SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=32, AC=x, ∠ABC=51°, ∠CAB=73°. Find the value of x.
问题：如图所示，AB=32，AC=x，∠ABC=51°，∠CAB=73°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),32)
Equal(LengthOfLine(AC),x)
Equal(MeasureOfAngle(ABC),51)
Equal(MeasureOfAngle(CAB),73)
Goal: Value(x)
Answer: (-2*sqrt(3*sqrt(5)+15)-2*sqrt(10)-2*sqrt(sqrt(5)+5)-2*sqrt(2)+2*sqrt(6)+2*sqrt(30)+2*sqrt(5*sqrt(5)+25)+2*sqrt(15*sqrt(5)+75))/sin(14*pi/45)
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,ABC)
Problem: As shown in the diagram, RS=x+9, RT=3*x-9, ST=2*x, △SRT is an equilateral △. Find the length of line RT.
问题：如图所示，RS=x+9，RT=3*x-9，ST=2*x，三角形SRT为等边三角形。求直线RT的长度。
Construction (CDL):
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(RS),x+9)
Equal(LengthOfLine(RT),3*x-9)
Equal(LengthOfLine(ST),2*x)
EquilateralTriangle(SRT)
Goal: Value(LengthOfLine(RT))
Answer: 18
Theorem sequence:
Problem: As shown in the diagram, ∠ABE=35°, ∠ABE=∠FBA, ∠CBF=140°, ∠EBC=x°. Find the value of x.
问题：如图所示，∠ABE=35°，∠ABE=∠FBA，∠CBF=140°，∠EBC=x°。求x的值。
Construction (CDL):
Shape(BCE,EB,BC)
Shape(BE,BEA,AB)
Shape(BA,BAF,FB)
Shape(BF,BFC,CB)
Cocircular(B,CEAF)
Conditions (CDL):
Equal(MeasureOfAngle(ABE),35)
Equal(MeasureOfAngle(ABE),MeasureOfAngle(FBA))
Equal(MeasureOfAngle(CBF),140)
Equal(MeasureOfAngle(EBC),x)
Goal: Value(x)
Answer: 150
Theorem sequence:
angle_addition(1,CBF,FBA)
angle_addition(1,CBA,ABE)
round_angle(1,EBC,CBE)
Problem: As shown in the diagram, AB=12, AC=12, ∠CAB=60°. Find the perimeter of △ABC.
问题：如图所示，AB=12，AC=12，∠CAB=60°。求△ABC的周长。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),12)
Equal(MeasureOfAngle(CAB),60)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 36
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
isosceles_triangle_judgment_angle_equal(1,BCA)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, CH=7, EF=11, FH=10, HE=6, quadrilateral ADEG is a parallelogram, △CHB is similar to △FHE.. Find the perimeter of △CHB.
问题：如图所示，CH=7，EF=11，FH=10，HE=6，四边形ADEG是平行四边形，三角形CHB相似于三角形FHE。求三角形CHB的周长。
Construction (CDL):
Shape(CD,DE,EH,HC)
Shape(HE,EF,FH)
Shape(CH,HB,BC)
Shape(HF,FB,BH)
Shape(FG,GA,AB,BF)
Collinear(CHF)
Collinear(BHE)
Collinear(ABCD)
Collinear(EFG)
Conditions (CDL):
Equal(LengthOfLine(CH),7)
Equal(LengthOfLine(EF),11)
Equal(LengthOfLine(FH),10)
Equal(LengthOfLine(HE),6)
Parallelogram(ADEG)
SimilarBetweenTriangle(CHB,FHE)
Goal: Value(PerimeterOfTriangle(CHB))
Answer: 189/10
Theorem sequence:
similar_triangle_property_line_ratio(1,CHB,FHE)
similar_triangle_property_line_ratio(1,HBC,HEF)
similar_triangle_property_line_ratio(1,BCH,EFH)
triangle_perimeter_formula(1,CHB)
Problem: As shown in the diagram, DF=6, EF=4, FB=8, FC=x. Find the value of x.
