SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠ADI=4*y+7°, ∠CBI=7*x-8°, ∠IAD=5*x°, ∠ICB=5*y-3°. Find the measure of ∠ICB.
问题：如图所示，∠ADI=4*y+7°，∠CBI=7*x-8°，∠IAD=5*x°，∠ICB=5*y-3°。求∠ICB的大小。
Construction (CDL):
Shape(IA,AD,DI)
Shape(IC,CB,BI)
Shape(JBA,AI,IB)
Shape(JAD,DA)
Shape(JDC,CI,ID)
Shape(JCB,BC)
Collinear(AIC)
Collinear(BID)
Cocircular(J,ADCB)
Conditions (CDL):
Equal(MeasureOfAngle(ADI),4*y+7)
Equal(MeasureOfAngle(CBI),7*x-8)
Equal(MeasureOfAngle(IAD),5*x)
Equal(MeasureOfAngle(ICB),5*y-3)
Goal: Value(MeasureOfAngle(ICB))
Answer: 47
Theorem sequence:
arc_property_circumference_angle_external(1,JBA,D)
arc_property_circumference_angle_external(1,JBA,C)
Problem: As shown in the diagram, ∠AXW=30°, WY⊥ZY, WYZX is a rectangle. Find the measure of ∠ZAY.
问题：如图所示，∠AXW=30°，WY⊥ZY，WYZX是矩形。求∠ZAY的大小。
Construction (CDL):
Shape(WY,YA,AW)
Shape(XW,WA,AX)
Shape(AY,YZ,ZA)
Shape(XA,AZ,ZX)
Collinear(WAZ)
Collinear(YAX)
Conditions (CDL):
Equal(MeasureOfAngle(AXW),30)
PerpendicularBetweenLine(WY,ZY)
Rectangle(WYZX)
Goal: Value(MeasureOfAngle(ZAY))
Answer: 120
Theorem sequence:
rectangle_property_diagonal_equal(1,WYZX)
parallelogram_property_diagonal_bisection(1,WYZX,A)
parallelogram_property_diagonal_bisection(1,YZXW,A)
line_addition(1,WA,AZ)
line_addition(1,YA,AX)
isosceles_triangle_judgment_line_equal(1,AXW)
isosceles_triangle_property_angle_equal(1,AXW)
triangle_property_angle_sum(1,AXW)
vertical_angle(1,WAX,ZAY)
Problem: As shown in the diagram, JN=3*y-5, LN=y+5, NK=2*z+7, NM=z+9, MJ and LK are opposite sides of the ▱ JMLK. Find the value of y.
问题：如图所示，JN=3*y-5，LN=y+5，NK=2*z+7，NM=z+9，JK和ML是▱JMLK的一组对边。求y的值。
Construction (CDL):
Shape(JM,MN,NJ)
Shape(JN,NK,KJ)
Shape(NM,ML,LN)
Shape(NL,LK,KN)
Collinear(JNL)
Collinear(MNK)
Conditions (CDL):
Equal(LengthOfLine(JN),3*y-5)
Equal(LengthOfLine(LN),y+5)
Equal(LengthOfLine(NK),2*z+7)
Equal(LengthOfLine(NM),z+9)
Parallelogram(JMLK)
Goal: Value(y)
Answer: 5
Theorem sequence:
parallelogram_property_diagonal_bisection(1,JMLK,N)
Problem: As shown in the diagram, AC=3, ∠BAC=55°, ∠CBA=55°. Find the length of line CB.
问题：如图所示，AC=3，∠BAC=55°，∠CBA=55°。求直线CB的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),3)
Equal(MeasureOfAngle(BAC),55)
Equal(MeasureOfAngle(CBA),55)
Goal: Value(LengthOfLine(CB))
Answer: 3
Theorem sequence:
isosceles_triangle_judgment_angle_equal(1,CBA)
Problem: As shown in the diagram, HP=25, KM=x, HJ is the midsegment of triangle MLK, JP is the midsegment of triangle KML, PH is the midsegment of triangle LKM. Find the value of x.
