SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, WX=x+4, ZY=2*x+3, ZW⊥XW, WXYZ is a rectangle. Find the length of line WX.
问题：如图所示，WX=x+4，ZY=2*x+3，ZW⊥XW，WXYZ是长方形。求直线WX的长度。
Construction (CDL):
Shape(ZW,WP,PZ)
Shape(PW,WX,XP)
Shape(PX,XY,YP)
Shape(ZP,PY,YZ)
Collinear(ZPX)
Collinear(WPY)
Conditions (CDL):
Equal(LengthOfLine(WX),x+4)
Equal(LengthOfLine(ZY),2*x+3)
PerpendicularBetweenLine(ZW,XW)
Rectangle(WXYZ)
Goal: Value(LengthOfLine(WX))
Answer: 5
Theorem sequence:
parallelogram_property_opposite_line_equal(1,WXYZ)
Problem: As shown in the diagram, CD=BD, ∠BCD=66°, ∠DAB=24°. Find the measure of ∠ABD.
问题：如图所示，CD=BD，∠BCD=66°，∠DAB=24°。求∠ABD的大小。
Construction (CDL):
Shape(DA,AB,BD)
Shape(CD,DB,BC)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(CD),LengthOfLine(BD))
Equal(MeasureOfAngle(BCD),66)
Equal(MeasureOfAngle(DAB),24)
Goal: Value(MeasureOfAngle(ABD))
Answer: 24
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DBC)
isosceles_triangle_property_angle_equal(1,DBC)
triangle_property_angle_sum(1,CDB)
flat_angle(1,CDA)
angle_addition(1,CDB,BDA)
triangle_property_angle_sum(1,BDA)
Problem: As shown in the diagram, ∠NMR=x°, the measure of ⌒CNR is 62, the measure of ⌒CPN is 108, C is the center of ⊙C, the tangent to circle C is MR. Find the value of x.
问题：如图所示，∠NMR=x°，⌒CNR的角度为62，⌒CPN的角度为108，C是⊙C的圆心，MR是圆O的切线。求x的值。
Construction (CDL):
Shape(NP,CPN)
Shape(NC,CP,PN)
Shape(PC,CR,CRP)
Shape(RC,CN,CNR)
Shape(NM,MR,CRN)
Shape(NM,MR,RC,CN)
Collinear(MNP)
Cocircular(C,NRP)
Conditions (CDL):
Equal(MeasureOfAngle(NMR),x)
Equal(MeasureOfArc(CNR),62)
Equal(MeasureOfArc(CPN),108)
IsCentreOfCircle(C,C)
IsTangentOfCircle(MR,C)
Goal: Value(x)
Answer: 64
Theorem sequence:
tangent_of_circle_property_perpendicular(2,MR,C,C)
radius_of_circle_property_length_equal(1,CN,C)
radius_of_circle_property_length_equal(1,CP,C)
arc_property_center_angle(1,CPN,C)
isosceles_triangle_judgment_line_equal(1,CPN)
isosceles_triangle_property_angle_equal(1,CPN)
triangle_property_angle_sum(1,CPN)
adjacent_complementary_angle(1,PNC,CNM)
arc_property_center_angle(1,CNR,C)
quadrilateral_property_angle_sum(1,NMRC)
Problem: As shown in the diagram, AC=b, CB=a, EA=y, ∠CAE=30°, ∠EBC=60°, a=10*sqrt(3), AE⊥CE, BC⊥AC. Find the length of line CE.
问题：如图所示，AC=b，CB=a，EA=y，∠CAE=30°，∠EBC=60°，a=10*sqrt(3)，AE垂直于CE，BC⊥AC。求直线CE的长度。
Construction (CDL):
Shape(BC,CE,EB)
Shape(EC,CA,AE)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(CB),a)
Equal(LengthOfLine(EA),y)
Equal(MeasureOfAngle(CAE),30)
Equal(MeasureOfAngle(EBC),60)
Equal(a,10*sqrt(3))
PerpendicularBetweenLine(AE,CE)
PerpendicularBetweenLine(BC,AC)
Goal: Value(LengthOfLine(CE))
Answer: 15
Theorem sequence:
adjacent_complementary_angle(1,AEC,CEB)
sine_theorem(1,CEB)
Problem: As shown in the diagram, ∠HUN=123°, AM∥QE, QI is parallel to UH. Find the measure of ∠EQI.
