SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, XY=15, ZX=25, ZY is perpendicular to XY. Find the measure of ∠XZY.
问题：如图所示，XY=15，ZX=25，ZY⊥XY。求∠XZY的大小。
Construction (CDL):
Shape(XZ,ZY,YX)
Conditions (CDL):
Equal(LengthOfLine(XY),15)
Equal(LengthOfLine(ZX),25)
PerpendicularBetweenLine(ZY,XY)
Goal: Value(MeasureOfAngle(XZY))
Answer: 180*acos(4/5)/pi
Theorem sequence:
right_triangle_judgment_angle(1,ZYX)
right_triangle_property_pythagorean(1,ZYX)
cosine_theorem(1,ZYX)
Problem: As shown in the diagram, the measure of arc GKJ is 194, the center of ⊙G is G, LK is the tangent to circle G. Find the measure of ∠JKL.
问题：如图所示，弧GKJ的角度为194，⊙G的圆心为G，圆O的切线为LK。求∠JKL的大小。
Construction (CDL):
Shape(GJ,JK,KG)
Shape(GKJ,JG,GK)
Shape(GJK,KJ)
Shape(GK,KL)
Cocircular(G,JK)
Conditions (CDL):
Equal(MeasureOfArc(GKJ),194)
IsCentreOfCircle(G,G)
IsTangentOfCircle(LK,G)
Goal: Value(MeasureOfAngle(JKL))
Answer: 97
Theorem sequence:
arc_property_center_angle(1,GKJ,G)
round_angle(1,JGK,KGJ)
radius_of_circle_property_length_equal(1,GJ,G)
radius_of_circle_property_length_equal(1,GK,G)
isosceles_triangle_judgment_line_equal(1,GJK)
isosceles_triangle_property_angle_equal(1,GJK)
triangle_property_angle_sum(1,GJK)
tangent_of_circle_property_perpendicular(1,LK,G,G)
angle_addition(1,JKG,GKL)
Problem: As shown in the diagram, AD=ED, AO=13, DE=9, OJ=10, x=Add(AreaOfTriangle(AED), OJEA is a rectangle. Find the value of x.
问题：如图所示，AD=ED，AO=13，DE=9，OJ=10，x=Add(AreaOfTriangle(AED)，四边形OJEA是矩形。求x的值。
Construction (CDL):
Shape(OJ,JE,EA,AO)
Shape(AE,ED,DA)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(ED))
Equal(LengthOfLine(AO),13)
Equal(LengthOfLine(DE),9)
Equal(LengthOfLine(OJ),10)
Equal(x,Add(AreaOfTriangle(AED),AreaOfQuadrilateral(OJEA)))
Rectangle(OJEA)
Goal: Value(x)
Answer: 10*sqrt(14)+130
Theorem sequence:
parallelogram_area_formula_sine(1,OJEA)
parallelogram_area_formula_sine(1,JEAO)
parallelogram_area_formula_sine(1,AOJE)
cosine_theorem(1,AED)
triangle_area_formula_sine(1,AED)
Problem: As shown in the diagram, FK=16, KA=9, AD is perpendicular to FD, DF is perpendicular to KF, FK is perpendicular to AK, KA is perpendicular to DA. Find the perimeter of quadrilateral DFKA.
问题：如图所示，FK=16，KA=9，AD⊥FD，DF垂直于KF，FK垂直于AK，KA⊥DA。求四边形DFKA的周长。
Construction (CDL):
Shape(AD,DF,FK,KA)
Conditions (CDL):
Equal(LengthOfLine(FK),16)
Equal(LengthOfLine(KA),9)
PerpendicularBetweenLine(AD,FD)
PerpendicularBetweenLine(DF,KF)
PerpendicularBetweenLine(FK,AK)
PerpendicularBetweenLine(KA,DA)
Goal: Value(PerimeterOfQuadrilateral(DFKA))
Answer: 50
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,DA,FK)
parallel_judgment_ipsilateral_internal_angle(1,FD,KA)
parallelogram_judgment_parallel_and_parallel(1,ADFK)
parallelogram_property_opposite_line_equal(1,ADFK)
parallelogram_property_opposite_line_equal(1,DFKA)
quadrilateral_perimeter_formula(1,ADFK)
Problem: As shown in the diagram, BA=9*x-1, CA=4*x+1, CA=BC, CB=5*x-1/2, △ABC is an isosceles △. Find the length of line AB.