问题：如图所示，HP=25，KM=x，HJ是△MLK的中位线，JP是三角形KML的中位线，PH是△LKM的中位线。求x的值。
Construction (CDL):
Shape(PK,KJ,JP)
Shape(PJ,JH,HP)
Shape(LP,PH,HL)
Shape(HJ,JM,MH)
Collinear(KPL)
Collinear(KJM)
Collinear(MHL)
Conditions (CDL):
Equal(LengthOfLine(HP),25)
Equal(LengthOfLine(KM),x)
IsMidsegmentOfTriangle(HJ,MLK)
IsMidsegmentOfTriangle(JP,KML)
IsMidsegmentOfTriangle(PH,LKM)
Goal: Value(x)
Answer: 50
Theorem sequence:
line_addition(1,MH,HL)
midsegment_of_triangle_property_parallel(1,PH,LKM)
parallel_property_corresponding_angle(1,PH,KM,L)
similar_triangle_judgment_aa(1,HLP,MLK)
similar_triangle_property_line_ratio(1,LPH,LKM)
similar_triangle_property_line_ratio(1,PHL,KML)
Problem: As shown in the diagram, AB=36, AD=22, ∠BCD=30°, BA and CD are opposite sides of the parallelogram BCDA, CE⊥BE. Find the area of BCDA.
问题：如图所示，AB=36，AD=22，∠BCD=30°，CB和DA是平行四边形BCDA的一组对边，CE垂直于BE。求BCDA的面积。
Construction (CDL):
Shape(BC,CE,ED,DA,AB)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AB),36)
Equal(LengthOfLine(AD),22)
Equal(MeasureOfAngle(BCD),30)
Parallelogram(BCDA)
PerpendicularBetweenLine(CE,BE)
Goal: Value(AreaOfQuadrilateral(BCDA))
Answer: 396
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,CDAB)
parallelogram_area_formula_sine(1,DABC)
Problem: As shown in the diagram, TR=18, WT=12, ∠TWV=60°, TR and WV are opposite sides of the parallelogram TWVR, WA⊥TA. Find the area of quadrilateral TWVR.
问题：如图所示，TR=18，WT=12，∠TWV=60°，WT和VR是▱TWVR的一组对边，WA⊥TA。求四边形TWVR的面积。
Construction (CDL):
Shape(TW,WA,AV,VR,RT)
Collinear(WAV)
Conditions (CDL):
Equal(LengthOfLine(TR),18)
Equal(LengthOfLine(WT),12)
Equal(MeasureOfAngle(TWV),60)
Parallelogram(TWVR)
PerpendicularBetweenLine(WA,TA)
Goal: Value(AreaOfQuadrilateral(TWVR))
Answer: 108*sqrt(3)
Theorem sequence:
parallelogram_property_opposite_line_equal(1,RTWV)
parallelogram_area_formula_sine(1,TWVR)
Problem: As shown in the diagram, AC=25, AN=21, OB=20, quadrilateral CANB is a ▱, NO⊥BO. Find the perimeter of quadrilateral CANB.
问题：如图所示，AC=25，AN=21，OB=20，CB和AN是平行四边形CANB的一组对边，NO垂直于BO。求CANB的周长。
Construction (CDL):
Shape(CA,AN,NB,BC)
Shape(BN,NO,OB)
Collinear(ANO)
Conditions (CDL):
Equal(LengthOfLine(AC),25)
Equal(LengthOfLine(AN),21)
Equal(LengthOfLine(OB),20)
Parallelogram(CANB)
PerpendicularBetweenLine(NO,BO)
Goal: Value(PerimeterOfQuadrilateral(CANB))
Answer: 92
Theorem sequence:
parallelogram_property_opposite_line_equal(1,CANB)
parallelogram_property_opposite_line_equal(1,ANBC)
quadrilateral_perimeter_formula(1,CANB)
Problem: As shown in the diagram, ∠AYB=y+25°, ∠BYD=3*x-15°, ∠GBE=105°, DA is parallel to EI. Find the value of x.
问题：如图所示，∠AYB=y+25°，∠BYD=3*x-15°，∠GBE=105°，DA∥EI。求x的值。
Construction (CDL):
Shape(DY,YH)
Shape(HY,YA)
Shape(BY,YD)
Shape(AY,YB)
Shape(EB,BY)
Shape(YB,BI)
Shape(GB,BE)
Shape(IB,BG)
Collinear(GBYH)
Collinear(DYA)
Collinear(EBI)
Conditions (CDL):
Equal(MeasureOfAngle(AYB),y+25)
Equal(MeasureOfAngle(BYD),3*x-15)
Equal(MeasureOfAngle(GBE),105)
ParallelBetweenLine(DA,EI)
Goal: Value(x)
Answer: 40
Theorem sequence:
parallel_property_collinear_extend(3,IE,AD,B)
parallel_property_collinear_extend(3,DA,EB,Y)
parallel_property_corresponding_angle(1,BE,YD,G)
Problem: As shown in the diagram, JK=5, LJ=13, LK=12, LK is perpendicular to JK. Find the value of cos(JL).