问题：如图所示，DF=6，EF=4，FB=8，FC=x。求x的值。
Construction (CDL):
Shape(ACE,EF,FC)
Shape(FE,AEB,BF)
Shape(ABD,DF,FB)
Shape(FD,ADC,CF)
Collinear(EFD)
Collinear(CFB)
Cocircular(A,CEBD)
Conditions (CDL):
Equal(LengthOfLine(DF),6)
Equal(LengthOfLine(EF),4)
Equal(LengthOfLine(FB),8)
Equal(LengthOfLine(FC),x)
Goal: Value(x)
Answer: 3
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,CFB,EFD,A)
Problem: As shown in the diagram, BC=6, ∠BAC=4*x°, ∠CBA=2*x°, △ACB is a right △. Find the length of line AB.
问题：如图所示，BC=6，∠BAC=4*x°，∠CBA=2*x°，AC和CB是直角三角形ACB的两个直角边。求直线AB的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
Equal(MeasureOfAngle(BAC),4*x)
Equal(MeasureOfAngle(CBA),2*x)
RightTriangle(ACB)
Goal: Value(LengthOfLine(AB))
Answer: 4*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,BAC)
Problem: As shown in the diagram, AB=x, AC=5, CD=y, ∠CBD=45°, AC⊥BC, CD⊥AD. Find the value of x.
问题：如图所示，AB=x，AC=5，CD=y，∠CBD=45°，AC垂直于BC，CD垂直于AD。求x的值。
Construction (CDL):
Shape(CB,BD,DC)
Shape(CD,DA,AC)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(CD),y)
Equal(MeasureOfAngle(CBD),45)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(CD,AD)
Goal: Value(x)
Answer: 5*sqrt(2)
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, AE=8, BE=6, ED=4, CD is parallel to BE. Find the length of line CD.
问题：如图所示，AE=8，BE=6，ED=4，CD∥BE。求直线CD的长度。
Construction (CDL):
Shape(BA,AE,EB)
Shape(CB,BE,ED,DC)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AE),8)
Equal(LengthOfLine(BE),6)
Equal(LengthOfLine(ED),4)
ParallelBetweenLine(CD,BE)
Goal: Value(LengthOfLine(CD))
Answer: 9
Theorem sequence:
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
similar_triangle_property_line_ratio(1,AEB,ADC)
similar_triangle_property_line_ratio(1,BAE,CAD)
line_addition(1,AE,ED)
Problem: As shown in the diagram, AB=18, AD=9, DC=6, AD is perpendicular to FD, DC⊥BC. Find the length of line AF.
问题：如图所示，AB=18，AD=9，DC=6，AD垂直于FD，DC⊥BC。求直线AF的长度。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CB,BF)
Collinear(ADC)
Collinear(AFB)
Conditions (CDL):
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(AD),9)
Equal(LengthOfLine(DC),6)
PerpendicularBetweenLine(AD,FD)
PerpendicularBetweenLine(DC,BC)
Goal: Value(LengthOfLine(AF))
Answer: 54/5
Theorem sequence:
similar_triangle_judgment_aa(1,FAD,BAC)
line_addition(1,AD,DC)
similar_triangle_property_line_ratio(1,FAD,BAC)
similar_triangle_property_line_ratio(1,DFA,CBA)
Problem: As shown in the diagram, ∠GFB=x°, the measure of ⌒DBH is 128, the measure of ⌒DHG is 148. Find the value of x.
问题：如图所示，∠GFB=x°，⌒DBH的角度为128，弧DHG的角度为148。求x的值。
Construction (CDL):
Shape(DGB,GF,FB)
Shape(DHG,GH)
Shape(DGB,DBH,HG)
Collinear(FGHC)
Collinear(FBE)
Cocircular(D,GBH)
Conditions (CDL):
Equal(MeasureOfAngle(GFB),x)
Equal(MeasureOfArc(DBH),128)
Equal(MeasureOfArc(DHG),148)
Goal: Value(x)
Answer: 22
Theorem sequence:
arc_addition_measure(1,DBH,DHG)
round_arc(1,DGB,DBG)
circle_property_circular_power_tangent_and_segment_angle(2,FB,FGH,D)
Problem: As shown in the diagram, ∠ADC=x°, ∠BAD=3*x°, CB is perpendicular to AB, DC is perpendicular to BC. Find the measure of ∠CBA.
