y.
问题：如图所示，AB=18，AD=y，BC=x，BD=z，∠ABC=45°，∠ADB=60°，BA⊥DA，BC⊥AC。求y的值。
Construction (CDL):
Shape(DB,BA,AD)
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(BD),z)
Equal(MeasureOfAngle(ABC),45)
Equal(MeasureOfAngle(ADB),60)
PerpendicularBetweenLine(BA,DA)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 6*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,DBA)
sine_theorem(1,ADB)
Problem: As shown in the diagram, the area of ADCB is 150, the area of quadrilateral FJHG is 54, CD=10, JH=x, quadrilateral ADCB is a ▱, FG and JH are opposite sides of the ▱ FJHG, ADCB is similar to FJHG. Find the value of x.
问题：如图所示，四边形ADCB的面积为150，四边形FJHG的面积为54，CD=10，JH=x，ADCB是平行四边形，四边形FJHG是▱，ADCB相似于FJHG。求x的值。
Construction (CDL):
Shape(FJ,JH,HG,GF)
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ADCB),150)
Equal(AreaOfQuadrilateral(FJHG),54)
Equal(LengthOfLine(CD),10)
Equal(LengthOfLine(JH),x)
Parallelogram(ADCB)
Parallelogram(FJHG)
SimilarBetweenQuadrilateral(ADCB,FJHG)
Goal: Value(x)
Answer: 6
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,DCBA,JHGF)
similar_quadrilateral_property_area_square_ratio(1,ADCB,FJHG)
Problem: As shown in the diagram, CA=10, C is the center of circle C, quadrilateral ABFD is a square. Find the area of the circle C minus the area of quadrilateral ABFD.
问题：如图所示，CA=10，圆C的圆心为C，四边形ABFD是正方形。求⊙C的面积减去四边形ABFD的面积。
Construction (CDL):
Shape(CBF,FB)
Shape(CFD,DF)
Shape(CDA,AD)
Shape(CAB,BA)
Shape(CB,BF,FC)
Shape(CA,AB,BC)
Shape(CD,DA,AC)
Shape(CF,FD,DC)
Collinear(FCA)
Collinear(DCB)
Cocircular(C,BFDA)
Conditions (CDL):
Equal(LengthOfLine(CA),10)
IsCentreOfCircle(C,C)
Square(ABFD)
Goal: Value(Sub(AreaOfCircle(C),AreaOfQuadrilateral(ABFD)))
Answer: -200+100*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,CA,C)
radius_of_circle_property_length_equal(1,CF,C)
line_addition(1,FC,CA)
rectangle_property_diagonal_equal(1,ABFD)
kite_area_formula_diagonal(1,ABFD)
circle_area_formula(1,C)
Problem: As shown in the diagram, AB=c, CA=b, CB=a, ∠ABC=60°, ∠CAB=30°, b=18, BC⊥AC. Find the value of a.
问题：如图所示，AB=c，CA=b，CB=a，∠ABC=60°，∠CAB=30°，b=18，BC⊥AC。求a的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),c)
Equal(LengthOfLine(CA),b)
Equal(LengthOfLine(CB),a)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),30)
Equal(b,18)
PerpendicularBetweenLine(BC,AC)
Goal: Value(a)
Answer: 6*sqrt(3)
Theorem sequence:
sine_theorem(1,BCA)
sine_theorem(1,ABC)
Problem: As shown in the diagram, AB=12, EB=9, ∠EBA=55°, quadrilateral ADCB is a rhombus. Find the measure of ∠ECB.
问题：如图所示，AB=12，EB=9，∠EBA=55°，ADCB是菱形。求∠ECB的大小。
Construction (CDL):
Shape(AE,EB,BA)
Shape(AD,DE,EA)
Shape(ED,DC,CE)
Shape(BE,EC,CB)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(EB),9)
Equal(MeasureOfAngle(EBA),55)
Rhombus(ADCB)
Goal: Value(MeasureOfAngle(ECB))
Answer: 180*asin(3/4)/pi
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,CBAD,E)
sine_theorem(1,BEC)
Problem: As shown in the diagram, RP=6, SQ=4, the center of ⊙R is R, S is the center of circle S, the tangent to circle S is PQ, QP is the tangent to ⊙R, MA is perpendicular to RA. Find the length of line PQ.