问题：如图所示，JK=5，LJ=13，LK=12，LK⊥JK。求cos(JL)的值。
Construction (CDL):
Shape(JL,LK,KJ)
Conditions (CDL):
Equal(LengthOfLine(JK),5)
Equal(LengthOfLine(LJ),13)
Equal(LengthOfLine(LK),12)
PerpendicularBetweenLine(LK,JK)
Goal: Value(Cos(MeasureOfAngle(JLK)))
Answer: 12/13
Theorem sequence:
cosine_theorem(1,LKJ)
Problem: As shown in the diagram, AD=8, AE=4, FC=10, ∠CFA=60°, the center of ⊙E is E, AB⊥FB, CB⊥AB, DCFA is a trapezoid. Find the sum of the area of ADCF and the area of the sector EFA.
问题：如图所示，AD=8，AE=4，FC=10，∠CFA=60°，⊙E的圆心为E，AB垂直于FB，CB⊥AB，DC和FA是梯形DCFA的腰。求ADCF的面积与扇形EFA的面积之和。
Construction (CDL):
Shape(AD,DC,CB,BA)
Shape(AB,BF,FA)
Shape(EFA,AE,EF)
Collinear(AEF)
Collinear(CBF)
Cocircular(E,FA)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(AE),4)
Equal(LengthOfLine(FC),10)
Equal(MeasureOfAngle(CFA),60)
IsCentreOfCircle(E,E)
PerpendicularBetweenLine(AB,FB)
PerpendicularBetweenLine(CB,AB)
Trapezoid(DCFA)
Goal: Value(Add(AreaOfQuadrilateral(ADCF),AreaOfSector(EFA)))
Answer: 8*pi+36*sqrt(3)
Theorem sequence:
line_addition(1,AE,EF)
sine_theorem(1,ABF)
radius_of_circle_property_length_equal(1,EA,E)
radius_of_circle_property_length_equal(1,EF,E)
flat_angle(1,AEF)
altitude_of_quadrilateral_judgment_right_vertex(2,AB,DCFA)
trapezoid_area_formula(1,DCFA)
arc_property_center_angle(1,EFA,E)
sector_area_formula(1,EFA)
Problem: As shown in the diagram, ∠ETV=9*x-11°, ∠REU=8*x+4°, DU∥CV. Find the measure of ∠REU.
问题：如图所示，∠ETV=9*x-11°，∠REU=8*x+4°，DU平行于CV。求∠REU的大小。
Construction (CDL):
Shape(DE,ER)
Shape(RE,EU)
Shape(TE,ED)
Shape(UE,ET)
Shape(CT,TE)
Shape(ET,TV)
Shape(VT,TB)
Shape(BT,TC)
Collinear(RETB)
Collinear(DEU)
Collinear(CTV)
Conditions (CDL):
Equal(MeasureOfAngle(ETV),9*x-11)
Equal(MeasureOfAngle(REU),8*x+4)
ParallelBetweenLine(DU,CV)
Goal: Value(MeasureOfAngle(REU))
Answer: 124
Theorem sequence:
parallel_property_collinear_extend(3,VC,UD,T)
parallel_property_collinear_extend(3,DU,TV,E)
parallel_property_corresponding_angle(1,EU,TV,R)
Problem: As shown in the diagram, DC=10, ∠BAC=40°, ∠CAD=40°, CD is perpendicular to ND, FB is perpendicular to CB. Find the length of line BC.
问题：如图所示，DC=10，∠BAC=40°，∠CAD=40°，CD⊥ND，FB⊥CB。求直线BC的长度。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AC,CB,BA)
Collinear(ADN)
Collinear(ABF)
Conditions (CDL):
Equal(LengthOfLine(DC),10)
Equal(MeasureOfAngle(BAC),40)
Equal(MeasureOfAngle(CAD),40)
PerpendicularBetweenLine(CD,ND)
PerpendicularBetweenLine(FB,CB)
Goal: Value(LengthOfLine(BC))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,ADC,CDN)
adjacent_complementary_angle(1,FBC,CBA)
mirror_congruent_triangle_judgment_aas(3,CBA,CAD)
mirror_congruent_triangle_property_line_equal(1,ACB,ADC)
Problem: As shown in the diagram, ∠AXC=118°, ∠BXA=x°, ∠CXB=135°. Find the value of x.