问题：如图所示，∠HUN=123°，AM平行于QE，QI∥UH。求∠EQI的大小。
Construction (CDL):
Shape(GK,FK)
Shape(FK,KA)
Shape(UK,KG)
Shape(AK,KU)
Shape(NU,UK)
Shape(KU,UQ)
Shape(HU,UN)
Shape(QU,UH)
Shape(KA,AJ)
Shape(JA,AM)
Shape(QA,AK)
Shape(MA,AQ)
Shape(UQ,QA)
Shape(AQ,QE)
Shape(IQ,QU)
Shape(EQ,QI)
Collinear(GKAM)
Collinear(NUQE)
Collinear(FKUH)
Collinear(JAQI)
Conditions (CDL):
Equal(MeasureOfAngle(HUN),123)
ParallelBetweenLine(AM,QE)
ParallelBetweenLine(QI,UH)
Goal: Value(MeasureOfAngle(EQI))
Answer: 57
Theorem sequence:
adjacent_complementary_angle(1,QUH,HUN)
parallel_property_corresponding_angle(1,QI,UH,E)
Problem: As shown in the diagram, AB=y, AD=12, BD=12, AD⊥BD. Find the value of y.
问题：如图所示，AB=y，AD=12，BD=12，AD⊥BD。求y的值。
Construction (CDL):
Shape(CA,AB,BC)
Shape(BA,AD,DB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(BD),12)
PerpendicularBetweenLine(AD,BD)
Goal: Value(y)
Answer: 12*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,ADB)
right_triangle_property_pythagorean(1,ADB)
Problem: As shown in the diagram, ∠FHE=4*x+7°, ∠KMP=5*x-13°, HG is parallel to MK. Find the value of x.
问题：如图所示，∠FHE=4*x+7°，∠KMP=5*x-13°，HG∥MK。求x的值。
Construction (CDL):
Shape(EH,HG)
Shape(GH,HM)
Shape(FH,HE)
Shape(MH,HF)
Shape(HM,MK)
Shape(DM,MH)
Shape(KM,MP)
Shape(PM,MD)
Collinear(GHF)
Collinear(EHMP)
Collinear(KMD)
Conditions (CDL):
Equal(MeasureOfAngle(FHE),4*x+7)
Equal(MeasureOfAngle(KMP),5*x-13)
ParallelBetweenLine(HG,MK)
Goal: Value(x)
Answer: 20
Theorem sequence:
vertical_angle(1,FHE,GHM)
parallel_property_corresponding_angle(2,HG,MK,P)
Problem: As shown in the diagram, ∠BPY=3*x°, ∠CPA=2*x+15°, ∠YPC=3*x-3°, the center of ⊙P is P. Find the measure of arc PBX.
问题：如图所示，∠BPY=3*x°，∠CPA=2*x+15°，∠YPC=3*x-3°，圆P的圆心为P。求⌒PBX的角度。
Construction (CDL):
Shape(XP,PB,PBX)
Shape(YP,PB,PYB)
Shape(YP,PC,PCY)
Shape(CP,PA,PAC)
Shape(AP,PX,PXA)
Cocircular(P,XACYB)
Collinear(YPX)
Collinear(BPA)
Conditions (CDL):
Equal(MeasureOfAngle(BPY),3*x)
Equal(MeasureOfAngle(CPA),2*x+15)
Equal(MeasureOfAngle(YPC),3*x-3)
IsCentreOfCircle(P,P)
Goal: Value(MeasureOfArc(PBX))
Answer: 117
Theorem sequence:
vertical_angle(1,APX,BPY)
angle_addition(1,BPY,YPC)
angle_addition(1,BPC,CPA)
flat_angle(1,BPA)
angle_addition(1,BPA,APX)
round_angle(1,XPB,BPX)
arc_property_center_angle(1,PBX,P)
Problem: As shown in the diagram, ∠DFE=4-5*x°, ∠IEG=7*x+100°, FD is parallel to EI. Find the value of x.
问题：如图所示，∠DFE=4-5*x°，∠IEG=7*x+100°，FD∥EI。求x的值。
Construction (CDL):
Shape(DF,FE)
Shape(AF,FD)
Shape(HF,FA)
Shape(EF,FH)
Shape(FE,EI)
Shape(CE,EF)
Shape(GE,EC)
Shape(IE,EG)
Collinear(DFH)
Collinear(IEC)
Collinear(AFEG)
Conditions (CDL):
Equal(MeasureOfAngle(DFE),4-5*x)
Equal(MeasureOfAngle(IEG),7*x+100)
ParallelBetweenLine(FD,EI)
Goal: Value(x)
Answer: -8
Theorem sequence:
parallel_property_corresponding_angle(2,FD,EI,G)
Problem: As shown in the diagram, QM=12, QP=13, Triangle QPM is mirror similar to triangle RQP., QM⊥RM, RQ⊥PQ. Find the perimeter of triangle RQP.