问题：如图所示，BA=9*x-1，CA=4*x+1，CA=BC，CB=5*x-1/2，△ABC为等腰△。求直线AB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(ABC)
Goal: Value(LengthOfLine(AB))
Answer: 25/2
Theorem sequence:
Problem: As shown in the diagram, AE=9, BE=3*x-7, DE=x+5, ADCB is a parallelogram. Find the value of x.
问题：如图所示，AE=9，BE=3*x-7，DE=x+5，四边形ADCB是平行四边形。求x的值。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DC,CE)
Shape(EC,CB,BE)
Shape(AE,EB,BA)
Collinear(AEC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(AE),9)
Equal(LengthOfLine(BE),3*x-7)
Equal(LengthOfLine(DE),x+5)
Parallelogram(ADCB)
Goal: Value(x)
Answer: 6
Theorem sequence:
parallelogram_property_diagonal_bisection(1,DCBA,E)
Problem: As shown in the diagram, ∠BAF=2*x+20°, ∠FAD=5*x-4°, quadrilateral BADC is a rhombus. Find the value of x.
问题：如图所示，∠BAF=2*x+20°，∠FAD=5*x-4°，四边形BADC是菱形。求x的值。
Construction (CDL):
Shape(BA,AF,FB)
Shape(FA,AD,DF)
Shape(FD,DC,CF)
Shape(BF,FC,CB)
Collinear(BFD)
Collinear(AFC)
Conditions (CDL):
Equal(MeasureOfAngle(BAF),2*x+20)
Equal(MeasureOfAngle(FAD),5*x-4)
Rhombus(BADC)
Goal: Value(x)
Answer: 8
Theorem sequence:
parallelogram_property_diagonal_bisection(1,BADC,F)
mirror_congruent_triangle_judgment_sss(1,ADF,AFB)
mirror_congruent_triangle_property_angle_equal(1,ADF,AFB)
Problem: As shown in the diagram, VW=3*x-6, WX=x+4, YW=5, ZW=6, YX is parallel to VZ. Find the length of line VW.
问题：如图所示，VW=3*x-6，WX=x+4，YW=5，ZW=6，YX∥VZ。求直线VW的长度。
Construction (CDL):
Shape(VZ,ZW,WV)
Shape(WX,XY,YW)
Collinear(VWX)
Collinear(ZWY)
Conditions (CDL):
Equal(LengthOfLine(VW),3*x-6)
Equal(LengthOfLine(WX),x+4)
Equal(LengthOfLine(YW),5)
Equal(LengthOfLine(ZW),6)
ParallelBetweenLine(YX,VZ)
Goal: Value(LengthOfLine(VW))
Answer: 12
Theorem sequence:
parallel_property_alternate_interior_angle(1,YX,VZ)
parallel_property_alternate_interior_angle(2,YX,VZ)
similar_triangle_judgment_aa(1,WVZ,WXY)
similar_triangle_property_line_ratio(1,ZWV,YWX)
similar_triangle_property_line_ratio(1,VZW,XYW)
Problem: As shown in the diagram, the length of arc KJL is 56, ∠LKJ=340°, K is the center of ⊙K. Find the radius of circle K.
问题：如图所示，弧KJL的长度为56，∠LKJ=340°，圆K的圆心为K。求⊙K的半径。
Construction (CDL):
Shape(KLJ,JK,KL)
Shape(KJL,LK,KJ)
Cocircular(K,JL)
Conditions (CDL):
Equal(LengthOfArc(KJL),56)
Equal(MeasureOfAngle(LKJ),340)
IsCentreOfCircle(K,K)
Goal: Value(RadiusOfCircle(K))
Answer: 504/(17*pi)
Theorem sequence:
arc_length_formula(1,KJL)
arc_property_center_angle(1,KJL,K)
Problem: As shown in the diagram, RS=TS, TR=14, SX⊥RX. Find the length of line XT.
问题：如图所示，RS=TS，TR=14，SX⊥RX。求直线XT的长度。
Construction (CDL):
Shape(ST,TX,XS)
Shape(SX,XR,RS)
Collinear(TXR)
Conditions (CDL):
Equal(LengthOfLine(RS),LengthOfLine(TS))
Equal(LengthOfLine(TR),14)
PerpendicularBetweenLine(SX,RX)
Goal: Value(LengthOfLine(XT))
Answer: 7
Theorem sequence:
adjacent_complementary_angle(1,TXS,SXR)
mirror_congruent_triangle_judgment_hl(1,TXS,RSX)
mirror_congruent_triangle_property_line_equal(1,STX,SXR)
line_addition(1,RX,XT)
Problem: As shown in the diagram, ∠SRQ=110°, RS∥QT. Find the measure of ∠RQT.
