SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=30, CD=30, the measure of arc XCZ is 40, X is the center of circle X, AM is perpendicular to YM, DN is perpendicular to ZN. Find the length of line ND.
问题：如图所示，AB=30，CD=30，⌒XCZ的角度为40，X是⊙X的圆心，AM⊥YM，DN垂直于ZN。求直线ND的长度。
Construction (CDL):
Shape(XYA,AM,MY)
Shape(XAC,CX,XM,MA)
Shape(XCZ,ZN,NC)
Shape(NX,XC,CN)
Shape(XN,ND,DX)
Shape(XZD,DN,NZ)
Shape(XDB,BM,MX,XD)
Shape(XBY,YM,MB)
Collinear(XMY)
Collinear(XNZ)
Collinear(AMB)
Collinear(CND)
Cocircular(X,ACZDBY)
Conditions (CDL):
Equal(LengthOfLine(AB),30)
Equal(LengthOfLine(CD),30)
Equal(MeasureOfArc(XCZ),40)
IsCentreOfCircle(X,X)
PerpendicularBetweenLine(AM,YM)
PerpendicularBetweenLine(DN,ZN)
Goal: Value(LengthOfLine(ND))
Answer: 15
Theorem sequence:
radius_of_circle_property_length_equal(1,XD,X)
radius_of_circle_property_length_equal(1,XC,X)
vertical_angle(1,CNX,DNZ)
adjacent_complementary_angle(1,XND,DNZ)
mirror_congruent_triangle_judgment_hl(1,CNX,DXN)
mirror_congruent_triangle_property_line_equal(1,XCN,XND)
line_addition(1,CN,ND)
Problem: As shown in the diagram, AB=15, CD=15, ∠DCB=30°, BD⊥CD, CA is perpendicular to BA, quadrilateral BDCA is a rectangle. Find the area of BDCA.
问题：如图所示，AB=15，CD=15，∠DCB=30°，BD垂直于CD，CA⊥BA，BDCA是矩形。求四边形BDCA的面积。
Construction (CDL):
Shape(BD,DC,CB)
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(CD),15)
Equal(MeasureOfAngle(DCB),30)
PerpendicularBetweenLine(BD,CD)
PerpendicularBetweenLine(CA,BA)
Rectangle(BDCA)
Goal: Value(AreaOfQuadrilateral(BDCA))
Answer: 75*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,BDC)
sine_theorem(1,DCB)
parallelogram_area_formula_sine(1,BDCA)
Problem: As shown in the diagram, YX=24, YZ=28, ∠XWZ=105°, quadrilateral WZYX is a parallelogram. Find the measure of ∠ZYX.
问题：如图所示，YX=24，YZ=28，∠XWZ=105°，WX和ZY是平行四边形WZYX的一组对边。求∠ZYX的大小。
Construction (CDL):
Shape(WZ,ZY,YX,XW)
Conditions (CDL):
Equal(LengthOfLine(YX),24)
Equal(LengthOfLine(YZ),28)
Equal(MeasureOfAngle(XWZ),105)
Parallelogram(WZYX)
Goal: Value(MeasureOfAngle(ZYX))
Answer: 105
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,WZYX)
Problem: As shown in the diagram, ∠PRZ=4*b-17°, ∠ZRQ=3*b-4°, RZ bisects ∠PRQ. Find the measure of ∠PRZ.
问题：如图所示，∠PRZ=4*b-17°，∠ZRQ=3*b-4°，RZ是∠PRQ的角平分线。求∠PRZ的大小。
Construction (CDL):
Shape(ZP,PR,RZ)
Shape(QZ,ZR,RQ)
Collinear(PZQ)
Conditions (CDL):
Equal(MeasureOfAngle(PRZ),4*b-17)
Equal(MeasureOfAngle(ZRQ),3*b-4)
IsBisectorOfAngle(RZ,PRQ)
Goal: Value(MeasureOfAngle(PRZ))
Answer: 35
Theorem sequence:
Problem: As shown in the diagram, AB=14, AC=10, BC=18, BD=x, ∠BAD=∠DAC. Find the value of x.
问题：如图所示，AB=14，AC=10，BC=18，BD=x，∠BAD=∠DAC。求x的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),14)
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(BC),18)
Equal(LengthOfLine(BD),x)
Equal(MeasureOfAngle(BAD),MeasureOfAngle(DAC))
Goal: Value(x)
Answer: 21/2
Theorem sequence:
line_addition(1,CD,DB)
bisector_of_angle_judgment_angle_equal(1,AD,BAC)
bisector_of_angle_property_line_ratio(1,AD,BAC)
Problem: As shown in the diagram, the area of triangle ACB is 357, AD=x, BC=34, CD is perpendicular to AD. Find the value of x.