问题：如图所示，∠AXC=118°，∠BXA=x°，∠CXB=135°。求x的值。
Construction (CDL):
Shape(XBC,CX,XB)
Shape(XCA,AX,XC)
Shape(XAB,BX,XA)
Cocircular(X,BCA)
Conditions (CDL):
Equal(MeasureOfAngle(AXC),118)
Equal(MeasureOfAngle(BXA),x)
Equal(MeasureOfAngle(CXB),135)
Goal: Value(x)
Answer: 107
Theorem sequence:
angle_addition(1,AXC,CXB)
round_angle(1,BXA,AXB)
Problem: As shown in the diagram, CA=20, CB=6, BC⊥AC. Find the measure of ∠CAB.
问题：如图所示，CA=20，CB=6，BC⊥AC。求∠CAB的大小。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(CA),20)
Equal(LengthOfLine(CB),6)
PerpendicularBetweenLine(BC,AC)
Goal: Value(MeasureOfAngle(CAB))
Answer: 180*acos(10*sqrt(109)/109)/pi
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,ABC)
Problem: As shown in the diagram, the diameter of circle A is 10, the diameter of ⊙B is 30, the diameter of ⊙C is 10, AZ=CW, CW=2, the center of circle A is A, B is the center of ⊙B, the center of ⊙C is C. Find the length of line BY.
问题：如图所示，⊙A的直径为10，圆B的直径为30，⊙C的直径为10，AZ=CW，CW=2，A是⊙A的圆心，B是圆B的圆心，C是圆C的圆心。求直线BY的长度。
Construction (CDL):
Shape(ADF,FA,AZ,BDZ)
Shape(AFE,BZE,ZA,ZF)
Shape(BDZ,ZX,AXD)
Shape(BZE,AEX,XZ)
Shape(BHD,AXD,XB,BY,CHY)
Shape(BEI,CYI,YB,BX,AEX)
Shape(CHY,YW,BWH)
Shape(CYI,BIW,WY)
Shape(CGH,BWH,WC,CG)
Shape(CIG,GC,CW,BIW)
Collinear(FAZXBYWCG)
Cocircular(A,DFEX)
Cocircular(B,DZEIWH)
Cocircular(C,HYIG)
Conditions (CDL):
Equal(DiameterOfCircle(A),10)
Equal(DiameterOfCircle(B),30)
Equal(DiameterOfCircle(C),10)
Equal(LengthOfLine(AZ),LengthOfLine(CW))
Equal(LengthOfLine(CW),2)
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
Goal: Value(LengthOfLine(BY))
Answer: 12
Theorem sequence:
circle_property_length_of_radius_and_diameter(1,B)
circle_property_length_of_radius_and_diameter(1,C)
radius_of_circle_property_length_equal(1,BW,B)
radius_of_circle_property_length_equal(1,CY,C)
line_addition(1,BY,YW)
line_addition(1,YW,WC)
Problem: As shown in the diagram, AE=9, ED=12, quadrilateral DEAC is a rectangle. Find the area of the circle B minus the area of quadrilateral DEAC.
问题：如图所示，AE=9，ED=12，四边形DEAC是长方形。求⊙B的面积减去四边形DEAC的面积。
Construction (CDL):
Shape(BCD,DC)
Shape(BDE,ED)
Shape(BEA,AE)
Shape(BAC,CA)
Shape(DE,EA,AD)
Shape(CD,DA,AC)
Cocircular(B,DEAC)
Conditions (CDL):
Equal(LengthOfLine(AE),9)
Equal(LengthOfLine(ED),12)
Rectangle(DEAC)
Goal: Value(Sub(AreaOfCircle(B),AreaOfQuadrilateral(DEAC)))
Answer: -108+225*pi/4
Theorem sequence:
right_triangle_judgment_angle(1,DEA)
right_triangle_property_pythagorean(1,DEA)
diameter_of_circle_judgment_right_angle(1,DEA,B)
diameter_of_circle_property_length_equal(1,DA,B)
circle_property_length_of_radius_and_diameter(1,B)
parallelogram_area_formula_sine(1,DEAC)
circle_area_formula(1,B)
Problem: As shown in the diagram, AB=x, AD=3*sqrt(2), BC=y, BD=sqrt(2), CD=z, AB is perpendicular to CB, CD is perpendicular to BD. Find the value of z.
问题：如图所示，AB=x，AD=3*sqrt(2)，BC=y，BD=sqrt(2)，CD=z，AB垂直于CB，CD⊥BD。求z的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),3*sqrt(2))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),sqrt(2))
Equal(LengthOfLine(CD),z)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(CD,BD)
Goal: Value(z)
Answer: sqrt(2)/3
Theorem sequence:
line_addition(1,AD,DC)
adjacent_complementary_angle(1,CDB,BDA)
right_triangle_judgment_angle(1,ABC)
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,BDA)
right_triangle_property_pythagorean(1,ABC)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,CDB)
Problem: As shown in the diagram, ∠BED=62°, ∠FEC=5*x+2°. Find the value of x.