问题：如图所示，∠SRQ=110°，RS平行于QT。求∠RQT的大小。
Construction (CDL):
Shape(RQ,QT,TS,SR)
Conditions (CDL):
Equal(MeasureOfAngle(SRQ),110)
ParallelBetweenLine(RS,QT)
Goal: Value(MeasureOfAngle(RQT))
Answer: 70
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,RS,QT)
Problem: As shown in the diagram, GF=3*x+10, HF=6*x+1, HG=9*x-8, △FHG is an equilateral △. Find the length of line HG.
问题：如图所示，GF=3*x+10，HF=6*x+1，HG=9*x-8，△FHG为等边△。求直线HG的长度。
Construction (CDL):
Shape(HG,GF,FH)
Conditions (CDL):
Equal(LengthOfLine(GF),3*x+10)
Equal(LengthOfLine(HF),6*x+1)
Equal(LengthOfLine(HG),9*x-8)
EquilateralTriangle(FHG)
Goal: Value(LengthOfLine(HG))
Answer: 19
Theorem sequence:
Problem: As shown in the diagram, ∠EBD=60°, the measure of ⌒CFG is 5*x, the measure of ⌒CJF is 105, BF is the tangent to ⊙C, BD⊥HD, FB is perpendicular to DB, HE is perpendicular to CE. Find the value of x.
问题：如图所示，∠EBD=60°，⌒CFG的角度为5*x，弧CJF的角度为105，圆O的切线为BF，BD⊥HD，FB垂直于DB，HE垂直于CE。求x的值。
Construction (CDL):
Shape(BD,DH,CGH,GB)
Shape(CGH,HE,EG)
Shape(CHJ,JE,EH)
Shape(CE,EJ,JC)
Shape(CJ,CJF,FC)
Shape(GE,EC,CG)
Shape(GC,CF,CFG)
Shape(FB,BG,CFG)
Collinear(BGEJA)
Collinear(FCE)
Collinear(DHE)
Cocircular(C,FGHJ)
Conditions (CDL):
Equal(MeasureOfAngle(EBD),60)
Equal(MeasureOfArc(CFG),5*x)
Equal(MeasureOfArc(CJF),105)
IsTangentOfCircle(BF,C)
PerpendicularBetweenLine(BD,HD)
PerpendicularBetweenLine(FB,DB)
PerpendicularBetweenLine(HE,CE)
Goal: Value(x)
Answer: 9
Theorem sequence:
angle_addition(1,FBG,GBD)
circle_property_circular_power_tangent_and_segment_angle(1,BF,BGJ,C)
Problem: As shown in the diagram, AC=8, CB=15, D is the center of ⊙D. Find the diameter of circle D.
问题：如图所示，AC=8，CB=15，⊙D的圆心为D。求圆D的直径。
Construction (CDL):
Shape(DBA,AD,DB)
Shape(DAC,CA)
Shape(DA,AC,CB,BD)
Shape(DCB,BC)
Collinear(ADB)
Cocircular(D,ACB)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(CB),15)
IsCentreOfCircle(D,D)
Goal: Value(DiameterOfCircle(D))
Answer: 17
Theorem sequence:
diameter_of_circle_judgment_pass_centre(1,ADB,D)
diameter_of_circle_property_right_angle(1,ACB,D)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
diameter_of_circle_property_length_equal(1,AB,D)
Problem: As shown in the diagram, QR=11, ST=y-4, TQ=8, QT and RS are opposite sides of the parallelogram TQRS. Find the value of y.
问题：如图所示，QR=11，ST=y-4，TQ=8，四边形TQRS是平行四边形。求y的值。
Construction (CDL):
Shape(TQ,QR,RS,ST)
Conditions (CDL):
Equal(LengthOfLine(QR),11)
Equal(LengthOfLine(ST),y-4)
Equal(LengthOfLine(TQ),8)
Parallelogram(TQRS)
Goal: Value(y)
Answer: 15
Theorem sequence:
parallelogram_property_opposite_line_equal(1,QRST)
Problem: As shown in the diagram, AD=15, CB=6, ES=28, SA=17, AD⊥ED, SB is perpendicular to CB. Find the sum of the area of △CSA and the area of △ASE.
