SunnyLin/FormalGeo7k_v1 huggingface dataset:
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 ∠FGB.
问题：如图所示，∠BGC=40°，∠DGF=53°，CB垂直于GB，FG⊥CG，GF垂直于DF。求∠FGB的大小。
Construction (CDL):
Shape(DG,GF,FD)
Shape(FG,GB,BA,AF)
Shape(BG,GC,CB)
Shape(AB,BC,CA)
Collinear(GBA)
Collinear(DFA)
Conditions (CDL):
Equal(MeasureOfAngle(BGC),40)
Equal(MeasureOfAngle(DGF),53)
PerpendicularBetweenLine(CB,GB)
PerpendicularBetweenLine(FG,CG)
PerpendicularBetweenLine(GF,DF)
Goal: Value(MeasureOfAngle(FGB))
Answer: 50
Theorem sequence:
angle_addition(1,FGA,AGC)
Problem: As shown in the diagram, FQ=10, FW=x, WQ=4. Find the value of x.
问题：如图所示，FQ=10，FW=x，WQ=4。求x的值。
Construction (CDL):
Shape(DF,FW,WB,BD)
Shape(ND,DB,BW,WQ,QC,CN)
Collinear(NDF)
Collinear(FWQ)
Conditions (CDL):
Equal(LengthOfLine(FQ),10)
Equal(LengthOfLine(FW),x)
Equal(LengthOfLine(WQ),4)
Goal: Value(x)
Answer: 6
Theorem sequence:
line_addition(1,FW,WQ)
Problem: As shown in the diagram, AB=2, triangle DBC is an equilateral triangle, A is the center of circle A. Find the area of the ⊙A minus the area of △DBC.
问题：如图所示，AB=2，△DBC为等边△，A是⊙A的圆心。求圆A的面积减去△DBC的面积。
Construction (CDL):
Shape(ADB,BD)
Shape(DB,BA,AD)
Shape(AB,BC,CD,DA)
Shape(ACD,DC)
Shape(ABC,CB)
Cocircular(A,DBC)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
EquilateralTriangle(DBC)
IsCentreOfCircle(A,A)
Goal: Value(Sub(AreaOfCircle(A),AreaOfTriangle(DBC)))
Answer: -3*sqrt(3)+4*pi
Theorem sequence:
equilateral_triangle_property_angle(1,CDB)
equilateral_triangle_property_angle(1,DBC)
arc_property_center_angle(1,ADB,A)
arc_property_circumference_angle_external(1,ADB,C)
radius_of_circle_property_length_equal(1,AB,A)
radius_of_circle_property_length_equal(1,AD,A)
cosine_theorem(1,ADB)
triangle_area_formula_sine(1,DBC)
circle_area_formula(1,A)
Problem: As shown in the diagram, ∠TOR=71°, ∠UOR=179°, O is the center of ⊙O, SR is the tangent to ⊙O. Find the measure of ∠RST.
问题：如图所示，∠TOR=71°，∠UOR=179°，圆O的圆心为O，SR是⊙O的切线。求∠RST的大小。
Construction (CDL):
Shape(OUR,RO,OU)
Shape(OTU,UT)
Shape(OT,TU,UO)
Shape(ORT,TO,OR)
Shape(ORT,RS,ST)
Collinear(UTS)
Cocircular(O,URT)
Conditions (CDL):
Equal(MeasureOfAngle(TOR),71)
Equal(MeasureOfAngle(UOR),179)
IsCentreOfCircle(O,O)
IsTangentOfCircle(SR,O)
Goal: Value(MeasureOfAngle(RST))
Answer: 55
Theorem sequence:
angle_addition(1,UOT,TOR)
radius_of_circle_property_length_equal(1,OU,O)
radius_of_circle_property_length_equal(1,OT,O)
isosceles_triangle_judgment_line_equal(1,OTU)
isosceles_triangle_property_angle_equal(1,OTU)
triangle_property_angle_sum(1,UOT)
flat_angle(1,STU)
angle_addition(1,STO,OTU)
tangent_of_circle_property_perpendicular(1,SR,O,O)
quadrilateral_property_angle_sum(1,ORST)
Problem: As shown in the diagram, AB=sqrt(481), AC=16, BC=15, AC is perpendicular to BC. Find the value of tan(BA).