问题：如图所示，∠BED=62°，∠FEC=5*x+2°。求x的值。
Construction (CDL):
Shape(BE,ED)
Shape(CE,EB)
Shape(FE,EC)
Shape(DE,EF)
Collinear(DEC)
Collinear(BEF)
Conditions (CDL):
Equal(MeasureOfAngle(BED),62)
Equal(MeasureOfAngle(FEC),5*x+2)
Goal: Value(x)
Answer: 12
Theorem sequence:
vertical_angle(1,BED,FEC)
Problem: As shown in the diagram, ∠DIA=110°, the center of ⊙I is I, BA is the tangent to circle I. Find the measure of ∠BAD.
问题：如图所示，∠DIA=110°，I是圆I的圆心，圆O的切线为BA。求∠BAD的大小。
Construction (CDL):
Shape(BA,AI)
Shape(IA,AD,DI)
Shape(IAD,DA)
Shape(IDA,AI,ID)
Cocircular(I,AD)
Conditions (CDL):
Equal(MeasureOfAngle(DIA),110)
IsCentreOfCircle(I,I)
IsTangentOfCircle(BA,I)
Goal: Value(MeasureOfAngle(BAD))
Answer: 125
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BA,I,I)
radius_of_circle_property_length_equal(1,IA,I)
radius_of_circle_property_length_equal(1,ID,I)
isosceles_triangle_judgment_line_equal(1,IAD)
triangle_property_angle_sum(1,IAD)
isosceles_triangle_property_angle_equal(1,IAD)
angle_addition(1,BAI,IAD)
Problem: As shown in the diagram, BO=15, BO=CA, CD=7, CO=13, CO=BA, OD⊥AD. Find the area of OCAB.
问题：如图所示，BO=15，BO=CA，CD=7，CO=13，CO=BA，OD⊥AD。求OCAB的面积。
Construction (CDL):
Shape(OC,CD,DO)
Shape(OD,DA,AB,BO)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(BO),15)
Equal(LengthOfLine(BO),LengthOfLine(CA))
Equal(LengthOfLine(CD),7)
Equal(LengthOfLine(CO),13)
Equal(LengthOfLine(CO),LengthOfLine(BA))
PerpendicularBetweenLine(OD,AD)
Goal: Value(AreaOfQuadrilateral(OCAB))
Answer: 30*sqrt(30)
Theorem sequence:
parallelogram_judgment_equal_and_equal(1,OCAB)
flat_angle(1,CDA)
angle_addition(1,CDO,ODA)
right_triangle_judgment_angle(1,CDO)
right_triangle_property_pythagorean(1,CDO)
altitude_of_quadrilateral_judgment_left_vertex(1,OD,OCAB)
parallelogram_area_formula_common(1,OCAB)
Problem: As shown in the diagram, BL=9, JK=18, JL=27, BA∥LK. Find the length of line JA.
问题：如图所示，BL=9，JK=18，JL=27，BA∥LK。求直线JA的长度。
Construction (CDL):
Shape(JB,BA,AJ)
Shape(BL,LK,KA,AB)
Collinear(JAK)
Collinear(JBL)
Conditions (CDL):
Equal(LengthOfLine(BL),9)
Equal(LengthOfLine(JK),18)
Equal(LengthOfLine(JL),27)
ParallelBetweenLine(BA,LK)
Goal: Value(LengthOfLine(JA))
Answer: 12
Theorem sequence:
line_addition(1,JB,BL)
parallel_property_corresponding_angle(1,BA,LK,J)
similar_triangle_judgment_aa(1,AJB,KJL)
similar_triangle_property_line_ratio(1,AJB,KJL)
similar_triangle_property_line_ratio(1,BAJ,LKJ)
Problem: As shown in the diagram, LE=x, LK=36, MK=15, ∠EML=66°, M is the center of ⊙M, A bisects segment EK, LE is the tangent to circle M, the tangent to circle M is LK. Find the value of x.