问题：如图所示，QM=12，QP=13，三角形QPM与三角形RQP是镜像相似三角形，QM⊥RM，RQ垂直于PQ。求三角形RQP的周长。
Construction (CDL):
Shape(QP,PM,MQ)
Shape(QM,MR,RQ)
Collinear(PMR)
Conditions (CDL):
Equal(LengthOfLine(QM),12)
Equal(LengthOfLine(QP),13)
MirrorSimilarBetweenTriangle(QPM,RQP)
PerpendicularBetweenLine(QM,RM)
PerpendicularBetweenLine(RQ,PQ)
Goal: Value(PerimeterOfTriangle(RQP))
Answer: 78
Theorem sequence:
adjacent_complementary_angle(1,PMQ,QMR)
right_triangle_judgment_angle(1,PMQ)
right_triangle_property_pythagorean(1,PMQ)
mirror_similar_triangle_property_line_ratio(1,QPM,RQP)
mirror_similar_triangle_property_line_ratio(1,PMQ,PRQ)
mirror_similar_triangle_property_line_ratio(1,MQP,QPR)
triangle_perimeter_formula(1,PRQ)
Problem: As shown in the diagram, HJ=12, HK=8, JG=x, KG=x, G is the center of ⊙G, the tangent to circle G is HJ. Find the value of x.
问题：如图所示，HJ=12，HK=8，JG=x，KG=x，⊙G的圆心为G，HJ是圆O的切线。求x的值。
Construction (CDL):
Shape(JG,GK,GKJ)
Shape(KG,GJ,GJK)
Shape(JH,HK,GKJ)
Collinear(GKH)
Cocircular(G,JK)
Shape(GJ,JH,HK,KG)
Conditions (CDL):
Equal(LengthOfLine(HJ),12)
Equal(LengthOfLine(HK),8)
Equal(LengthOfLine(JG),x)
Equal(LengthOfLine(KG),x)
IsCentreOfCircle(G,G)
IsTangentOfCircle(HJ,G)
Goal: Value(x)
Answer: 5
Theorem sequence:
line_addition(1,GK,KH)
tangent_of_circle_property_perpendicular(1,HJ,G,G)
right_triangle_judgment_angle(1,GJH)
right_triangle_property_pythagorean(1,GJH)
Problem: As shown in the diagram, GH=10, the center of ⊙G is G, the center of circle J is J, the center of ⊙K is K. Find the length of line JK.
问题：如图所示，GH=10，⊙G的圆心为G，⊙J的圆心为J，⊙K的圆心为K。求直线JK的长度。
Construction (CDL):
Shape(BG,GH,GHB)
Shape(FG,GB,GBF)
Shape(HG,JGK,GLH)
Shape(GF,GFL,JGL)
Shape(JG,JGL,KJL)
Shape(GJ,KLJ,JLG)
Shape(JK,KL,KLJ)
Shape(LK,KJ,KJL)
Collinear(HGF)
Collinear(BGJKL)
Cocircular(K,JL)
Cocircular(J,GL)
Cocircular(G,FLHB)
Conditions (CDL):
Equal(LengthOfLine(GH),10)
IsCentreOfCircle(G,G)
IsCentreOfCircle(J,J)
IsCentreOfCircle(K,K)
Goal: Value(LengthOfLine(JK))
Answer: 5/2
Theorem sequence:
radius_of_circle_property_length_equal(1,GH,G)
radius_of_circle_property_length_equal(1,GL,G)
radius_of_circle_property_length_equal(1,JG,J)
radius_of_circle_property_length_equal(1,JL,J)
line_addition(1,GJ,JL)
radius_of_circle_property_length_equal(1,KJ,K)
radius_of_circle_property_length_equal(1,KL,K)
line_addition(1,JK,KL)
Problem: As shown in the diagram, ∠WZY=a+2°, ∠XWZ=a/2+8°, ∠YXW=a°, ∠ZYX=a-28°. Find the measure of ∠YXW.
问题：如图所示，∠WZY=a+2°，∠XWZ=a/2+8°，∠YXW=a°，∠ZYX=a-28°。求∠YXW的大小。
Construction (CDL):
Shape(XW,WZ,ZY,YX)
Conditions (CDL):
Equal(MeasureOfAngle(WZY),a+2)
Equal(MeasureOfAngle(XWZ),a/2+8)
Equal(MeasureOfAngle(YXW),a)
Equal(MeasureOfAngle(ZYX),a-28)
Goal: Value(MeasureOfAngle(YXW))
Answer: 108
Theorem sequence:
quadrilateral_property_angle_sum(1,XWZY)
Problem: As shown in the diagram, ∠BGC=40°, ∠DGF=53°, CB is perpendicular to GB, FG is perpendicular to CG, GF⊥DF. Find the measure of ∠FDG.
