SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠BCE=5*x+62°, ∠BDC=3*x+47°, ∠CBD=37°. Find the value of x.
问题：如图所示，∠BCE=5*x+62°，∠BDC=3*x+47°，∠CBD=37°。求x的值。
Construction (CDL):
Shape(BD,DC,CB)
Shape(BC,CE)
Collinear(DCE)
Conditions (CDL):
Equal(MeasureOfAngle(BCE),5*x+62)
Equal(MeasureOfAngle(BDC),3*x+47)
Equal(MeasureOfAngle(CBD),37)
Goal: Value(x)
Answer: 11
Theorem sequence:
adjacent_complementary_angle(1,DCB,BCE)
triangle_property_angle_sum(1,BDC)
Problem: As shown in the diagram, ∠DOB=∠EOD, ∠EOD=∠FOE, ∠FOE=∠AOF, O is the center of circle O, CO is perpendicular to AO. Find the measure of arc OCA.
问题：如图所示，∠DOB=∠EOD，∠EOD=∠FOE，∠FOE=∠AOF，⊙O的圆心为O，CO垂直于AO。求弧OCA的角度。
Construction (CDL):
Shape(FO,OE,OEF)
Shape(AO,OF,OFA)
Shape(CO,OA,OAC)
Shape(BO,OC,OCB)
Shape(DO,OB,OBD)
Shape(EO,OD,ODE)
Collinear(AOB)
Collinear(EOC)
Cocircular(O,ACBDEF)
Conditions (CDL):
Equal(MeasureOfAngle(DOB),MeasureOfAngle(EOD))
Equal(MeasureOfAngle(EOD),MeasureOfAngle(FOE))
Equal(MeasureOfAngle(FOE),MeasureOfAngle(AOF))
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CO,AO)
Goal: Value(MeasureOfArc(OCA))
Answer: 270
Theorem sequence:
arc_property_center_angle(1,OCA,O)
round_angle(1,AOC,COA)
Problem: As shown in the diagram, ∠NMQ=10*x°, ∠PNM=20*x°, ∠PNM=∠MQP, ∠QPN=∠NMQ, QM and PN are opposite sides of the parallelogram MQPN. Find the measure of ∠PNM.
问题：如图所示，∠NMQ=10*x°，∠PNM=20*x°，∠PNM=∠MQP，∠QPN=∠NMQ，MN和QP是平行四边形MQPN的一组对边。求∠PNM的大小。
Construction (CDL):
Shape(MQ,QP,PN,NM)
Conditions (CDL):
Equal(MeasureOfAngle(NMQ),10*x)
Equal(MeasureOfAngle(PNM),20*x)
Equal(MeasureOfAngle(PNM),MeasureOfAngle(MQP))
Equal(MeasureOfAngle(QPN),MeasureOfAngle(NMQ))
Parallelogram(MQPN)
Goal: Value(MeasureOfAngle(PNM))
Answer: 120
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,NP,MQ)
Problem: As shown in the diagram, DA=6, ∠BDC=x°, ∠EDA=x°, AB is parallel to EC. Find the length of line DB.
问题：如图所示，DA=6，∠BDC=x°，∠EDA=x°，AB平行于EC。求直线DB的长度。
Construction (CDL):
Shape(ED,DA,AE)
Shape(AD,DB,BA)
Shape(DC,CB,BD)
Collinear(EDC)
Conditions (CDL):
Equal(LengthOfLine(DA),6)
Equal(MeasureOfAngle(BDC),x)
Equal(MeasureOfAngle(EDA),x)
ParallelBetweenLine(AB,EC)
Goal: Value(LengthOfLine(DB))
Answer: 6
Theorem sequence:
parallel_property_collinear_extend(3,CE,BA,D)
parallel_property_alternate_interior_angle(1,AB,ED)
parallel_property_alternate_interior_angle(2,AB,DC)
isosceles_triangle_judgment_angle_equal(1,DBA)
Problem: As shown in the diagram, AD=13, BA=3*y-8, BC=2*x+7, CD=10, ∠ADF=59°, ∠BFC=49°, ∠FBA=20°, quadrilateral ADCB is a parallelogram. Find the value of x.