问题：如图所示，△ACB的面积为357，AD=x，BC=34，CD垂直于AD。求x的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(AreaOfTriangle(ACB),357)
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BC),34)
PerpendicularBetweenLine(CD,AD)
Goal: Value(x)
Answer: 21
Theorem sequence:
altitude_of_triangle_judgment(1,AD,ACB)
triangle_area_formula_common(1,ACB)
Problem: As shown in the diagram, ∠ABC=6*y-57°, ∠BCD=3*x+9°, ∠CDA=3*y+36°, ∠DAB=5*x-19°, BCDA is a ▱. Find the value of y.
问题：如图所示，∠ABC=6*y-57°，∠BCD=3*x+9°，∠CDA=3*y+36°，∠DAB=5*x-19°，四边形BCDA是▱。求y的值。
Construction (CDL):
Shape(BC,CD,DA,AB)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),6*y-57)
Equal(MeasureOfAngle(BCD),3*x+9)
Equal(MeasureOfAngle(CDA),3*y+36)
Equal(MeasureOfAngle(DAB),5*x-19)
Parallelogram(BCDA)
Goal: Value(y)
Answer: 31
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BCDA)
Problem: As shown in the diagram, AH=BH, BH=2*x+3, EN=2*y, HA=6-x, IJ=4/3*y+1, AE is parallel to BI, FC is parallel to MN, OL is parallel to FC, OL is parallel to MN. Find the value of x.
问题：如图所示，AH=BH，BH=2*x+3，EN=2*y，HA=6-x，IJ=4/3*y+1，AE平行于BI，FC∥MN，OL平行于FC，OL平行于MN。求x的值。
Construction (CDL):
Shape(BH,HJ,JI,IB)
Shape(HA,AE,EJ,JH)
Collinear(OBIL)
Collinear(EHJC)
Collinear(MAEN)
Collinear(GBHAK)
Collinear(DIJEP)
Conditions (CDL):
Equal(LengthOfLine(AH),LengthOfLine(BH))
Equal(LengthOfLine(BH),2*x+3)
Equal(LengthOfLine(EN),2*y)
Equal(LengthOfLine(HA),6-x)
Equal(LengthOfLine(IJ),4/3*y+1)
ParallelBetweenLine(AE,BI)
ParallelBetweenLine(FC,MN)
ParallelBetweenLine(OL,FC)
ParallelBetweenLine(OL,MN)
Goal: Value(x)
Answer: 1
Theorem sequence:
Problem: As shown in the diagram, BC=13, BE=5, DC=15, CBAD is a ▱, BE is perpendicular to CE. Find the area of quadrilateral CBAD.
问题：如图所示，BC=13，BE=5，DC=15，BC和AD是▱CBAD的一组对边，BE垂直于CE。求CBAD的面积。
Construction (CDL):
Shape(CB,BE,EC)
Shape(CE,EA,AD,DC)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(BC),13)
Equal(LengthOfLine(BE),5)
Equal(LengthOfLine(DC),15)
Parallelogram(CBAD)
PerpendicularBetweenLine(BE,CE)
Goal: Value(AreaOfQuadrilateral(CBAD))
Answer: 180
Theorem sequence:
right_triangle_judgment_angle(1,BEC)
right_triangle_property_pythagorean(1,BEC)
altitude_of_quadrilateral_judgment_left_vertex(1,CE,CBAD)
parallelogram_property_opposite_line_equal(1,DCBA)
parallelogram_area_formula_common(1,CBAD)
Problem: As shown in the diagram, AB=CD, AC=3, AC=AB, CD=BD, quadrilateral DCAB is a parallelogram, DC is perpendicular to AC. Find the area of DCAB.
问题：如图所示，AB=CD，AC=3，AC=AB，CD=BD，四边形DCAB是▱，DC⊥AC。求四边形DCAB的面积。
Construction (CDL):
Shape(DC,CA,AB,BD)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(CD))
Equal(LengthOfLine(AC),3)
Equal(LengthOfLine(AC),LengthOfLine(AB))
Equal(LengthOfLine(CD),LengthOfLine(BD))
Parallelogram(DCAB)
PerpendicularBetweenLine(DC,AC)
Goal: Value(AreaOfQuadrilateral(DCAB))
Answer: 9
Theorem sequence:
parallelogram_area_formula_sine(1,DCAB)
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, K is the center of circle K. Find the length of line CK.