问题：如图所示，∠BGC=40°，∠DGF=53°，CB垂直于GB，FG垂直于CG，GF⊥DF。求∠FDG的大小。
Construction (CDL):
Shape(DG,GF,FD)
Shape(FG,GA,AF)
Shape(AB,BC,CA)
Shape(BG,GC,CB)
Collinear(DFA)
Collinear(ABG)
Conditions (CDL):
Equal(MeasureOfAngle(BGC),40)
Equal(MeasureOfAngle(DGF),53)
PerpendicularBetweenLine(CB,GB)
PerpendicularBetweenLine(FG,CG)
PerpendicularBetweenLine(GF,DF)
Goal: Value(MeasureOfAngle(FDG))
Answer: 37
Theorem sequence:
triangle_property_angle_sum(1,GFD)
Problem: As shown in the diagram, RT=13, XY=7, ∠ZXY=124°, XZ is the midsegment of △ SRT, YX is the midsegment of triangle RTS. Find the length of line XZ.
问题：如图所示，RT=13，XY=7，∠ZXY=124°，XZ是三角形SRT的中位线，YX是三角形RTS的中位线。求直线XZ的长度。
Construction (CDL):
Shape(SX,XZ,ZS)
Shape(XR,RY,YX)
Shape(XY,YT,TZ,ZX)
Collinear(SXR)
Collinear(SZT)
Collinear(RYT)
Conditions (CDL):
Equal(LengthOfLine(RT),13)
Equal(LengthOfLine(XY),7)
Equal(MeasureOfAngle(ZXY),124)
IsMidsegmentOfTriangle(XZ,SRT)
IsMidsegmentOfTriangle(YX,RTS)
Goal: Value(LengthOfLine(XZ))
Answer: 13/2
Theorem sequence:
line_addition(1,SX,XR)
midsegment_of_triangle_property_parallel(1,XZ,SRT)
parallel_property_corresponding_angle(1,XZ,RT,S)
similar_triangle_judgment_aa(1,ZSX,TSR)
similar_triangle_property_line_ratio(1,ZSX,TSR)
similar_triangle_property_line_ratio(1,SXZ,SRT)
Problem: As shown in the diagram, AB=y, AF=z, AL=sqrt(3), FB=x, FL=2*sqrt(3), FB⊥AB, LA⊥FA. Find the value of y.
问题：如图所示，AB=y，AF=z，AL=sqrt(3)，FB=x，FL=2*sqrt(3)，FB垂直于AB，LA垂直于FA。求y的值。
Construction (CDL):
Shape(FB,BA,AF)
Shape(BL,LA,AB)
Collinear(LBF)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AF),z)
Equal(LengthOfLine(AL),sqrt(3))
Equal(LengthOfLine(FB),x)
Equal(LengthOfLine(FL),2*sqrt(3))
PerpendicularBetweenLine(FB,AB)
PerpendicularBetweenLine(LA,FA)
Goal: Value(y)
Answer: 3/2
Theorem sequence:
right_triangle_judgment_angle(1,LAF)
right_triangle_property_pythagorean(1,LAF)
altitude_of_triangle_judgment(1,AB,AFL)
triangle_area_formula_sine(1,AFL)
triangle_area_formula_common(1,AFL)
Problem: As shown in the diagram, AE=x, BC=4, BE=2, the tangent to ⊙D is BC. Find the value of x.
问题：如图所示，AE=x，BC=4，BE=2，圆O的切线为BC。求x的值。
Construction (CDL):
Shape(EA,DAC,DCE)
Shape(CB,BE,DEC)
Shape(AE,DEA)
Collinear(AEB)
Cocircular(D,ACE)
Conditions (CDL):
Equal(LengthOfLine(AE),x)
Equal(LengthOfLine(BC),4)
Equal(LengthOfLine(BE),2)
IsTangentOfCircle(BC,D)
Goal: Value(x)
Answer: 6
Theorem sequence:
line_addition(1,AE,EB)
circle_property_circular_power_tangent_and_segment_line(1,BC,BEA,D)
Problem: As shown in the diagram, KN=9, LN=16, PM=Mul(LengthOfLine(KP),2), LK∥RP, KN⊥MN, LM is perpendicular to KM. Find the length of line MR.