问题：如图所示，∠ADC=x°，∠BAD=3*x°，CB垂直于AB，DC垂直于BC。求∠CBA的大小。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),x)
Equal(MeasureOfAngle(BAD),3*x)
PerpendicularBetweenLine(CB,AB)
PerpendicularBetweenLine(DC,BC)
Goal: Value(MeasureOfAngle(CBA))
Answer: 90
Theorem sequence:
Problem: As shown in the diagram, PY=2*c-1, YR=4*c-11, ZP=a+5, ZQ=3*a-11, ∠PRZ=4*b-17°, ∠QYR=7*b+6°, ∠RXP=2*a+10°, ∠ZRQ=3*b-4°, PX is the altitude of △ PRQ. Find the value of a.
问题：如图所示，PY=2*c-1，YR=4*c-11，ZP=a+5，ZQ=3*a-11，∠PRZ=4*b-17°，∠QYR=7*b+6°，∠RXP=2*a+10°，∠ZRQ=3*b-4°，PX是△PRQ的高。求a的值。
Construction (CDL):
Shape(QZ,ZM,MN,NQ)
Shape(ZP,PM,MZ)
Shape(MP,PY,YA,AM)
Shape(NM,MA,AN)
Shape(AY,YR,RA)
Shape(XN,NA,AR,RX)
Shape(QN,NX,XQ)
Collinear(PZQ)
Collinear(PMNX)
Collinear(PYR)
Collinear(ZMAR)
Collinear(QNAY)
Collinear(QXR)
Conditions (CDL):
Equal(LengthOfLine(PY),2*c-1)
Equal(LengthOfLine(YR),4*c-11)
Equal(LengthOfLine(ZP),a+5)
Equal(LengthOfLine(ZQ),3*a-11)
Equal(MeasureOfAngle(PRZ),4*b-17)
Equal(MeasureOfAngle(QYR),7*b+6)
Equal(MeasureOfAngle(RXP),2*a+10)
Equal(MeasureOfAngle(ZRQ),3*b-4)
IsAltitudeOfTriangle(PX,PRQ)
Goal: Value(a)
Answer: 40
Theorem sequence:
Problem: As shown in the diagram, ∠EDF=x°, ∠FDA=140°. Find the value of x.
问题：如图所示，∠EDF=x°，∠FDA=140°。求x的值。
Construction (CDL):
Shape(BFD,DF)
Shape(BDF,FD)
Shape(ED,DC)
Shape(CD,DA)
Collinear(EDA)
Collinear(DFC)
Cocircular(B,FD)
Conditions (CDL):
Equal(MeasureOfAngle(EDF),x)
Equal(MeasureOfAngle(FDA),140)
Goal: Value(x)
Answer: 40
Theorem sequence:
adjacent_complementary_angle(1,EDC,CDA)
Problem: As shown in the diagram, BC=10, ∠BCA=30°, ∠CAB=30°. Find the length of line AB.
问题：如图所示，BC=10，∠BCA=30°，∠CAB=30°。求直线AB的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(BC),10)
Equal(MeasureOfAngle(BCA),30)
Equal(MeasureOfAngle(CAB),30)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
isosceles_triangle_judgment_angle_equal(1,BCA)
Problem: As shown in the diagram, CK=8, KJ=10, JMLK is a rhombus. Find the length of line JC.
问题：如图所示，CK=8，KJ=10，四边形JMLK是菱形。求直线JC的长度。
Construction (CDL):
Shape(JM,MC,CJ)
Shape(CM,ML,LC)
Shape(CL,LK,KC)
Shape(JC,CK,KJ)
Collinear(JCL)
Collinear(MCK)
Conditions (CDL):
Equal(LengthOfLine(CK),8)
Equal(LengthOfLine(KJ),10)
Rhombus(JMLK)
Goal: Value(LengthOfLine(JC))
Answer: 6
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,JMLK,C)
right_triangle_judgment_angle(1,MCJ)
right_triangle_property_pythagorean(1,MCJ)
Problem: As shown in the diagram, JK=3*x+6, JK=LK, JL=5*x-6, LK=4*x, LK=JL. Find the length of line JK.