问题：如图所示，AD=15，CB=6，ES=28，SA=17，AD垂直于ED，SB⊥CB。求三角形CSA的面积与三角形ASE的面积之和。
Construction (CDL):
Shape(CS,SB,BC)
Shape(CB,BA,AC)
Shape(AS,SD,DA)
Shape(AD,DE,EA)
Collinear(SBA)
Collinear(SDE)
Conditions (CDL):
Equal(LengthOfLine(AD),15)
Equal(LengthOfLine(CB),6)
Equal(LengthOfLine(ES),28)
Equal(LengthOfLine(SA),17)
PerpendicularBetweenLine(AD,ED)
PerpendicularBetweenLine(SB,CB)
Goal: Value(Add(AreaOfTriangle(CSA),AreaOfTriangle(ASE)))
Answer: 261
Theorem sequence:
adjacent_complementary_angle(1,SDA,ADE)
altitude_of_triangle_judgment(1,CB,CSA)
altitude_of_triangle_judgment(1,AD,ASE)
triangle_area_formula_common(1,CSA)
triangle_area_formula_common(1,ASE)
Problem: As shown in the diagram, AB=15, AD=x, BC=10, CD=4, ∠ABD=∠DBC. Find the value of x.
问题：如图所示，AB=15，AD=x，BC=10，CD=4，∠ABD=∠DBC。求x的值。
Construction (CDL):
Shape(CD,DB,BC)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BC),10)
Equal(LengthOfLine(CD),4)
Equal(MeasureOfAngle(ABD),MeasureOfAngle(DBC))
Goal: Value(x)
Answer: 6
Theorem sequence:
bisector_of_angle_judgment_angle_equal(1,BD,ABC)
bisector_of_angle_property_line_ratio(1,BD,ABC)
Problem: As shown in the diagram, ∠BFE=30°, ∠BGE=52°, the measure of ⌒OBA is 108, the measure of arc OFE is 118, the center of circle O is O. Find the measure of ∠BDE.
问题：如图所示，∠BFE=30°，∠BGE=52°，⌒OBA的角度为108，⌒OFE的角度为118，O是圆O的圆心。求∠BDE的大小。
Construction (CDL):
Shape(DC,OAC,AD)
Shape(OCF,FG,GC)
Shape(GF,FE,EG)
Shape(OFE,EF)
Shape(OEB,BG,GE)
Shape(OAC,CO,OA)
Shape(CG,GO,OC)
Shape(AO,OB,BO)
Shape(OBA,AB)
Collinear(DAB)
Collinear(DCGE)
Collinear(BOGF)
Cocircular(O,ACFEB)
Conditions (CDL):
Equal(MeasureOfAngle(BFE),30)
Equal(MeasureOfAngle(BGE),52)
Equal(MeasureOfArc(OBA),108)
Equal(MeasureOfArc(OFE),118)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(BDE))
Answer: 16
Theorem sequence:
flat_angle(1,BGF)
angle_addition(1,BGE,EGF)
triangle_property_angle_sum(1,GFE)
arc_property_circumference_angle_external(1,OCF,E)
arc_property_circumference_angle_external(1,OEB,F)
flat_angle(1,FOB)
arc_property_center_angle(1,OAC,O)
arc_property_center_angle(1,OBA,O)
arc_property_center_angle(1,OCF,O)
angle_addition(1,FOC,COB)
angle_addition(1,COA,AOB)
circle_property_circular_power_segment_and_segment_angle(1,DAB,DCE,O)
Problem: As shown in the diagram, SU=20, YW=20, the measure of arc RYX is 45, the center of circle R is R, RV is perpendicular to SV, YZ⊥RZ. Find the measure of ⌒RUS.
问题：如图所示，SU=20，YW=20，弧RYX的角度为45，⊙R的圆心为R，RV垂直于SV，YZ垂直于RZ。求⌒RUS的角度。
Construction (CDL):
Shape(RTS,SV,VT)
Shape(VS,SR,RV)
Shape(RSY,YR,RS)
Shape(RY,YZ,ZR)
Shape(RYX,XZ,ZY)
Shape(RXW,WZ,ZX)
Shape(RZ,ZW,RWU,UR)
Shape(VR,RU,UV)
Shape(RUT,TV,VU)
Collinear(SVU)
Collinear(YZW)
Collinear(RZX)
Collinear(RVT)
Cocircular(R,TSYXWU)
Conditions (CDL):
Equal(LengthOfLine(SU),20)
Equal(LengthOfLine(YW),20)
Equal(MeasureOfArc(RYX),45)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(RV,SV)
PerpendicularBetweenLine(YZ,RZ)
Goal: Value(MeasureOfArc(RUS))
Answer: 90
Theorem sequence:
radius_of_circle_property_length_equal(1,RS,R)
radius_of_circle_property_length_equal(1,RY,R)
radius_of_circle_property_length_equal(1,RU,R)
arc_property_center_angle(1,RYX,R)
adjacent_complementary_angle(1,UVR,RVS)
circle_property_chord_perpendicular_bisect_chord(1,R,RV,US)
circle_property_chord_perpendicular_bisect_chord(1,R,RZ,YW)
line_addition(1,SV,VU)
line_addition(1,YZ,ZW)
mirror_congruent_triangle_judgment_hl(2,YZR,SRV)
congruent_triangle_judgment_hl(1,YZR,UVR)
mirror_congruent_triangle_property_angle_equal(1,RYZ,RVS)
congruent_triangle_property_angle_equal(1,RYZ,RUV)
angle_addition(1,SRV,VRU)
arc_property_center_angle(1,RUS,R)
Problem: As shown in the diagram, AC=3*x+2, AF=4*x-6, EC=ED, AF is perpendicular to DF, CA is perpendicular to EA. Find the value of x.
问题：如图所示，AC=3*x+2，AF=4*x-6，EC=ED，AF垂直于DF，CA⊥EA。求x的值。
Construction (CDL):
Shape(CA,AE,EC)
Shape(EA,AF,FD,DE)
Collinear(CAF)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AC),3*x+2)
Equal(LengthOfLine(AF),4*x-6)
Equal(LengthOfLine(EC),LengthOfLine(ED))
PerpendicularBetweenLine(AF,DF)
PerpendicularBetweenLine(CA,EA)
Goal: Value(x)
Answer: 8
Theorem sequence:
similar_triangle_judgment_aa(1,ECA,DCF)
line_addition(1,CE,ED)
line_addition(1,CA,AF)
similar_triangle_property_line_ratio(1,ECA,DCF)
similar_triangle_property_line_ratio(1,AEC,FDC)
Problem: As shown in the diagram, AB=z, AC=x, AD=y, BD=6, CD=4, AD is perpendicular to BD, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=z，AC=x，AD=y，BD=6，CD=4，AD垂直于BD，BA⊥CA。求x的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BD),6)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 2*sqrt(10)
Theorem sequence:
adjacent_complementary_angle(1,CDA,ADB)
right_triangle_judgment_angle(1,ADB)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,ADB)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,BAC)
line_addition(1,CD,DB)
Problem: As shown in the diagram, ∠AGC=40°, FG is perpendicular to CG. Find the measure of ∠FGA.
问题：如图所示，∠AGC=40°，FG⊥CG。求∠FGA的大小。
Construction (CDL):
Shape(FG,GA,AF)
Shape(AG,GC,CA)
Collinear(ABG)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),40)
PerpendicularBetweenLine(FG,CG)
Goal: Value(MeasureOfAngle(FGA))
Answer: 50
Theorem sequence:
angle_addition(1,FGA,AGC)
Problem: As shown in the diagram, RQ=RS, ∠RQS=75°. Find the measure of ∠SRQ.
问题：如图所示，RQ=RS，∠RQS=75°。求∠SRQ的大小。
Construction (CDL):
Shape(RQ,QS,SR)
Conditions (CDL):
Equal(LengthOfLine(RQ),LengthOfLine(RS))
Equal(MeasureOfAngle(RQS),75)
Goal: Value(MeasureOfAngle(SRQ))
Answer: 30
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,RQS)
isosceles_triangle_property_angle_equal(1,RQS)
triangle_property_angle_sum(1,RQS)
Problem: As shown in the diagram, AD=5, EA=3, EX=x, XB=10. Find the value of x.
问题：如图所示，AD=5，EA=3，EX=x，XB=10。求x的值。
Construction (CDL):
Shape(CXB,BX)
Shape(XB,CBD,DA,CAX)
Shape(CDA,AD)
Shape(CAX,AE,EX)
Collinear(BXE)
Collinear(DAE)
Cocircular(C,XBDA)
Conditions (CDL):
Equal(LengthOfLine(AD),5)
Equal(LengthOfLine(EA),3)
Equal(LengthOfLine(EX),x)
Equal(LengthOfLine(XB),10)
Goal: Value(x)
Answer: 2
Theorem sequence:
line_addition(1,BX,XE)
line_addition(1,DA,AE)
circle_property_circular_power_segment_and_segment_line(1,EXB,EAD,C)
Problem: As shown in the diagram, UR=16, US=x, VR=11, VS=7, RS is the angle bisector of ∠URV. Find the value of x.
问题：如图所示，UR=16，US=x，VR=11，VS=7，RS平分∠URV。求x的值。
Construction (CDL):
Shape(VS,SR,RV)
Shape(RS,SU,UR)
Collinear(VSU)
Conditions (CDL):
Equal(LengthOfLine(UR),16)
Equal(LengthOfLine(US),x)
Equal(LengthOfLine(VR),11)
Equal(LengthOfLine(VS),7)
IsBisectorOfAngle(RS,URV)
Goal: Value(x)
Answer: 112/11
Theorem sequence:
bisector_of_angle_property_line_ratio(1,RS,URV)
Problem: As shown in the diagram, AX=x+4, AY=x+9, WA=x, ZA=x+3. Find the value of x.
问题：如图所示，AX=x+4，AY=x+9，WA=x，ZA=x+3。求x的值。
Construction (CDL):
Shape(PZY,YA,AZ)
Shape(PYX,XA,AY)
Shape(PXW,WA,AX)
Shape(PWZ,ZA,AW)
Collinear(ZAX)
Collinear(YAW)
Cocircular(P,ZYXW)
Conditions (CDL):
Equal(LengthOfLine(AX),x+4)
Equal(LengthOfLine(AY),x+9)
Equal(LengthOfLine(WA),x)
Equal(LengthOfLine(ZA),x+3)
Goal: Value(x)
Answer: 6
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,ZAX,YAW,P)
Problem: As shown in the diagram, ∠BTS=3*x-5°, ∠VUS=2*x+15°. Find the measure of ∠VTS.
问题：如图所示，∠BTS=3*x-5°，∠VUS=2*x+15°。求∠VTS的大小。
Construction (CDL):
Shape(ATS,ST)
Shape(TS,SB,BT)
Shape(ASV,VB,BS)
Shape(BV,VU,UB)
Shape(AVU,UV)
Shape(AUT,TB,BU)
Collinear(TBV)
Collinear(UBS)
Cocircular(A,UTSV)
Conditions (CDL):
Equal(MeasureOfAngle(BTS),3*x-5)
Equal(MeasureOfAngle(VUS),2*x+15)
Goal: Value(MeasureOfAngle(VTS))
Answer: 55
Theorem sequence:
arc_property_circumference_angle_external(1,ASV,U)
arc_property_circumference_angle_external(1,ASV,T)
Problem: As shown in the diagram, ∠AGC=60°, DG is perpendicular to AG. Find the measure of ∠DGA.
问题：如图所示，∠AGC=60°，DG⊥AG。求∠DGA的大小。
Construction (CDL):
Shape(CG,GB,GBC)
Shape(GC,GCA,AG)
Shape(DG,GA,GAD)
Shape(GD,GDE,EG)
Shape(GE,GEB,BG)
Collinear(AGB)
Collinear(CGE)
Cocircular(G,CADEB)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),60)
PerpendicularBetweenLine(DG,AG)
Goal: Value(MeasureOfAngle(DGA))
Answer: 90
Theorem sequence:
Problem: As shown in the diagram, the center of circle L is L. Find the measure of arc LHK.
问题：如图所示，⊙L的圆心为L。求⌒LHK的角度。
Construction (CDL):
Shape(LHK,KH)
Shape(LKH,HK)
Collinear(HLK)
Cocircular(L,HK)
Conditions (CDL):
IsCentreOfCircle(L,L)
Goal: Value(MeasureOfArc(LHK))
Answer: 180
Theorem sequence:
flat_angle(1,KLH)
arc_property_center_angle(1,LHK,L)
Problem: As shown in the diagram, ∠HBF=38°, NO∥MH. Find the measure of ∠OCB.
问题：如图所示，∠HBF=38°，NO∥MH。求∠OCB的大小。
Construction (CDL):
Shape(OC,CB)
Shape(HB,BF)
Shape(CK,KB,BC)
Collinear(JCBF)
Collinear(ICKG)
Collinear(NCO)
Collinear(MKBH)
Conditions (CDL):
Equal(MeasureOfAngle(HBF),38)
ParallelBetweenLine(NO,MH)
Goal: Value(MeasureOfAngle(OCB))
Answer: 38
Theorem sequence:
parallel_property_collinear_extend(3,NO,MH,C)
parallel_property_collinear_extend(3,HM,OC,B)
parallel_property_corresponding_angle(2,CO,BH,F)
Problem: As shown in the diagram, the measure of ⌒AJH is 114, A is the center of circle A. Find the measure of ∠AJH.
问题：如图所示，弧AJH的角度为114，圆A的圆心为A。求∠AJH的大小。
Construction (CDL):
Shape(AHJ,JA,AH)
Shape(AJ,JH,HA)
Shape(AJH,HJ)
Cocircular(A,JH)
Conditions (CDL):
Equal(MeasureOfArc(AJH),114)
IsCentreOfCircle(A,A)
Goal: Value(MeasureOfAngle(AJH))
Answer: 33
Theorem sequence:
arc_property_center_angle(1,AJH,A)
triangle_property_angle_sum(1,AJH)
radius_of_circle_property_length_equal(1,AJ,A)
radius_of_circle_property_length_equal(1,AH,A)
isosceles_triangle_judgment_line_equal(1,AJH)
isosceles_triangle_property_angle_equal(1,AJH)
Problem: As shown in the diagram, ∠RQS=33°, ∠TSQ=3*z°, quadrilateral QTSR is a parallelogram. Find the value of z.
问题：如图所示，∠RQS=33°，∠TSQ=3*z°，QTSR是平行四边形。求z的值。
Construction (CDL):
Shape(QT,TP,PQ)
Shape(PT,TS,SP)
Shape(PS,SR,RP)
Shape(PR,RQ,QP)
Collinear(QPS)
Collinear(RPT)
Conditions (CDL):
Equal(MeasureOfAngle(RQS),33)
Equal(MeasureOfAngle(TSQ),3*z)
Parallelogram(QTSR)
Goal: Value(z)
Answer: 11
Theorem sequence:
parallel_property_alternate_interior_angle(1,QR,TS)
Problem: As shown in the diagram, ∠MLJ=30°, ∠RLK=70°, JMLK is a ▱. Find the measure of ∠JML.
问题：如图所示，∠MLJ=30°，∠RLK=70°，四边形JMLK是平行四边形。求∠JML的大小。
Construction (CDL):
Shape(JM,MR,RJ)
Shape(RM,ML,LR)
Shape(RL,LK,KR)
Shape(JR,RK,KJ)
Collinear(JRL)
Collinear(MRK)
Conditions (CDL):
Equal(MeasureOfAngle(MLJ),30)
Equal(MeasureOfAngle(RLK),70)
Parallelogram(JMLK)
Goal: Value(MeasureOfAngle(JML))
Answer: 80
Theorem sequence:
angle_addition(1,MLR,RLK)
parallel_property_ipsilateral_internal_angle(1,MJ,LK)
Problem: As shown in the diagram, ∠AFD=50°, the center of circle F is F, CF⊥BF. Find the measure of ⌒FEC.
问题：如图所示，∠AFD=50°，F是⊙F的圆心，CF垂直于BF。求⌒FEC的角度。
Construction (CDL):
Shape(FCD,DF,FC)
Shape(FDA,AF,FD)
Shape(FAE,EF,FA)
Shape(FEB,BF,FE)
Shape(FBC,CF,FB)
Collinear(AFB)
Collinear(DFE)
Cocircular(F,CDAEB)
Conditions (CDL):
Equal(MeasureOfAngle(AFD),50)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(CF,BF)
Goal: Value(MeasureOfArc(FEC))
Answer: 140
Theorem sequence:
vertical_angle(1,AFD,BFE)
angle_addition(1,CFB,BFE)
arc_property_center_angle(1,FEC,F)
Problem: As shown in the diagram, quadrilateral EDCB is congruent to quadrilateral TURS, CB=4*w-7, CD=11, ED=3*z+10, RS=2*w+13, RU=12, UT=z+16, ∠CBE=2*x+9°, ∠EDC=2*y-31°, ∠STU=y+11°, ∠URS=49°. Find the value of z.
问题：如图所示，四边形EDCB与四边形TURS全等，CB=4*w-7，CD=11，ED=3*z+10，RS=2*w+13，RU=12，UT=z+16，∠CBE=2*x+9°，∠EDC=2*y-31°，∠STU=y+11°，∠URS=49°。求z的值。
Construction (CDL):
Shape(BE,ED,DC,CB)
Shape(UR,RS,ST,TU)
Conditions (CDL):
CongruentBetweenQuadrilateral(EDCB,TURS)
Equal(LengthOfLine(CB),4*w-7)
Equal(LengthOfLine(CD),11)
Equal(LengthOfLine(ED),3*z+10)
Equal(LengthOfLine(RS),2*w+13)
Equal(LengthOfLine(RU),12)
Equal(LengthOfLine(UT),z+16)
Equal(MeasureOfAngle(CBE),2*x+9)
Equal(MeasureOfAngle(EDC),2*y-31)
Equal(MeasureOfAngle(STU),y+11)
Equal(MeasureOfAngle(URS),49)
Goal: Value(z)
Answer: 3
Theorem sequence:
congruent_quadrilateral_property_line_equal(1,EDCB,TURS)
Problem: As shown in the diagram, AH=4, BC=6, HC=BH, AH⊥BH. Find the area of triangle BAC.
问题：如图所示，AH=4，BC=6，HC=BH，AH垂直于BH。求三角形BAC的面积。
Construction (CDL):
Shape(CH,HA,AC)
Shape(AH,HB,BA)
Collinear(CHB)
Conditions (CDL):
Equal(LengthOfLine(AH),4)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(HC),LengthOfLine(BH))
PerpendicularBetweenLine(AH,BH)
Goal: Value(AreaOfTriangle(BAC))
Answer: 12
Theorem sequence:
adjacent_complementary_angle(1,CHA,AHB)
altitude_of_triangle_judgment(1,AH,ACB)
triangle_area_formula_common(1,ACB)
Problem: As shown in the diagram, AB=17, AC=7, BC=x, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AB=17，AC=7，BC=x，CA垂直于BA。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),17)
Equal(LengthOfLine(AC),7)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 13*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, ∠BFC=72°, ∠CEB=3*y°, ∠EBF=4*x°, ∠FCE=112°, BE is parallel to FC. Find the value of x.
问题：如图所示，∠BFC=72°，∠CEB=3*y°，∠EBF=4*x°，∠FCE=112°，BE平行于FC。求x的值。
Construction (CDL):
Shape(AF,FB)
Shape(BF,FC,CE,EB)
Collinear(AFC)
Conditions (CDL):
Equal(MeasureOfAngle(BFC),72)
Equal(MeasureOfAngle(CEB),3*y)
Equal(MeasureOfAngle(EBF),4*x)
Equal(MeasureOfAngle(FCE),112)
ParallelBetweenLine(BE,FC)
Goal: Value(x)
Answer: 27
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,BE,FC)
Problem: As shown in the diagram, ∠BED=3*x°, ∠DCB=5*x-12°, B is the center of circle B, the diameter of circle B is EC. Find the measure of ∠DCB.
问题：如图所示，∠BED=3*x°，∠DCB=5*x-12°，圆B的圆心为B，⊙B的直径为EC。求∠DCB的大小。
Construction (CDL):
Shape(BDC,CD)
Shape(BC,BCE,EB)
Shape(DC,CB,BE,ED)
Shape(BED,DE)
Collinear(CBE)
Cocircular(B,DCE)
Conditions (CDL):
Equal(MeasureOfAngle(BED),3*x)
Equal(MeasureOfAngle(DCB),5*x-12)
IsCentreOfCircle(B,B)
IsDiameterOfCircle(EC,B)
Goal: Value(MeasureOfAngle(DCB))
Answer: 207/4
Theorem sequence:
diameter_of_circle_property_right_angle(1,EDC,B)
triangle_property_angle_sum(1,DCE)
Problem: As shown in the diagram, AB=z, AC=x, AY=17, BC=y, CY=6, AB⊥AB, AY is perpendicular to BY. Find the value of x.
问题：如图所示，AB=z，AC=x，AY=17，BC=y，CY=6，AB垂直于AB，AY垂直于BY。求x的值。
Construction (CDL):
Shape(AC,CY,YA)
Shape(AY,YB,BA)
Collinear(CYB)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AY),17)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CY),6)
PerpendicularBetweenLine(AB,AC)
PerpendicularBetweenLine(AY,BY)
Goal: Value(x)
Answer: 5*sqrt(13)
Theorem sequence:
adjacent_complementary_angle(1,CYA,AYB)
right_triangle_judgment_angle(1,CYA)
right_triangle_property_pythagorean(1,CYA)