问题：如图所示，AD=13，BA=3*y-8，BC=2*x+7，CD=10，∠ADF=59°，∠BFC=49°，∠FBA=20°，ADCB是平行四边形。求x的值。
Construction (CDL):
Shape(AD,DF,FA)
Shape(FD,DC,CF)
Shape(FC,CB,BF)
Shape(FB,BA,AF)
Collinear(AFC)
Collinear(DFB)
Conditions (CDL):
Equal(LengthOfLine(AD),13)
Equal(LengthOfLine(BA),3*y-8)
Equal(LengthOfLine(BC),2*x+7)
Equal(LengthOfLine(CD),10)
Equal(MeasureOfAngle(ADF),59)
Equal(MeasureOfAngle(BFC),49)
Equal(MeasureOfAngle(FBA),20)
Parallelogram(ADCB)
Goal: Value(x)
Answer: 3
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ADCB)
Problem: As shown in the diagram, AB=9.6, AC=CD, BC=y, CD=BA, CD=BD, ∠BCD=x°, CD is perpendicular to BD. Find the value of x.
问题：如图所示，AB=9.6，AC=CD，BC=y，CD=BA，CD=BD，∠BCD=x°，CD⊥BD。求x的值。
Construction (CDL):
Shape(CD,DB,BC)
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),9.6)
Equal(LengthOfLine(AC),LengthOfLine(CD))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CD),LengthOfLine(BA))
Equal(LengthOfLine(CD),LengthOfLine(BD))
Equal(MeasureOfAngle(BCD),x)
PerpendicularBetweenLine(CD,BD)
Goal: Value(x)
Answer: 45
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DBC)
isosceles_triangle_property_angle_equal(1,DBC)
triangle_property_angle_sum(1,DBC)
Problem: As shown in the diagram, AB=x, AC=7, BC=15, ∠BCA=18°. Find the value of x.
问题：如图所示，AB=x，AC=7，BC=15，∠BCA=18°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),7)
Equal(LengthOfLine(BC),15)
Equal(MeasureOfAngle(BCA),18)
Goal: Value(x)
Answer: 2**(3/4)*sqrt(-105*sqrt(sqrt(5)+5)+274*sqrt(2))/2
Theorem sequence:
cosine_theorem(1,CAB)
Problem: As shown in the diagram, ∠NAP=51°, the measure of arc OPN is 74. Find the measure of ⌒ORQ.
问题：如图所示，∠NAP=51°，⌒OPN的角度为74。求弧ORQ的角度。
Construction (CDL):
Shape(ONR,RA,AN)
Shape(OPN,NA,AP)
Shape(ORQ,QA,AR)
Shape(AQ,QP,PA)
Shape(PQ,OQP)
Collinear(NAQ)
Collinear(RAP)
Cocircular(O,RQPN)
Conditions (CDL):
Equal(MeasureOfAngle(NAP),51)
Equal(MeasureOfArc(OPN),74)
Goal: Value(MeasureOfArc(ORQ))
Answer: 28
Theorem sequence:
arc_property_circumference_angle_external(1,OPN,Q)
adjacent_complementary_angle(1,NAP,PAQ)
triangle_property_angle_sum(1,PAQ)
arc_property_circumference_angle_external(1,ORQ,P)
Problem: As shown in the diagram, BD=4*x+9, CD=7*x-6, ∠CDA=2*y-6°, AD is the altitude of triangle ACB. Find the value of y.
问题：如图所示，BD=4*x+9，CD=7*x-6，∠CDA=2*y-6°，AD是△ACB的高。求y的值。
Construction (CDL):
Shape(BA,AD,DB)
Shape(DA,AC,CD)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(BD),4*x+9)
Equal(LengthOfLine(CD),7*x-6)
Equal(MeasureOfAngle(CDA),2*y-6)
IsAltitudeOfTriangle(AD,ACB)
Goal: Value(y)
Answer: 48
Theorem sequence:
Problem: As shown in the diagram, AB=6, AE=9, BC=4, CD∥BE. Find the length of line ED.
问题：如图所示，AB=6，AE=9，BC=4，CD∥BE。求直线ED的长度。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BE,ED,DC,CB)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AE),9)
Equal(LengthOfLine(BC),4)
ParallelBetweenLine(CD,BE)
Goal: Value(LengthOfLine(ED))
Answer: 6
Theorem sequence:
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
line_addition(1,AE,ED)
line_addition(1,AB,BC)
similar_triangle_property_line_ratio(1,BAE,CAD)
similar_triangle_property_line_ratio(1,EBA,DCA)
Problem: As shown in the diagram, AB=72, AC=78, BC=x, AB⊥CB. Find the value of x.
问题：如图所示，AB=72，AC=78，BC=x，AB⊥CB。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),72)
Equal(LengthOfLine(AC),78)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(AB,CB)
Goal: Value(x)
Answer: 30
Theorem sequence:
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,ABC)
Problem: As shown in the diagram, ∠UEN=64°, IW∥KA, JL∥BN. Find the measure of ∠AUL.