问题：如图所示，KN=9，LN=16，PM=Mul(LengthOfLine(KP),2)，LK∥RP，KN垂直于MN，LM垂直于KM。求直线MR的长度。
Construction (CDL):
Shape(LR,RQ,QN,NL)
Shape(KN,NQ,QP,PK)
Shape(RM,MQ,QR)
Shape(QM,MP,PQ)
Collinear(LNK)
Collinear(RQP)
Collinear(LRM)
Collinear(KPM)
Collinear(NQM)
Conditions (CDL):
Equal(LengthOfLine(KN),9)
Equal(LengthOfLine(LN),16)
Equal(LengthOfLine(PM),Mul(LengthOfLine(KP),2))
ParallelBetweenLine(LK,RP)
PerpendicularBetweenLine(KN,MN)
PerpendicularBetweenLine(LM,KM)
Goal: Value(LengthOfLine(MR))
Answer: 40/3
Theorem sequence:
parallel_property_corresponding_angle(1,PR,KL,M)
parallel_property_corresponding_angle(2,LK,RP,M)
line_addition(1,LN,NK)
line_addition(1,KP,PM)
mirror_similar_triangle_judgment_aa(1,LMK,MKN)
mirror_similar_triangle_property_line_ratio(1,LMK,MKN)
mirror_similar_triangle_property_line_ratio(1,MKL,NMK)
right_triangle_judgment_angle(1,LMK)
right_triangle_property_pythagorean(1,LMK)
similar_triangle_judgment_aa(1,MPR,MKL)
similar_triangle_property_line_ratio(1,PRM,KLM)
similar_triangle_property_line_ratio(1,RMP,LMK)
Problem: As shown in the diagram, ∠FOE=45°, O is the center of circle O, CO is perpendicular to AO, EO⊥BO. Find the measure of arc OBE.
问题：如图所示，∠FOE=45°，⊙O的圆心为O，CO⊥AO，EO⊥BO。求弧OBE的角度。
Construction (CDL):
Shape(FO,OE,OEF)
Shape(AO,OF,OFA)
Shape(CO,OA,OAC)
Shape(BO,OC,OCB)
Shape(EO,OB,OBE)
Collinear(EOC)
Collinear(AOB)
Cocircular(O,EFACB)
Conditions (CDL):
Equal(MeasureOfAngle(FOE),45)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CO,AO)
PerpendicularBetweenLine(EO,BO)
Goal: Value(MeasureOfArc(OBE))
Answer: 90
Theorem sequence:
arc_property_center_angle(1,OBE,O)
Problem: As shown in the diagram, AB=15, CA=25, CB=20, CB is perpendicular to AB. Find the value of sin(AC).
问题：如图所示，AB=15，CA=25，CB=20，CB⊥AB。求sin(AC)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(CA),25)
Equal(LengthOfLine(CB),20)
PerpendicularBetweenLine(CB,AB)
Goal: Value(Sin(MeasureOfAngle(ACB)))
Answer: 3/5
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, ∠GFJ=4*x+2°, ∠JHG=9*x-3°, J is the center of circle J, the diameter of circle J is HF. Find the measure of ∠GFH.
问题：如图所示，∠GFJ=4*x+2°，∠JHG=9*x-3°，圆J的圆心为J，圆J的直径为HF。求∠GFH的大小。
Construction (CDL):
Shape(FJ,JH,HG,GF)
Shape(Hj,JF,JFH)
Shape(FG,JGF)
Shape(GH,JHG)
Collinear(FJH)
Cocircular(J,FHG)
Conditions (CDL):
Equal(MeasureOfAngle(GFJ),4*x+2)
Equal(MeasureOfAngle(JHG),9*x-3)
IsCentreOfCircle(J,J)
IsDiameterOfCircle(HF,J)
Goal: Value(MeasureOfAngle(GFH))
Answer: 30
Theorem sequence:
diameter_of_circle_property_right_angle(1,HGF,J)
triangle_property_angle_sum(1,FHG)
Problem: As shown in the diagram, BA=x+2, CB=6, DA=8, DE=5, EB∥DC. Find the length of line AB.
问题：如图所示，BA=x+2，CB=6，DA=8，DE=5，EB∥DC。求直线AB的长度。
Construction (CDL):
Shape(AE,EB,BA)
Shape(ED,DC,CB,BE)
Collinear(AED)
Collinear(ABC)
Conditions (CDL):
Equal(LengthOfLine(BA),x+2)
Equal(LengthOfLine(CB),6)
Equal(LengthOfLine(DA),8)
Equal(LengthOfLine(DE),5)
ParallelBetweenLine(EB,DC)
Goal: Value(LengthOfLine(AB))
Answer: 18/5
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, ∠ABC=30°, the measure of arc DCA is x, the center of circle D is D, the tangent to ⊙D is BA, BC is the tangent to ⊙D. Find the value of x.