问题：如图所示，BC=5，圆J的半径为10，圆K的半径为8，⊙J的圆心为J，⊙K的圆心为K。求直线CK的长度。
Construction (CDL):
Shape(JB,KEB,JEA,AJ)
Shape(JA,JAF,KBF,BJ)
Shape(JCE,KEB,BC)
Shape(CB,KBF,JFC)
Shape(KD,KDE,JCE,CK)
Shape(KC,JFC,KFD,DK)
Collinear(AJBCKD)
Cocircular(J,CEAF)
Cocircular(K,DEBF)
Conditions (CDL):
Equal(LengthOfLine(BC),5)
Equal(RadiusOfCircle(J),10)
Equal(RadiusOfCircle(K),8)
IsCentreOfCircle(J,J)
IsCentreOfCircle(K,K)
Goal: Value(LengthOfLine(CK))
Answer: 3
Theorem sequence:
radius_of_circle_property_length_equal(1,KB,K)
line_addition(1,BC,CK)
Problem: As shown in the diagram, ∠AXW=30°, WY is perpendicular to ZY, quadrilateral WYZX is a rectangle. Find the measure of ∠AYZ.
问题：如图所示，∠AXW=30°，WY垂直于ZY，WYZX是长方形。求∠AYZ的大小。
Construction (CDL):
Shape(WA,AX,XW)
Shape(WY,YA,AW)
Shape(AY,YZ,ZA)
Shape(AZ,ZX,XA)
Collinear(WAZ)
Collinear(XAY)
Conditions (CDL):
Equal(MeasureOfAngle(AXW),30)
PerpendicularBetweenLine(WY,ZY)
Rectangle(WYZX)
Goal: Value(MeasureOfAngle(AYZ))
Answer: 30
Theorem sequence:
parallel_property_alternate_interior_angle(2,WX,YZ)
Problem: As shown in the diagram, AB=y, AC=12, BC=x, ∠CAB=45°, AB is perpendicular to CB. Find the value of y.
问题：如图所示，AB=y，AC=12，BC=x，∠CAB=45°，AB⊥CB。求y的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(AB,CB)
Goal: Value(y)
Answer: 6*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,BCA)
sine_theorem(1,ABC)
Problem: As shown in the diagram, AE=3, DB=24, EC=18, DE is parallel to BC, ED∥CB. Find the length of line AD.
问题：如图所示，AE=3，DB=24，EC=18，DE平行于BC，ED平行于CB。求直线AD的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CE,ED)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AE),3)
Equal(LengthOfLine(DB),24)
Equal(LengthOfLine(EC),18)
ParallelBetweenLine(DE,BC)
ParallelBetweenLine(ED,CB)
Goal: Value(LengthOfLine(AD))
Answer: 4
Theorem sequence:
line_addition(1,AD,DB)
line_addition(1,AE,EC)
parallel_property_corresponding_angle(1,DE,BC,A)
parallel_property_corresponding_angle(2,CB,ED,A)
similar_triangle_judgment_aa(1,ADE,ABC)
similar_triangle_property_line_ratio(1,DEA,BCA)
similar_triangle_property_line_ratio(1,EAD,CAB)
Problem: As shown in the diagram, BA=5*x, BC=3*y-4, DA=29, DC=25, ∠ADB=34°, ∠DCF=54°, ∠FAD=49°, BC and AD are opposite sides of the ▱ BADC. Find the measure of ∠FDC.
问题：如图所示，BA=5*x，BC=3*y-4，DA=29，DC=25，∠ADB=34°，∠DCF=54°，∠FAD=49°，四边形BADC是▱。求∠FDC的大小。
Construction (CDL):
Shape(BF,FC,CB)
Shape(BA,AF,FB)
Shape(FA,AD,DF)
Shape(CF,FD,DC)
Collinear(BFD)
Collinear(CFA)
Conditions (CDL):
Equal(LengthOfLine(BA),5*x)
Equal(LengthOfLine(BC),3*y-4)
Equal(LengthOfLine(DA),29)
Equal(LengthOfLine(DC),25)
Equal(MeasureOfAngle(ADB),34)
Equal(MeasureOfAngle(DCF),54)
Equal(MeasureOfAngle(FAD),49)
Parallelogram(BADC)
Goal: Value(MeasureOfAngle(FDC))
Answer: 43
Theorem sequence:
triangle_property_angle_sum(1,FAD)
flat_angle(1,CFA)
angle_addition(1,CFD,DFA)
triangle_property_angle_sum(1,CFD)
Problem: As shown in the diagram, ∠ADC=x°, ∠BAD=x°, ∠CBA=2*x°, ∠DCB=2*x°. Find the measure of ∠CBA.