问题：如图所示，∠UEN=64°，IW平行于KA，JL平行于BN。求∠AUL的大小。
Construction (CDL):
Shape(JD,DW)
Shape(WD,DU)
Shape(DU,UA)
Shape(AU,UL)
Shape(HD,DJ)
Shape(BH,HD)
Shape(IH,HB)
Shape(DH,HE,EU,UD)
Shape(EH,HI)
Shape(KE,EH)
Shape(LU,UE)
Shape(UE,EN)
Shape(NE,EK)
Collinear(WDHI)
Collinear(AUEK)
Collinear(JDUL)
Collinear(BHEN)
Conditions (CDL):
Equal(MeasureOfAngle(UEN),64)
ParallelBetweenLine(IW,KA)
ParallelBetweenLine(JL,BN)
Goal: Value(MeasureOfAngle(AUL))
Answer: 64
Theorem sequence:
parallel_property_collinear_extend(3,JL,BN,U)
parallel_property_collinear_extend(3,NB,LU,E)
parallel_property_corresponding_angle(1,UL,EN,A)
Problem: As shown in the diagram, R is the center of ⊙R, RS⊥XS, quadrilateral WZYX is a square. Find the measure of ∠XRY.
问题：如图所示，R是圆R的圆心，RS垂直于XS，WZYX是正方形。求∠XRY的大小。
Construction (CDL):
Shape(WZ,ZY,YR,RW)
Shape(WR,RX,XW)
Shape(XR,RS,SX)
Shape(SR,RY,YS)
Shape(WX,RXW)
Shape(RWZ,ZW)
Shape(RZY,YZ)
Shape(RYX,XY)
Collinear(XSY)
Collinear(YRW)
Cocircular(R,WZYX)
Conditions (CDL):
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(RS,XS)
Square(WZYX)
Goal: Value(MeasureOfAngle(XRY))
Answer: 90
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,XWY)
isosceles_triangle_property_angle_equal(1,XWY)
triangle_property_angle_sum(1,XWY)
arc_property_circumference_angle_external(1,RYX,W)
arc_property_center_angle(1,RYX,R)
Problem: As shown in the diagram, BA=2*x+4, BC=4*x-8, CA=3*x-1, △ABC is an isosceles △. Find the perimeter of triangle ABC.
问题：如图所示，BA=2*x+4，BC=4*x-8，CA=3*x-1，三角形ABC为等腰三角形。求三角形ABC的周长。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(BA),2*x+4)
Equal(LengthOfLine(BC),4*x-8)
Equal(LengthOfLine(CA),3*x-1)
IsoscelesTriangle(ABC)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 40
Theorem sequence:
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, ∠BAD=58°, ∠CBE=26°, ∠EBA=47°, AB∥EC. Find the measure of ∠ECB.
问题：如图所示，∠BAD=58°，∠CBE=26°，∠EBA=47°，AB∥EC。求∠ECB的大小。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BE,EC,CB)
Shape(CE,ED,DC)
Collinear(AED)
Collinear(BCD)
Conditions (CDL):
Equal(MeasureOfAngle(BAD),58)
Equal(MeasureOfAngle(CBE),26)
Equal(MeasureOfAngle(EBA),47)
ParallelBetweenLine(AB,EC)
Goal: Value(MeasureOfAngle(ECB))
Answer: 107
Theorem sequence:
angle_addition(1,CBE,EBA)
parallel_property_ipsilateral_internal_angle(1,CE,BA)
Problem: As shown in the diagram, RT=11, ∠RPS=130°, ∠TPQ=112°, P is the center of circle P, the diameter of ⊙P is RT. Find the length of ⌒PQR.
问题：如图所示，RT=11，∠RPS=130°，∠TPQ=112°，圆P的圆心为P，⊙P的直径为RT。求⌒PQR的长度。
Construction (CDL):
Shape(QP,PR,PRQ)
Shape(TP,PQ,PQT)
Shape(SP,PT,PTS)
Shape(RP,PS,PSR)
Collinear(RPT)
Cocircular(P,RQTS)
Conditions (CDL):
Equal(LengthOfLine(RT),11)
Equal(MeasureOfAngle(RPS),130)
Equal(MeasureOfAngle(TPQ),112)
IsCentreOfCircle(P,P)
IsDiameterOfCircle(RT,P)
Goal: Value(LengthOfArc(PQR))
Answer: 803*pi/90
Theorem sequence:
flat_angle(1,RPT)
angle_addition(1,RPT,TPQ)
arc_length_formula(1,PQR)
arc_property_center_angle(1,PQR,P)
diameter_of_circle_property_length_equal(1,RT,P)
circle_property_length_of_radius_and_diameter(1,P)
Problem: As shown in the diagram, JK=8, JS=10, ∠SLK=25°, ∠SPM=28°, S is the incenter of triangle JPL, PM is perpendicular to SM, SK⊥JK, SQ⊥PQ. Find the measure of ∠SJP.