问题：如图所示，AB=sqrt(481)，AC=16，BC=15，AC⊥BC。求tan(BA)的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),sqrt(481))
Equal(LengthOfLine(AC),16)
Equal(LengthOfLine(BC),15)
PerpendicularBetweenLine(AC,BC)
Goal: Value(Tan(MeasureOfAngle(BAC)))
Answer: 15/16
Theorem sequence:
cosine_theorem(1,ACB)
Problem: As shown in the diagram, GF=5*x-9, GH=x+7, GF is the tangent to circle J, the tangent to circle J is GH. Find the value of x.
问题：如图所示，GF=5*x-9，GH=x+7，圆O的切线为GF，圆O的切线为GH。求x的值。
Construction (CDL):
Shape(JHF,HG,GF)
Shape(JHF,JFH)
Cocircular(J,FH)
Conditions (CDL):
Equal(LengthOfLine(GF),5*x-9)
Equal(LengthOfLine(GH),x+7)
IsTangentOfCircle(GF,J)
IsTangentOfCircle(GH,J)
Goal: Value(x)
Answer: 4
Theorem sequence:
tangent_of_circle_property_length_equal(1,GF,GH,J)
Problem: As shown in the diagram, ZX=3*x+5, ZY=5*x-7, ∠ZWX=∠YWZ, WX⊥ZX, ZY is perpendicular to WY. Find the length of line XZ.
问题：如图所示，ZX=3*x+5，ZY=5*x-7，∠ZWX=∠YWZ，WX垂直于ZX，ZY垂直于WY。求直线XZ的长度。
Construction (CDL):
Shape(WX,XZ,ZW)
Shape(WZ,ZY,YW)
Conditions (CDL):
Equal(LengthOfLine(ZX),3*x+5)
Equal(LengthOfLine(ZY),5*x-7)
Equal(MeasureOfAngle(ZWX),MeasureOfAngle(YWZ))
PerpendicularBetweenLine(WX,ZX)
PerpendicularBetweenLine(ZY,WY)
Goal: Value(LengthOfLine(XZ))
Answer: 23
Theorem sequence:
mirror_congruent_triangle_judgment_aas(3,ZYW,ZWX)
mirror_congruent_triangle_property_line_equal(1,WZY,WXZ)
Problem: As shown in the diagram, SR=5, TR=3, TS=4, RT⊥ST. Find the value of cos(SR).
问题：如图所示，SR=5，TR=3，TS=4，RT⊥ST。求cos(SR)的值。
Construction (CDL):
Shape(TS,SR,RT)
Conditions (CDL):
Equal(LengthOfLine(SR),5)
Equal(LengthOfLine(TR),3)
Equal(LengthOfLine(TS),4)
PerpendicularBetweenLine(RT,ST)
Goal: Value(Cos(MeasureOfAngle(SRT)))
Answer: 3/5
Theorem sequence:
cosine_theorem(1,RTS)
Problem: As shown in the diagram, AC=x, AD=8, BD=y, ∠BCA=43°, CA is perpendicular to BA, DB⊥CB. Find the value of y.
问题：如图所示，AC=x，AD=8，BD=y，∠BCA=43°，CA垂直于BA，DB⊥CB。求y的值。
Construction (CDL):
Shape(DB,BA,AD)
Shape(AB,BC,CA)
Collinear(DAC)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BD),y)
Equal(MeasureOfAngle(BCA),43)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(DB,CB)
Goal: Value(y)
Answer: 8/sin(43*pi/180)
Theorem sequence:
adjacent_complementary_angle(1,CAB,BAD)
triangle_property_angle_sum(1,DBA)
triangle_property_angle_sum(1,DBC)
sine_theorem(1,DBA)
Problem: As shown in the diagram, ∠BDF=47°, ∠CIF=112°, ∠DFB=65°. Find the measure of ∠IAF.
问题：如图所示，∠BDF=47°，∠CIF=112°，∠DFB=65°。求∠IAF的大小。
Construction (CDL):
Shape(BD,DF,FB)
Shape(FI,IA,AF)
Shape(DB,BG)
Shape(BF,FA)
Shape(CI,IF)
Collinear(GBFI)
Collinear(AFD)
Collinear(AIC)
Conditions (CDL):
Equal(MeasureOfAngle(BDF),47)
Equal(MeasureOfAngle(CIF),112)
Equal(MeasureOfAngle(DFB),65)
Goal: Value(MeasureOfAngle(IAF))
Answer: 47
Theorem sequence:
vertical_angle(1,DFB,AFI)
adjacent_complementary_angle(1,CIF,FIA)
triangle_property_angle_sum(1,FIA)
Problem: As shown in the diagram, ∠ACE=y°, ∠BDA=68°, ∠EAC=2*x°, ∠EBD=3*x-15°, EA∥BD. Find the value of y.
