SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠GID=4*x°, ∠HIM=8*x-12°, ∠KHA=6*y+10°, HO is parallel to IM. Find the value of y.
问题：如图所示，∠GID=4*x°，∠HIM=8*x-12°，∠KHA=6*y+10°，HO平行于IM。求y的值。
Construction (CDL):
Shape(KH,HA)
Shape(AH,HO)
Shape(DI,IH)
Shape(HI,IM)
Shape(IH,HK)
Shape(OH,HI)
Shape(GI,ID)
Shape(MI,IG)
Collinear(KHO)
Collinear(DIM)
Collinear(AHIG)
Conditions (CDL):
Equal(MeasureOfAngle(GID),4*x)
Equal(MeasureOfAngle(HIM),8*x-12)
Equal(MeasureOfAngle(KHA),6*y+10)
ParallelBetweenLine(HO,IM)
Goal: Value(y)
Answer: 79/3
Theorem sequence:
vertical_angle(1,AIM,GID)
parallel_property_ipsilateral_internal_angle(1,HO,IM)
vertical_angle(1,KHA,OHI)
Problem: As shown in the diagram, ∠FOE=118°, ∠JDA=104°, OD is parallel to FI. Find the measure of ∠IFJ.
问题：如图所示，∠FOE=118°，∠JDA=104°，OD∥FI。求∠IFJ的大小。
Construction (CDL):
Shape(JD,DA)
Shape(FO,OE)
Shape(IF,FJ)
Shape(JF,FK)
Collinear(JDIC)
Collinear(HOFJ)
Collinear(ADOE)
Collinear(BIFK)
Conditions (CDL):
Equal(MeasureOfAngle(FOE),118)
Equal(MeasureOfAngle(JDA),104)
ParallelBetweenLine(OD,FI)
Goal: Value(MeasureOfAngle(IFJ))
Answer: 62
Theorem sequence:
parallel_property_collinear_extend(1,OD,FI,E)
parallel_property_collinear_extend(2,IF,OE,K)
parallel_property_corresponding_angle(1,FK,OE,J)
flat_angle(1,IFK)
angle_addition(1,IFJ,JFK)
Problem: As shown in the diagram, DC=DA, ∠ACD=66°, ∠DBA=24°. Find the measure of ∠DAC.
问题：如图所示，DC=DA，∠ACD=66°，∠DBA=24°。求∠DAC的大小。
Construction (CDL):
Shape(DB,BA,AD)
Shape(CD,DA,AC)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(DC),LengthOfLine(DA))
Equal(MeasureOfAngle(ACD),66)
Equal(MeasureOfAngle(DBA),24)
Goal: Value(MeasureOfAngle(DAC))
Answer: 66
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DAC)
isosceles_triangle_property_angle_equal(1,DAC)
Problem: As shown in the diagram, ∠BTS=3*x°, ∠BVU=y+16°, ∠TSB=3*y°, ∠VUB=2*x+15°. Find the measure of ∠TSB.
问题：如图所示，∠BTS=3*x°，∠BVU=y+16°，∠TSB=3*y°，∠VUB=2*x+15°。求∠TSB的大小。
Construction (CDL):
Shape(ATS,ST)
Shape(AVU,UV)
Shape(AUT,TB,BU)
Shape(TS,SB,BT)
Shape(UB,BV,VU)
Shape(ASV,VA,AS)
Shape(AV,VB,BS,SA)
Collinear(TBV)
Collinear(UBS)
Cocircular(A,UTSV)
Conditions (CDL):
Equal(MeasureOfAngle(BTS),3*x)
Equal(MeasureOfAngle(BVU),y+16)
Equal(MeasureOfAngle(TSB),3*y)
Equal(MeasureOfAngle(VUB),2*x+15)
Goal: Value(MeasureOfAngle(TSB))
Answer: 24
Theorem sequence:
arc_property_circumference_angle_external(1,AUT,S)
arc_property_circumference_angle_external(1,AUT,V)
Problem: As shown in the diagram, ∠CDE=57°, ∠DBE=42°, ∠EBA=36°, ∠ECD=28°. Find the measure of ∠BAE.
问题：如图所示，∠CDE=57°，∠DBE=42°，∠EBA=36°，∠ECD=28°。求∠BAE的大小。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BE,ED,DB)
Shape(EC,CD,DE)
Collinear(AED)
Collinear(CEB)
Conditions (CDL):
Equal(MeasureOfAngle(CDE),57)
Equal(MeasureOfAngle(DBE),42)
Equal(MeasureOfAngle(EBA),36)
Equal(MeasureOfAngle(ECD),28)
Goal: Value(MeasureOfAngle(BAE))
Answer: 49
Theorem sequence:
triangle_property_angle_sum(1,ADB)
triangle_property_angle_sum(1,BCD)
angle_addition(1,DBE,EBA)
angle_addition(1,CDE,EDB)
Problem: As shown in the diagram, SR=5, ST=4, TR=3, ST⊥RT. Find the value of tan(SR).