问题：如图所示，LE=x，LK=36，MK=15，∠EML=66°，圆M的圆心为M，A平分线段EK，圆O的切线为LE，圆O的切线为LK。求x的值。
Construction (CDL):
Shape(EM,MA,AE)
Shape(AM,MK,KA)
Shape(MPE,EA,AP)
Shape(MKP,PA,AK)
Shape(LE,MPE,PL)
Shape(LP,MKP,KL)
Shape(MEK,KM,ME)
Collinear(MAPL)
Collinear(KAE)
Cocircular(M,EKP)
Conditions (CDL):
Equal(LengthOfLine(LE),x)
Equal(LengthOfLine(LK),36)
Equal(LengthOfLine(MK),15)
Equal(MeasureOfAngle(EML),66)
IsCentreOfCircle(M,M)
IsMidpointOfLine(A,EK)
IsTangentOfCircle(LE,M)
IsTangentOfCircle(LK,M
Goal: Value(x)
Answer: 36
Theorem sequence:
tangent_of_circle_property_perpendicular(2,LE,M,M)
tangent_of_circle_property_perpendicular(1,LK,M,M)
right_triangle_judgment_angle(1,LEM)
right_triangle_judgment_angle(1,MKL)
radius_of_circle_property_length_equal(1,ME,M)
radius_of_circle_property_length_equal(1,MK,M)
right_triangle_property_pythagorean(1,LEM)
right_triangle_property_pythagorean(1,MKL)
Problem: As shown in the diagram, ∠DBH=x°, the measure of ⌒FDH is 34, the measure of arc FED is 106, the measure of ⌒FHA is 160, F is the center of ⊙F. Find the value of x.
问题：如图所示，∠DBH=x°，弧FDH的角度为34，⌒FED的角度为106，弧FHA的角度为160，⊙F的圆心为F。求x的值。
Construction (CDL):
Shape(FDH,DB,BH)
Shape(FDH,HA,AF,FE,ED)
Shape(FAE,EF,FA)
Shape(FED,DE)
Shape(FHA,AH)
Collinear(BHA)
Collinear(BDE)
Cocircular(F,DHAE)
Conditions (CDL):
Equal(MeasureOfAngle(DBH),x)
Equal(MeasureOfArc(FDH),34)
Equal(MeasureOfArc(FED),106)
Equal(MeasureOfArc(FHA),160)
IsCentreOfCircle(F,F)
Goal: Value(x)
Answer: 13
Theorem sequence:
arc_addition_measure(1,FDH,FHA)
arc_addition_measure(1,FED,FDA)
round_angle(1,EFA,AFE)
arc_property_center_angle(1,FEA,F)
arc_property_center_angle(1,FAE,F)
circle_property_circular_power_segment_and_segment_angle(1,BDE,BHA,F)
Problem: As shown in the diagram, BC=5, the radius of circle J is 10, the radius of circle K is 8, J is the center of ⊙J, the center of ⊙K is K, AC is the diameter of ⊙J. Find the length of line AB.
问题：如图所示，BC=5，⊙J的半径为10，圆K的半径为8，J是⊙J的圆心，K是圆K的圆心，圆J的直径为AC。求直线AB的长度。
Construction (CDL):
Shape(JEA,AJ,JB,KEB)
Shape(JCE,KEB,BC)
Shape(KDE,JCE,CK,KD)
Shape(JAF,KBF,BJ,JA)
Shape(JFC,CB,KBF)
Shape(KFD,DK,KC,JFC)
Collinear(AJBCKD)
Cocircular(J,AFCE)
Cocircular(K,EBFD)
Conditions (CDL):
Equal(LengthOfLine(BC),5)
Equal(RadiusOfCircle(J),10)
Equal(RadiusOfCircle(K),8)
IsCentreOfCircle(J,J)
IsCentreOfCircle(K,K)
IsDiameterOfCircle(AC,J)
Goal: Value(LengthOfLine(AB))
Answer: 15
Theorem sequence:
circle_property_length_of_radius_and_diameter(1,J)
diameter_of_circle_property_length_equal(1,AC,J)
line_addition(1,AB,BC)
Problem: As shown in the diagram, ∠ABC=38°, ∠ADC=72°, ∠BCA=42°. Find the measure of ∠DAB.
问题：如图所示，∠ABC=38°，∠ADC=72°，∠BCA=42°。求∠DAB的大小。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AB,BD,DA)
Collinear(CDB)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),38)
Equal(MeasureOfAngle(ADC),72)
Equal(MeasureOfAngle(BCA),42)
Goal: Value(MeasureOfAngle(DAB))
Answer: 34
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADC)
triangle_property_angle_sum(1,ABD)
Problem: As shown in the diagram, ∠CFH=70°, ∠HEA=40°, AB is parallel to XY, XY∥CD. Find the measure of ∠EHF.