问题：如图所示，∠ACE=y°，∠BDA=68°，∠EAC=2*x°，∠EBD=3*x-15°，EA∥BD。求y的值。
Construction (CDL):
Shape(BD,DA,AE,EB)
Shape(EA,AC,CE)
Collinear(BEC)
Collinear(DAC)
Conditions (CDL):
Equal(MeasureOfAngle(ACE),y)
Equal(MeasureOfAngle(BDA),68)
Equal(MeasureOfAngle(EAC),2*x)
Equal(MeasureOfAngle(EBD),3*x-15)
ParallelBetweenLine(EA,BD)
Goal: Value(y)
Answer: 25
Theorem sequence:
parallel_property_corresponding_angle(2,DB,AE,C)
triangle_property_angle_sum(1,BDC)
Problem: As shown in the diagram, AB=10, AE=25/4, BC=x+2, DE=x-1, △ABE is similar to △ACD.. Find the length of line BC.
问题：如图所示，AB=10，AE=25/4，BC=x+2，DE=x-1，三角形ABE相似于三角形ACD。求直线BC的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(BC,CD,DE,EB)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AE),25/4)
Equal(LengthOfLine(BC),x+2)
Equal(LengthOfLine(DE),x-1)
SimilarBetweenTriangle(ABE,ACD)
Goal: Value(LengthOfLine(BC))
Answer: 8
Theorem sequence:
similar_triangle_property_line_ratio(1,EAB,DAC)
similar_triangle_property_line_ratio(1,BEA,CDA)
line_addition(1,AB,BC)
line_addition(1,AE,ED)
Problem: As shown in the diagram, AD=5+x, BD=x, BF=5, CF=5+x. Find the value of x.
问题：如图所示，AD=5+x，BD=x，BF=5，CF=5+x。求x的值。
Construction (CDL):
Shape(EDF,DB,BF)
Shape(EAD,DA)
Shape(EDF,FC,ECA,AD)
Shape(EFC,CF)
Collinear(BDA)
Collinear(BFC)
Cocircular(E,DFCA)
Conditions (CDL):
Equal(LengthOfLine(AD),5+x)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(BF),5)
Equal(LengthOfLine(CF),5+x)
Goal: Value(x)
Answer: 5
Theorem sequence:
line_addition(1,BD,DA)
line_addition(1,BF,FC)
circle_property_circular_power_segment_and_segment_line(1,BDA,BFC,E)
Problem: As shown in the diagram, AG=12, the center of ⊙A is A. Find the length of line LA.
问题：如图所示，AG=12，圆A的圆心为A。求直线LA的长度。
Construction (CDL):
Shape(APD,DA,AP)
Shape(ADL,LA,AD)
Shape(ALG,GA,AL)
Shape(AGF,FA,AG)
Shape(AFP,PA,AF)
Cocircular(A,PDLGF)
Conditions (CDL):
Equal(LengthOfLine(AG),12)
IsCentreOfCircle(A,A)
Goal: Value(LengthOfLine(LA))
Answer: 12
Theorem sequence:
radius_of_circle_property_length_equal(1,AG,A)
radius_of_circle_property_length_equal(1,AL,A)
Problem: As shown in the diagram, ∠FJH=82°, GFJH is a rhombus. Find the measure of ∠JHK.
问题：如图所示，∠FJH=82°，GFJH是菱形。求∠JHK的大小。
Construction (CDL):
Shape(GF,FK,KG)
Shape(GK,KH,HG)
Shape(FJ,JK,KF)
Shape(KJ,JH,HK)
Collinear(FKH)
Collinear(GKJ)
Conditions (CDL):
Equal(MeasureOfAngle(FJH),82)
Rhombus(GFJH)
Goal: Value(MeasureOfAngle(JHK))
Answer: 49
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,JHGF,K)
altitude_of_triangle_judgment(1,JK,JHF)
isosceles_triangle_judgment_line_equal(1,JHF)
isosceles_triangle_property_line_coincidence(1,JHF,K)
angle_addition(1,FJK,KJH)
triangle_property_angle_sum(1,JHK)
Problem: As shown in the diagram, AB=13, AC=7, CB=10, FE=14, △ACB is mirror similar to △DFE.. Find the perimeter of △DFE.