问题：如图所示，SR=5，ST=4，TR=3，ST垂直于RT。求tan(SR)的值。
Construction (CDL):
Shape(TS,SR,RT)
Conditions (CDL):
Equal(LengthOfLine(SR),5)
Equal(LengthOfLine(ST),4)
Equal(LengthOfLine(TR),3)
PerpendicularBetweenLine(ST,RT)
Goal: Value(Tan(MeasureOfAngle(SRT)))
Answer: 4/3
Theorem sequence:
cosine_theorem(1,RTS)
Problem: As shown in the diagram, ∠ADC=105°, ∠BAD=74°, ∠DCB=x°, quadrilateral ADCB is a kite. Find the measure of ∠DCB.
问题：如图所示，∠ADC=105°，∠BAD=74°，∠DCB=x°，四边形ADCB是风筝形。求∠DCB的大小。
Construction (CDL):
Shape(BA,AD,DC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),105)
Equal(MeasureOfAngle(BAD),74)
Equal(MeasureOfAngle(DCB),x)
Kite(ADCB)
Goal: Value(MeasureOfAngle(DCB))
Answer: 76
Theorem sequence:
kite_property_opposite_angle_equal(1,ADCB)
quadrilateral_property_angle_sum(1,ADCB)
Problem: As shown in the diagram, CE=24, DE=24, EA=12, EB=x. Find the length of line AB.
问题：如图所示，CE=24，DE=24，EA=12，EB=x。求直线AB的长度。
Construction (CDL):
Shape(FAD,DE,EA)
Shape(FCA,AE,EC)
Shape(FDB,BF,FE,ED)
Shape(FBC,CE,EF,FB)
Collinear(DEC)
Collinear(AEFB)
Cocircular(F,ADBC)
Conditions (CDL):
Equal(LengthOfLine(CE),24)
Equal(LengthOfLine(DE),24)
Equal(LengthOfLine(EA),12)
Equal(LengthOfLine(EB),x)
Goal: Value(LengthOfLine(AB))
Answer: 60
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,AEB,DEC,F)
line_addition(1,AE,EB)
Problem: As shown in the diagram, RS=x, RT=y+3, TS=49, VU=21, WU=29, WV=20, ∠RST=∠UWV, ∠STR=∠WVU, ∠TRS=∠VUW. Find the value of x.
问题：如图所示，RS=x，RT=y+3，TS=49，VU=21，WU=29，WV=20，∠RST=∠UWV，∠STR=∠WVU，∠TRS=∠VUW。求x的值。
Construction (CDL):
Shape(RS,ST,TR)
Shape(UW,WV,VU)
Conditions (CDL):
Equal(LengthOfLine(RS),x)
Equal(LengthOfLine(RT),y+3)
Equal(LengthOfLine(TS),49)
Equal(LengthOfLine(VU),21)
Equal(LengthOfLine(WU),29)
Equal(LengthOfLine(WV),20)
Equal(MeasureOfAngle(RST),MeasureOfAngle(UWV))
Equal(MeasureOfAngle(STR),MeasureOfAngle(WVU))
Equal(MeasureOfAngle(TRS),MeasureOfAngle(VUW))
Goal: Value(x)
Answer: 1421/20
Theorem sequence:
similar_triangle_judgment_aa(1,RST,UWV)
similar_triangle_property_line_ratio(1,RST,UWV)
similar_triangle_property_line_ratio(1,TRS,VUW)
Problem: As shown in the diagram, DA=8, DH=x, HF=6, ∠ABC=45°, ∠DAB=30°, DC∥AB, AH is perpendicular to DH, CF is perpendicular to BF. Find the value of x.