问题：如图所示，∠ABC=30°，⌒DCA的角度为x，D是圆D的圆心，圆O的切线为BA，BC是⊙O的切线。求x的值。
Construction (CDL):
Shape(CD,DA,DAC)
Shape(AD,DC,DCA)
Shape(AB,BC,DCA)
Shape(CD,DA,AB,BC)
Cocircular(D,AC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),30)
Equal(MeasureOfArc(DCA),x)
IsCentreOfCircle(D,D)
IsTangentOfCircle(BA,D)
IsTangentOfCircle(BC,D)
Goal: Value(x)
Answer: 210
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BA,D,D)
tangent_of_circle_property_perpendicular(2,BC,D,D)
quadrilateral_property_angle_sum(1,CDAB)
round_angle(1,CDA,ADC)
arc_property_center_angle(1,DCA,D)
Problem: As shown in the diagram, ID=6, IG=6, ∠DGI=24°, the center of ⊙I is I. Find the measure of ⌒IDG.
问题：如图所示，ID=6，IG=6，∠DGI=24°，圆I的圆心为I。求弧IDG的角度。
Construction (CDL):
Shape(ID,DG,GI)
Shape(GD,IDG)
Shape(DI,IG,IGD)
Cocircular(I,DG)
Conditions (CDL):
Equal(LengthOfLine(ID),6)
Equal(LengthOfLine(IG),6)
Equal(MeasureOfAngle(DGI),24)
IsCentreOfCircle(I,I)
Goal: Value(MeasureOfArc(IDG))
Answer: 132
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,IDG)
isosceles_triangle_property_angle_equal(1,IDG)
triangle_property_angle_sum(1,IDG)
arc_property_center_angle(1,IDG,I)
Problem: As shown in the diagram, ∠BDF=78°, ∠EGF=120°, ∠FBD=50°, ∠GEC=56°. Find the measure of ∠GAF.
问题：如图所示，∠BDF=78°，∠EGF=120°，∠FBD=50°，∠GEC=56°。求∠GAF的大小。
Construction (CDL):
Shape(BD,DF,FB)
Shape(AF,FG,GA)
Shape(GE,EC,CG)
Shape(EG,GF)
Collinear(BFGCH)
Collinear(AFD)
Collinear(AGE)
Conditions (CDL):
Equal(MeasureOfAngle(BDF),78)
Equal(MeasureOfAngle(EGF),120)
Equal(MeasureOfAngle(FBD),50)
Equal(MeasureOfAngle(GEC),56)
Goal: Value(MeasureOfAngle(GAF))
Answer: 68
Theorem sequence:
triangle_property_angle_sum(1,BDF)
vertical_angle(1,DFB,AFG)
flat_angle(1,EGA)
angle_addition(1,EGF,FGA)
triangle_property_angle_sum(1,AFG)
Problem: As shown in the diagram, ∠AEC=74°, ∠BAE=15°, EC is perpendicular to AC. Find the measure of ∠BEA.
问题：如图所示，∠AEC=74°，∠BAE=15°，EC⊥AC。求∠BEA的大小。
Construction (CDL):
Shape(BE,EA,AB)
Shape(EC,CA,AE)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(AEC),74)
Equal(MeasureOfAngle(BAE),15)
PerpendicularBetweenLine(EC,AC)
Goal: Value(MeasureOfAngle(BEA))
Answer: 106
Theorem sequence:
flat_angle(1,BEC)
angle_addition(1,BEA,AEC)
Problem: As shown in the diagram, AB=y, AC=14, BC=x, ∠ABC=60°, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=y，AC=14，BC=x，∠ABC=60°，BC⊥AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ABC),60)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 14*sqrt(3)/3
Theorem sequence:
triangle_property_angle_sum(1,BCA)
sine_theorem(1,BCA)
sine_theorem(1,ABC)
Problem: As shown in the diagram, ∠AED=x°, ∠BEA=110°, ∠CEB=x°, ∠DEC=110°. Find the value of x.
问题：如图所示，∠AED=x°，∠BEA=110°，∠CEB=x°，∠DEC=110°。求x的值。
Construction (CDL):
Shape(CD,DE,EC)
Shape(EA,AB,BE)
Shape(DC,ECD)
Shape(BA,DAB)
Shape(AE,ED,EDA)
Shape(CE,EB,EBC)
Collinear(DEB)
Collinear(AEC)
Cocircular(E,DABC)
Conditions (CDL):
Equal(MeasureOfAngle(AED),x)
Equal(MeasureOfAngle(BEA),110)
Equal(MeasureOfAngle(CEB),x)
Equal(MeasureOfAngle(DEC),110)
Goal: Value(x)
Answer: 70
Theorem sequence:
adjacent_complementary_angle(1,DEC,CEB)
Problem: As shown in the diagram, ∠PRQ=33°, ∠SQP=51°, RQ⊥PQ. Find the measure of ∠RQS.