问题：如图所示，AB=13，AC=7，CB=10，FE=14，三角形ACB镜像相似于三角形DFE。求三角形DFE的周长。
Construction (CDL):
Shape(AC,CB,BA)
Shape(DF,FE,ED)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),7)
Equal(LengthOfLine(CB),10)
Equal(LengthOfLine(FE),14)
MirrorSimilarBetweenTriangle(ACB,DFE)
Goal: Value(PerimeterOfTriangle(DFE))
Answer: 42
Theorem sequence:
triangle_perimeter_formula(1,ACB)
mirror_similar_triangle_property_line_ratio(1,ACB,DFE)
mirror_similar_triangle_property_perimeter_ratio(1,ACB,DFE)
Problem: As shown in the diagram, ∠CPD=3*x-15°, DP is perpendicular to AP, quadrilateral ADCB is a rhombus. Find the value of x.
问题：如图所示，∠CPD=3*x-15°，DP垂直于AP，ADCB是菱形。求x的值。
Construction (CDL):
Shape(AD,DP,PA)
Shape(AP,PB,BA)
Shape(PD,DC,CP)
Shape(PC,CB,BP)
Collinear(APC)
Collinear(DPB)
Conditions (CDL):
Equal(MeasureOfAngle(CPD),3*x-15)
PerpendicularBetweenLine(DP,AP)
Rhombus(ADCB)
Goal: Value(x)
Answer: 35
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,CBAD,P)
Problem: As shown in the diagram, ∠KFD=x°, the measure of ⌒ECG is 45, the measure of arc EDC is 130, E is the center of circle E. Find the value of x.
问题：如图所示，∠KFD=x°，⌒ECG的角度为45，⌒EDC的角度为130，圆E的圆心为E。求x的值。
Construction (CDL):
Shape(EDC,CD)
Shape(DC,CE,ED)
Shape(ECG,GE,EC)
Shape(EGK,KE,EG)
Shape(EKD,DE,EK)
Shape(EKD,KF,FD)
Collinear(GEKF)
Collinear(CDF)
Cocircular(E,CGKD)
Conditions (CDL):
Equal(MeasureOfAngle(KFD),x)
Equal(MeasureOfArc(ECG),45)
Equal(MeasureOfArc(EDC),130)
IsCentreOfCircle(E,E)
Goal: Value(x)
Answer: 20
Theorem sequence:
arc_property_center_angle(1,EDC,E)
arc_property_center_angle(1,ECG,E)
radius_of_circle_property_length_equal(1,EC,E)
radius_of_circle_property_length_equal(1,ED,E)
isosceles_triangle_judgment_line_equal(1,EDC)
isosceles_triangle_property_angle_equal(1,EDC)
triangle_property_angle_sum(1,DCE)
flat_angle(1,FDC)
angle_addition(1,FDE,EDC)
flat_angle(1,GEK)
angle_addition(1,GEC,CED)
angle_addition(1,GED,DEF)
triangle_property_angle_sum(1,DEF)
Problem: As shown in the diagram, AB=c, AC=b, CB=a, a=14, b=48, c=50, BC⊥AC. Find the value of cos(AB).
问题：如图所示，AB=c，AC=b，CB=a，a=14，b=48，c=50，BC⊥AC。求cos(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),c)
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(CB),a)
Equal(a,14)
Equal(b,48)
Equal(c,50)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Cos(MeasureOfAngle(ABC)))
Answer: 7/25
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AF=x+4, BD=x, BF=8, DC=2*x. Find the value of x.
问题：如图所示，AF=x+4，BD=x，BF=8，DC=2*x。求x的值。
Construction (CDL):
Shape(EAF,FA)
Shape(ECA,AF,EFD,DC)
Shape(EDC,CD)
Shape(EFD,FB,BD)
Collinear(CDB)
Collinear(AFB)
Cocircular(E,AFDC)
Conditions (CDL):
Equal(LengthOfLine(AF),x+4)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(BF),8)
Equal(LengthOfLine(DC),2*x)
Goal: Value(x)
Answer: 4/3+4*sqrt(19)/3
Theorem sequence:
line_addition(1,CD,DB)
line_addition(1,AF,FB)
circle_property_circular_power_segment_and_segment_line(1,BFA,BDC,E)
Problem: As shown in the diagram, ∠BFD=165°, ∠CFB=x°, ∠DFC=145°. Find the value of x.
问题：如图所示，∠BFD=165°，∠CFB=x°，∠DFC=145°。求x的值。
Construction (CDL):
Shape(FDB,BF,FD)
Shape(FBC,CF,FB)
Shape(FCD,DF,FC)
Cocircular(F,DBC)
Conditions (CDL):
Equal(MeasureOfAngle(BFD),165)
Equal(MeasureOfAngle(CFB),x)
Equal(MeasureOfAngle(DFC),145)
Goal: Value(x)
Answer: 50
Theorem sequence:
angle_addition(1,BFD,DFC)
round_angle(1,CFB,BFC)
Problem: As shown in the diagram, KL=AJ, the measure of arc CKL is 5*x, the measure of arc DJA is 3*x+54, the radius of ⊙C is equal to the radius of circle D, C is the center of circle C, D is the center of circle D. Find the value of x.