问题：如图所示，DA=8，DH=x，HF=6，∠ABC=45°，∠DAB=30°，DC∥AB，AH垂直于DH，CF垂直于BF。求x的值。
Construction (CDL):
Shape(DA,AH,HD)
Shape(DH,HF,FC,CD)
Shape(CF,FB,BC)
Collinear(AHFB)
Conditions (CDL):
Equal(LengthOfLine(DA),8)
Equal(LengthOfLine(DH),x)
Equal(LengthOfLine(HF),6)
Equal(MeasureOfAngle(ABC),45)
Equal(MeasureOfAngle(DAB),30)
ParallelBetweenLine(DC,AB)
PerpendicularBetweenLine(AH,DH)
PerpendicularBetweenLine(CF,BF)
Goal: Value(x)
Answer: 4
Theorem sequence:
sine_theorem(1,DAH)
Problem: As shown in the diagram, BD=7, FE=6, OE=9, OB and OD are one pair of adjacent sides of the kite FBOD. Find the area of quadrilateral FBOD.
问题：如图所示，BD=7，FE=6，OE=9，四边形FBOD是风筝形。求四边形FBOD的面积。
Construction (CDL):
Shape(FB,BE,EF)
Shape(FE,ED,DF)
Shape(BO,OE,EB)
Shape(EO,OD,DE)
Collinear(FEO)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(BD),7)
Equal(LengthOfLine(FE),6)
Equal(LengthOfLine(OE),9)
Kite(FBOD)
Goal: Value(AreaOfQuadrilateral(FBOD))
Answer: 105/2
Theorem sequence:
line_addition(1,FE,EO)
kite_area_formula_diagonal(1,FBOD)
Problem: As shown in the diagram, ∠KGL=x°, the measure of ⌒AJH is 47, the measure of ⌒ALK is 116. Find the value of x.
问题：如图所示，∠KGL=x°，弧AJH的角度为47，⌒ALK的角度为116。求x的值。
Construction (CDL):
Shape(LG,GH,AHL)
Shape(HG,GJ,AJH)
Shape(KG,GL,ALK)
Shape(JG,GK,KJ)
Shape(JK,AKJ)
Collinear(LGJ)
Collinear(HGK)
Cocircular(A,HLKJ)
Conditions (CDL):
Equal(MeasureOfAngle(KGL),x)
Equal(MeasureOfArc(AJH),47)
Equal(MeasureOfArc(ALK),116)
Goal: Value(x)
Answer: 163/2
Theorem sequence:
arc_property_circumference_angle_external(1,ALK,J)
arc_property_circumference_angle_external(1,AJH,K)
triangle_property_angle_sum(1,JGK)
adjacent_complementary_angle(1,JGK,KGL)
Problem: As shown in the diagram, ∠APC=72+x°, CP is the altitude of △ CAB, CQ is the angle bisector of ∠BCA, R bisects segment AB. Find the value of x.
问题：如图所示，∠APC=72+x°，CP是△CAB的高，CQ是∠BCA的角平分线，R平分线段AB。求x的值。
Construction (CDL):
Shape(CA,AP,PC)
Shape(CP,PQ,QC)
Shape(CQ,QR,RC)
Shape(CR,RB,BC)
Collinear(APQRB)
Conditions (CDL):
Equal(MeasureOfAngle(APC),72+x)
IsAltitudeOfTriangle(CP,CAB)
IsBisectorOfAngle(CQ,BCA)
IsMidpointOfLine(R,AB)
Goal: Value(x)
Answer: 18
Theorem sequence:
Problem: As shown in the diagram, AZ=y, QZ=z, RQ=12, RS=10, RZ=x, ∠AQR=30°, ∠SPA=45°, SR is parallel to AZ, PA is perpendicular to SA, RZ⊥QZ. Find the value of y.
问题：如图所示，AZ=y，QZ=z，RQ=12，RS=10，RZ=x，∠AQR=30°，∠SPA=45°，SR平行于AZ，PA垂直于SA，RZ⊥QZ。求y的值。
Construction (CDL):
Shape(SP,PA,AS)
Shape(SA,AZ,ZR,RS)
Shape(RZ,ZQ,QR)
Collinear(PAZQ)
Conditions (CDL):
Equal(LengthOfLine(AZ),y)
Equal(LengthOfLine(QZ),z)
Equal(LengthOfLine(RQ),12)
Equal(LengthOfLine(RS),10)
Equal(LengthOfLine(RZ),x)
Equal(MeasureOfAngle(AQR),30)
Equal(MeasureOfAngle(SPA),45)
ParallelBetweenLine(SR,AZ)
PerpendicularBetweenLine(PA,SA)
PerpendicularBetweenLine(RZ,QZ)
Goal: Value(y)
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,PAS,SAZ)
adjacent_complementary_angle(1,PZR,RZQ)
perpendicular_judgment_angle(1,SA,ZA)
perpendicular_judgment_angle(1,AZ,RZ)
parallel_judgment_per_per(1,AS,ZR)
parallelogram_judgment_parallel_and_parallel(1,SAZR)
parallelogram_property_opposite_line_equal(1,AZRS)
Problem: As shown in the diagram, AC=x-3, BA=2*x-7, BC=4*x-21, BA and BC are the legs of the isosceles △ BAC. Find the length of line AC.