问题：如图所示，RP=6，SQ=4，圆R的圆心为R，S是⊙S的圆心，PQ是⊙O的切线，QP是圆O的切线，MA垂直于RA。求直线PQ的长度。
Construction (CDL):
Shape(RPC,CR,RA,AP)
Shape(RCM,MA,AR,RC)
Shape(RMP,PA,AM)
Shape(RCM,SNC,NM)
Shape(SNC,CS,SN)
Shape(SQN,NS,SQ)
Shape(SCQ,QS,SC)
Shape(RMP,MN,SQN,QP)
Collinear(RAP)
Collinear(RCS)
Collinear(AMNS)
Cocircular(R,CMP)
Cocircular(S,QNC)
Conditions (CDL):
Equal(LengthOfLine(RP),6)
Equal(LengthOfLine(SQ),4)
IsCentreOfCircle(R,R)
IsCentreOfCircle(S,S)
IsTangentOfCircle(PQ,S)
IsTangentOfCircle(QP,R)
PerpendicularBetweenLine(MA,RA)
Goal: Value(LengthOfLine(PQ))
Answer: 4*sqrt(6)
Theorem sequence:
radius_of_circle_property_length_equal(1,RP,R)
radius_of_circle_property_length_equal(1,RC,R)
radius_of_circle_property_length_equal(1,SQ,S)
radius_of_circle_property_length_equal(1,SC,S)
line_addition(1,RC,CS)
tangent_of_circle_property_perpendicular(1,PQ,S,S)
tangent_of_circle_property_perpendicular(2,QP,R,R)
parallel_judgment_corresponding_angle(2,PQ,AS,R)
parallel_judgment_per_per(1,QS,PR)
parallel_property_collinear_extend(3,RP,SQ,A)
parallelogram_judgment_parallel_and_parallel(1,PASQ)
parallelogram_property_opposite_line_equal(1,PASQ)
parallelogram_property_opposite_line_equal(1,ASQP)
line_addition(1,PA,AR)
right_triangle_judgment_angle(1,SAR)
right_triangle_property_pythagorean(1,SAR)
Problem: As shown in the diagram, AN=12, BN=6, ∠NBE=45°, ANBC is a ▱, AE⊥BE. Find the area of ANBC.
问题：如图所示，AN=12，BN=6，∠NBE=45°，四边形ANBC是平行四边形，AE⊥BE。求ANBC的面积。
Construction (CDL):
Shape(BC,CA,AE,EB)
Shape(BE,EN,NB)
Collinear(AEN)
Conditions (CDL):
Equal(LengthOfLine(AN),12)
Equal(LengthOfLine(BN),6)
Equal(MeasureOfAngle(NBE),45)
Parallelogram(ANBC)
PerpendicularBetweenLine(AE,BE)
Goal: Value(AreaOfQuadrilateral(ANBC))
Answer: 36*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,BEN)
sine_theorem(1,BEN)
altitude_of_quadrilateral_judgment_right_vertex(1,BE,CANB)
parallelogram_area_formula_common(1,CANB)
Problem: As shown in the diagram, AB=h, AC=15, ∠ACB=60°, BA⊥CA. Find the value of h.
问题：如图所示，AB=h，AC=15，∠ACB=60°，BA垂直于CA。求h的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),h)
Equal(LengthOfLine(AC),15)
Equal(MeasureOfAngle(ACB),60)
PerpendicularBetweenLine(BA,CA)
Goal: Value(h)
Answer: 15*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,BAC)
sine_theorem(1,BAC)
sine_theorem(1,CBA)
Problem: As shown in the diagram, XY=8, XZ=8, ∠YXZ=60°. Find the measure of ∠ZYX.
问题：如图所示，XY=8，XZ=8，∠YXZ=60°。求∠ZYX的大小。
Construction (CDL):
Shape(XZ,ZY,YX)
Conditions (CDL):
Equal(LengthOfLine(XY),8)
Equal(LengthOfLine(XZ),8)
Equal(MeasureOfAngle(YXZ),60)
Goal: Value(MeasureOfAngle(ZYX))
Answer: 60
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,XZY)
isosceles_triangle_property_angle_equal(1,XZY)
triangle_property_angle_sum(1,XZY)
Problem: As shown in the diagram, ∠FEI=2*x+70°, ∠HGL=2*y-20°, ∠RFX=3*x+40°, GF∥HE, HG is parallel to EF. Find the value of y.