问题：如图所示，∠PRQ=33°，∠SQP=51°，RQ⊥PQ。求∠RQS的大小。
Construction (CDL):
Shape(QP,PS,SQ)
Shape(QS,SR,RQ)
Collinear(PSR)
Conditions (CDL):
Equal(MeasureOfAngle(PRQ),33)
Equal(MeasureOfAngle(SQP),51)
PerpendicularBetweenLine(RQ,PQ)
Goal: Value(MeasureOfAngle(RQS))
Answer: 39
Theorem sequence:
angle_addition(1,RQS,SQP)
Problem: As shown in the diagram, FH=17, GF=9*x-6, GF=GH, GH=7*x+4. Find the length of line FG.
问题：如图所示，FH=17，GF=9*x-6，GF=GH，GH=7*x+4。求直线FG的长度。
Construction (CDL):
Shape(FG,GH,HF)
Conditions (CDL):
Equal(LengthOfLine(FH),17)
Equal(LengthOfLine(GF),9*x-6)
Equal(LengthOfLine(GF),LengthOfLine(GH))
Equal(LengthOfLine(GH),7*x+4)
Goal: Value(LengthOfLine(FG))
Answer: 39
Theorem sequence:
Problem: As shown in the diagram, AP=3*y+11, PH=7*y-5, ∠HWQ=4*x-16°, ∠HWX=x+12°, ∠QAP=3*x-2°, WP bisects ∠HWA, WP is the median of triangle WAH. Find the value of x.
问题：如图所示，AP=3*y+11，PH=7*y-5，∠HWQ=4*x-16°，∠HWX=x+12°，∠QAP=3*x-2°，WP是∠HWA的角平分线，WP是△WAH的中线。求x的值。
Construction (CDL):
Shape(HW,WX,XH)
Shape(HW,WP,PH)
Shape(XW,WQ,QX)
Shape(PX,XQ,QA,AP)
Collinear(HXQ)
Collinear(HPA)
Collinear(WXP)
Collinear(WQA)
Conditions (CDL):
Equal(LengthOfLine(AP),3*y+11)
Equal(LengthOfLine(PH),7*y-5)
Equal(MeasureOfAngle(HWQ),4*x-16)
Equal(MeasureOfAngle(HWX),x+12)
Equal(MeasureOfAngle(QAP),3*x-2)
IsBisectorOfAngle(WP,HWA)
IsMedianOfTriangle(WP,WAH)
Goal: Value(x)
Answer: 20
Theorem sequence:
angle_addition(1,HWX,XWQ)
Problem: As shown in the diagram, QP=17, QR=15, RP=8, QR⊥PR. Find the value of tan(RP).
问题：如图所示，QP=17，QR=15，RP=8，QR⊥PR。求tan(RP)的值。
Construction (CDL):
Shape(QR,RP,PQ)
Conditions (CDL):
Equal(LengthOfLine(QP),17)
Equal(LengthOfLine(QR),15)
Equal(LengthOfLine(RP),8)
PerpendicularBetweenLine(QR,PR)
Goal: Value(Tan(MeasureOfAngle(RPQ)))
Answer: 15/8
Theorem sequence:
sine_theorem(1,QRP)
Problem: As shown in the diagram, MP=11, OP=27, NP is perpendicular to OP, ON⊥MN. Find the height of triangle NMO.
问题：如图所示，MP=11，OP=27，NP⊥OP，ON垂直于MN。求△NMO的高。
Construction (CDL):
Shape(NM,MP,PN)
Shape(NP,PO,ON)
Collinear(MPO)
Conditions (CDL):
Equal(LengthOfLine(MP),11)
Equal(LengthOfLine(OP),27)
PerpendicularBetweenLine(NP,OP)
PerpendicularBetweenLine(ON,MN)
Goal: Value(HeightOfTriangle(NMO))
Answer: 3*sqrt(33)
Theorem sequence:
adjacent_complementary_angle(1,MPN,NPO)
angle_addition(1,ONP,PNM)
triangle_property_angle_sum(1,NPO)
similar_triangle_judgment_aa(1,MPN,NPO)
similar_triangle_property_line_ratio(1,MPN,NPO)
similar_triangle_property_line_ratio(1,NMP,ONP)
altitude_of_triangle_judgment(1,NP,NMO)
Problem: As shown in the diagram, ∠CBD=110°, ∠DBE=x°, ∠EBA=30°, AB⊥CB. Find the value of x.