问题：如图所示，JK=3*x+6，JK=LK，JL=5*x-6，LK=4*x，LK=JL。求直线JK的长度。
Construction (CDL):
Shape(KJ,JL,LK)
Conditions (CDL):
Equal(LengthOfLine(JK),3*x+6)
Equal(LengthOfLine(JK),LengthOfLine(LK))
Equal(LengthOfLine(JL),5*x-6)
Equal(LengthOfLine(LK),4*x)
Equal(LengthOfLine(LK),LengthOfLine(JL))
Goal: Value(LengthOfLine(JK))
Answer: 24
Theorem sequence:
Problem: As shown in the diagram, ∠FOE=118°, ∠LDA=104°, OD is parallel to FI. Find the measure of ∠FID.
问题：如图所示，∠FOE=118°，∠LDA=104°，OD平行于FI。求∠FID的大小。
Construction (CDL):
Shape(DO,OF,FI,ID)
Shape(LD,DA)
Shape(BL,LC)
Shape(EO,OH)
Shape(KF,FJ)
Collinear(LDIC)
Collinear(HOFJ)
Collinear(ADOE)
Collinear(BIFK)
Conditions (CDL):
Equal(MeasureOfAngle(FOE),118)
Equal(MeasureOfAngle(LDA),104)
ParallelBetweenLine(OD,FI)
Goal: Value(MeasureOfAngle(FID))
Answer: 76
Theorem sequence:
vertical_angle(1,LDA,IDO)
parallel_property_ipsilateral_internal_angle(1,IF,DO)
Problem: As shown in the diagram, DE=9/2, FB=9, the center of ⊙A is O, O is the center of circle C, DE is the diameter of ⊙A, the diameter of ⊙C is FB. Find the area of the circle C minus the area of the ⊙A.
问题：如图所示，DE=9/2，FB=9，圆A的圆心为O，⊙C的圆心为O，圆A的直径为DE，⊙C的直径为FB。求圆C的面积减去⊙A的面积。
Construction (CDL):
Shape(A)
Shape(C)
Collinear(FDOEB)
Cocircular(C,FB)
Cocircular(A,DE)
Conditions (CDL):
Equal(LengthOfLine(DE),9/2)
Equal(LengthOfLine(FB),9)
IsCentreOfCircle(O,A)
IsCentreOfCircle(O,C)
IsDiameterOfCircle(DE,A)
IsDiameterOfCircle(FB,C)
Goal: Value(Sub(AreaOfCircle(C),AreaOfCircle(A)))
Answer: 243*pi/16
Theorem sequence:
diameter_of_circle_property_length_equal(1,FB,C)
diameter_of_circle_property_length_equal(1,DE,A)
circle_property_length_of_radius_and_diameter(1,C)
circle_property_length_of_radius_and_diameter(1,A)
circle_area_formula(1,C)
circle_area_formula(1,A)
Problem: As shown in the diagram, AB=11*sqrt(2), AC=11*sqrt(2), BC=x, ∠ACB=45°, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=11*sqrt(2)，AC=11*sqrt(2)，BC=x，∠ACB=45°，BA垂直于CA。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),11*sqrt(2))
Equal(LengthOfLine(AC),11*sqrt(2))
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ACB),45)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 22
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, AB=y, AC=9, AD=x, BC=z, BD=8, CA⊥BA, DB is perpendicular to CB. Find the value of z.
问题：如图所示，AB=y，AC=9，AD=x，BC=z，BD=8，CA⊥BA，DB⊥CB。求z的值。
Construction (CDL):
Shape(DB,BA,AD)
Shape(AB,BC,CA)
Collinear(DAC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),9)
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BD),8)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(DB,CB)
Goal: Value(z)
Answer: 3*sqrt(9/2+sqrt(337)/2)
Theorem sequence:
line_addition(1,DA,AC)
adjacent_complementary_angle(1,CAB,BAD)
right_triangle_judgment_angle(1,CAB)
right_triangle_judgment_angle(1,BAD)
right_triangle_judgment_angle(1,DBC)
right_triangle_property_pythagorean(1,CAB)
right_triangle_property_pythagorean(1,BAD)
right_triangle_property_pythagorean(1,DBC)
Problem: As shown in the diagram, QR=x, SP=2, SR=4, PQ is perpendicular to RQ, RS is perpendicular to QS. Find the value of x.