问题：如图所示，JK=8，JS=10，∠SLK=25°，∠SPM=28°，三角形JPL的内切圆圆心为S，PM⊥SM，SK垂直于JK，SQ垂直于PQ。求∠SJP的大小。
Construction (CDL):
Shape(JQ,QS,SJ)
Shape(QP,PS,SQ)
Shape(PM,MS,SP)
Shape(SM,ML,LS)
Shape(SL,LK,KS)
Shape(SK,KJ,JS)
Collinear(JQP)
Collinear(PML)
Collinear(LKJ)
Conditions (CDL):
Equal(LengthOfLine(JK),8)
Equal(LengthOfLine(JS),10)
Equal(MeasureOfAngle(SLK),25)
Equal(MeasureOfAngle(SPM),28)
IsIncenterOfTriangle(S,JPL)
PerpendicularBetweenLine(PM,SM)
PerpendicularBetweenLine(SK,JK)
PerpendicularBetweenLine(SQ,PQ)
Goal: Value(MeasureOfAngle(SJP))
Answer: 37
Theorem sequence:
angle_addition(1,MLS,SLK)
angle_addition(1,KJS,SJQ)
angle_addition(1,QPS,SPM)
triangle_property_angle_sum(1,JPL)
Problem: As shown in the diagram, ∠WZX=24°, ∠XWZ=23°, ∠YXZ=105°, ∠ZYX=∠XZY. Find the measure of ∠ZXW.
问题：如图所示，∠WZX=24°，∠XWZ=23°，∠YXZ=105°，∠ZYX=∠XZY。求∠ZXW的大小。
Construction (CDL):
Shape(WZ,ZX,XW)
Shape(XZ,ZY,YX)
Conditions (CDL):
Equal(MeasureOfAngle(WZX),24)
Equal(MeasureOfAngle(XWZ),23)
Equal(MeasureOfAngle(YXZ),105)
Equal(MeasureOfAngle(ZYX),MeasureOfAngle(XZY))
Goal: Value(MeasureOfAngle(ZXW))
Answer: 133
Theorem sequence:
triangle_property_angle_sum(1,WZX)
Problem: As shown in the diagram, CD=22, the radius of circle A is 14, the center of ⊙A is A, AE⊥DE. Find the length of line CE.
问题：如图所示，CD=22，圆A的半径为14，A是⊙A的圆心，AE⊥DE。求直线CE的长度。
Construction (CDL):
Shape(ADC,CE,ED)
Shape(EC,ACB,BE)
Shape(ABD,DE,EB)
Shape(AE,ED)
Shape(CE,EA)
Collinear(AEB)
Collinear(CED)
Cocircular(A,CBD)
Conditions (CDL):
Equal(LengthOfLine(CD),22)
Equal(RadiusOfCircle(A),14)
IsCentreOfCircle(A,A)
PerpendicularBetweenLine(AE,DE)
Goal: Value(LengthOfLine(CE))
Answer: 11
Theorem sequence:
adjacent_complementary_angle(1,CEA,AED)
circle_property_chord_perpendicular_bisect_chord(1,A,AE,CD)
line_addition(1,CE,ED)
Problem: As shown in the diagram, ∠ABP=3*x-60°, ∠MCL=2*x+15°, ∠PBC=y°, OB∥NC, OP∥NM. Find the measure of ∠BCM.
问题：如图所示，∠ABP=3*x-60°，∠MCL=2*x+15°，∠PBC=y°，OB∥NC，OP平行于NM。求∠BCM的大小。
Construction (CDL):
Shape(OR,RA)
Shape(AR,RP)
Shape(CB,BO)
Shape(PB,BC)
Shape(NC,CB)
Shape(LC,CN)
Shape(BC,CM)
Shape(MC,CL)
Collinear(OBP)
Collinear(ABCL)
Collinear(NCM)
Conditions (CDL):
Equal(MeasureOfAngle(ABP),3*x-60)
Equal(MeasureOfAngle(MCL),2*x+15)
Equal(MeasureOfAngle(PBC),y)
ParallelBetweenLine(OB,NC)
ParallelBetweenLine(OP,NM)
Goal: Value(MeasureOfAngle(BCM))
Answer: 75
Theorem sequence:
parallel_property_collinear_extend(2,OB,NC,P)
parallel_property_collinear_extend(1,CN,PB,M)
parallel_property_corresponding_angle(2,BP,CM,L)
adjacent_complementary_angle(1,ABP,PBC)
adjacent_complementary_angle(1,BCM,MCL)
Problem: As shown in the diagram, AC=24, CB=x, ∠ACB=86°, ∠BAC=33°. Find the value of x.