问题：如图所示，∠CBD=110°，∠DBE=x°，∠EBA=30°，AB垂直于CB。求x的值。
Construction (CDL):
Shape(EB,BA,BAE)
Shape(AB,BC,BCA)
Shape(CB,BD,BDC)
Shape(DB,BE,BED)
Cocircular(B,EACD)
Conditions (CDL):
Equal(MeasureOfAngle(CBD),110)
Equal(MeasureOfAngle(DBE),x)
Equal(MeasureOfAngle(EBA),30)
PerpendicularBetweenLine(AB,CB)
Goal: Value(x)
Answer: 130
Theorem sequence:
angle_addition(1,EBA,ABC)
angle_addition(1,EBC,CBD)
round_angle(1,EBD,DBE)
Problem: As shown in the diagram, AB=3, AC=x, BC=y, ∠ACB=45°, BA is perpendicular to CA. Find the value of x.
问题：如图所示，AB=3，AC=x，BC=y，∠ACB=45°，BA垂直于CA。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(ACB),45)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 3
Theorem sequence:
triangle_property_angle_sum(1,BAC)
sine_theorem(1,BAC)
sine_theorem(1,CBA)
Problem: As shown in the diagram, AC=20, BM=30, CM=12, CM is perpendicular to AM. Find the area of triangle AMB.
问题：如图所示，AC=20，BM=30，CM=12，CM垂直于AM。求△AMB的面积。
Construction (CDL):
Shape(AC,CM,MA)
Shape(AM,MB,BA)
Collinear(CMB)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BM),30)
Equal(LengthOfLine(CM),12)
PerpendicularBetweenLine(CM,AM)
Goal: Value(AreaOfTriangle(AMB))
Answer: 240
Theorem sequence:
right_triangle_judgment_angle(1,CMA)
right_triangle_property_pythagorean(1,CMA)
adjacent_complementary_angle(1,CMA,AMB)
triangle_area_formula_sine(1,MBA)
Problem: As shown in the diagram, AB=z, AN=x, AY=5, YB=14, YN=y, AN is perpendicular to BN, BA is perpendicular to YA. Find the value of x.
问题：如图所示，AB=z，AN=x，AY=5，YB=14，YN=y，AN⊥BN，BA垂直于YA。求x的值。
Construction (CDL):
Shape(AY,YN,NA)
Shape(AN,NB,BA)
Collinear(YNB)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AN),x)
Equal(LengthOfLine(AY),5)
Equal(LengthOfLine(YB),14)
Equal(LengthOfLine(YN),y)
PerpendicularBetweenLine(AN,BN)
PerpendicularBetweenLine(BA,YA)
Goal: Value(x)
Answer: 15*sqrt(19)/14
Theorem sequence:
adjacent_complementary_angle(1,YNA,ANB)
right_triangle_judgment_angle(1,BAY)
right_triangle_property_pythagorean(1,BAY)
altitude_of_triangle_judgment(1,AN,AYB)
triangle_area_formula_sine(1,AYB)
triangle_area_formula_common(1,AYB)
Problem: As shown in the diagram, GF=4, GH=4, ∠FOG=x°, the measure of ⌒OFH is 220, the center of circle O is O. Find the value of x.
问题：如图所示，GF=4，GH=4，∠FOG=x°，⌒OFH的角度为220，⊙O的圆心为O。求x的值。
Construction (CDL):
Shape(FO,OG,GF)
Shape(OH,HG,GO)
Shape(HO,OF,OFH)
Shape(FG,OGF)
Shape(GH,OHG)
Cocircular(O,FHG)
Conditions (CDL):
Equal(LengthOfLine(GF),4)
Equal(LengthOfLine(GH),4)
Equal(MeasureOfAngle(FOG),x)
Equal(MeasureOfArc(OFH),220)
IsCentreOfCircle(O,O)
Goal: Value(x)
Answer: 70
Theorem sequence:
arc_property_center_angle(1,OFH,O)
radius_of_circle_property_length_equal(1,OF,O)
radius_of_circle_property_length_equal(1,OG,O)
radius_of_circle_property_length_equal(1,OH,O)
congruent_triangle_judgment_sss(1,OGF,OHG)
congruent_triangle_property_angle_equal(1,OGF,OHG)
angle_addition(1,FOG,GOH)
round_angle(1,FOH,HOF)
Problem: As shown in the diagram, ∠CBA=76°, AD is the angle bisector of ∠BAC, CD bisects ∠ACB. Find the measure of ∠CDA.
问题：如图所示，∠CBA=76°，AD是∠BAC的角平分线，CD是∠ACB的角平分线。求∠CDA的大小。
Construction (CDL):
Shape(BA,AD,DC,CB)
Shape(DA,AC,CD)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),76)
IsBisectorOfAngle(AD,BAC)
IsBisectorOfAngle(CD,ACB)
Goal: Value(MeasureOfAngle(CDA))
Answer: 128
Theorem sequence:
triangle_property_angle_sum(1,DAC)
triangle_property_angle_sum(1,BAC)
angle_addition(1,BAD,DAC)
angle_addition(1,ACD,DCB)