问题：如图所示，QR=x，SP=2，SR=4，PQ⊥RQ，RS垂直于QS。求x的值。
Construction (CDL):
Shape(RS,SQ,QR)
Shape(PQ,QS,SP)
Collinear(PSR)
Conditions (CDL):
Equal(LengthOfLine(QR),x)
Equal(LengthOfLine(SP),2)
Equal(LengthOfLine(SR),4)
PerpendicularBetweenLine(PQ,RQ)
PerpendicularBetweenLine(RS,QS)
Goal: Value(x)
Answer: 2*sqrt(6)
Theorem sequence:
line_addition(1,RS,SP)
adjacent_complementary_angle(1,RSQ,QSP)
right_triangle_judgment_angle(1,RSQ)
right_triangle_judgment_angle(1,QSP)
right_triangle_judgment_angle(1,PQR)
right_triangle_property_pythagorean(1,RSQ)
right_triangle_property_pythagorean(1,QSP)
right_triangle_property_pythagorean(1,PQR)
Problem: As shown in the diagram, ∠ZYW=53°, XWZY is a rhombus. Find the measure of ∠WYX.
问题：如图所示，∠ZYW=53°，四边形XWZY是菱形。求∠WYX的大小。
Construction (CDL):
Shape(XW,WV,VX)
Shape(VW,WZ,ZV)
Shape(VZ,ZY,YV)
Shape(XV,VY,YX)
Collinear(XVZ)
Collinear(WVY)
Conditions (CDL):
Equal(MeasureOfAngle(ZYW),53)
Rhombus(XWZY)
Goal: Value(MeasureOfAngle(WYX))
Answer: 53
Theorem sequence:
mirror_congruent_triangle_judgment_sss(1,XWY,ZYW)
mirror_congruent_triangle_property_angle_equal(1,YXW,YWZ)
Problem: As shown in the diagram, QS=3*a-14, SN=2*a+1, SQ=SN, ∠MSQ=7*a+1°, MS is the median of △ MQN. Find the value of a.
问题：如图所示，QS=3*a-14，SN=2*a+1，SQ=SN，∠MSQ=7*a+1°，MS是△MQN的中线。求a的值。
Construction (CDL):
Shape(QS,SR,RQ)
Shape(RS,SM,MR)
Shape(MS,SN,NM)
Collinear(QRM)
Collinear(QSN)
Conditions (CDL):
Equal(LengthOfLine(QS),3*a-14)
Equal(LengthOfLine(SN),2*a+1)
Equal(LengthOfLine(SQ),LengthOfLine(SN))
Equal(MeasureOfAngle(MSQ),7*a+1)
IsMedianOfTriangle(MS,MQN)
Goal: Value(a)
Answer: 15
Theorem sequence:
Problem: As shown in the diagram, the measure of arc RQP is 115, R is the center of circle R, NR is perpendicular to PR. Find the measure of arc RQM.
问题：如图所示，弧RQP的角度为115，R是圆R的圆心，NR垂直于PR。求弧RQM的角度。
Construction (CDL):
Shape(RPN,NR,RP)
Shape(RNM,MR,RN)
Shape(RMQ,QR,RM)
Shape(RQP,PR,RQ)
Collinear(MRP)
Cocircular(R,NMQP)
Conditions (CDL):
Equal(MeasureOfArc(RQP),115)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(NR,PR)
Goal: Value(MeasureOfArc(RQM))
Answer: 295
Theorem sequence:
flat_angle(1,MRP)
angle_addition(1,MRP,PRQ)
arc_property_center_angle(1,RQM,R)
arc_property_center_angle(1,RQP,R)
Problem: As shown in the diagram, AC=6*y-8, BD=4*y+6, CB=2*x+6, DA=3*x-2, ACBD is a parallelogram. Find the value of y.
问题：如图所示，AC=6*y-8，BD=4*y+6，CB=2*x+6，DA=3*x-2，AD和CB是平行四边形ACBD的一组对边。求y的值。
Construction (CDL):
Shape(AC,CB,BD,DA)
Conditions (CDL):
Equal(LengthOfLine(AC),6*y-8)
Equal(LengthOfLine(BD),4*y+6)
Equal(LengthOfLine(CB),2*x+6)
Equal(LengthOfLine(DA),3*x-2)
Parallelogram(ACBD)
Goal: Value(y)
Answer: 7
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ACBD)
Problem: As shown in the diagram, AC=z, AD=4, BC=x, BD=12, CD=y, AC⊥BC, CD⊥AD. Find the value of y.