问题：如图所示，AC=x-3，BA=2*x-7，BC=4*x-21，三角形BAC为等腰三角形。求直线AC的长度。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),x-3)
Equal(LengthOfLine(BA),2*x-7)
Equal(LengthOfLine(BC),4*x-21)
IsoscelesTriangle(BAC)
Goal: Value(LengthOfLine(AC))
Answer: 4
Theorem sequence:
Problem: As shown in the diagram, BC=z, BX=6*x, CA=y, CX=36, XA=x, AC is perpendicular to BC, BX⊥CX. Find the value of y.
问题：如图所示，BC=z，BX=6*x，CA=y，CX=36，XA=x，AC垂直于BC，BX垂直于CX。求y的值。
Construction (CDL):
Shape(CB,BX,XC)
Shape(CX,XA,AC)
Collinear(BXA)
Conditions (CDL):
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BX),6*x)
Equal(LengthOfLine(CA),y)
Equal(LengthOfLine(CX),36)
Equal(LengthOfLine(XA),x)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BX,CX)
Goal: Value(y)
Answer: 6*sqrt(42)
Theorem sequence:
adjacent_complementary_angle(1,BXC,CXA)
angle_addition(1,ACX,XCB)
triangle_property_angle_sum(1,CBX)
similar_triangle_judgment_aa(1,CBX,ACX)
similar_triangle_property_line_ratio(1,CBX,ACX)
similar_triangle_property_line_ratio(1,BXC,CXA)
right_triangle_judgment_angle(1,CXA)
right_triangle_property_pythagorean(1,CXA)
Problem: As shown in the diagram, ∠GOI=3*y+1°, ∠HBI=3*x+11°, ∠OIE=4*x-5°, GE∥OI, HG∥BE, OI∥HB. Find the value of y.
问题：如图所示，∠GOI=3*y+1°，∠HBI=3*x+11°，∠OIE=4*x-5°，GE∥OI，HG∥BE，OI∥HB。求y的值。
Construction (CDL):
Shape(GO,OI,IE,EG)
Shape(OH,HB,BI,IO)
Collinear(GOH)
Collinear(EIB)
Conditions (CDL):
Equal(MeasureOfAngle(GOI),3*y+1)
Equal(MeasureOfAngle(HBI),3*x+11)
Equal(MeasureOfAngle(OIE),4*x-5)
ParallelBetweenLine(GE,OI)
ParallelBetweenLine(HG,BE)
ParallelBetweenLine(OI,HB)
Goal: Value(y)
Answer: 40
Theorem sequence:
parallel_property_corresponding_angle(2,BH,IO,E)
parallel_property_collinear_extend(3,HG,BE,O)
parallel_property_collinear_extend(3,EB,GO,I)
parallel_property_ipsilateral_internal_angle(1,OG,IE)
Problem: As shown in the diagram, ∠EAB=40°, AB⊥DB, ABDC is a rectangle. Find the measure of ∠DEB.
问题：如图所示，∠EAB=40°，AB⊥DB，四边形ABDC是长方形。求∠DEB的大小。
Construction (CDL):
Shape(AE,EC,CA)
Shape(BE,EA,AB)
Shape(EB,BD,DE)
Shape(CE,ED,DC)
Collinear(AED)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(EAB),40)
PerpendicularBetweenLine(AB,DB)
Rectangle(ABDC)
Goal: Value(MeasureOfAngle(DEB))
Answer: 80
Theorem sequence:
rectangle_property_diagonal_equal(1,ABDC)
parallelogram_property_diagonal_bisection(1,ABDC,E)
parallelogram_property_diagonal_bisection(1,BDCA,E)
line_addition(1,AE,ED)
line_addition(1,BE,EC)
isosceles_triangle_judgment_line_equal(1,EBD)
isosceles_triangle_property_angle_equal(1,EBD)
triangle_property_angle_sum(1,ABD)
triangle_property_angle_sum(1,EBD)
Problem: As shown in the diagram, AB=5, AC=7, BC=6, DE=3, △ABC is similar to △DEF.. Find the perimeter of △DEF.
