SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, BS=x-1, RB=5, RC=x+1, SD=x-1, TC=3*x-2, TD=10, the center of ⊙A is A, the tangent to circle A is RB, the tangent to ⊙A is RC, SB is the tangent to circle A, SD is the tangent to circle A, the tangent to circle A is TC, the tangent to circle A is TD. Find the perimeter of △RST.
问题：如图所示，BS=x-1，RB=5，RC=x+1，SD=x-1，TC=3*x-2，TD=10，⊙A的圆心为A，⊙O的切线为RB，⊙O的切线为RC，⊙O的切线为SB，SD是圆O的切线，⊙O的切线为TC，圆O的切线为TD。求△RST的周长。
Construction (CDL):
Shape(RB,ACB,CR)
Shape(ACB,BA,AC)
Shape(BS,SD,ABD)
Shape(ABD,DA,AB)
Shape(ADE,EA,AD)
Shape(ADE,DT,TE)
Shape(AEC,CA,AE)
Shape(AEC,ET,TC)
Collinear(RBS)
Collinear(SDT)
Collinear(RCT)
Collinear(AET)
Cocircular(A,BDEC)
Conditions (CDL):
Equal(LengthOfLine(BS),x-1)
Equal(LengthOfLine(RB),5)
Equal(LengthOfLine(RC),x+1)
Equal(LengthOfLine(SD),x-1)
Equal(LengthOfLine(TC),3*x-2)
Equal(LengthOfLine(TD),10)
IsCentreOfCircle(A,A)
IsTangentOfCircle(RB,A)
IsTangentOfCircle(RC,A)
IsTangentOfCircle(SB,A)
IsTangentOfCircle(SD,A)
IsTangentOfCircle(TC,A)
IsTangentOfCircle(TD,A)
Goal: Value(PerimeterOfTriangle(RST))
Answer: 36
Theorem sequence:
tangent_of_circle_property_perpendicular(2,TC,A,A)
tangent_of_circle_property_perpendicular(1,TD,A,A)
radius_of_circle_property_length_equal(1,AC,A)
radius_of_circle_property_length_equal(1,AD,A)
mirror_congruent_triangle_judgment_hl(2,ADT,ATC)
mirror_congruent_triangle_property_line_equal(1,ADT,ATC)
line_addition(1,RB,BS)
line_addition(1,SD,DT)
line_addition(1,RC,CT)
triangle_perimeter_formula(1,RST)
Problem: As shown in the diagram, AF=3, DH=x, the perimeter of ADCB is 32, AE is the tangent to ⊙P, the tangent to circle P is AF, BF is the tangent to circle P, BG is the tangent to circle P, the tangent to circle P is CG, CH is the tangent to ⊙P, the tangent to circle P is DE, DH is the tangent to ⊙P, quadrilateral ADCB is a rhombus. Find the value of x.
问题：如图所示，AF=3，DH=x，ADCB的周长为32，AE是圆O的切线，AF是⊙O的切线，BF是⊙O的切线，BG是⊙O的切线，圆O的切线为CG，圆O的切线为CH，⊙O的切线为DE，DH是⊙O的切线，四边形ADCB是菱形。求x的值。
Construction (CDL):
Shape(PFE,FA,AE)
Shape(PEH,ED,DH)
Shape(PHG,HC,CG)
Shape(PGF,GB,BF)
Shape(PGF,PFE,PEH,PHG)
Collinear(AED)
Collinear(BFA)
Collinear(DHC)
Collinear(CGB)
Cocircular(P,FEHG)
Conditions (CDL):
Equal(LengthOfLine(AF),3)
Equal(LengthOfLine(DH),x)
Equal(PerimeterOfQuadrilateral(ADCB),32)
IsTangentOfCircle(AE,P)
IsTangentOfCircle(AF,P)
IsTangentOfCircle(BF,P)
IsTangentOfCircle(BG,P)
IsTangentOfCircle(CG,P)
IsTangentOfCircle(CH,P)
IsTangentOfCircle(DE,P)
IsTangentOfCircle(DH,P)
Rhombus(ADCB)
Goal: Value(x)
Answer: 5
Theorem sequence:
quadrilateral_perimeter_formula(1,ADCB)
tangent_of_circle_property_length_equal(1,AF,AE,P)
tangent_of_circle_property_length_equal(1,DE,DH,P)
line_addition(1,AE,ED)
Problem: As shown in the diagram, ∠FGM=82°, the tangent to circle C is HG, the tangent to ⊙C is MG. Find the measure of arc CFG.