问题：如图所示，KL=AJ，弧CKL的角度为5*x，弧DJA的角度为3*x+54，⊙C的半径与⊙D的半径相等，C是圆C的圆心，⊙D的圆心为D。求x的值。
Construction (CDL):
Shape(CKL,LK)
Shape(KL,LC,CK)
Shape(CLK,KC,CL)
Shape(DJA,AJ)
Shape(DJ,JA,AD)
Shape(DAJ,JD,DA)
Cocircular(C,KL)
Cocircular(D,JA)
Conditions (CDL):
Equal(LengthOfLine(KL),LengthOfLine(AJ))
Equal(MeasureOfArc(CKL),5*x)
Equal(MeasureOfArc(DJA),3*x+54)
Equal(RadiusOfCircle(C),RadiusOfCircle(D))
IsCentreOfCircle(C,C)
IsCentreOfCircle(D,D)
Goal: Value(x)
Answer: 27
Theorem sequence:
arc_property_center_angle(1,CKL,C)
arc_property_center_angle(1,DJA,D)
radius_of_circle_property_length_equal(1,CK,C)
radius_of_circle_property_length_equal(1,CL,C)
radius_of_circle_property_length_equal(1,DJ,D)
radius_of_circle_property_length_equal(1,DA,D)
congruent_triangle_judgment_sss(1,KLC,JAD)
congruent_triangle_property_angle_equal(1,CKL,DJA)
Problem: As shown in the diagram, AB=c, AC=b, CB=a, a=14, b=48, c=50, BC is perpendicular to AC. Find the value of tan(CA).
问题：如图所示，AB=c，AC=b，CB=a，a=14，b=48，c=50，BC⊥AC。求tan(CA)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),c)
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(CB),a)
Equal(a,14)
Equal(b,48)
Equal(c,50)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Tan(MeasureOfAngle(CAB)))
Answer: 7/24
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, ∠FEC=a°, ∠GDA=b°, DH⊥EH. Find the value of a+b-90.
问题：如图所示，∠FEC=a°，∠GDA=b°，DH⊥EH。求a+b-90的值。
Construction (CDL):
Shape(DH,HE,ED)
Collinear(ADH)
Collinear(HEF)
Collinear(GDEC)
Shape(GD,DA)
Shape(FE,EC)
Conditions (CDL):
Equal(MeasureOfAngle(FEC),a)
Equal(MeasureOfAngle(GDA),b)
PerpendicularBetweenLine(DH,EH)
Goal: Value(a+b-90)
Answer: 0
Theorem sequence:
vertical_angle(1,HEG,FEC)
vertical_angle(1,GDA,EDH)
triangle_property_angle_sum(1,DHE)
Problem: As shown in the diagram, ∠UZY=2*x+24°, ∠VZU=4*x°, ∠XZW=∠YZX, Z is the center of circle Z. Find the measure of ⌒ZYU.
问题：如图所示，∠UZY=2*x+24°，∠VZU=4*x°，∠XZW=∠YZX，Z是⊙Z的圆心。求⌒ZYU的角度。
Construction (CDL):
Shape(ZVW,WZ,ZV)
Shape(ZWX,XZ,ZW)
Shape(ZXY,YZ,ZX)
Shape(ZYU,UZ,ZY)
Shape(ZUV,VZ,ZU)
Collinear(VZY)
Collinear(WZU)
Cocircular(Z,VWXYU)
Conditions (CDL):
Equal(MeasureOfAngle(UZY),2*x+24)
Equal(MeasureOfAngle(VZU),4*x)
Equal(MeasureOfAngle(XZW),MeasureOfAngle(YZX))
IsCentreOfCircle(Z,Z)
Goal: Value(MeasureOfArc(ZYU))
Answer: 76
Theorem sequence:
flat_angle(1,VZY)
angle_addition(1,VZU,UZY)
arc_property_center_angle(1,ZYU,Z)
Problem: As shown in the diagram, ∠ABG=47°, ∠ACG=136°, ∠BED=63°, ∠DFB=∠BDF, ∠EBA=69°. Find the measure of ∠BCA.