问题：如图所示，AB=5，AC=7，BC=6，DE=3，三角形ABC与三角形DEF是相似三角形。求三角形DEF的周长。
Construction (CDL):
Shape(AB,BC,CA)
Shape(DE,EF,FD)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),7)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(DE),3)
SimilarBetweenTriangle(ABC,DEF)
Goal: Value(PerimeterOfTriangle(DEF))
Answer: 54/5
Theorem sequence:
triangle_perimeter_formula(1,DEF)
similar_triangle_property_line_ratio(1,ABC,DEF)
similar_triangle_property_line_ratio(1,CAB,FDE)
similar_triangle_property_line_ratio(1,BCA,EFD)
Problem: As shown in the diagram, AB=x, BC=y, CA=14, ∠CAB=30°, AB⊥CB. Find the value of x.
问题：如图所示，AB=x，BC=y，CA=14，∠CAB=30°，AB垂直于CB。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CA),14)
Equal(MeasureOfAngle(CAB),30)
PerpendicularBetweenLine(AB,CB)
Goal: Value(x)
Answer: 7*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,ABC)
Problem: As shown in the diagram, ∠BCF=120°, ∠DBC=5*y°, ∠DFE=108°, ∠FDB=2*x°, BD is parallel to CE. Find the value of y.
问题：如图所示，∠BCF=120°，∠DBC=5*y°，∠DFE=108°，∠FDB=2*x°，BD∥CE。求y的值。
Construction (CDL):
Shape(BC,CF,FD,DB)
Collinear(CFE)
Conditions (CDL):
Equal(MeasureOfAngle(BCF),120)
Equal(MeasureOfAngle(DBC),5*y)
Equal(MeasureOfAngle(DFE),108)
Equal(MeasureOfAngle(FDB),2*x)
ParallelBetweenLine(BD,CE)
Goal: Value(y)
Answer: 12
Theorem sequence:
parallel_property_collinear_extend(3,EC,DB,F)
parallel_property_ipsilateral_internal_angle(1,FC,DB)
quadrilateral_property_angle_sum(1,BCFD)
Problem: As shown in the diagram, AF=10*z-40, CF=18-6*x, DA=2*z, DB=12*y-4, EB=4*y, EC=3*x, AD is the tangent to circle G, the tangent to ⊙G is AF, the tangent to ⊙G is BD, BE is the tangent to circle G, the tangent to circle G is CE, the tangent to circle G is CF. Find the perimeter of △ABC.
问题：如图所示，AF=10*z-40，CF=18-6*x，DA=2*z，DB=12*y-4，EB=4*y，EC=3*x，AD是⊙O的切线，⊙O的切线为AF，BD是⊙O的切线，BE是⊙O的切线，CE是圆O的切线，CF是⊙O的切线。求△ABC的周长。
Construction (CDL):
Shape(GFD,FA,AD)
Shape(GDE,DB,BE)
Shape(GEF,EC,CF)
Shape(GFD,GDE,GEF)
Collinear(AFC)
Collinear(CEB)
Collinear(BDA)
Cocircular(G,FDE)
Conditions (CDL):
Equal(LengthOfLine(AF),10*z-40)
Equal(LengthOfLine(CF),18-6*x)
Equal(LengthOfLine(DA),2*z)
Equal(LengthOfLine(DB),12*y-4)
Equal(LengthOfLine(EB),4*y)
Equal(LengthOfLine(EC),3*x)
IsTangentOfCircle(AD,G)
IsTangentOfCircle(AF,G)
IsTangentOfCircle(BD,G)
IsTangentOfCircle(BE,G)
IsTangentOfCircle(CE,G)
IsTangentOfCircle(CF,G)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 36
Theorem sequence:
tangent_of_circle_property_length_equal(1,AF,AD,G)
tangent_of_circle_property_length_equal(1,CE,CF,G)
tangent_of_circle_property_length_equal(1,BD,BE,G)
line_addition(1,AF,FC)
line_addition(1,AD,DB)
line_addition(1,BE,EC)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, BC=y, BD=x, CD=2*sqrt(3), DA=2, AC⊥BC, BD⊥CD. Find the value of x.
问题：如图所示，BC=y，BD=x，CD=2*sqrt(3)，DA=2，AC垂直于BC，BD垂直于CD。求x的值。
Construction (CDL):
Shape(CB,BD,DC)
Shape(CD,DA,AC)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CD),2*sqrt(3))
Equal(LengthOfLine(DA),2)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BD,CD)
Goal: Value(x)
Answer: 6
Theorem sequence:
adjacent_complementary_angle(1,BDC,CDA)
angle_addition(1,ACD,DCB)
triangle_property_angle_sum(1,CBD)
similar_triangle_judgment_aa(1,CBD,ACD)
similar_triangle_property_line_ratio(1,CBD,ACD)
similar_triangle_property_line_ratio(1,BDC,CDA)
Problem: As shown in the diagram, ∠CEB=40°, ∠EFD=2*x°, ∠EFD=∠FDE. Find the value of x.