问题：如图所示，∠FGM=82°，HG是⊙O的切线，MG是⊙O的切线。求弧CFG的角度。
Construction (CDL):
Shape(FG,GM)
Shape(HG,GF)
Shape(CGF,FG)
Shape(CFJ,CJG,GF)
Collinear(MGH)
Cocircular(C,GFJ)
Conditions (CDL):
Equal(MeasureOfAngle(FGM),82)
IsTangentOfCircle(HG,C)
IsTangentOfCircle(MG,C)
Goal: Value(MeasureOfArc(CFG))
Answer: 196
Theorem sequence:
adjacent_complementary_angle(1,HGF,FGM)
circle_property_angle_of_osculation(2,CFG,H)
Problem: As shown in the diagram, AB=33, BE=24, CD=32, EB∥DC. Find the length of line BC.
问题：如图所示，AB=33，BE=24，CD=32，EB平行于DC。求直线BC的长度。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BE,ED,DC,CB)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),33)
Equal(LengthOfLine(BE),24)
Equal(LengthOfLine(CD),32)
ParallelBetweenLine(EB,DC)
Goal: Value(LengthOfLine(BC))
Answer: 11
Theorem sequence:
line_addition(1,AB,BC)
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
similar_triangle_property_line_ratio(1,EBA,DCA)
similar_triangle_property_line_ratio(1,AEB,ADC)
Problem: As shown in the diagram, HJ=5, LM=x, HJ is the midsegment of triangle KML. Find the value of x.
问题：如图所示，HJ=5，LM=x，HJ是三角形KML的中位线。求x的值。
Construction (CDL):
Shape(JK,KH,HJ)
Shape(JH,HM,MX,XJ)
Shape(JX,XL,LJ)
Collinear(KHM)
Collinear(KJL)
Collinear(LXM)
Conditions (CDL):
Equal(LengthOfLine(HJ),5)
Equal(LengthOfLine(LM),x)
IsMidsegmentOfTriangle(HJ,KML)
Goal: Value(x)
Answer: 10
Theorem sequence:
midsegment_of_triangle_property_parallel(1,HJ,KML)
parallel_property_corresponding_angle(1,HJ,ML,K)
similar_triangle_judgment_aa(1,JKH,LKM)
line_addition(1,KH,HM)
similar_triangle_property_line_ratio(1,KHJ,KML)
similar_triangle_property_line_ratio(1,JKH,LKM)
Problem: As shown in the diagram, AF=24, AH=25, BH=11, ∠ABH=30°, ∠GCH=28°, H is the center of the inscribed circle of triangle ABC, BG⊥HG, HD⊥BD, HF is perpendicular to AF. Find the measure of ∠GHD.
问题：如图所示，AF=24，AH=25，BH=11，∠ABH=30°，∠GCH=28°，△ABC的内切圆圆心为H，BG垂直于HG，HD⊥BD，HF⊥AF。求∠GHD的大小。
Construction (CDL):
Shape(AD,DH,HA)
Shape(DB,BH,HD)
Shape(BG,GH,HB)
Shape(HG,GC,CH)
Shape(HC,CF,FH)
Shape(HF,FA,AH)
Collinear(ADB)
Collinear(BGC)
Collinear(CFA)
Conditions (CDL):
Equal(LengthOfLine(AF),24)
Equal(LengthOfLine(AH),25)
Equal(LengthOfLine(BH),11)
Equal(MeasureOfAngle(ABH),30)
Equal(MeasureOfAngle(GCH),28)
IsIncenterOfTriangle(H,ABC)
PerpendicularBetweenLine(BG,HG)
PerpendicularBetweenLine(HD,BD)
PerpendicularBetweenLine(HF,AF)
Goal: Value(MeasureOfAngle(GHD))
Answer: 120
Theorem sequence:
triangle_property_angle_sum(1,DBH)
triangle_property_angle_sum(1,HBG)
angle_addition(1,GHB,BHD)
Problem: As shown in the diagram, BF=4, BG=3, BG=HeightOfTriangle(ABC), CE=1, EH=HeightOfTriangle(DFE), FC=2, Triangle ABC is mirror similar to triangle DFE., AB⊥FB, BG⊥AG, CE is perpendicular to DE, EH⊥FH. Find the length of line EH.