问题：如图所示，∠ADC=x°，∠BAD=x°，∠CBA=2*x°，∠DCB=2*x°。求∠CBA的大小。
Construction (CDL):
Shape(BA,AD,DC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),x)
Equal(MeasureOfAngle(BAD),x)
Equal(MeasureOfAngle(CBA),2*x)
Equal(MeasureOfAngle(DCB),2*x)
Goal: Value(MeasureOfAngle(CBA))
Answer: 120
Theorem sequence:
quadrilateral_property_angle_sum(1,BADC)
Problem: As shown in the diagram, ∠ABC=1/2*y°, ∠BDC=3*y-4°, ∠CAB=4*x-8°, ∠DCB=x-12°, CD and AB are opposite sides of the parallelogram CABD. Find the value of x.
问题：如图所示，∠ABC=1/2*y°，∠BDC=3*y-4°，∠CAB=4*x-8°，∠DCB=x-12°，CD和AB是平行四边形CABD的一组对边。求x的值。
Construction (CDL):
Shape(CB,BD,DC)
Shape(CA,AB,BC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),1/2*y)
Equal(MeasureOfAngle(BDC),3*y-4)
Equal(MeasureOfAngle(CAB),4*x-8)
Equal(MeasureOfAngle(DCB),x-12)
Parallelogram(CABD)
Goal: Value(x)
Answer: 34
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,DCAB)
parallel_property_alternate_interior_angle(1,CD,AB)
Problem: As shown in the diagram, AB=33, AC=x, BC=66, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=33，AC=x，BC=66，BA垂直于CA。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),33)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),66)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 33*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, ∠POM=43°, BF is parallel to MQ, KC is parallel to GJ. Find the measure of ∠HNO.
问题：如图所示，∠POM=43°，BF∥MQ，KC平行于GJ。求∠HNO的大小。
Construction (CDL):
Shape(BP,PO,OM,MB)
Shape(PH,HN,NO,OP)
Shape(HF,FQ,QN,NH)
Shape(GF,FH,HG)
Shape(KG,GH,HP,PK)
Shape(MO,OC,CM)
Shape(NQ,QJ,JN)
Shape(BK,KP,PB)
Shape(ON,NJ,JC,CO)
Collinear(BPHF)
Collinear(MONQ)
Collinear(KPOC)
Collinear(GHNJ)
Conditions (CDL):
Equal(MeasureOfAngle(POM),43)
ParallelBetweenLine(BF,MQ)
ParallelBetweenLine(KC,GJ)
Goal: Value(MeasureOfAngle(HNO))
Answer: 43
Theorem sequence:
flat_angle(1,POC)
angle_addition(1,POM,MOC)
flat_angle(1,HNJ)
parallel_property_collinear_extend(3,KC,GJ,O)
parallel_property_collinear_extend(3,JG,CO,N)
parallel_property_corresponding_angle(1,OC,NJ,M)
angle_addition(1,HNO,ONJ)
Problem: As shown in the diagram, ∠DEF=25°, ∠EFD=65°, EG⊥DG. Find the measure of ∠GDE.
问题：如图所示，∠DEF=25°，∠EFD=65°，EG垂直于DG。求∠GDE的大小。
Construction (CDL):
Shape(DE,EG,GD)
Shape(DG,GF,FD)
Collinear(EGF)
Conditions (CDL):
Equal(MeasureOfAngle(DEF),25)
Equal(MeasureOfAngle(EFD),65)
PerpendicularBetweenLine(EG,DG)
Goal: Value(MeasureOfAngle(GDE))
Answer: 65
Theorem sequence:
triangle_property_angle_sum(1,DEG)
Problem: As shown in the diagram, ∠ABD=3*x-60°, ∠DBC=y°, ∠ECF=2*x+15°, BD∥CE. Find the measure of ∠ECF.
问题：如图所示，∠ABD=3*x-60°，∠DBC=y°，∠ECF=2*x+15°，BD∥CE。求∠ECF的大小。
Construction (CDL):
Shape(IB,BP,PA)
Shape(BC,CM,MP,PB)
Shape(CL,LM,MC)
Shape(IO,OB,BI)
Shape(ON,NC,CB,BO)
Shape(NL,LC,CN)
Collinear(OBDSP)
Collinear(NCETM)
Collinear(IABCFL)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),3*x-60)
Equal(MeasureOfAngle(DBC),y)
Equal(MeasureOfAngle(ECF),2*x+15)
ParallelBetweenLine(BD,CE)
Goal: Value(MeasureOfAngle(ECF))
Answer: 105
Theorem sequence:
parallel_property_corresponding_angle(1,BD,CE,A)
flat_angle(1,BCF)
angle_addition(1,BCE,ECF)
Problem: As shown in the diagram, LJ=2*x+1, LN=10, PJ=3*x-1, LJMK is a ▱, PNML is a rectangle. Find the value of x.
问题：如图所示，LJ=2*x+1，LN=10，PJ=3*x-1，JL和MK是▱LJMK的一组对边，四边形PNML是矩形。求x的值。
Construction (CDL):
Shape(PJ,JL,LP)
Shape(PN,NJ,JP)
Shape(JN,NM,MJ)
Shape(LJ,JM,ML)
Shape(LM,MK.KL)
Collinear(PJM)
Collinear(LJN)
Conditions (CDL):
Equal(LengthOfLine(LJ),2*x+1)
Equal(LengthOfLine(LN),10)
Equal(LengthOfLine(PJ),3*x-1)
Parallelogram(LJMK)
Rectangle(PNML)
Goal: Value(x)
Answer: 2
Theorem sequence:
parallelogram_property_diagonal_bisection(1,LPNM,J)
line_addition(1,LJ,JN)
Problem: As shown in the diagram, KM=3*x, KN=9, KP=x, LN=16, PM=2*x, the ratio of mirror similar triangles LMK and MKN is 5/3, PR∥KL, KN is perpendicular to MN, RM⊥PM. Find the value of x.
问题：如图所示，KM=3*x，KN=9，KP=x，LN=16，PM=2*x，△LMK和△MKN的镜像相似比为5/3，PR∥KL，KN垂直于MN，RM⊥PM。求x的值。
Construction (CDL):
Shape(NL,LR,RQ,QN)
Shape(KN,NQ,QP,PK)
Shape(QR,RM,MQ)
Shape(PQ,QM,MP)
Collinear(LNK)
Collinear(RQP)
Collinear(LRM)
Collinear(KPM)
Collinear(NQM)
Conditions (CDL):
Equal(LengthOfLine(KM),3*x)
Equal(LengthOfLine(KN),9)
Equal(LengthOfLine(KP),x)
Equal(LengthOfLine(LN),16)
Equal(LengthOfLine(PM),2*x)
Equal(RatioOfMirrorSimilarTriangle(LMK,MKN),5/3)
ParallelBetweenLine(PR,KL)
PerpendicularBetweenLine(KN,MN)
PerpendicularBetweenLine(RM,PM)
Goal: Value(x)
Answer: 5
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,LMK,MKN)
mirror_similar_triangle_property_line_ratio(1,LMK,MKN)
Problem: As shown in the diagram, AB=2*x, AC=12, BC=20, DE=y+4, EF=15, FD=12, ∠ACB=∠DEF, ∠BAC=∠FDE, ∠CBA=∠EFD, triangle ACB is similar to triangle DEF.. Find the value of y.
问题：如图所示，AB=2*x，AC=12，BC=20，DE=y+4，EF=15，FD=12，∠ACB=∠DEF，∠BAC=∠FDE，∠CBA=∠EFD，△ACB相似于△DEF。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Shape(DE,EF,FD)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(BC),20)
Equal(LengthOfLine(DE),y+4)
Equal(LengthOfLine(EF),15)
Equal(LengthOfLine(FD),12)
Equal(MeasureOfAngle(ACB),MeasureOfAngle(DEF))
Equal(MeasureOfAngle(BAC),MeasureOfAngle(FDE))
Equal(MeasureOfAngle(CBA),MeasureOfAngle(EFD))
SimilarBetweenTriangle(ACB,DEF)
Goal: Value(y)
Answer: 5
Theorem sequence:
similar_triangle_property_line_ratio(1,ACB,DEF)
similar_triangle_property_line_ratio(1,BAC,FDE)
Problem: As shown in the diagram, RS=2*z-15, RT=9, TS=7, ∠STR=∠RST. Find the value of z.
问题：如图所示，RS=2*z-15，RT=9，TS=7，∠STR=∠RST。求z的值。
Construction (CDL):
Shape(RS,ST,TR)
Conditions (CDL):
Equal(LengthOfLine(RS),2*z-15)
Equal(LengthOfLine(RT),9)
Equal(LengthOfLine(TS),7)
Equal(MeasureOfAngle(STR),MeasureOfAngle(RST))
Goal: Value(z)
Answer: 12
Theorem sequence:
isosceles_triangle_judgment_angle_equal(1,RST)
Problem: As shown in the diagram, triangle LMN is congruent to triangle QRS, ML=2*x+11, QR=3*x-9, ∠LMN=y+10°, ∠QRS=2*y-40°. Find the value of x.