问题：如图所示，AC=24，CB=x，∠ACB=86°，∠BAC=33°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),24)
Equal(LengthOfLine(CB),x)
Equal(MeasureOfAngle(ACB),86)
Equal(MeasureOfAngle(BAC),33)
Goal: Value(x)
Answer: (-3*sqrt(5)*sqrt(5-sqrt(5))-3*sqrt(6)-3*sqrt(5-sqrt(5))-3*sqrt(2)+3*sqrt(3)*sqrt(5-sqrt(5))+3*sqrt(10)+3*sqrt(30)+3*sqrt(15)*sqrt(5-sqrt(5)))/(2*sin(61*pi/180))
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,CBA)
Problem: As shown in the diagram, ∠GCH=2*x°, ∠HCD=6*x+28°, C is the center of circle C, FC⊥GC. Find the measure of arc CDH.
问题：如图所示，∠GCH=2*x°，∠HCD=6*x+28°，⊙C的圆心为C，FC垂直于GC。求弧CDH的角度。
Construction (CDL):
Shape(GC,CH,CHG)
Shape(FC,CG,CGF)
Shape(EC,CF,CFE)
Shape(DC,CE,CED)
Shape(HC,CD,CDH)
Collinear(HCE)
Collinear(GCD)
Cocircular(C,HGFED)
Conditions (CDL):
Equal(MeasureOfAngle(GCH),2*x)
Equal(MeasureOfAngle(HCD),6*x+28)
IsCentreOfCircle(C,C)
PerpendicularBetweenLine(FC,GC)
Goal: Value(MeasureOfArc(CDH))
Answer: 142
Theorem sequence:
adjacent_complementary_angle(1,GCH,HCD)
arc_property_center_angle(1,CDH,C)
Problem: As shown in the diagram, AB=11, AC=x, BC=12, ∠CBA=67°. Find the value of x.
问题：如图所示，AB=11，AC=x，BC=12，∠CBA=67°。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),11)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),12)
Equal(MeasureOfAngle(CBA),67)
Goal: Value(x)
Answer: sqrt(265-264*cos(67*pi/180))
Theorem sequence:
cosine_theorem(1,BAC)
Problem: As shown in the diagram, ∠ADC=x+34°, ∠BAD=82°, ∠DCB=83°, ∠PRS=3*y-13°, ∠QPR=97°, ∠RSQ=98°, quadrilateral ADCB is similar to quadrilateral RSQP. Find the value of y.
问题：如图所示，∠ADC=x+34°，∠BAD=82°，∠DCB=83°，∠PRS=3*y-13°，∠QPR=97°，∠RSQ=98°，四边形ADCB相似于四边形RSQP。求y的值。
Construction (CDL):
Shape(AD,DC,CB,BA)
Shape(PR,RS,SQ,QP)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),x+34)
Equal(MeasureOfAngle(BAD),82)
Equal(MeasureOfAngle(DCB),83)
Equal(MeasureOfAngle(PRS),3*y-13)
Equal(MeasureOfAngle(QPR),97)
Equal(MeasureOfAngle(RSQ),98)
SimilarBetweenQuadrilateral(ADCB,RSQP)
Goal: Value(y)
Answer: 95/3
Theorem sequence:
similar_quadrilateral_property_angle_equal(1,ADCB,RSQP)
Problem: As shown in the diagram, RS=x+9, RT=3*x-9, ST=2*x, triangle SRT is an equilateral triangle. Find the length of line RS.
问题：如图所示，RS=x+9，RT=3*x-9，ST=2*x，△SRT为等边△。求直线RS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(RS))
Answer: 18
Theorem sequence:
Problem: As shown in the diagram, the diameter of ⊙A is 10, the diameter of ⊙B is 30, the diameter of circle C is 10, AZ=CW, CW=2. Find the length of line AZ.