问题：如图所示，BF=4，BG=3，BG=HeightOfTriangle(ABC)，CE=1，EH=HeightOfTriangle(DFE)，FC=2，三角形ABC与三角形DFE是镜像相似三角形，AB垂直于FB，BG⊥AG，CE⊥DE，EH⊥FH。求直线EH的长度。
Construction (CDL):
Shape(AB,BG,GA)
Shape(GB,BF,FI,IG)
Shape(IF,FC,CI)
Shape(IC,CE,EH,HI)
Shape(HE,ED,DH)
Collinear(AGIC)
Collinear(FIHD)
Collinear(BFCE)
Conditions (CDL):
Equal(LengthOfLine(BF),4)
Equal(LengthOfLine(BG),3)
Equal(LengthOfLine(BG),HeightOfTriangle(ABC))
Equal(LengthOfLine(CE),1)
Equal(LengthOfLine(EH),HeightOfTriangle(DFE))
Equal(LengthOfLine(FC),2)
MirrorSimilarBetweenTriangle(ABC,DFE)
PerpendicularBetweenLine(AB,FB)
PerpendicularBetweenLine(BG,AG)
PerpendicularBetweenLine(CE,DE)
PerpendicularBetweenLine(EH,FH)
Goal: Value(LengthOfLine(EH))
Answer: 3/2
Theorem sequence:
line_addition(1,BF,FC)
line_addition(1,FC,CE)
mirror_similar_triangle_property_angle_equal(1,CAB,FED)
adjacent_complementary_angle(1,IGB,BGA)
mirror_similar_triangle_judgment_aa(1,BCG,EHF)
mirror_similar_triangle_property_line_ratio(1,CGB,FEH)
mirror_similar_triangle_property_line_ratio(1,GBC,HFE)
Problem: As shown in the diagram, ∠BGC=20*x°, ∠DGB=100+20*x°. Find the measure of ∠DGB.
问题：如图所示，∠BGC=20*x°，∠DGB=100+20*x°。求∠DGB的大小。
Construction (CDL):
Shape(BG,GC)
Shape(DG,GB)
Shape(EG,GD)
Shape(CG,GE)
Collinear(BGE)
Collinear(CGD)
Conditions (CDL):
Equal(MeasureOfAngle(BGC),20*x)
Equal(MeasureOfAngle(DGB),100+20*x)
Goal: Value(MeasureOfAngle(DGB))
Answer: 140
Theorem sequence:
adjacent_complementary_angle(1,DGB,BGC)
Problem: As shown in the diagram, ∠DBC=2*x°, the measure of ⌒ECA is 9*x-1, the measure of arc EDC is 94, the tangent to circle E is BC. Find the value of x.
问题：如图所示，∠DBC=2*x°，弧ECA的角度为9*x-1，⌒EDC的角度为94，BC是⊙O的切线。求x的值。
Construction (CDL):
Shape(EDC,ECA,AD)
Shape(EDC,DB,BC)
Shape(EAD,DA)
Collinear(BDA)
Cocircular(E,DCA)
Conditions (CDL):
Equal(MeasureOfAngle(DBC),2*x)
Equal(MeasureOfArc(ECA),9*x-1)
Equal(MeasureOfArc(EDC),94)
IsTangentOfCircle(BC,E)
Goal: Value(x)
Answer: 19
Theorem sequence:
circle_property_circular_power_tangent_and_segment_angle(2,BC,BDA,E)
Problem: As shown in the diagram, AC=14, AD=21, BC=8, EB∥DC. Find the length of line ED.
问题：如图所示，AC=14，AD=21，BC=8，EB∥DC。求直线ED的长度。
Construction (CDL):
Shape(BA,AE,EB)
Shape(CB,BE,ED,DC)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(AD),21)
Equal(LengthOfLine(BC),8)
ParallelBetweenLine(EB,DC)
Goal: Value(LengthOfLine(ED))
Answer: 12
Theorem sequence:
line_addition(1,AB,BC)
line_addition(1,AE,ED)
parallel_property_corresponding_angle(2,CD,BE,A)
similar_triangle_judgment_aa(1,EBA,DCA)
similar_triangle_property_line_ratio(1,EBA,DCA)
similar_triangle_property_line_ratio(1,BAE,CAD)
Problem: As shown in the diagram, AB=12, AD=10, BC=y, BD=x, CD=z, AB is perpendicular to CB, BD⊥AD. Find the value of z.
问题：如图所示，AB=12，AD=10，BC=y，BD=x，CD=z，AB⊥CB，BD⊥AD。求z的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CD),z)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(BD,AD)
Goal: Value(z)
Answer: 22/5
Theorem sequence:
line_addition(1,CD,DA)
adjacent_complementary_angle(1,CDB,BDA)
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,CDB)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,ABC)
Problem: As shown in the diagram, ∠HEL=55°, ∠IHF=110°, IC is parallel to HF. Find the measure of ∠HIA.
问题：如图所示，∠HEL=55°，∠IHF=110°，IC∥HF。求∠HIA的大小。
Construction (CDL):
Shape(DI,IC)
Shape(CI,IH)
Shape(IH,HF)
Shape(FH,HB)
Shape(AI,ID)
Shape(HI,IA,AE,EH)
Shape(BH,HE)
Shape(JA,AI)
Shape(HE,EL)
Shape(KA,AJ)
Shape(EA,AK)
Shape(GE,EA)
Shape(LE,EG)
Collinear(JAEL)
Collinear(DIHB)
Collinear(CIAK)
Collinear(FHEG)
Conditions (CDL):
Equal(MeasureOfAngle(HEL),55)
Equal(MeasureOfAngle(IHF),110)
ParallelBetweenLine(IC,HF)
Goal: Value(MeasureOfAngle(HIA))
Answer: 110
Theorem sequence:
parallel_property_corresponding_angle(1,IC,HF,D)
vertical_angle(1,DIC,HIA)
Problem: As shown in the diagram, JP=9, KR=RJ, KS=SL, LT=TJ, RP=7/2. Find the length of line PL.
问题：如图所示，JP=9，KR=RJ，KS=SL，LT=TJ，RP=7/2。求直线PL的长度。
Construction (CDL):
Shape(RJ,JP,PR)
Shape(PJ,JT,TP)
Shape(PT,TL,LP)
Shape(PL,LS,SP)
Shape(PS,SK,KP)
Shape(PK,KR,RP)
Collinear(JRK)
Collinear(KSL)
Collinear(JTL)
Collinear(JPS)
Collinear(LPR)
Collinear(KPT)
Conditions (CDL):
Equal(LengthOfLine(JP),9)
Equal(LengthOfLine(KR),LengthOfLine(RJ))
Equal(LengthOfLine(KS),LengthOfLine(SL))
Equal(LengthOfLine(LT),LengthOfLine(TJ))
Equal(LengthOfLine(RP),7/2)
Goal: Value(LengthOfLine(PL))
Answer: 7
Theorem sequence:
median_of_triangle_judgment(1,LR,LKJ)
median_of_triangle_judgment(1,JS,JLK)
centroid_of_triangle_judgment_intersection(1,P,KJL,R,S)
centroid_of_triangle_property_line_ratio(1,P,LKJ,R)
Problem: As shown in the diagram, DE=x-2, FD=14, DE is the tangent to ⊙B, the tangent to ⊙B is DF. Find the value of x.
问题：如图所示，DE=x-2，FD=14，⊙O的切线为DE，DF是⊙O的切线。求x的值。
Construction (CDL):
Shape(BEF,ED,DF)
Shape(BFE)
Cocircular(B,EF)
Conditions (CDL):
Equal(LengthOfLine(DE),x-2)
Equal(LengthOfLine(FD),14)
IsTangentOfCircle(DE,B)
IsTangentOfCircle(DF,B)
Goal: Value(x)
Answer: 16
Theorem sequence:
tangent_of_circle_property_length_equal(1,DF,DE,B)
Problem: As shown in the diagram, ∠CBA=28°, ∠DAE=68°, ∠ECB=82°. Find the measure of ∠BEC.
问题：如图所示，∠CBA=28°，∠DAE=68°，∠ECB=82°。求∠BEC的大小。
Construction (CDL):
Shape(BE,EC,CB)
Shape(ED,DA,AE)
Collinear(BEA)
Collinear(CED)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),28)
Equal(MeasureOfAngle(DAE),68)
Equal(MeasureOfAngle(ECB),82)
Goal: Value(MeasureOfAngle(BEC))
Answer: 70
Theorem sequence:
triangle_property_angle_sum(1,BEC)
Problem: As shown in the diagram, BH=Mul(LengthOfLine(EB),2), EB=HA, HE=9, quadrilateral CBAD is a square, EHGF is a square. Find the area of ADCB.
问题：如图所示，BH=Mul(LengthOfLine(EB),2)，EB=HA，HE=9，CBAD是正方形，EHGF是正方形。求四边形ADCB的面积。
Construction (CDL):
Shape(EB,BC,CE)
Shape(BH,HA,AB)
Shape(AG,GD,DA)
Shape(CD,DF,FC)
Shape(AD,DC,CB,BA)
Collinear(EBH)
Collinear(ECF)
Collinear(FDG)
Collinear(HAG)
Conditions (CDL):
Equal(LengthOfLine(BH),Mul(LengthOfLine(EB),2))
Equal(LengthOfLine(EB),LengthOfLine(HA))
Equal(LengthOfLine(HE),9)
Square(CBAD)
Square(EHGF)
Goal: Value(AreaOfQuadrilateral(ADCB))
Answer: 45
Theorem sequence:
line_addition(1,EB,BH)
right_triangle_judgment_angle(1,BHA)
right_triangle_property_pythagorean(1,BHA)
parallelogram_area_formula_sine(1,ADCB)
Problem: As shown in the diagram, AF=24, AH=25, BH=11, ∠ABH=30°, ∠GCH=28°, H is the center of the inscribed circle of △ABC, BG⊥HG, HD⊥BD, HF is perpendicular to AF. Find the length of line BD.
问题：如图所示，AF=24，AH=25，BH=11，∠ABH=30°，∠GCH=28°，H是△ABC内切圆的圆心，BG⊥HG，HD垂直于BD，HF垂直于AF。求直线BD的长度。
Construction (CDL):
Shape(AD,DH,HA)
Shape(HF,FA,AH)
Shape(HD,DB,BH)
Shape(HG,GC,CH)
Shape(HC,CF,FH)
Shape(HB,BG,GH)
Collinear(ADB)
Collinear(BGC)
Collinear(AFC)
Conditions (CDL):
Equal(LengthOfLine(AF),24)
Equal(LengthOfLine(AH),25)
Equal(LengthOfLine(BH),11)
Equal(MeasureOfAngle(ABH),30)
Equal(MeasureOfAngle(GCH),28)
IsIncenterOfTriangle(H,ABC)
PerpendicularBetweenLine(BG,HG)
PerpendicularBetweenLine(HD,BD)
PerpendicularBetweenLine(HF,AF)
Goal: Value(LengthOfLine(BD))
Answer: 11*sqrt(3)/2
Theorem sequence:
triangle_property_angle_sum(1,HDB)
sine_theorem(1,BHD)
Problem: As shown in the diagram, ∠FEG=57°, quadrilateral EHGF is a rectangle. Find the measure of ∠GEH.
问题：如图所示，∠FEG=57°，四边形EHGF是长方形。求∠GEH的大小。
Construction (CDL):
Shape(HK,KE,EH)
Shape(KH,HG,GK)
Shape(KG,GF,FK)
Shape(EK,KF,FE)
Collinear(EKG)
Collinear(HKF)
Conditions (CDL):
Equal(MeasureOfAngle(FEG),57)
Rectangle(EHGF)
Goal: Value(MeasureOfAngle(GEH))
Answer: 33
Theorem sequence:
angle_addition(1,FEK,KEH)
Problem: As shown in the diagram, the area of quadrilateral EFHG is 525, the area of quadrilateral XBAC is 1575, AB=18, FH=x, XBAC is similar to EFHG. Find the value of x.
问题：如图所示，EFHG的面积为525，XBAC的面积为1575，AB=18，FH=x，四边形XBAC与四边形EFHG相似。求x的值。
Construction (CDL):
Shape(XB,BA,AC,CX)
Shape(EF,FH,HG,GE)
Conditions (CDL):
Equal(AreaOfQuadrilateral(EFHG),525)
Equal(AreaOfQuadrilateral(XBAC),1575)
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(FH),x)
SimilarBetweenQuadrilateral(XBAC,EFHG)
Goal: Value(x)
Answer: 6*sqrt(3)
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,XBAC,EFHG)
similar_quadrilateral_property_line_ratio(1,BACX,FHGE)
Problem: As shown in the diagram, ML=w, NM=2*y+5, NQ=3*x+2, QL=12, RM=4*x-2, RQ=3*y, MN and RQ are opposite sides of the ▱ NMRQ. Find the length of line QR.
问题：如图所示，ML=w，NM=2*y+5，NQ=3*x+2，QL=12，RM=4*x-2，RQ=3*y，NQ和MR是▱NMRQ的一组对边。求直线QR的长度。
Construction (CDL):
Shape(NM,ML,LN)
Shape(LM,MR,RL)
Shape(LR,RQ,QL)
Shape(NL,LQ,QN)
Collinear(NLR)
Collinear(MLQ)
Conditions (CDL):
Equal(LengthOfLine(ML),w)
Equal(LengthOfLine(NM),2*y+5)
Equal(LengthOfLine(NQ),3*x+2)
Equal(LengthOfLine(QL),12)
Equal(LengthOfLine(RM),4*x-2)
Equal(LengthOfLine(RQ),3*y)
Parallelogram(NMRQ)
Goal: Value(LengthOfLine(QR))
Answer: 15
Theorem sequence:
parallelogram_property_diagonal_bisection(1,NMRQ,L)
parallelogram_property_diagonal_bisection(1,MRQN,L)
vertical_angle(1,MLN,QLR)
congruent_triangle_judgment_sas(1,LNM,LRQ)
congruent_triangle_property_line_equal(1,LNM,LRQ)
Problem: As shown in the diagram, AB=x, AC=8, BC=15, AC⊥BC. Find the value of x.
问题：如图所示，AB=x，AC=8，BC=15，AC⊥BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),15)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 17
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, O is the center of ⊙O. Find the ratio of the measure of ∠ACB to the measure of ∠AOB.
问题：如图所示，O是⊙O的圆心。求∠ACB的大小与∠AOB的大小之比。
Construction (CDL):
Shape(CA,OAC)
Shape(OA,AC,CB,BO)
Shape(BC,OCB)
Shape(OB,OBA,AO)
Cocircular(O,BAC)
Conditions (CDL):
IsCentreOfCircle(O,O)
Goal: Value(Div(MeasureOfAngle(ACB),MeasureOfAngle(AOB)))
Answer: 1/2
Theorem sequence:
arc_property_circumference_angle_external(1,OBA,C)
arc_property_center_angle(1,OBA,O)
Problem: As shown in the diagram, ∠BDC=3*y+10°, ∠BNC=2*x+1°, ∠DBN=5*y°, ∠END=5*x+5°, quadrilateral NDCB is a rectangle. Find the value of y.
问题：如图所示，∠BDC=3*y+10°，∠BNC=2*x+1°，∠DBN=5*y°，∠END=5*x+5°，NDCB是矩形。求y的值。
Construction (CDL):
Shape(EB,BN,NE)
Shape(DE,EN,ND)
Shape(ED,DC,CE)
Shape(EC,CB,BE)
Collinear(NEC)
Collinear(DEB)
Conditions (CDL):
Equal(MeasureOfAngle(BDC),3*y+10)
Equal(MeasureOfAngle(BNC),2*x+1)
Equal(MeasureOfAngle(DBN),5*y)
Equal(MeasureOfAngle(END),5*x+5)
Rectangle(NDCB)
Goal: Value(y)
Answer: 5
Theorem sequence:
parallel_property_alternate_interior_angle(2,NB,DC)
Problem: As shown in the diagram, WZ=4*a, XW=x+11, XY=3*a+7, YZ=2, XWZY is a parallelogram. Find the value of a.
问题：如图所示，WZ=4*a，XW=x+11，XY=3*a+7，YZ=2，WX和ZY是▱XWZY的一组对边。求a的值。
Construction (CDL):
Shape(XW,WZ,ZY,YX)
Conditions (CDL):
Equal(LengthOfLine(WZ),4*a)
Equal(LengthOfLine(XW),x+11)
Equal(LengthOfLine(XY),3*a+7)
Equal(LengthOfLine(YZ),2)
Parallelogram(XWZY)
Goal: Value(a)
Answer: 7
Theorem sequence:
parallelogram_property_opposite_line_equal(1,WZYX)
Problem: As shown in the diagram, QP=11, QR=5, RQ=5, TP=16, TQ=15, △QSR is similar to △QPT.. Find the perimeter of triangle QSR.
问题：如图所示，QP=11，QR=5，RQ=5，TP=16，TQ=15，三角形QSR与三角形QPT是相似三角形。求三角形QSR的周长。
Construction (CDL):
Shape(QP,PT,TQ)
Shape(QS,SR,RQ)
Collinear(PQS)
Collinear(TQR)
Conditions (CDL):
Equal(LengthOfLine(QP),11)
Equal(LengthOfLine(QR),5)
Equal(LengthOfLine(RQ),5)
Equal(LengthOfLine(TP),16)
Equal(LengthOfLine(TQ),15)
SimilarBetweenTriangle(QSR,QPT)
Goal: Value(PerimeterOfTriangle(QSR))
Answer: 14
Theorem sequence:
triangle_perimeter_formula(1,QPT)
similar_triangle_property_line_ratio(1,SRQ,PTQ)
similar_triangle_property_perimeter_ratio(1,RQS,TQP)
Problem: As shown in the diagram, ∠ABC=32°, ∠CAD=62°, ∠CDS=125°. Find the measure of ∠SAB.
问题：如图所示，∠ABC=32°，∠CAD=62°，∠CDS=125°。求∠SAB的大小。
Construction (CDL):
Shape(CA,AD,DC)
Shape(DA,AB,BD)
Shape(SD,DB)
Collinear(CDB)
Collinear(ADS)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),32)
Equal(MeasureOfAngle(CAD),62)
Equal(MeasureOfAngle(CDS),125)
Goal: Value(MeasureOfAngle(SAB))
Answer: 23
Theorem sequence:
adjacent_complementary_angle(1,CDS,SDB)
adjacent_complementary_angle(1,SDB,BDA)
triangle_property_angle_sum(1,DAB)
Problem: As shown in the diagram, EB=3, ∠CBE=32°, the center of ⊙B is B. Find the area of the sector BEC.
问题：如图所示，EB=3，∠CBE=32°，B是⊙B的圆心。求扇形BEC的面积。
Construction (CDL):
Shape(CB,BE,BEC)
Shape(EB,BC,BCE)
Cocircular(B,EC)
Conditions (CDL):
Equal(LengthOfLine(EB),3)
Equal(MeasureOfAngle(CBE),32)
IsCentreOfCircle(B,B)
Goal: Value(AreaOfSector(BEC))
Answer: 4*pi/5
Theorem sequence:
radius_of_circle_property_length_equal(1,BE,B)
arc_property_center_angle(1,BEC,B)
sector_area_formula(1,BEC)
Problem: As shown in the diagram, AB=13, AC=10, BC=13, BD⊥DD. Find the perimeter of △ACB.
问题：如图所示，AB=13，AC=10，BC=13，BD垂直于DD。求三角形ACB的周长。
Construction (CDL):
Shape(AD,DB,BA)
Shape(BD,DC,CB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(BC),13)
PerpendicularBetweenLine(BD,DC)
Goal: Value(PerimeterOfTriangle(ACB))
Answer: 36
Theorem sequence:
triangle_perimeter_formula(1,ACB)
Problem: As shown in the diagram, BC=CD, CB=2*x+4, CD=10, DB=x+2. Find the length of line BC.
问题：如图所示，BC=CD，CB=2*x+4，CD=10，DB=x+2。求直线BC的长度。
Construction (CDL):
Shape(BD,DC,CB)
Conditions (CDL):
Equal(LengthOfLine(BC),LengthOfLine(CD))
Equal(LengthOfLine(CB),2*x+4)
Equal(LengthOfLine(CD),10)
Equal(LengthOfLine(DB),x+2)
Goal: Value(LengthOfLine(BC))
Answer: 10
Theorem sequence:
Problem: As shown in the diagram, ∠TWV=3*x-4°, ∠UTW=x°, ∠VUT=3*x-4°, ∠WVU=x°, TU and WV are opposite sides of the parallelogram UTWV. Find the measure of ∠TWV.
问题：如图所示，∠TWV=3*x-4°，∠UTW=x°，∠VUT=3*x-4°，∠WVU=x°，UV和TW是▱UTWV的一组对边。求∠TWV的大小。
Construction (CDL):
Shape(UT,TW,WV,VU)
Conditions (CDL):
Equal(MeasureOfAngle(TWV),3*x-4)
Equal(MeasureOfAngle(UTW),x)
Equal(MeasureOfAngle(VUT),3*x-4)
Equal(MeasureOfAngle(WVU),x)
Parallelogram(UTWV)
Goal: Value(MeasureOfAngle(TWV))
Answer: 134
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,UV,TW)
Problem: As shown in the diagram, AB=x, AC=y, BC=14, ∠CBA=30°, BA⊥CA. Find the value of y.
问题：如图所示，AB=x，AC=y，BC=14，∠CBA=30°，BA⊥CA。求y的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(BC),14)
Equal(MeasureOfAngle(CBA),30)
PerpendicularBetweenLine(BA,CA)
Goal: Value(y)
Answer: 7
Theorem sequence:
sine_theorem(1,CBA)
Problem: As shown in the diagram, ∠FBC=50°, the measure of arc AFC is x, the center of circle A is A, the tangent to ⊙A is BC, BF is the tangent to circle A. Find the value of x.
问题：如图所示，∠FBC=50°，弧AFC的角度为x，圆A的圆心为A，⊙O的切线为BC，圆O的切线为BF。求x的值。
Construction (CDL):
Shape(BC,AFC,FB)
Shape(CA,AF,AFC)
Shape(AC,ACF,FA)
Cocircular(A,FC)
Conditions (CDL):
Equal(MeasureOfAngle(FBC),50)
Equal(MeasureOfArc(AFC),x)
IsCentreOfCircle(A,A)
IsTangentOfCircle(BC,A)
IsTangentOfCircle(BF,A)
Goal: Value(x)
Answer: 130
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BF,A,A)
tangent_of_circle_property_perpendicular(2,BC,A,A)
quadrilateral_property_angle_sum(1,FBCA)
arc_property_center_angle(1,AFC,A)
Problem: As shown in the diagram, CF=60, DE=EC, the center of ⊙D is D, AC is the diameter of circle D, CE is perpendicular to DE, DE⊥EE. Find the measure of ∠BAC.
问题：如图所示，CF=60，DE=EC，D是圆D的圆心，圆D的直径为AC，CE⊥DE，DE垂直于EE。求∠BAC的大小。
Construction (CDL):
Shape(AB,DBA)
Shape(AD,DE,EB,BA)
Shape(ED,DC,CE)
Shape(BC,DCB)
Shape(AF,FC,CA)
Shape(FA,DAF)
Shape(CF,DFC)
Collinear(ADC)
Collinear(BEC)
Cocircular(D,AFCB)
Conditions (CDL):
Equal(LengthOfLine(CF),60)
Equal(LengthOfLine(DE),LengthOfLine(EC))
IsCentreOfCircle(D,D)
IsDiameterOfCircle(AC,D)
PerpendicularBetweenLine(CE,DE)
PerpendicularBetweenLine(DE,EC)
Goal: Value(MeasureOfAngle(BAC))
Answer: 45
Theorem sequence:
diameter_of_circle_property_right_angle(1,CBA,D)
isosceles_triangle_judgment_line_equal(1,EDC)
isosceles_triangle_property_angle_equal(1,EDC)
triangle_property_angle_sum(1,EDC)
triangle_property_angle_sum(1,BAC)
Problem: As shown in the diagram, AB=6, AC=3.6, DC=2, BD⊥CD. Find the area of △BAC.
问题：如图所示，AB=6，AC=3.6，DC=2，BD垂直于CD。求△BAC的面积。
Construction (CDL):
Shape(CB,BD,DC)
Shape(CD,DA,AC)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AC),3.6)
Equal(LengthOfLine(DC),2)
PerpendicularBetweenLine(BD,CD)
Goal: Value(AreaOfTriangle(BAC))
Answer: 6
Theorem sequence:
altitude_of_triangle_judgment(1,CD,CBA)
triangle_area_formula_common(1,CBA)
Problem: As shown in the diagram, GK=GH, HK=KJ, ∠HGK=42°. Find the measure of ∠KJH.
问题：如图所示，GK=GH，HK=KJ，∠HGK=42°。求∠KJH的大小。
Construction (CDL):
Shape(HG,GK,KH)
Shape(HK,KJ,JH)
Collinear(GKJ)
Conditions (CDL):
Equal(LengthOfLine(GK),LengthOfLine(GH))
Equal(LengthOfLine(HK),LengthOfLine(KJ))
Equal(MeasureOfAngle(HGK),42)
Goal: Value(MeasureOfAngle(KJH))
Answer: 69/2
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,GKH)
isosceles_triangle_property_angle_equal(1,GKH)
triangle_property_angle_sum(1,GKH)
adjacent_complementary_angle(1,GKH,HKJ)
isosceles_triangle_judgment_line_equal(1,KJH)
isosceles_triangle_property_angle_equal(1,KJH)
triangle_property_angle_sum(1,KJH)
Problem: As shown in the diagram, PS=y+2, PT=y-3, QS=16, RS=12, RT is parallel to QP. Find the value of y.
问题：如图所示，PS=y+2，PT=y-3，QS=16，RS=12，RT平行于QP。求y的值。
Construction (CDL):
Shape(QP,PT,TR,RQ)
Shape(RT,TS,SR)
Collinear(PTS)
Collinear(QRS)
Conditions (CDL):
Equal(LengthOfLine(PS),y+2)
Equal(LengthOfLine(PT),y-3)
Equal(LengthOfLine(QS),16)
Equal(LengthOfLine(RS),12)
ParallelBetweenLine(RT,QP)
Goal: Value(y)
Answer: 14/3
Theorem sequence:
line_addition(1,PT,TS)
parallel_property_corresponding_angle(1,RT,QP,S)
similar_triangle_judgment_aa(1,TSR,PSQ)
similar_triangle_property_line_ratio(1,TSR,PSQ)
similar_triangle_property_line_ratio(1,RTS,QPS)
Problem: As shown in the diagram, AB=x+3, BC=2*x-8, DB=3, EB=5, EA∥CD. Find the length of line AB.
问题：如图所示，AB=x+3，BC=2*x-8，DB=3，EB=5，EA平行于CD。求直线AB的长度。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BC,CD,DB)
Collinear(ABC)
Collinear(EBD)
Conditions (CDL):
Equal(LengthOfLine(AB),x+3)
Equal(LengthOfLine(BC),2*x-8)
Equal(LengthOfLine(DB),3)
Equal(LengthOfLine(EB),5)
ParallelBetweenLine(EA,CD)
Goal: Value(LengthOfLine(AB))
Answer: 10
Theorem sequence:
parallel_property_alternate_interior_angle(1,EA,CD)
vertical_angle(1,EBA,DBC)
similar_triangle_judgment_aa(1,AEB,CDB)
similar_triangle_property_line_ratio(1,EBA,DBC)
similar_triangle_property_line_ratio(1,AEB,CDB)