问题：如图所示，∠ABG=47°，∠ACG=136°，∠BED=63°，∠DFB=∠BDF，∠EBA=69°。求∠BCA的大小。
Construction (CDL):
Shape(ED,DB,BE)
Shape(BD,DF,FB)
Shape(AB,BC,CA)
Collinear(DBCG)
Collinear(EBF)
Shape(EB,BA)
Shape(AC,CG)
Conditions (CDL):
Equal(MeasureOfAngle(ABG),47)
Equal(MeasureOfAngle(ACG),136)
Equal(MeasureOfAngle(BED),63)
Equal(MeasureOfAngle(DFB),MeasureOfAngle(BDF))
Equal(MeasureOfAngle(EBA),69)
Goal: Value(MeasureOfAngle(BCA))
Answer: 44
Theorem sequence:
flat_angle(1,BCG)
angle_addition(1,BCA,ACG)
Problem: As shown in the diagram, ∠ADC=60°, D is the center of circle D. Find the measure of ∠ABC.
问题：如图所示，∠ADC=60°，D是⊙D的圆心。求∠ABC的大小。
Construction (CDL):
Shape(DAB,BA)
Shape(DBC,CB)
Shape(DCA,AD,DC)
Shape(DA,AB,BC,CD)
Cocircular(D,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),60)
IsCentreOfCircle(D,D)
Goal: Value(MeasureOfAngle(ABC))
Answer: 30
Theorem sequence:
arc_property_center_angle(1,DCA,D)
arc_property_circumference_angle_external(1,DCA,B)
Problem: As shown in the diagram, DC=6-x, DG=2, JL=4, JM=x, DG is the altitude of triangle DDC, JM is the altitude of triangle JJL, DG is perpendicular to CG, JM is perpendicular to LM, triangle KLJ is similar to triangle ECD.. Find the length of line DC.
问题：如图所示，DC=6-x，DG=2，JL=4，JM=x，DG是△DDC的高，JM是三角形JJL的高，DG⊥CG，JM垂直于LM，△KLJ相似于△ECD。求直线DC的长度。
Construction (CDL):
Shape(JK,KM,MJ)
Shape(JM,ML,LJ)
Shape(DE,EG,GD)
Shape(DG,GC,CD)
Collinear(KML)
Collinear(EGC)
Conditions (CDL):
Equal(LengthOfLine(DC),6-x)
Equal(LengthOfLine(DG),2)
Equal(LengthOfLine(JL),4)
Equal(LengthOfLine(JM),x)
IsAltitudeOfTriangle(DG,EDC)
IsAltitudeOfTriangle(JM,KJL)
PerpendicularBetweenLine(DG,CG)
PerpendicularBetweenLine(JM,LM)
SimilarBetweenTriangle(KLJ,ECD)
Goal: Value(LengthOfLine(DC))
Answer: 2
Theorem sequence:
similar_triangle_property_angle_equal(1,LJK,CDE)
similar_triangle_judgment_aa(1,JML,DGC)
similar_triangle_property_line_ratio(1,LJM,CDG)
similar_triangle_property_line_ratio(1,MLJ,GCD)
Problem: As shown in the diagram, BA=3, BD=x-1, CE=x+2, EF=8, ∠GAB=∠EFG, AB⊥CB, DE is perpendicular to FE. Find the length of line EC.
问题：如图所示，BA=3，BD=x-1，CE=x+2，EF=8，∠GAB=∠EFG，AB垂直于CB，DE⊥FE。求直线EC的长度。
Construction (CDL):
Shape(AB,BC,CG,GA)
Shape(GC,CD,DG)
Shape(GD,DE,EF,FG)
Collinear(BCDE)
Collinear(AGD)
Collinear(CGF)
Conditions (CDL):
Equal(LengthOfLine(BA),3)
Equal(LengthOfLine(BD),x-1)
Equal(LengthOfLine(CE),x+2)
Equal(LengthOfLine(EF),8)
Equal(MeasureOfAngle(GAB),MeasureOfAngle(EFG))
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(DE,FE)
Goal: Value(LengthOfLine(EC))
Answer: 24/5
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,DAB,CEF)
mirror_similar_triangle_property_line_ratio(1,DAB,CEF)
mirror_similar_triangle_property_line_ratio(1,ABD,FCE)
Problem: As shown in the diagram, AC=5, BC=3, AC is perpendicular to BC. Find the length of line AB.
问题：如图所示，AC=5，BC=3，AC⊥BC。求直线AB的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),3)
PerpendicularBetweenLine(AC,BC)
Goal: Value(LengthOfLine(AB))
Answer: sqrt(34)
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, the area of △CEA is 52, CD=b+5, EA=b, ED⊥CD. Find the value of b.
问题：如图所示，△CEA的面积为52，CD=b+5，EA=b，ED垂直于CD。求b的值。
Construction (CDL):
Shape(CE,ED,DC)
Shape(CD,DA,AC)
Collinear(EDA)
Conditions (CDL):
Equal(AreaOfTriangle(CEA),52)
Equal(LengthOfLine(CD),b+5)
Equal(LengthOfLine(EA),b)
PerpendicularBetweenLine(ED,CD)
Goal: Value(b)
Answer: 8
Theorem sequence:
altitude_of_triangle_judgment(1,CD,CEA)
triangle_area_formula_common(1,CEA)
Problem: As shown in the diagram, AI=18, FH=37, GB=9, GB is perpendicular to HB, IA is perpendicular to FA. Find the sum of the area of △IHF and the area of △FHG.
问题：如图所示，AI=18，FH=37，GB=9，GB垂直于HB，IA⊥FA。求三角形IHF的面积与△FHG的面积之和。
Construction (CDL):
Shape(GF,FA,AB,BG)
Shape(FI,IA,AF)
Shape(AI,IH,HB,BA)
Shape(GB,BH,HG)
Collinear(FABH)
Conditions (CDL):
Equal(LengthOfLine(AI),18)
Equal(LengthOfLine(FH),37)
Equal(LengthOfLine(GB),9)
PerpendicularBetweenLine(GB,HB)
PerpendicularBetweenLine(IA,FA)
Goal: Value(Add(AreaOfTriangle(IHF),AreaOfTriangle(FHG)))
Answer: 999/2
Theorem sequence:
adjacent_complementary_angle(1,BAI,IAF)
adjacent_complementary_angle(1,FBG,GBH)
altitude_of_triangle_judgment(1,GB,GFH)
altitude_of_triangle_judgment(1,IA,IHF)
triangle_area_formula_common(1,GFH)
triangle_area_formula_common(1,IHF)
Problem: As shown in the diagram, AC=9, AY=18, PO=x, QP=y, RO=14, YB=21, quadrilateral RQPO is similar to quadrilateral BCAY. Find the value of y.
问题：如图所示，AC=9，AY=18，PO=x，QP=y，RO=14，YB=21，四边形RQPO相似于四边形BCAY。求y的值。
Construction (CDL):
Shape(RQ,QP,PO,OR)
Shape(AY,YB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),9)
Equal(LengthOfLine(AY),18)
Equal(LengthOfLine(PO),x)
Equal(LengthOfLine(QP),y)
Equal(LengthOfLine(RO),14)
Equal(LengthOfLine(YB),21)
SimilarBetweenQuadrilateral(RQPO,BCAY)
Goal: Value(y)
Answer: 6
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,QPOR,CAYB)
similar_quadrilateral_property_line_ratio(1,ORQP,YBCA)
Problem: As shown in the diagram, RT=2*x+6, VT=10, WR=x+6, WS=8, ∠SWR=∠VTR. Find the length of line RT.
问题：如图所示，RT=2*x+6，VT=10，WR=x+6，WS=8，∠SWR=∠VTR。求直线RT的长度。
Construction (CDL):
Shape(SW,WR,RS)
Shape(VT,TR,RV)
Collinear(SRV)
Collinear(WRT)
Conditions (CDL):
Equal(LengthOfLine(RT),2*x+6)
Equal(LengthOfLine(VT),10)
Equal(LengthOfLine(WR),x+6)
Equal(LengthOfLine(WS),8)
Equal(MeasureOfAngle(SWR),MeasureOfAngle(VTR))
Goal: Value(LengthOfLine(RT))
Answer: 10
Theorem sequence:
vertical_angle(1,WRS,TRV)
similar_triangle_judgment_aa(1,SWR,VTR)
similar_triangle_property_line_ratio(1,SWR,VTR)
similar_triangle_property_line_ratio(1,RSW,RVT)
Problem: As shown in the diagram, ∠EFA=63°, the center of circle F is F, DF is perpendicular to EF. Find the measure of arc FDA.
问题：如图所示，∠EFA=63°，F是圆F的圆心，DF⊥EF。求⌒FDA的角度。
Construction (CDL):
Shape(FCB,BF,FC)
Shape(FBA,AF,FB)
Shape(FAE,EF,FA)
Shape(FED,DF,FE)
Shape(FDC,CF,FD)
Cocircular(F,AEDCB)
Conditions (CDL):
Equal(MeasureOfAngle(EFA),63)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(DF,EF)
Goal: Value(MeasureOfArc(FDA))
Answer: 207
Theorem sequence:
angle_addition(1,DFE,EFA)
round_angle(1,DFA,AFD)
arc_property_center_angle(1,FDA,F)
Problem: As shown in the diagram, AB=y, BC=x, CA=7*sqrt(2), ∠CAB=45°, BC is perpendicular to AC. Find the value of y.
问题：如图所示，AB=y，BC=x，CA=7*sqrt(2)，∠CAB=45°，BC⊥AC。求y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CA),7*sqrt(2))
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 14
Theorem sequence:
triangle_property_angle_sum(1,BCA)
sine_theorem(1,ABC)
Problem: As shown in the diagram, ∠FBE=4*x°, the measure of ⌒ACD is 9*x+26, the measure of arc AFE is 35. Find the value of x.
问题：如图所示，∠FBE=4*x°，弧ACD的角度为9*x+26，⌒AFE的角度为35。求x的值。
Construction (CDL):
Shape(AEC,CE)
Shape(ACD,DF,AFE,EC)
Shape(ADF,FD)
Shape(BE,AFE,FB)
Collinear(CEB)
Collinear(DFB)
Cocircular(A,ECDF)
Conditions (CDL):
Equal(MeasureOfAngle(FBE),4*x)
Equal(MeasureOfArc(ACD),9*x+26)
Equal(MeasureOfArc(AFE),35)
Goal: Value(x)
Answer: 9
Theorem sequence:
circle_property_circular_power_segment_and_segment_angle(1,BFD,BEC,A)
Problem: As shown in the diagram, BD=22, DN=18, ∠NAD=40°, AD and CB are opposite sides of the parallelogram ACBD, DN⊥AN. Find the area of ACBD.
问题：如图所示，BD=22，DN=18，∠NAD=40°，ACBD是▱，DN垂直于AN。求ACBD的面积。
Construction (CDL):
Shape(CB,BD,DA,AC)
Shape(AD,DN,NA)
Collinear(BDN)
Conditions (CDL):
Equal(LengthOfLine(BD),22)
Equal(LengthOfLine(DN),18)
Equal(MeasureOfAngle(NAD),40)
Parallelogram(ACBD)
PerpendicularBetweenLine(DN,AN)
Goal: Value(AreaOfQuadrilateral(ACBD))
Answer: 396/tan(2*pi/9)
Theorem sequence:
triangle_property_angle_sum(1,ADN)
sine_theorem(1,NAD)
altitude_of_quadrilateral_judgment_right_vertex(5,AN,CBDA)
parallelogram_area_formula_common(1,CBDA)
Problem: As shown in the diagram, AC=8, AD=4, CB=10, AC⊥BC, DA is perpendicular to CA. Find the area of ACBD.
问题：如图所示，AC=8，AD=4，CB=10，AC⊥BC，DA垂直于CA。求四边形ACBD的面积。
Construction (CDL):
Shape(BD,DA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(CB),10)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(DA,CA)
Goal: Value(AreaOfQuadrilateral(ACBD))
Answer: 56
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,AD,CB)
trapezoid_judgment_parallel(1,ACBD)
right_trapezoid_judgment_right_angle(1,ACBD)
right_trapezoid_area_formular(1,ACBD)
Problem: As shown in the diagram, AB=13, AC=12, BC=15. Find the measure of ∠BAC.
问题：如图所示，AB=13，AC=12，BC=15。求∠BAC的大小。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(BC),15)
Goal: Value(MeasureOfAngle(BAC))
Answer: 180*acos(11/39)/pi
Theorem sequence:
cosine_theorem(1,ACB)
Problem: As shown in the diagram, BE=15, CB=12, CE=x, the center of ⊙E is E, BC is the tangent to ⊙E. Find the value of x.
问题：如图所示，BE=15，CB=12，CE=x，E是⊙E的圆心，BC是圆O的切线。求x的值。
Construction (CDL):
Shape(ECF,FE,EC)
Shape(EFC,CE,EF)
Shape(ECF,CB,BF)
Collinear(BFE)
Cocircular(E,CF)
Conditions (CDL):
Equal(LengthOfLine(BE),15)
Equal(LengthOfLine(CB),12)
Equal(LengthOfLine(CE),x)
IsCentreOfCircle(E,E)
IsTangentOfCircle(BC,E)
Goal: Value(x)
Answer: 9
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BC,E,E)
right_triangle_judgment_angle(1,ECB)
right_triangle_property_pythagorean(1,ECB)