问题：如图所示，∠FEI=2*x+70°，∠HGL=2*y-20°，∠RFX=3*x+40°，GF平行于HE，HG平行于EF。求y的值。
Construction (CDL):
Shape(RF,FX)
Shape(XF,FE)
Shape(FE,EI)
Shape(IE,EO)
Shape(GF.FR)
Shape(EF,FG)
Shape(HE,EF)
Shape(OE,EH)
Shape(QG,GF)
Shape(FG,GH)
Shape(GH,HE)
Shape(EH,HM)
Shape(LG,GQ)
Shape(HG,GL)
Shape(JH,HG)
Shape(MH,HJ)
Collinear(RFEO)
Collinear(QGHM)
Collinear(XFGL)
Collinear(IEHJ)
Conditions (CDL):
Equal(MeasureOfAngle(FEI),2*x+70)
Equal(MeasureOfAngle(HGL),2*y-20)
Equal(MeasureOfAngle(RFX),3*x+40)
ParallelBetweenLine(GF,HE)
ParallelBetweenLine(HG,EF)
Goal: Value(y)
Answer: 75
Theorem sequence:
vertical_angle(1,RFX,EFG)
vertical_angle(1,FEI,OEH)
parallel_property_corresponding_angle(1,EH,FG,O)
parallel_property_corresponding_angle(2,FE,GH,L)
Problem: As shown in the diagram, RT=6, S is the center of ⊙S, PT⊥QT. Find the length of line PR.
问题：如图所示，RT=6，S是圆S的圆心，PT垂直于QT。求直线PR的长度。
Construction (CDL):
Shape(SRQ,QT,TR)
Shape(SQP,PT,TQ)
Shape(SPR,RT,TS,ST,TP)
Collinear(PTR)
Collinear(QTS)
Cocircular(S,RQP)
Conditions (CDL):
Equal(LengthOfLine(RT),6)
IsCentreOfCircle(S,S)
PerpendicularBetweenLine(PT,QT)
Goal: Value(LengthOfLine(PR))
Answer: 12
Theorem sequence:
vertical_angle(1,PTQ,RTS)
circle_property_chord_perpendicular_bisect_chord(1,S,ST,RP)
line_addition(1,PT,TR)
Problem: As shown in the diagram, ∠BGE=52°, ∠GFE=30°, the measure of ⌒OBA is 108, the measure of arc OFE is 118, O is the center of ⊙O. Find the measure of ⌒OAC.
问题：如图所示，∠BGE=52°，∠GFE=30°，⌒OBA的角度为108，弧OFE的角度为118，圆O的圆心为O。求弧OAC的角度。
Construction (CDL):
Shape(OEB,BE)
Shape(OE,EB,BO)
Shape(OBA,AB)
Shape(OAC,CB,BA)
Shape(GB,BC,CG)
Shape(OCF,FG,GC)
Shape(GE,EO,OB,BG)
Shape(OFE,EF)
Shape(GF,FE,EG)
Shape(OAC,AD,DC)
Collinear(DAB)
Collinear(DCGE)
Collinear(FGB)
Cocircular(O,ACFEB)
Conditions (CDL):
Equal(MeasureOfAngle(BGE),52)
Equal(MeasureOfAngle(GFE),30)
Equal(MeasureOfArc(OBA),108)
Equal(MeasureOfArc(OFE),118)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfArc(OAC))
Answer: 30
Theorem sequence:
arc_property_circumference_angle_external(1,OEB,C)
flat_angle(1,CGE)
angle_addition(1,CGB,BGE)
triangle_property_angle_sum(1,BCG)
arc_property_circumference_angle_external(1,OCF,B)
arc_property_center_angle(1,OEB,O)
arc_property_center_angle(1,OBE,O)
arc_addition_measure(1,OAC,OCE)
arc_addition_measure(1,OBA,OAE)
arc_addition_measure(1,OCF,OFE)
round_angle(1,BOE,EOB)
Problem: As shown in the diagram, BA=6, CF=7, DA=9, ∠BAF=32°, ∠CBF=40°, ∠FAD=20°, DA and CB are opposite sides of the parallelogram ADCB. Find the length of line DF.