问题：如图所示，圆A的直径为10，圆B的直径为30，⊙C的直径为10，AZ=CW，CW=2。求直线AZ的长度。
Construction (CDL):
Shape(GA,AZ,BJZ,AJG)
Shape(ZA,AG,AGK,BZK)
Shape(AXJ,BJZ,ZX)
Shape(AKX,XZ,BZK)
Shape(XB,BY,CMY,BMJ,AXJ)
Shape(YB,BX,AKX,BKN,CYN)
Shape(CMY,YW,BWM)
Shape(BNW,WY,CYN)
Shape(WC,CF,CFM,BWM)
Shape(FC,CW,BNW,CNF)
Collinear(GAZXBYWCF)
Cocircular(A,JGKX)
Cocircular(B,JZKNWM)
Cocircular(C,MYNF)
Conditions (CDL):
Equal(DiameterOfCircle(A),10)
Equal(DiameterOfCircle(B),30)
Equal(DiameterOfCircle(C),10)
Equal(LengthOfLine(AZ),LengthOfLine(CW))
Equal(LengthOfLine(CW),2)
Goal: Value(LengthOfLine(AZ))
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, ∠XZW=5*x-12°, ∠YZX=3*x+6°, ZY and WX are opposite sides of the parallelogram ZWXY, ZW is perpendicular to XW. Find the measure of ∠ZXY.
问题：如图所示，∠XZW=5*x-12°，∠YZX=3*x+6°，ZWXY是▱，ZW垂直于XW。求∠ZXY的大小。
Construction (CDL):
Shape(ZW,WP,PZ)
Shape(PW,WX,XP)
Shape(YP,PX,XY)
Shape(ZP,PY,YZ)
Collinear(ZPX)
Collinear(WPY)
Conditions (CDL):
Equal(MeasureOfAngle(XZW),5*x-12)
Equal(MeasureOfAngle(YZX),3*x+6)
Parallelogram(ZWXY)
PerpendicularBetweenLine(ZW,XW)
Goal: Value(MeasureOfAngle(ZXY))
Answer: 48
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,ZY,WX)
angle_addition(1,YZX,XZW)
parallel_property_alternate_interior_angle(2,YX,ZW)
Problem: As shown in the diagram, BA=9*x-1, CA=4*x+1, CA=BC, CB=5*x-1/2, △CBA is an isosceles △. Find the length of line CB.
问题：如图所示，BA=9*x-1，CA=4*x+1，CA=BC，CB=5*x-1/2，三角形CBA为等腰三角形。求直线CB的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(BA),9*x-1)
Equal(LengthOfLine(CA),4*x+1)
Equal(LengthOfLine(CA),LengthOfLine(BC))
Equal(LengthOfLine(CB),5*x-1/2)
IsoscelesTriangle(CBA)
Goal: Value(LengthOfLine(CB))
Answer: 7
Theorem sequence:
Problem: As shown in the diagram, XY=7, ZX=14, ZY=x, XY is perpendicular to ZY. Find the value of x.
问题：如图所示，XY=7，ZX=14，ZY=x，XY垂直于ZY。求x的值。
Construction (CDL):
Shape(XY,YZ,ZX)
Conditions (CDL):
Equal(LengthOfLine(XY),7)
Equal(LengthOfLine(ZX),14)
Equal(LengthOfLine(ZY),x)
PerpendicularBetweenLine(XY,ZY)
Goal: Value(x)
Answer: 7*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,XYZ)
right_triangle_property_pythagorean(1,XYZ)
Problem: As shown in the diagram, QR=5, QS=RS, RS=5*x+3, SQ=10*x-7, ∠PSR=15*x+42°, PS is the median of triangle PQR. Find the value of x.
问题：如图所示，QR=5，QS=RS，RS=5*x+3，SQ=10*x-7，∠PSR=15*x+42°，PS是△PQR的中线。求x的值。
Construction (CDL):
Shape(PQ,QS,SP)
Shape(PS,SR,RP)
Collinear(QSR)
Conditions (CDL):
Equal(LengthOfLine(QR),5)
Equal(LengthOfLine(QS),LengthOfLine(RS))
Equal(LengthOfLine(RS),5*x+3)
Equal(LengthOfLine(SQ),10*x-7)
Equal(MeasureOfAngle(PSR),15*x+42)
IsMedianOfTriangle(PS,PQR)
Goal: Value(x)
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, TW=13, ZV=1, the center of circle T is T, TV is perpendicular to XV. Find the length of line XV.
问题：如图所示，TW=13，ZV=1，T是圆T的圆心，TV垂直于XV。求直线XV的长度。
Construction (CDL):
Shape(TZX,XV,VZ)
Shape(TYZ,ZV,VY)
Shape(TV,VX,XT)
Shape(WT,TX,TXW)
Shape(VT,TW,TWY,YV)
Collinear(ZVTW)
Collinear(XVY)
Cocircular(T,ZXWY)
Conditions (CDL):
Equal(LengthOfLine(TW),13)
Equal(LengthOfLine(ZV),1)
IsCentreOfCircle(T,T)
PerpendicularBetweenLine(TV,XV)
Goal: Value(LengthOfLine(XV))
Answer: 5
Theorem sequence:
radius_of_circle_property_length_equal(1,TW,T)
radius_of_circle_property_length_equal(1,TX,T)
radius_of_circle_property_length_equal(1,TZ,T)
line_addition(1,ZV,VT)
right_triangle_judgment_angle(1,TVX)
right_triangle_property_pythagorean(1,TVX)
Problem: As shown in the diagram, ∠CDB=20°, ∠DBC=25°, ∠ECB=x°. Find the value of x.
问题：如图所示，∠CDB=20°，∠DBC=25°，∠ECB=x°。求x的值。
Construction (CDL):
Shape(BE,EC,CB)
Shape(BC,CD,DB)
Collinear(ECD)
Conditions (CDL):
Equal(MeasureOfAngle(CDB),20)
Equal(MeasureOfAngle(DBC),25)
Equal(MeasureOfAngle(ECB),x)
Goal: Value(x)
Answer: 45
Theorem sequence:
triangle_property_angle_sum(1,BCD)
adjacent_complementary_angle(1,ECB,BCD)
Problem: As shown in the diagram, AB=x, AC=19, AD=15, BC=z, BD=y, CA⊥BA, DB⊥CB. Find the value of x.
问题：如图所示，AB=x，AC=19，AD=15，BC=z，BD=y，CA垂直于BA，DB垂直于CB。求x的值。
Construction (CDL):
Shape(DB,BA,AD)
Shape(AB,BC,CA)
Collinear(DAC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),19)
Equal(LengthOfLine(AD),15)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BD),y)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(DB,CB)
Goal: Value(x)
Answer: sqrt(285)
Theorem sequence:
adjacent_complementary_angle(1,CAB,BAD)
mirror_similar_triangle_judgment_aa(1,BAD,CDB)
mirror_similar_triangle_judgment_aa(1,BCA,DBC)
mirror_similar_triangle_property_line_ratio(1,BAD,CDB)
mirror_similar_triangle_property_line_ratio(1,DBA,DBC)
mirror_similar_triangle_property_line_ratio(1,BCA,DBC)
mirror_similar_triangle_property_line_ratio(1,CAB,CDB)
Problem: As shown in the diagram, the area of quadrilateral NMQP is 375, MP=25, MQPN is a rhombus. Find the length of line NQ.
问题：如图所示，四边形NMQP的面积为375，MP=25，MQPN是菱形。求直线NQ的长度。
Construction (CDL):
Shape(MA,AN,NM)
Shape(NA,AP,PN)
Shape(MQ,QA,AM)
Shape(AQ,QP,PA)
Collinear(MAP)
Collinear(NAQ)
Conditions (CDL):
Equal(AreaOfQuadrilateral(NMQP),375)
Equal(LengthOfLine(MP),25)
Rhombus(MQPN)
Goal: Value(LengthOfLine(NQ))
Answer: 30
Theorem sequence:
kite_area_formula_diagonal(1,NMQP)
Problem: As shown in the diagram, GH=10, G is the center of ⊙G, J is the center of ⊙J, the center of circle K is K. Find the length of line GL.
问题：如图所示，GH=10，⊙G的圆心为G，J是圆J的圆心，K是圆K的圆心。求直线GL的长度。
Construction (CDL):
Shape(BG,GH,GHB)
Shape(FG,GB,GBF)
Shape(GF,GFL,JGL)
Shape(JGL,KJL,JG)
Shape(KJL,LK,KJ)
Shape(JK,KL,KLJ)
Shape(KLJ,JLG,GJ)
Shape(JLG,GKH,HG)
Collinear(BGJKL)
Collinear(HGF)
Cocircular(G,HBFL)
Cocircular(J,GL)
Cocircular(K,JL)
Conditions (CDL):
Equal(LengthOfLine(GH),10)
IsCentreOfCircle(G,G)
IsCentreOfCircle(J,J)
IsCentreOfCircle(K,K)
Goal: Value(LengthOfLine(GL))
Answer: 10
Theorem sequence:
radius_of_circle_property_length_equal(1,GH,G)
radius_of_circle_property_length_equal(1,GL,G)
Problem: As shown in the diagram, ∠PNO=56°, the measure of ⌒BMN is 70. Find the measure of ∠NPM.
问题：如图所示，∠PNO=56°，弧BMN的角度为70。求∠NPM的大小。
Construction (CDL):
Shape(MP,BPM)
Shape(NP,PM,BMN)
Shape(ON,BNO)
Shape(PN,NO,BOP)
Cocircular(B,NOPM)
Conditions (CDL):
Equal(MeasureOfAngle(PNO),56)
Equal(MeasureOfArc(BMN),70)
Goal: Value(MeasureOfAngle(NPM))
Answer: 35
Theorem sequence:
arc_property_circumference_angle_external(1,BMN,P)
Problem: As shown in the diagram, ∠BCH=8*y-12°, ∠BHD=1/4*x°, ∠HBC=y-8°, ∠HDB=4*x-8°, BD and CH are opposite sides of the ▱ DBCH. Find the value of y.
问题：如图所示，∠BCH=8*y-12°，∠BHD=1/4*x°，∠HBC=y-8°，∠HDB=4*x-8°，DBCH是平行四边形。求y的值。
Construction (CDL):
Shape(DB,BH,HD)
Shape(HB,BC,CH)
Conditions (CDL):
Equal(MeasureOfAngle(BCH),8*y-12)
Equal(MeasureOfAngle(BHD),1/4*x)
Equal(MeasureOfAngle(HBC),y-8)
Equal(MeasureOfAngle(HDB),4*x-8)
Parallelogram(DBCH)
Goal: Value(y)
Answer: 31/2
Theorem sequence:
parallel_property_alternate_interior_angle(2,DH,BC)
parallelogram_property_opposite_angle_equal(1,DBCH)
Problem: As shown in the diagram, CA=x, CB=8, ∠ABC=60°, ∠CAB=78°. Find the value of x.
问题：如图所示，CA=x，CB=8，∠ABC=60°，∠CAB=78°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(CA),x)
Equal(LengthOfLine(CB),8)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),78)
Goal: Value(x)
Answer: 32*sqrt(6)/(-sqrt(2)+sqrt(10)+2*sqrt(3)*sqrt(sqrt(5)+5))
Theorem sequence:
sine_theorem(1,CAB)
Problem: As shown in the diagram, AB=6, AF=8, BC=x, CD=y, DE=2*y-3, FE=x+10/3, BC∥FD, BF is parallel to CD. Find the length of line FE.
问题：如图所示，AB=6，AF=8，BC=x，CD=y，DE=2*y-3，FE=x+10/3，BC∥FD，BF平行于CD。求直线FE的长度。
Construction (CDL):
Shape(AB,BF,FA)
Shape(BC,CD,DF,FB)
Shape(FD,DE,EF)
Collinear(ABC)
Collinear(AFE)
Collinear(CDE)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AF),8)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CD),y)
Equal(LengthOfLine(DE),2*y-3)
Equal(LengthOfLine(FE),x+10/3)
ParallelBetweenLine(BC,FD)
ParallelBetweenLine(BF,CD)
Goal: Value(LengthOfLine(FE))
Answer: 40/3
Theorem sequence:
line_addition(1,AB,BC)
line_addition(1,AF,FE)
parallel_property_collinear_extend(1,DC,FB,E)
parallel_property_corresponding_angle(1,BF,CE,A)
similar_triangle_judgment_aa(1,FAB,EAC)
similar_triangle_property_line_ratio(1,FAB,EAC)
similar_triangle_property_line_ratio(1,BFA,CEA)
Problem: As shown in the diagram, BC=30, BC=AB, BD=x, ∠BCA=42°, BD is perpendicular to AD. Find the value of x.
问题：如图所示，BC=30，BC=AB，BD=x，∠BCA=42°，BD⊥AD。求x的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(BC),30)
Equal(LengthOfLine(BC),LengthOfLine(AB))
Equal(LengthOfLine(BD),x)
Equal(MeasureOfAngle(BCA),42)
PerpendicularBetweenLine(BD,AD)
Goal: Value(x)
Answer: -15*sqrt(5)/4+15/4+15*sqrt(6*sqrt(5)+30)/4
Theorem sequence:
sine_theorem(1,BDA)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AC=4*x-2, EB=6*x-10, the measure of arc DCA is 133, the measure of arc DEB is 133. Find the value of x.
问题：如图所示，AC=4*x-2，EB=6*x-10，⌒DCA的角度为133，弧DEB的角度为133。求x的值。
Construction (CDL):
Shape(AC,DCA)
Shape(CA,DAE,EB,DBC)
Shape(BE,DEB)
Cocircular(D,AEBC)
Conditions (CDL):
Equal(LengthOfLine(AC),4*x-2)
Equal(LengthOfLine(EB),6*x-10)
Equal(MeasureOfArc(DCA),133)
Equal(MeasureOfArc(DEB),133)
Goal: Value(x)
Answer: 4
Theorem sequence:
congruent_arc_judgment_measure_equal(1,DEB,DCA)
congruent_arc_property_chord_equal(1,DEB,DCA)