问题：如图所示，∠CFH=70°，∠HEA=40°，AB∥XY，XY平行于CD。求∠EHF的大小。
Construction (CDL):
Shape(AE,ED)
Shape(XH,HE)
Shape(CF,FH)
Shape(DE,EB)
Shape(EH,HY)
Shape(HF,FD)
Shape(HE,EA)
Shape(FH,HX)
Shape(IF,FC)
Shape(BE,EH)
Shape(YH,HF)
Shape(DF,FI)
Collinear(AEB)
Collinear(XHY)
Collinear(CFD)
Collinear(HED)
Collinear(HFI)
Conditions (CDL):
Equal(MeasureOfAngle(CFH),70)
Equal(MeasureOfAngle(HEA),40)
ParallelBetweenLine(AB,XY)
ParallelBetweenLine(XY,CD)
Goal: Value(MeasureOfAngle(EHF))
Answer: 110
Theorem sequence:
parallel_property_collinear_extend(3,YX,BA,H)
parallel_property_collinear_extend(3,DC,YX,F)
parallel_property_collinear_extend(3,AB,HY,E)
parallel_property_collinear_extend(3,XY,CF,H)
parallel_property_alternate_interior_angle(1,HY,CF)
parallel_property_alternate_interior_angle(2,AE,HY)
angle_addition(1,EHY,YHF)
Problem: As shown in the diagram, MR=Mul(LengthOfLine(MN),1/4), QT=TP, MQPN is a parallelogram. Find the ratio of similar triangles RMS and TPS.
问题：如图所示，MR=Mul(LengthOfLine(MN),1/4)，QT=TP，MQPN是平行四边形。求△RMS和△TPS的相似比。
Construction (CDL):
Shape(RM,MS,SR)
Shape(SM,MQ,QT,TS)
Shape(ST,TP,PS)
Shape(RS,SP,PN,NR)
Collinear(RST)
Collinear(MSP)
Collinear(MRN)
Collinear(QTP)
Conditions (CDL):
Equal(LengthOfLine(MR),Mul(LengthOfLine(MN),1/4))
Equal(LengthOfLine(QT),LengthOfLine(TP))
Parallelogram(MQPN)
Goal: Value(RatioOfSimilarTriangle(RMS,TPS))
Answer: 1/2
Theorem sequence:
parallelogram_property_opposite_line_equal(1,NMQP)
line_addition(1,QT,TP)
parallel_property_collinear_extend(3,MN,QP,R)
parallel_property_collinear_extend(3,PQ,RM,T)
parallel_property_alternate_interior_angle(2,PT,RM)
vertical_angle(1,MSR,PST)
similar_triangle_judgment_aa(1,MSR,PST)
similar_triangle_property_line_ratio(1,SRM,STP)
Problem: As shown in the diagram, RA=x+9, RS=x+9, RT=3*x-9, ST=2*x, TA=2*x, triangle RTS is an equilateral triangle. Find the length of line ST.
问题：如图所示，RA=x+9，RS=x+9，RT=3*x-9，ST=2*x，TA=2*x，三角形RTS为等边三角形。求直线ST的长度。
Construction (CDL):
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(RA),x+9)
Equal(LengthOfLine(RS),x+9)
Equal(LengthOfLine(RT),3*x-9)
Equal(LengthOfLine(ST),2*x)
Equal(LengthOfLine(TA),2*x)
EquilateralTriangle(RTS)
Goal: Value(LengthOfLine(ST))
Answer: 18
Theorem sequence:
Problem: As shown in the diagram, JA=x, LK=10, the radius of circle J is 10, the center of circle J is J, LA⊥JA. Find the value of x.
问题：如图所示，JA=x，LK=10，圆J的半径为10，J是⊙J的圆心，LA垂直于JA。求x的值。
Construction (CDL):
Shape(JLK,KA,AL)
Shape(LA,AJ,JL)
Shape(JKR,RJ,JA,AK)
Shape(JRL,LA,AJ,JR)
Collinear(LAK)
Cocircular(J,LKR)
Conditions (CDL):
Equal(LengthOfLine(JA),x)
Equal(LengthOfLine(LK),10)
Equal(RadiusOfCircle(J),10)
IsCentreOfCircle(J,J)
PerpendicularBetweenLine(LA,JA)
Goal: Value(x)
Answer: 5*sqrt(3)
Theorem sequence:
radius_of_circle_property_length_equal(1,JL,J)
circle_property_chord_perpendicular_bisect_chord(1,J,JA,LK)
right_triangle_judgment_angle(1,LAJ)
line_addition(1,LA,AK)
right_triangle_property_pythagorean(1,LAJ)
Problem: As shown in the diagram, AB=AC, BC=4, ∠BCA=48°. Find the area of △ABC.
问题：如图所示，AB=AC，BC=4，∠BCA=48°。求三角形ABC的面积。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(BC),4)
Equal(MeasureOfAngle(BCA),48)
Goal: Value(AreaOfTriangle(ABC))
Answer: (-4*sqrt(3*sqrt(5)+15)-4*sqrt(10)+28*sqrt(2)+4*sqrt(15*sqrt(5)+75))/(sqrt(6)+2*sqrt(5-sqrt(5))+sqrt(30))
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
sine_theorem(1,BCA)
triangle_area_formula_sine(1,BCA)
Problem: As shown in the diagram, AC=13, AO=12, BA=5, OB and AC are opposite sides of the parallelogram BOAC, OA is perpendicular to BA. Find the area of quadrilateral BOAC.
问题：如图所示，AC=13，AO=12，BA=5，四边形BOAC是平行四边形，OA垂直于BA。求四边形BOAC的面积。
Construction (CDL):
Shape(BO,OA,AB)
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),13)
Equal(LengthOfLine(AO),12)
Equal(LengthOfLine(BA),5)
Parallelogram(BOAC)
PerpendicularBetweenLine(OA,BA)
Goal: Value(AreaOfQuadrilateral(BOAC))
Answer: 60
Theorem sequence:
altitude_of_quadrilateral_judgment_diagonal(1,BOAC)
parallelogram_area_formula_common(1,BOAC)
Problem: As shown in the diagram, AB=x, AC=14, BC=y, ∠CBA=30°, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=x，AC=14，BC=y，∠CBA=30°，BA⊥CA。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(CBA),30)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 14*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,BAC)
sine_theorem(1,BAC)
sine_theorem(1,CBA)
Problem: As shown in the diagram, VW=17, WY=10, XW=17, M is the center of ⊙M. Find the diameter of circle M.
问题：如图所示，VW=17，WY=10，XW=17，圆M的圆心为M。求圆M的直径。
Construction (CDL):
Shape(MYV,VW,WY)
Shape(MW,WV,VM)
Shape(MVZ,ZM,MV)
Shape(MZX,XW,WM,MZ)
Shape(MXY,YW,WX)
Collinear(VWX)
Collinear(YWMZ)
Cocircular(M,YVZX)
Conditions (CDL):
Equal(LengthOfLine(VW),17)
Equal(LengthOfLine(WY),10)
Equal(LengthOfLine(XW),17)
IsCentreOfCircle(M,M)
Goal: Value(DiameterOfCircle(M))
Answer: 389/10
Theorem sequence:
radius_of_circle_property_length_equal(1,MV,M)
radius_of_circle_property_length_equal(1,MY,M)
line_addition(1,YW,WM)
circle_property_chord_perpendicular_bisect_chord(2,M,MW,XV)
right_triangle_judgment_angle(1,MWV)
right_triangle_property_pythagorean(1,MWV)
circle_property_length_of_radius_and_diameter(1,M)
Problem: As shown in the diagram, ∠AFN=∠ECA, ∠ANE=30°, ∠CEK=110°, ∠ENJ=130°. Find the measure of ∠AEC.
问题：如图所示，∠AFN=∠ECA，∠ANE=30°，∠CEK=110°，∠ENJ=130°。求∠AEC的大小。
Construction (CDL):
Shape(DA,AG)
Shape(FA,AD)
Shape(GA,AC)
Shape(AF,FN,NA)
Shape(MN,NF)
Shape(JN,NM)
Shape(EN,NJ)
Shape(KE,EN)
Shape(CE,EK)
Shape(CA,AE,EC)
Shape(EA,AN,NE)
Collinear(FNJ)
Collinear(CEN)
Collinear(DAEK)
Collinear(GANM)
Conditions (CDL):
Equal(MeasureOfAngle(AFN),MeasureOfAngle(ECA))
Equal(MeasureOfAngle(ANE),30)
Equal(MeasureOfAngle(CEK),110)
Equal(MeasureOfAngle(ENJ),130)
Goal: Value(MeasureOfAngle(AEC))
Answer: 70
Theorem sequence:
adjacent_complementary_angle(1,AEC,CEK)
Problem: As shown in the diagram, ∠AEC=74°, ∠EAB=15°, EC is perpendicular to AC. Find the measure of ∠ABE.
问题：如图所示，∠AEC=74°，∠EAB=15°，EC垂直于AC。求∠ABE的大小。
Construction (CDL):
Shape(CA,AE,EC)
Shape(EA,AB,BE)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(AEC),74)
Equal(MeasureOfAngle(EAB),15)
PerpendicularBetweenLine(EC,AC)
Goal: Value(MeasureOfAngle(ABE))
Answer: 59
Theorem sequence:
adjacent_complementary_angle(1,BEA,AEC)
triangle_property_angle_sum(1,BEA)