问题：如图所示，三角形LMN与三角形QRS是全等三角形，ML=2*x+11，QR=3*x-9，∠LMN=y+10°，∠QRS=2*y-40°。求x的值。
Construction (CDL):
Shape(LM,MN,NL)
Shape(RS,SQ,QR)
Conditions (CDL):
CongruentBetweenTriangle(LMN,QRS)
Equal(LengthOfLine(ML),2*x+11)
Equal(LengthOfLine(QR),3*x-9)
Equal(MeasureOfAngle(LMN),y+10)
Equal(MeasureOfAngle(QRS),2*y-40)
Goal: Value(x)
Answer: 20
Theorem sequence:
congruent_triangle_property_line_equal(1,NLM,SQR)
Problem: As shown in the diagram, the diameter of ⊙B is 30, the diameter of ⊙C is 10, CW=2, the center of circle B is B, the center of circle C is C. Find the length of line YW.
问题：如图所示，⊙B的直径为30，⊙C的直径为10，CW=2，圆B的圆心为B，C是⊙C的圆心。求直线YW的长度。
Construction (CDL):
Shape(BY,CEY,BEZ,ZB)
Shape(BZ,BZF,CYF,YB)
Shape(BWE,CEY,YW)
Shape(WY,CYF,BFW)
Shape(CD,CDE,BWE,WC)
Shape(CW,BFW,CFD,DC)
Collinear(ZXBYWCD)
Cocircular(B,WZ)
Cocircular(C,DY)
Conditions (CDL):
Equal(DiameterOfCircle(B),30)
Equal(DiameterOfCircle(C),10)
Equal(LengthOfLine(CW),2)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
Goal: Value(LengthOfLine(YW))
Answer: 3
Theorem sequence:
circle_property_length_of_radius_and_diameter(1,C)
radius_of_circle_property_length_equal(1,CY,C)
line_addition(1,YW,WC)
Problem: As shown in the diagram, ∠LMN=51°, ∠NLM=65°, ∠XZY=∠NLM, ∠YXZ=4*x°, ∠ZYX=∠LMN. Find the value of x.
问题：如图所示，∠LMN=51°，∠NLM=65°，∠XZY=∠NLM，∠YXZ=4*x°，∠ZYX=∠LMN。求x的值。
Construction (CDL):
Shape(LM,MN,NL)
Shape(XZ,ZY,YX)
Conditions (CDL):
Equal(MeasureOfAngle(LMN),51)
Equal(MeasureOfAngle(NLM),65)
Equal(MeasureOfAngle(XZY),MeasureOfAngle(NLM))
Equal(MeasureOfAngle(YXZ),4*x)
Equal(MeasureOfAngle(ZYX),MeasureOfAngle(LMN))
Goal: Value(x)
Answer: 16
Theorem sequence:
triangle_property_angle_sum(1,XZY)
Problem: As shown in the diagram, YX=5, ZX=9, ZY⊥XY. Find the measure of ∠XZY.
问题：如图所示，YX=5，ZX=9，ZY垂直于XY。求∠XZY的大小。
Construction (CDL):
Shape(ZY,YX,XZ)
Conditions (CDL):
Equal(LengthOfLine(YX),5)
Equal(LengthOfLine(ZX),9)
PerpendicularBetweenLine(ZY,XY)
Goal: Value(MeasureOfAngle(XZY))
Answer: 180*asin(5/9)/pi
Theorem sequence:
sine_theorem(1,XZY)
Problem: As shown in the diagram, MN=3*x-4, NQ=15.4, PN=2*y+5, PQ=11.1, RM=17.9, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, MN and RP are opposite sides of the parallelogram MRPN. Find the value of x.
问题：如图所示，MN=3*x-4，NQ=15.4，PN=2*y+5，PQ=11.1，RM=17.9，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，MN和RP是平行四边形MRPN的一组对边。求x的值。
Construction (CDL):
Shape(MQ,QN,NM)
Shape(MR,RQ,QM)
Shape(QR,RP,PQ)
Shape(NQ,QP,PN)
Collinear(MQP)
Collinear(NQR)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(NQ),15.4)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),11.1)
Equal(LengthOfLine(RM),17.9)
Equal(LengthOfLine(RP),20)
Equal(LengthOfLine(RQ),3*z-3)
Equal(MeasureOfAngle(MRQ),38)
Equal(MeasureOfAngle(NQP),83)
Equal(MeasureOfAngle(QNM),33)
Parallelogram(MRPN)
Goal: Value(x)
Answer: 8
Theorem sequence:
parallelogram_property_opposite_line_equal(1,NMRP)
Problem: As shown in the diagram, EF=11, GF=x, RT=15, SR=20, ST=10, ∠EFG=110°, ∠GEF=27°, ∠RTS=110°, ∠SRT=27°, ∠TSR=43°, △SRT is similar to △GEF.. Find the length of line GF.