问题：如图所示，AC=z，AD=4，BC=x，BD=12，CD=y，AC⊥BC，CD垂直于AD。求y的值。
Construction (CDL):
Shape(BD,DC,CB)
Shape(CD,DA,AC)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(AC),z)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(BD),12)
Equal(LengthOfLine(CD),y)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(CD,AD)
Goal: Value(y)
Answer: 4*sqrt(3)
Theorem sequence:
line_addition(1,BD,DA)
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, ∠ACB=25*x°, ∠BCD=40°, ∠CBA=10*y°, ∠DBC=100°, CA and DB are opposite sides of the parallelogram CDBA. Find the value of y.
问题：如图所示，∠ACB=25*x°，∠BCD=40°，∠CBA=10*y°，∠DBC=100°，四边形CDBA是▱。求y的值。
Construction (CDL):
Shape(CD,DB,BC)
Shape(CB,BA,AC)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),25*x)
Equal(MeasureOfAngle(BCD),40)
Equal(MeasureOfAngle(CBA),10*y)
Equal(MeasureOfAngle(DBC),100)
Parallelogram(CDBA)
Goal: Value(y)
Answer: 4
Theorem sequence:
parallel_property_alternate_interior_angle(2,DC,BA)
Problem: As shown in the diagram, ∠CBD=55°, ∠GBF=35°, the center of circle B is B. Find the measure of arc BDC.
问题：如图所示，∠CBD=55°，∠GBF=35°，B是圆B的圆心。求⌒BDC的角度。
Construction (CDL):
Shape(AB,BC,BCA)
Shape(CB,BD,BDC)
Shape(DB,BF,BFD)
Shape(FB,BG,BGF)
Shape(BA,BGA,GB)
Collinear(CBG)
Collinear(ABD)
Cocircular(B,AGFDC)
Conditions (CDL):
Equal(MeasureOfAngle(CBD),55)
Equal(MeasureOfAngle(GBF),35)
IsCentreOfCircle(B,B)
Goal: Value(MeasureOfArc(BDC))
Answer: 55
Theorem sequence:
arc_property_center_angle(1,BDC,B)
Problem: As shown in the diagram, BD=15, CA=b, CB=28, CD=27, ∠ADB=∠CDA. Find the value of b.
问题：如图所示，BD=15，CA=b，CB=28，CD=27，∠ADB=∠CDA。求b的值。
Construction (CDL):
Shape(DB,BA,AD)
Shape(DA,AC,CD)
Collinear(BAC)
Conditions (CDL):
Equal(LengthOfLine(BD),15)
Equal(LengthOfLine(CA),b)
Equal(LengthOfLine(CB),28)
Equal(LengthOfLine(CD),27)
Equal(MeasureOfAngle(ADB),MeasureOfAngle(CDA))
Goal: Value(b)
Answer: 18
Theorem sequence:
line_addition(1,BA,AC)
bisector_of_angle_judgment_angle_equal(1,DA,CDB)
bisector_of_angle_property_line_ratio(1,DA,CDB)
Problem: As shown in the diagram, JK=3*y-2, LK=5*y-8, ∠PMJ=3*x-6°, PK is the median of △ PLJ. Find the length of line LK.
问题：如图所示，JK=3*y-2，LK=5*y-8，∠PMJ=3*x-6°，PK是三角形PLJ的中线。求直线LK的长度。
Construction (CDL):
Shape(LK,KB,BM,ML)
Shape(KJ,JB,BK)
Shape(BJ,JP,PB)
Shape(MB,BP,PM)
Collinear(JKL)
Collinear(JBM)
Collinear(PBK)
Collinear(PML)
Conditions (CDL):
Equal(LengthOfLine(JK),3*y-2)
Equal(LengthOfLine(LK),5*y-8)
Equal(MeasureOfAngle(PMJ),3*x-6)
IsMedianOfTriangle(PK,PLJ)
Goal: Value(LengthOfLine(LK))
Answer: 7
Theorem sequence:
Problem: As shown in the diagram, CD=8, DB=15, the center of circle A is A, the diameter of ⊙A is BC, CD is perpendicular to BD. Find the circumference of the circle A.