问题：如图所示，BA=6，CF=7，DA=9，∠BAF=32°，∠CBF=40°，∠FAD=20°，AB和DC是平行四边形ADCB的一组对边。求直线DF的长度。
Construction (CDL):
Shape(AD,DF,FA)
Shape(AF,FB,BA)
Shape(FD,DC,CF)
Shape(BF,FC,CB)
Collinear(AFC)
Collinear(DFB)
Conditions (CDL):
Equal(LengthOfLine(BA),6)
Equal(LengthOfLine(CF),7)
Equal(LengthOfLine(DA),9)
Equal(MeasureOfAngle(BAF),32)
Equal(MeasureOfAngle(CBF),40)
Equal(MeasureOfAngle(FAD),20)
Parallelogram(ADCB)
Goal: Value(LengthOfLine(DF))
Answer: sqrt(130-126*cos(pi/9))
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ADCB,F)
cosine_theorem(1,ADF)
Problem: As shown in the diagram, JH=4*x+7, JK=25, KH=6*x-2, NQ=12, PN=8, PQ=20, ∠JHK=∠QNP, ∠KJH=∠PQN. Find the length of line HK.
问题：如图所示，JH=4*x+7，JK=25，KH=6*x-2，NQ=12，PN=8，PQ=20，∠JHK=∠QNP，∠KJH=∠PQN。求直线HK的长度。
Construction (CDL):
Shape(JH,HK,KJ)
Shape(PQ,QN,NP)
Conditions (CDL):
Equal(LengthOfLine(JH),4*x+7)
Equal(LengthOfLine(JK),25)
Equal(LengthOfLine(KH),6*x-2)
Equal(LengthOfLine(NQ),12)
Equal(LengthOfLine(PN),8)
Equal(LengthOfLine(PQ),20)
Equal(MeasureOfAngle(JHK),MeasureOfAngle(QNP))
Equal(MeasureOfAngle(KJH),MeasureOfAngle(PQN))
Goal: Value(LengthOfLine(HK))
Answer: 10
Theorem sequence:
similar_triangle_judgment_aa(1,KJH,PQN)
similar_triangle_property_line_ratio(1,JHK,QNP)
similar_triangle_property_line_ratio(1,KJH,PQN)
Problem: As shown in the diagram, AF=c, DF=9, DG=d, DH=11, HE=15, the tangent to ⊙C is DG. Find the value of c.
问题：如图所示，AF=c，DF=9，DG=d，DH=11，HE=15，DG是⊙O的切线。求c的值。
Construction (CDL):
Shape(CFG,CGA,AF)
Shape(CAE,EH,CHF,FA)
Shape(CEH,HE)
Shape(CFG,FD,DG)
Shape(CHF,HD,DF)
Collinear(AFD)
Collinear(EHD)
Cocircular(C,GAEHF)
Conditions (CDL):
Equal(LengthOfLine(AF),c)
Equal(LengthOfLine(DF),9)
Equal(LengthOfLine(DG),d)
Equal(LengthOfLine(DH),11)
Equal(LengthOfLine(HE),15)
IsTangentOfCircle(DG,C)
Goal: Value(c)
Answer: 205/9
Theorem sequence:
line_addition(1,AF,FD)
line_addition(1,EH,HD)
circle_property_circular_power_segment_and_segment_line(1,DFA,DHE,C)
Problem: As shown in the diagram, AD=x+2, AY=AC, CB=3/2*x+11, DB=3*y-9, DB=YD, YD=2*y+6. Find the value of y.
问题：如图所示，AD=x+2，AY=AC，CB=3/2*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)
Conditions (CDL):
Equal(LengthOfLine(AD),x+2)
Equal(LengthOfLine(AY),LengthOfLine(AC))
Equal(LengthOfLine(CB),3/2*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, ∠BAC=2*x°, ∠CAD=3*x°, BA⊥DA. Find the value of x.
问题：如图所示，∠BAC=2*x°，∠CAD=3*x°，BA⊥DA。求x的值。
Construction (CDL):
Shape(BA,AC)
Shape(CA,AD)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),2*x)
Equal(MeasureOfAngle(CAD),3*x)
PerpendicularBetweenLine(BA,DA)
Goal: Value(x)
Answer: 18
Theorem sequence:
angle_addition(1,BAC,CAD)