问题：如图所示，EF=11，GF=x，RT=15，SR=20，ST=10，∠EFG=110°，∠GEF=27°，∠RTS=110°，∠SRT=27°，∠TSR=43°，△SRT相似于△GEF。求直线GF的长度。
Construction (CDL):
Shape(SR,RT,TS)
Shape(GE,EF,FG)
Conditions (CDL):
Equal(LengthOfLine(EF),11)
Equal(LengthOfLine(GF),x)
Equal(LengthOfLine(RT),15)
Equal(LengthOfLine(SR),20)
Equal(LengthOfLine(ST),10)
Equal(MeasureOfAngle(EFG),110)
Equal(MeasureOfAngle(GEF),27)
Equal(MeasureOfAngle(RTS),110)
Equal(MeasureOfAngle(SRT),27)
Equal(MeasureOfAngle(TSR),43)
SimilarBetweenTriangle(SRT,GEF)
Goal: Value(LengthOfLine(GF))
Answer: 22/3
Theorem sequence:
similar_triangle_property_line_ratio(1,SRT,GEF)
similar_triangle_property_line_ratio(1,RTS,EFG)
Problem: As shown in the diagram, AC=12, CB=13, DE=5, AD and CB are opposite sides of the ▱ DACB, DE⊥CE. Find the perimeter of quadrilateral DACB.
问题：如图所示，AC=12，CB=13，DE=5，DB和AC是▱DACB的一组对边，DE垂直于CE。求四边形DACB的周长。
Construction (CDL):
Shape(DA,AE,ED)
Shape(DE,EC,CB,BD)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(CB),13)
Equal(LengthOfLine(DE),5)
Parallelogram(DACB)
PerpendicularBetweenLine(DE,CE)
Goal: Value(PerimeterOfQuadrilateral(DACB))
Answer: 50
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DACB)
parallelogram_property_opposite_line_equal(1,BDAC)
quadrilateral_perimeter_formula(1,DACB)
Problem: As shown in the diagram, AE=6, the radius of ⊙E is 6, the center of ⊙E is E, quadrilateral DACB is a square. Find the length of line AD.
问题：如图所示，AE=6，圆E的半径为6，⊙E的圆心为E，四边形DACB是正方形。求直线AD的长度。
Construction (CDL):
Shape(DA,AE,ED)
Shape(DE,EB,BD)
Shape(AC,CE,EA)
Shape(EC,CB,BE)
Shape(AD,EDA)
Shape(DB,EBD)
Shape(BC,ECB)
Shape(CA,EAC)
Collinear(AEB)
Collinear(DEC)
Cocircular(E,DACB)
Conditions (CDL):
Equal(LengthOfLine(AE),6)
Equal(RadiusOfCircle(E),6)
IsCentreOfCircle(E,E)
Square(DACB)
Goal: Value(LengthOfLine(AD))
Answer: 6*sqrt(2)
Theorem sequence:
radius_of_circle_property_length_equal(1,ED,E)
kite_property_diagonal_perpendicular_bisection(1,DACB,E)
right_triangle_judgment_angle(1,AED)
right_triangle_property_pythagorean(1,AED)
Problem: As shown in the diagram, AB=y, AC=12, AD=BD, ∠CAB=x°, DB⊥CB, ADBC is a square. Find the value of y.
问题：如图所示，AB=y，AC=12，AD=BD，∠CAB=x°，DB⊥CB，四边形ADBC是正方形。求y的值。
Construction (CDL):
Shape(AD,DB,BA)
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(AD),LengthOfLine(BD))
Equal(MeasureOfAngle(CAB),x)
PerpendicularBetweenLine(DB,CB)
Square(ADBC)
Goal: Value(y)
Answer: 12*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠BDC=∠CBA, ∠BDC=∠DCB, ∠CDA=x°. Find the value of x.