问题：如图所示，CD=8，DB=15，⊙A的圆心为A，⊙A的直径为BC，CD⊥BD。求⊙A的周长。
Construction (CDL):
Shape(CA,AB,ABC)
Shape(BA,AC,CD,DB)
Shape(BD,ADB)
Shape(ACD,DC)
Collinear(CAB)
Cocircular(A,BCD)
Conditions (CDL):
Equal(LengthOfLine(CD),8)
Equal(LengthOfLine(DB),15)
IsCentreOfCircle(A,A) 
IsDiameterOfCircle(BC,A)
PerpendicularBetweenLine(CD,BD)
Goal: Value(PerimeterOfCircle(A))
Answer: 17*pi
Theorem sequence:
right_triangle_judgment_angle(1,CDB)
right_triangle_property_pythagorean(1,CDB)
diameter_of_circle_property_length_equal(1,BC,A)
circle_property_length_of_radius_and_diameter(1,A)
circle_perimeter_formula(1,A)
Problem: As shown in the diagram, JM=MN, MJ=2*x-5, MN=3*x-9, NJ=x-2, MJ and MN are the legs of the isosceles △ MJN. Find the length of line MN.
问题：如图所示，JM=MN，MJ=2*x-5，MN=3*x-9，NJ=x-2，∠MJN和∠JNM是等腰三角形MJN的底角。求直线MN的长度。
Construction (CDL):
Shape(JN,NM,MJ)
Conditions (CDL):
Equal(LengthOfLine(JM),LengthOfLine(MN))
Equal(LengthOfLine(MJ),2*x-5)
Equal(LengthOfLine(MN),3*x-9)
Equal(LengthOfLine(NJ),x-2)
IsoscelesTriangle(MJN)
Goal: Value(LengthOfLine(MN))
Answer: 3
Theorem sequence:
Problem: As shown in the diagram, AB=y, AD=9, BC=z, BD=x, CD=4, BD is perpendicular to CD, CB is perpendicular to AB. Find the value of x.
问题：如图所示，AB=y，AD=9，BC=z，BD=x，CD=4，BD垂直于CD，CB⊥AB。求x的值。
Construction (CDL):
Shape(AD,DB,BA)
Shape(BD,DC,CB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AD),9)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(BD,CD)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 6
Theorem sequence:
line_addition(1,AD,DC)
adjacent_complementary_angle(1,ADB,BDC)
right_triangle_judgment_angle(1,ADB)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,ADB)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, the area of △DEF is 40, BM=7, EF=5, △DEF is similar to △AMB.. Find the area of △AMB.
问题：如图所示，△DEF的面积为40，BM=7，EF=5，三角形DEF相似于三角形AMB。求△AMB的面积。
Construction (CDL):
Shape(DE,EF,FD)
Shape(MB,BA,AM)
Conditions (CDL):
Equal(AreaOfTriangle(DEF),40)
Equal(LengthOfLine(BM),7)
Equal(LengthOfLine(EF),5)
SimilarBetweenTriangle(DEF,AMB)
Goal: Value(AreaOfTriangle(AMB))
Answer: 392/5
Theorem sequence:
similar_triangle_property_line_ratio(1,DEF,AMB)
similar_triangle_property_area_square_ratio(1,DEF,AMB)
Problem: As shown in the diagram, AB=z, AC=y, AD=20, BD=x, CD=10, BD is perpendicular to AD, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AB=z，AC=y，AD=20，BD=x，CD=10，BD垂直于AD，CA垂直于BA。求x的值。
Construction (CDL):
Shape(BD,DA,AB)
Shape(DC,CA,AD)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AD),20)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CD),10)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 40
Theorem sequence:
line_addition(1,CD,DB)
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,ADC)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,ADC)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, WX=9*x, WY=6*x+3, WY=YX, XW=WY, YX=4*x+5. Find the length of line WY.