问题：如图所示，∠CEB=40°，∠EFD=2*x°，∠EFD=∠FDE。求x的值。
Construction (CDL):
Shape(FD,DE,EF)
Shape(EB,BC,CE)
Collinear(DEC)
Collinear(FEB)
Conditions (CDL):
Equal(MeasureOfAngle(CEB),40)
Equal(MeasureOfAngle(EFD),2*x)
Equal(MeasureOfAngle(EFD),MeasureOfAngle(FDE))
Goal: Value(x)
Answer: 35
Theorem sequence:
vertical_angle(1,DEF,CEB)
triangle_property_angle_sum(1,FDE)
Problem: As shown in the diagram, ∠BGE=75°, ∠FCG=100°. Find the measure of ∠EGD.
问题：如图所示，∠BGE=75°，∠FCG=100°。求∠EGD的大小。
Construction (CDL):
Shape(DG,GC,CD)
Shape(DC,CA)
Shape(AC,CF)
Shape(FC,CG)
Shape(CG,GB)
Shape(BG,GE)
Shape(EG,GD)
Collinear(DCF)
Collinear(DGB)
Collinear(ACGE)
Conditions (CDL):
Equal(MeasureOfAngle(BGE),75)
Equal(MeasureOfAngle(FCG),100)
Goal: Value(MeasureOfAngle(EGD))
Answer: 105
Theorem sequence:
adjacent_complementary_angle(1,BGE,EGD)
Problem: As shown in the diagram, FD=25, the center of ⊙O is O, the diameter of ⊙O is EH, the tangent to circle O is DH, the tangent to ⊙O is FE, EF⊥NF, HA⊥KA, KG is perpendicular to EG, ND is perpendicular to HD. Find the circumference of the circle O.
问题：如图所示，FD=25，圆O的圆心为O，EH是圆O的直径，圆O的切线为DH，圆O的切线为FE，EF垂直于NF，HA垂直于KA，KG⊥EG，ND垂直于HD。求圆O的周长。
Construction (CDL):
Shape(KG,GE,OKE)
Shape(EF,FN,OEN)
Shape(ND,DH,ONH)
Shape(HA,AK,OHK)
Shape(EO,OH,OHK,OKE)
Shape(HO,OE,OEN,ONH)
Shape(OHK,OKE,OEN,ONH)
Collinear(GKA)
Collinear(GEF)
Collinear(FND)
Collinear(AHD)
Collinear(EOH)
Cocircular(O,KENH)
Conditions (CDL):
Equal(LengthOfLine(FD),25)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(EH,O)
IsTangentOfCircle(DH,O)
IsTangentOfCircle(FE,O)
PerpendicularBetweenLine(EF,NF)
PerpendicularBetweenLine(HA,KA)
PerpendicularBetweenLine(KG,EG)
PerpendicularBetweenLine(ND,HD)
Goal: Value(PerimeterOfCircle(O))
Answer: 25*pi
Theorem sequence:
tangent_of_circle_property_perpendicular(1,FE,O,O)
tangent_of_circle_property_perpendicular(2,DH,O,O)
parallel_judgment_ipsilateral_internal_angle(1,EO,FD)
parallel_property_collinear_extend(2,EO,FD,H)
parallel_judgment_ipsilateral_internal_angle(1,HD,EF)
parallelogram_judgment_parallel_and_parallel(1,EFDH)
parallelogram_property_opposite_line_equal(1,FDHE)
diameter_of_circle_property_length_equal(1,EH,O)
circle_property_length_of_radius_and_diameter(1,O)
circle_perimeter_formula(1,O)
Problem: As shown in the diagram, AC=x, AD=8, CB=y, DB=25, DC=z, AC is perpendicular to BC, BD is perpendicular to CD. Find the value of x.
问题：如图所示，AC=x，AD=8，CB=y，DB=25，DC=z，AC垂直于BC，BD垂直于CD。求x的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(DC,CB,BD)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(CB),y)
Equal(LengthOfLine(DB),25)
Equal(LengthOfLine(DC),z)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BD,CD)
Goal: Value(x)
Answer: 2*sqrt(66)
Theorem sequence:
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,CDA)
triangle_property_angle_sum(1,ACD)
triangle_property_angle_sum(1,ACB)
similar_triangle_judgment_aa(1,ACD,CBD)
similar_triangle_property_line_ratio(1,CDA,BDC)
similar_triangle_property_line_ratio(1,ACD,CBD)
right_triangle_property_pythagorean(1,CDA)
Problem: As shown in the diagram, ∠BCJ=105°, ∠CIE=x°, ∠KEF=125°, BD is parallel to FH. Find the value of x.
问题：如图所示，∠BCJ=105°，∠CIE=x°，∠KEF=125°，BD∥FH。求x的值。
Construction (CDL):
Shape(BC,CJ)
Shape(JC,CD)
Shape(CI,IE,EC)
Shape(EI,IC)
Shape(FE,EI)
Shape(CE,EH)
Shape(KE,EF)
Shape(HE,EK)
Shape(IC,CB)
Shape(DC,CE)
Collinear(BCD)
Collinear(JCI)
Collinear(IEK)
Collinear(FEH)
Conditions (CDL):
Equal(MeasureOfAngle(BCJ),105)
Equal(MeasureOfAngle(CIE),x)
Equal(MeasureOfAngle(KEF),125)
ParallelBetweenLine(BD,FH)
Goal: Value(x)
Answer: 130
Theorem sequence:
adjacent_complementary_angle(1,BCJ,JCD)
adjacent_complementary_angle(1,HEK,KEF)
adjacent_complementary_angle(1,JCD,DCI)
adjacent_complementary_angle(1,IEH,HEK)
parallel_property_collinear_extend(3,BD,FH,C)
parallel_property_collinear_extend(3,HF,DC,E)
parallel_property_ipsilateral_internal_angle(1,CD,EH)
angle_addition(1,DCE,ECI)
angle_addition(1,IEC,CEH)
triangle_property_angle_sum(1,CIE)
Problem: As shown in the diagram, AB=26, AC=24, CB=10, BC is perpendicular to AC. Find the value of sin(AB).
问题：如图所示，AB=26，AC=24，CB=10，BC垂直于AC。求sin(AB)的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AC),24)
Equal(LengthOfLine(CB),10)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Sin(MeasureOfAngle(ABC)))
Answer: 12/13
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AB=15, PB=12, ∠DBA=24°, ADCB is a rhombus. Find the measure of ∠ACB.
问题：如图所示，AB=15，PB=12，∠DBA=24°，ADCB是菱形。求∠ACB的大小。
Construction (CDL):
Shape(AD,DP,PA)
Shape(PD,DC,CP)
Shape(PC,CB,BP)
Shape(AP,PB,BA)
Collinear(DPB)
Collinear(APC)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(PB),12)
Equal(MeasureOfAngle(DBA),24)
Rhombus(ADCB)
Goal: Value(MeasureOfAngle(ACB))
Answer: 66
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,BADC,P)
isosceles_triangle_judgment_line_equal(1,BAC)
altitude_of_triangle_judgment(1,BP,BAC)
isosceles_triangle_property_line_coincidence(1,BAC,P)
triangle_property_angle_sum(1,BPC)
Problem: As shown in the diagram, BA=h, CA=3, ∠BAC=45°, AC⊥BC. Find the value of h.
问题：如图所示，BA=h，CA=3，∠BAC=45°，AC⊥BC。求h的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(BA),h)
Equal(LengthOfLine(CA),3)
Equal(MeasureOfAngle(BAC),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(h)
Answer: 3*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,ACB)
Problem: As shown in the diagram, AB=2, AE=x-1, CD=5, ED=x+5, BA∥DC. Find the length of line AE.
问题：如图所示，AB=2，AE=x-1，CD=5，ED=x+5，BA平行于DC。求直线AE的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(ED,DC,CE)
Collinear(BEC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),2)
Equal(LengthOfLine(AE),x-1)
Equal(LengthOfLine(CD),5)
Equal(LengthOfLine(ED),x+5)
ParallelBetweenLine(BA,DC)
Goal: Value(LengthOfLine(AE))
Answer: 4
Theorem sequence:
vertical_angle(1,BEA,CED)
parallel_property_alternate_interior_angle(1,BA,DC)
similar_triangle_judgment_aa(1,ABE,DCE)
similar_triangle_property_line_ratio(1,EAB,EDC)
similar_triangle_property_line_ratio(1,BEA,CED)
Problem: As shown in the diagram, AB=26, AC=24, CB=10, BC is perpendicular to AC. Find the value of tan(AB).
问题：如图所示，AB=26，AC=24，CB=10，BC⊥AC。求tan(AB)的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AC),24)
Equal(LengthOfLine(CB),10)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Tan(MeasureOfAngle(ABC)))
Answer: 12/5
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, CB=23, DE=5, ∠EDB=45°, CADB is a ▱, AE is perpendicular to BE. Find the perimeter of quadrilateral CADB.
问题：如图所示，CB=23，DE=5，∠EDB=45°，AC和DB是▱CADB的一组对边，AE⊥BE。求CADB的周长。
Construction (CDL):
Shape(CA,AE,EB,BC)
Shape(BE,ED,DB)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(CB),23)
Equal(LengthOfLine(DE),5)
Equal(MeasureOfAngle(EDB),45)
Parallelogram(CADB)
PerpendicularBetweenLine(AE,BE)
Goal: Value(PerimeterOfQuadrilateral(CADB))
Answer: 10*sqrt(2)+46
Theorem sequence:
altitude_of_quadrilateral_judgment_right_vertex(1,BE,CADB)
triangle_property_angle_sum(1,EDB)
isosceles_triangle_judgment_angle_equal(1,EDB)
right_triangle_judgment_angle(1,BED)
right_triangle_property_pythagorean(1,BED)
parallelogram_property_opposite_line_equal(1,ADBC)
parallelogram_property_opposite_line_equal(1,CADB)
quadrilateral_perimeter_formula(1,CADB)
Problem: As shown in the diagram, ∠BFC=6*x°, ∠CFD=3*x°, ∠EFA=12*y-10°, EF⊥AF. Find the value of y.
问题：如图所示，∠BFC=6*x°，∠CFD=3*x°，∠EFA=12*y-10°，EF⊥AF。求y的值。
Construction (CDL):
Shape(AF,FB)
Shape(BF,FC)
Shape(CF,FD)
Shape(DF,FE)
Shape(EF,FA)
Collinear(AFD)
Collinear(BFE)
Conditions (CDL):
Equal(MeasureOfAngle(BFC),6*x)
Equal(MeasureOfAngle(CFD),3*x)
Equal(MeasureOfAngle(EFA),12*y-10)
PerpendicularBetweenLine(EF,AF)
Goal: Value(y)
Answer: 25/3
Theorem sequence:
Problem: As shown in the diagram, MN=3*x-4, NQ=15.4, PN=2*y+5, PQ=11.1, RM=17.9, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, quadrilateral MRPN is a parallelogram. Find the measure of ∠RQM.
问题：如图所示，MN=3*x-4，NQ=15.4，PN=2*y+5，PQ=11.1，RM=17.9，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，RM和PN是▱MRPN的一组对边。求∠RQM的大小。
Construction (CDL):
Shape(MQ,QN,NM)
Shape(MR,RQ,QM)
Shape(QR,RP,PQ)
Shape(QP,ON,NQ)
Collinear(MQP)
Collinear(RQN)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(NQ),15.4)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),11.1)
Equal(LengthOfLine(RM),17.9)
Equal(LengthOfLine(RP),20)
Equal(LengthOfLine(RQ),3*z-3)
Equal(MeasureOfAngle(MRQ),38)
Equal(MeasureOfAngle(NQP),83)
Equal(MeasureOfAngle(QNM),33)
Parallelogram(MRPN)
Goal: Value(MeasureOfAngle(RQM))
Answer: 83
Theorem sequence:
vertical_angle(1,RQM,NQP)
Problem: As shown in the diagram, AG=1/5*x+3, CJ=2*y+1, CJ=JE, EG=4*x-35, JE=5*y-8, AC is parallel to GJ. Find the value of x.
问题：如图所示，AG=1/5*x+3，CJ=2*y+1，CJ=JE，EG=4*x-35，JE=5*y-8，AC∥GJ。求x的值。
Construction (CDL):
Shape(AG,GJ,JC,CA)
Shape(GE,EJ,JG)
Collinear(AGE)
Collinear(EJC)
Conditions (CDL):
Equal(LengthOfLine(AG),1/5*x+3)
Equal(LengthOfLine(CJ),2*y+1)
Equal(LengthOfLine(CJ),LengthOfLine(JE))
Equal(LengthOfLine(EG),4*x-35)
Equal(LengthOfLine(JE),5*y-8)
ParallelBetweenLine(AC,GJ)
Goal: Value(x)
Answer: 10
Theorem sequence:
line_addition(1,AG,GE)
line_addition(1,EJ,JC)
parallel_property_corresponding_angle(2,AC,GJ,E)
similar_triangle_judgment_aa(1,JGE,CAE)
similar_triangle_property_line_ratio(1,GEJ,AEC)
similar_triangle_property_line_ratio(1,JGE,CAE)