问题：如图所示，∠BDC=∠CBA，∠BDC=∠DCB，∠CDA=x°。求x的值。
Construction (CDL):
Shape(CB,BD,DC)
Shape(CD,DA)
Collinear(BDA)
Conditions (CDL):
Equal(MeasureOfAngle(BDC),MeasureOfAngle(CBA))
Equal(MeasureOfAngle(BDC),MeasureOfAngle(DCB))
Equal(MeasureOfAngle(CDA),x)
Goal: Value(x)
Answer: 120
Theorem sequence:
triangle_property_angle_sum(1,CBD)
flat_angle(1,BDA)
angle_addition(1,BDC,CDA)
Problem: As shown in the diagram, AC=16, DB=6, EB=12, AE is perpendicular to BE, CBDA is a trapezoid. Find the area of CBDA.
问题：如图所示，AC=16，DB=6，EB=12，AE垂直于BE，CA和BD是梯形CBDA的平行边。求CBDA的面积。
Construction (CDL):
Shape(CB,BE,EC)
Shape(EB,BD,DA,AE)
Collinear(CEA)
Conditions (CDL):
Equal(LengthOfLine(AC),16)
Equal(LengthOfLine(DB),6)
Equal(LengthOfLine(EB),12)
PerpendicularBetweenLine(AE,BE)
Trapezoid(CBDA)
Goal: Value(AreaOfQuadrilateral(CBDA))
Answer: 132
Theorem sequence:
altitude_of_quadrilateral_judgment_right_vertex(2,BE,DACB)
trapezoid_area_formula(1,DACB)
Problem: As shown in the diagram, XW=2*x+9, YX=3*x+6, the center of circle Z is Z, XW is the tangent to circle Z, the tangent to ⊙Z is XY. Find the value of x.
问题：如图所示，XW=2*x+9，YX=3*x+6，Z是圆Z的圆心，⊙O的切线为XW，XY是圆O的切线。求x的值。
Construction (CDL):
Shape(XW,ZYW,YX)
Cocircular(Z,YW)
Conditions (CDL):
Equal(LengthOfLine(XW),2*x+9)
Equal(LengthOfLine(YX),3*x+6)
IsCentreOfCircle(Z,Z)
IsTangentOfCircle(XW,Z)
IsTangentOfCircle(XY,Z)
Goal: Value(x)
Answer: 3
Theorem sequence:
tangent_of_circle_property_length_equal(1,XW,XY,Z)
Problem: As shown in the diagram, AB=x, BC=13, ∠ACB=52°, CB⊥AB. Find the value of x.
问题：如图所示，AB=x，BC=13，∠ACB=52°，CB垂直于AB。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),13)
Equal(MeasureOfAngle(ACB),52)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 13/tan(19*pi/90)
Theorem sequence:
triangle_property_angle_sum(1,CBA)
sine_theorem(1,BAC)
Problem: As shown in the diagram, Div(LengthOfLine(JK)=LengthOfLine(LK)), KJ=5, LJ=13, LK=12, LK is perpendicular to JK. Find the value of x.
问题：如图所示，Div(LengthOfLine(JK)=LengthOfLine(LK))，KJ=5，LJ=13，LK=12，LK垂直于JK。求x的值。
Construction (CDL):
Shape(JL,LK,KJ)
Conditions (CDL):
Equal(Div(LengthOfLine(JK),LengthOfLine(LK)),x)
Equal(LengthOfLine(KJ),5)
Equal(LengthOfLine(LJ),13)
Equal(LengthOfLine(LK),12)
PerpendicularBetweenLine(LK,JK)
Goal: Value(x)
Answer: 5/12
Theorem sequence:
Problem: As shown in the diagram, AB=21, AC=17, EC=8, quadrilateral BDCA is a ▱, AE is perpendicular to CE. Find the area of BDCA.
问题：如图所示，AB=21，AC=17，EC=8，BA和DC是▱BDCA的一组对边，AE垂直于CE。求四边形BDCA的面积。
Construction (CDL):
Shape(BD,DE,EA,AB)
Shape(AE,EC,CA)
Collinear(DEC)
Conditions (CDL):
Equal(LengthOfLine(AB),21)
Equal(LengthOfLine(AC),17)
Equal(LengthOfLine(EC),8)
Parallelogram(BDCA)
PerpendicularBetweenLine(AE,CE)
Goal: Value(AreaOfQuadrilateral(BDCA))
Answer: 315
Theorem sequence:
right_triangle_judgment_angle(1,AEC)
right_triangle_property_pythagorean(1,AEC)
flat_angle(1,DEC)
angle_addition(1,DEA,AEC)
parallelogram_property_opposite_line_equal(1,ABDC)
altitude_of_quadrilateral_judgment_right_vertex(1,AE,BDCA)
parallelogram_area_formula_common(1,BDCA)