问题：如图所示，WX=9*x，WY=6*x+3，WY=YX，XW=WY，YX=4*x+5。求直线WY的长度。
Construction (CDL):
Shape(XW,WY,YX)
Conditions (CDL):
Equal(LengthOfLine(WX),9*x)
Equal(LengthOfLine(WY),6*x+3)
Equal(LengthOfLine(WY),LengthOfLine(YX))
Equal(LengthOfLine(XW),LengthOfLine(WY))
Equal(LengthOfLine(YX),4*x+5)
Goal: Value(LengthOfLine(WY))
Answer: 9
Theorem sequence:
Problem: As shown in the diagram, ∠CAE=42°, ∠CFE=77°, △CDE is an equilateral △, AB and AC are the legs of the isosceles triangle ABC, △CFG is an isosceles △. Find the measure of ∠ECG.
问题：如图所示，∠CAE=42°，∠CFE=77°，三角形CDE为等边三角形，三角形ABC为等腰三角形，∠CFG和∠FGC是等腰△CFG的底角。求∠ECG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(CAE),42)
Equal(MeasureOfAngle(CFE),77)
EquilateralTriangle(CDE)
IsoscelesTriangle(ABC)
IsoscelesTriangle(CFG)
Goal: Value(MeasureOfAngle(ECG))
Answer: 17
Theorem sequence:
isosceles_triangle_property_angle_equal(1,CFG)
adjacent_complementary_angle(1,FGC,CGE)
equilateral_triangle_property_angle(1,ECD)
triangle_property_angle_sum(1,CGE)
Problem: As shown in the diagram, AP=6*r+4, PH=22+3*r, ∠AHW=8*q+17°, ∠HWP=10+q°, WP perpendicular bisects AH. Find the value of r.
问题：如图所示，AP=6*r+4，PH=22+3*r，∠AHW=8*q+17°，∠HWP=10+q°，WP是AH的垂直平分线。求r的值。
Construction (CDL):
Shape(HW,WX,XH)
Shape(XW,WQ,QX)
Shape(XQ,QA,AP,PX)
Shape(HX,XP,PH)
Collinear(HXQ)
Collinear(HPA)
Collinear(WXP)
Collinear(WQA)
Conditions (CDL):
Equal(LengthOfLine(AP),6*r+4)
Equal(LengthOfLine(PH),22+3*r)
Equal(MeasureOfAngle(AHW),8*q+17)
Equal(MeasureOfAngle(HWP),10+q)
IsPerpendicularBisectorOfLine(WP,AH)
Goal: Value(r)
Answer: 6
Theorem sequence:
Problem: As shown in the diagram, CA=32, ES=12, ∠SCE=30°, AC and BS are opposite sides of the ▱ ABSC, BE is perpendicular to CE. Find the area of ABSC.
问题：如图所示，CA=32，ES=12，∠SCE=30°，ABSC是▱，BE垂直于CE。求ABSC的面积。
Construction (CDL):
Shape(AB,BE,EC,CA)
Shape(CE,ES,SC)
Collinear(BES)
Conditions (CDL):
Equal(LengthOfLine(CA),32)
Equal(LengthOfLine(ES),12)
Equal(MeasureOfAngle(SCE),30)
Parallelogram(ABSC)
PerpendicularBetweenLine(BE,CE)
Goal: Value(AreaOfQuadrilateral(ABSC))
Answer: 384*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,BEC,CES)
parallelogram_property_opposite_angle_equal(1,ABSC)
parallelogram_property_opposite_angle_equal(1,BSCA)
cosine_theorem(1,CES)
right_triangle_judgment_angle(1,CES)
right_triangle_property_pythagorean(1,CES)
altitude_of_quadrilateral_judgment_right_vertex(1,CE,ABSC)
parallelogram_area_formula_sine(1,ABSC)
parallelogram_area_formula_sine(1,BSCA)
parallelogram_area_formula_sine(1,SCAB)
parallelogram_area_formula_sine(1,CABS)
parallelogram_area_formula_common(1,ABSC)
Problem: As shown in the diagram, AC=27, BA=2*x+5, BA=BC, BC=3*x-4. Find the value of x.
问题：如图所示，AC=27，BA=2*x+5，BA=BC，BC=3*x-4。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),27)
Equal(LengthOfLine(BA),2*x+5)
Equal(LengthOfLine(BA),LengthOfLine(BC))
Equal(LengthOfLine(BC),3*x-4)
Goal: Value(x)
Answer: 9
Theorem sequence: