SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠JHL=76°, ∠LJH=x°, ∠MHP=44°, HP is the midsegment of triangle MLK, JH is the midsegment of △ LKM, PJ is the midsegment of △ KML. Find the value of x.
问题：如图所示，∠JHL=76°，∠LJH=x°，∠MHP=44°，HP是△MLK的中位线，JH是三角形LKM的中位线，PJ是三角形KML的中位线。求x的值。
Construction (CDL):
Shape(KP,PJ,JK)
Shape(PM,MH,HP)
Shape(JP,PH,HJ)
Shape(JH,HL,LJ)
Collinear(KJL)
Collinear(KPM)
Collinear(LHM)
Conditions (CDL):
Equal(MeasureOfAngle(JHL),76)
Equal(MeasureOfAngle(LJH),x)
Equal(MeasureOfAngle(MHP),44)
IsMidsegmentOfTriangle(HP,MLK)
IsMidsegmentOfTriangle(JH,LKM)
IsMidsegmentOfTriangle(PJ,KML)
Goal: Value(x)
Answer: 60
Theorem sequence:
adjacent_complementary_angle(1,MHJ,JHL)
angle_addition(1,MHP,PHJ)
midsegment_of_triangle_property_parallel(1,HP,MLK)
parallel_property_collinear_extend(3,KL,PH,J)
parallel_property_alternate_interior_angle(1,JL,PH)
Problem: As shown in the diagram, AM=9, BC=14, ME=15, BA is the tangent to ⊙O, the tangent to circle O is BC, the tangent to ⊙O is EC, the tangent to ⊙O is EF, the tangent to ⊙O is MA, the tangent to circle O is MF. Find the perimeter of triangle BEM.
问题：如图所示，AM=9，BC=14，ME=15，圆O的切线为BA，BC是⊙O的切线，EC是圆O的切线，EF是圆O的切线，MA是圆O的切线，圆O的切线为MF。求△BEM的周长。
Construction (CDL):
Shape(BC,OAC,AB)
Shape(CE,EF,OCF)
Shape(FM,MA,OFA)
Shape(OAC,OCF,OFA)
Collinear(BCE)
Collinear(EFM)
Collinear(MAB)
Cocircular(O,ACF)
Conditions (CDL):
Equal(LengthOfLine(AM),9)
Equal(LengthOfLine(BC),14)
Equal(LengthOfLine(ME),15)
IsTangentOfCircle(BA,O)
IsTangentOfCircle(BC,O)
IsTangentOfCircle(EC,O)
IsTangentOfCircle(EF,O)
IsTangentOfCircle(MA,O)
IsTangentOfCircle(MF,O)
Goal: Value(PerimeterOfTriangle(BEM))
Answer: 58
Theorem sequence:
tangent_of_circle_property_length_equal(1,EF,EC,O)
tangent_of_circle_property_length_equal(1,BC,BA,O)
tangent_of_circle_property_length_equal(1,MA,MF,O)
line_addition(1,BC,CE)
line_addition(1,EF,FM)
line_addition(1,BA,AM)
triangle_perimeter_formula(1,BEM)
Problem: As shown in the diagram, AB=26, AD=12, DG=9/2, EF=8, GF=14, ∠AGF=108°. Find the length of line AG.
问题：如图所示，AB=26，AD=12，DG=9/2，EF=8，GF=14，∠AGF=108°。求直线AG的长度。
Construction (CDL):
Shape(AG,GF,FE,EA)
Shape(FD,DC,CB,BE,EF,FG)
Collinear(AGD)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DG),9/2)
Equal(LengthOfLine(EF),8)
Equal(LengthOfLine(GF),14)
Equal(MeasureOfAngle(AGF),108)
Goal: Value(LengthOfLine(AG))
Answer: 15/2
Theorem sequence:
line_addition(1,AG,GD)
Problem: As shown in the diagram, SV=17, TU=26, S is the midpoint of segment RT, V bisects segment QU, QUTR is a trapezoid. Find the length of line QR.
问题：如图所示，SV=17，TU=26，S平分线段RT，V平分线段QU，QU和TR是梯形QUTR的腰。求直线QR的长度。
Construction (CDL):
Shape(QV,VS,SR,RQ)
Shape(VU,UT,TS,SV)
Collinear(QVU)
Collinear(RST)
Conditions (CDL):
Equal(LengthOfLine(SV),17)
Equal(LengthOfLine(TU),26)
IsMidpointOfLine(S,RT)
IsMidpointOfLine(V,QU)
Trapezoid(QUTR)
Goal: Value(LengthOfLine(QR))
Answer: 8
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,VS,QUTR)
midsegment_of_quadrilateral_property_length(1,VS,QUTR)
Problem: As shown in the diagram, EL=4, ES=9, GO=x, LG=3. Find the value of x.
问题：如图所示，EL=4，ES=9，GO=x，LG=3。求x的值。
Construction (CDL):
Shape(BGO,OG)
Shape(EL,LG,BEG)
Shape(BSE,ES)
Shape(BEG,GO,BOS,SE)
Collinear(OGL)
Collinear(LES)
Cocircular(B,OSEG)
Conditions (CDL):
Equal(LengthOfLine(EL),4)
Equal(LengthOfLine(ES),9)
Equal(LengthOfLine(GO),x)
Equal(LengthOfLine(LG),3)
Goal: Value(x)
Answer: 43/3
Theorem sequence:
line_addition(1,LG,GO)
line_addition(1,LE,ES)
circle_property_circular_power_segment_and_segment_line(1,LES,LGO,B)
Problem: As shown in the diagram, AB=x+15, BC=2*x-5, the tangent to circle D is BA, BC is the tangent to circle D. Find the value of x.
问题：如图所示，AB=x+15，BC=2*x-5，BA是⊙O的切线，BC是圆O的切线。求x的值。
Construction (CDL):
Shape(DAC,DCA)
Shape(DCA,CB,BA)
Cocircular(D,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),x+15)
Equal(LengthOfLine(BC),2*x-5)
IsTangentOfCircle(BA,D)
IsTangentOfCircle(BC,D)
Goal: Value(x)
Answer: 20
Theorem sequence:
tangent_of_circle_property_length_equal(1,BA,BC,D)
Problem: As shown in the diagram, FG=3*x+10, FH=6*x+1, GH=9*x-8, △HGF is an equilateral △. Find the length of line FG.
问题：如图所示，FG=3*x+10，FH=6*x+1，GH=9*x-8，△HGF为等边△。求直线FG的长度。
Construction (CDL):
Shape(HG,GF,FH)
Conditions (CDL):
Equal(LengthOfLine(FG),3*x+10)
Equal(LengthOfLine(FH),6*x+1)
Equal(LengthOfLine(GH),9*x-8)
EquilateralTriangle(HGF)
Goal: Value(LengthOfLine(FG))
Answer: 19
Theorem sequence:
Problem: As shown in the diagram, ∠GKJ=x+21°, ∠HGK=4*y-11°, ∠JHG=2*x°, ∠KJH=3*y+9°. Find the measure of ∠HGK.
问题：如图所示，∠GKJ=x+21°，∠HGK=4*y-11°，∠JHG=2*x°，∠KJH=3*y+9°。求∠HGK的大小。
Construction (CDL):
Shape(JH,HG,GK,KJ)
Shape(HJ,OJH)
Shape(GH,OHG)
Shape(KG,OGK)
Shape(JK,OKJ)
Cocircular(O,JHGK)
Conditions (CDL):
Equal(MeasureOfAngle(GKJ),x+21)
Equal(MeasureOfAngle(HGK),4*y-11)
Equal(MeasureOfAngle(JHG),2*x)
Equal(MeasureOfAngle(KJH),3*y+9)
Goal: Value(MeasureOfAngle(HGK))
Answer: 93
Theorem sequence:
arc_property_circumference_angle_external(1,OHK,J)
arc_property_circumference_angle_internal(1,OHK,G)
Problem: As shown in the diagram, MD=7, M is the center of circle M, the diameter of circle M is RH. Find the length of line RH.
问题：如图所示，MD=7，⊙M的圆心为M，⊙M的直径为RH。求直线RH的长度。
Construction (CDL):
Shape(AM,MD,MDA)
Shape(DM,MH,MHD)
Shape(RM,MA,MAR)
Shape(HR,MRS,SU,MUH)
Shape(MSU,US)
Collinear(RMH)
Cocircular(M,DARSUH)
Conditions (CDL):
Equal(LengthOfLine(MD),7)
IsCentreOfCircle(M,M)
IsDiameterOfCircle(RH,M)
Goal: Value(LengthOfLine(RH))
Answer: 14
Theorem sequence:
radius_of_circle_property_length_equal(1,MD,M)
circle_property_length_of_radius_and_diameter(1,M)
diameter_of_circle_property_length_equal(1,RH,M)
Problem: As shown in the diagram, ∠GKH=122°, K is the center of ⊙K. Find the measure of ⌒KHG.
问题：如图所示，∠GKH=122°，K是圆K的圆心。求弧KHG的角度。
Construction (CDL):
Shape(GK,KH,KHG)
Shape(JG,KGL,KLJ)
Shape(HK,KJ,KJH)
Collinear(GKJ)
Cocircular(K,GLJH)
Conditions (CDL):
Equal(MeasureOfAngle(GKH),122)
IsCentreOfCircle(K,K)
Goal: Value(MeasureOfArc(KHG))
Answer: 122
Theorem sequence:
arc_property_center_angle(1,KHG,K)
Problem: As shown in the diagram, BA=16, DA=9, DC=15, CA is perpendicular to DA. Find the perimeter of △CBA.
问题：如图所示，BA=16，DA=9，DC=15，CA⊥DA。求三角形CBA的周长。
Construction (CDL):
Shape(CB,BA,AC)
Shape(CA,AD,DC)
Collinear(BAD)
Conditions (CDL):
Equal(LengthOfLine(BA),16)
Equal(LengthOfLine(DA),9)
Equal(LengthOfLine(DC),15)
PerpendicularBetweenLine(CA,DA)
Goal: Value(PerimeterOfTriangle(CBA))
Answer: 48
Theorem sequence:
adjacent_complementary_angle(1,BAC,CAD)
right_triangle_judgment_angle(1,BAC)
right_triangle_judgment_angle(1,CAD)
right_triangle_property_pythagorean(1,BAC)
right_triangle_property_pythagorean(1,CAD)
triangle_perimeter_formula(1,CBA)
Problem: As shown in the diagram, NS=9, the center of ⊙J is J, SB is the diameter of ⊙J, quadrilateral SNBP is a square. Find the circumference of the circle J.
问题：如图所示，NS=9，⊙J的圆心为J，SB是⊙J的直径，四边形SNBP是正方形。求⊙J的周长。
Construction (CDL):
Shape(SP,JPS)
Shape(NS,JSN)
Shape(BN,SNB)
Shape(PB,JBP)
Shape(BP,PS,SB)
Shape(BS,SN,NB)
Collinear(SJB)
Cocircular(J,SNBP)
Conditions (CDL):
Equal(LengthOfLine(NS),9)
IsCentreOfCircle(J,J)
IsDiameterOfCircle(SB,J)
Square(SNBP)
Goal: Value(PerimeterOfCircle(J))
Answer: 9*sqrt(2)*pi
Theorem sequence:
right_triangle_judgment_angle(1,SNB)
right_triangle_property_pythagorean(1,SNB)
diameter_of_circle_property_length_equal(1,SB,J)
circle_property_length_of_radius_and_diameter(1,J)
circle_perimeter_formula(1,J)
Problem: As shown in the diagram, DC=9, FA=12, FD=x, VZ=3*y-1, YW=6, YZ=10, AFDC is similar to VZYW. Find the value of x.
问题：如图所示，DC=9，FA=12，FD=x，VZ=3*y-1，YW=6，YZ=10，四边形AFDC与四边形VZYW相似。求x的值。
Construction (CDL):
Shape(AF,FD,DC,CA)
Shape(YW,WV,VZ,ZY)
Conditions (CDL):
Equal(LengthOfLine(DC),9)
Equal(LengthOfLine(FA),12)
Equal(LengthOfLine(FD),x)
Equal(LengthOfLine(VZ),3*y-1)
Equal(LengthOfLine(YW),6)
Equal(LengthOfLine(YZ),10)
SimilarBetweenQuadrilateral(AFDC,VZYW)
Goal: Value(x)
Answer: 15
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,FDCA,ZYWV)
similar_quadrilateral_property_line_ratio(1,DCAF,YWVZ)
Problem: As shown in the diagram, ∠ADC=32°, ∠DCB=23°, ∠FAB=22°. Find the measure of ∠EBA.
问题：如图所示，∠ADC=32°，∠DCB=23°，∠FAB=22°。求∠EBA的大小。
Construction (CDL):
Shape(CF,FD,DC)
Shape(FA,AB,BF)
Shape(EB,BA)
Collinear(CFBE)
Collinear(DFA)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),32)
Equal(MeasureOfAngle(DCB),23)
Equal(MeasureOfAngle(FAB),22)
Goal: Value(MeasureOfAngle(EBA))
Answer: 147
Theorem sequence:
triangle_property_angle_sum(1,CFD)
triangle_property_angle_sum(1,FAB)
vertical_angle(1,CFD,BFA)
adjacent_complementary_angle(1,EBA,ABF)
Problem: As shown in the diagram, ∠EAB=40°, AB⊥DB, quadrilateral ABDC is a rectangle. Find the measure of ∠CAE.
问题：如图所示，∠EAB=40°，AB垂直于DB，ABDC是矩形。求∠CAE的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BD,DE)
Shape(ED,DC,CE)
Shape(EC,CA,AE)
Collinear(AED)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(EAB),40)
PerpendicularBetweenLine(AB,DB)
Rectangle(ABDC)
Goal: Value(MeasureOfAngle(CAE))
Answer: 50
Theorem sequence:
angle_addition(1,CAE,EAB)
Problem: As shown in the diagram, ∠DCB=54°, DP is perpendicular to AP, quadrilateral ADCB is a rhombus. Find the measure of ∠BAC.
问题：如图所示，∠DCB=54°，DP垂直于AP，ADCB是菱形。求∠BAC的大小。
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(DCB),54)
PerpendicularBetweenLine(DP,AP)
Rhombus(ADCB)
Goal: Value(MeasureOfAngle(BAC))
Answer: 27
Theorem sequence:
mirror_congruent_triangle_judgment_sss(1,DCA,BAC)
mirror_congruent_triangle_property_angle_equal(1,CAD,CBA)
isosceles_triangle_judgment_line_equal(1,BAC)
isosceles_triangle_property_angle_equal(1,BAC)
angle_addition(1,DCA,ACB)
Problem: As shown in the diagram, SX=x+2, SZ=x+6, WS=x, YS=x+12. Find the value of x.
问题：如图所示，SX=x+2，SZ=x+6，WS=x，YS=x+12。求x的值。
Construction (CDL):
Shape(ZS,SW,SWZ)
Shape(YS,SZ,SZY)
Shape(WS,SX,SXW)
Shape(XS,SY,SYX)
Collinear(WSY)
Collinear(ZSX)
Cocircular(B,WZYX)
Conditions (CDL):
Equal(LengthOfLine(SX),x+2)
Equal(LengthOfLine(SZ),x+6)
Equal(LengthOfLine(WS),x)
Equal(LengthOfLine(YS),x+12)
Goal: Value(x)
Answer: 3
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,ZSX,WSY,B)
Problem: As shown in the diagram, AB=15, ∠ABC=30°, BC is perpendicular to AC. Find the length of line AC.
问题：如图所示，AB=15，∠ABC=30°，BC⊥AC。求直线AC的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(BC,AC)
Goal: Value(LengthOfLine(AC))
Answer: 15/2
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, AC=13, CD=6, DB=29, CD⊥AD. Find the perimeter of triangle ADB.
问题：如图所示，AC=13，CD=6，DB=29，CD垂直于AD。求△ADB的周长。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AC),13)
Equal(LengthOfLine(CD),6)
Equal(LengthOfLine(DB),29)
PerpendicularBetweenLine(CD,AD)
Goal: Value(PerimeterOfTriangle(ADB))
Answer: sqrt(133)+29+sqrt(974)
Theorem sequence:
adjacent_complementary_angle(1,CDA,ADB)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,ADB)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,ADB)
triangle_perimeter_formula(1,ADB)
Problem: As shown in the diagram, ⌒WSR is congruent to ⌒WVT, SR=5*x-9, TV=2*x+3. Find the length of line RS.
问题：如图所示，⌒WSR全等于⌒WVT，SR=5*x-9，TV=2*x+3。求直线RS的长度。
Construction (CDL):
Shape(RS,WSR)
Shape(SR,WRV,VT,WTS)
Shape(TV,WVT)
Cocircular(W,RVTS)
Conditions (CDL):
CongruentBetweenArc(WSR,WVT)
Equal(LengthOfLine(SR),5*x-9)
Equal(LengthOfLine(TV),2*x+3)
Goal: Value(LengthOfLine(RS))
Answer: 11
Theorem sequence:
congruent_arc_property_chord_equal(1,WSR,WVT)
Problem: As shown in the diagram, ∠JKL=x+10°, ∠KLM=x°, ∠LMJ=2*x-8°, ∠MJK=3*x-6°. Find the measure of ∠KLM.
问题：如图所示，∠JKL=x+10°，∠KLM=x°，∠LMJ=2*x-8°，∠MJK=3*x-6°。求∠KLM的大小。
Construction (CDL):
Shape(JK,KL,LM,MJ)
Conditions (CDL):
Equal(MeasureOfAngle(JKL),x+10)
Equal(MeasureOfAngle(KLM),x)
Equal(MeasureOfAngle(LMJ),2*x-8)
Equal(MeasureOfAngle(MJK),3*x-6)
Goal: Value(MeasureOfAngle(KLM))
Answer: 52
Theorem sequence:
quadrilateral_property_angle_sum(1,JKLM)
Problem: As shown in the diagram, ∠DAG=7*y+6°, ∠RGA=4*y+30°, FR is parallel to DS. Find the value of y.
问题：如图所示，∠DAG=7*y+6°，∠RGA=4*y+30°，FR平行于DS。求y的值。
Construction (CDL):
Shape(EG,GB)
Shape(BG,GR)
Shape(AG,GF)
Shape(RG,GA)
Shape(DA,AG)
Shape(HA,AD)
Shape(SA,AH)
Shape(GA,AS)
Collinear(BGAH)
Collinear(FGR)
Collinear(DAS)
Conditions (CDL):
Equal(MeasureOfAngle(DAG),7*y+6)
Equal(MeasureOfAngle(RGA),4*y+30)
ParallelBetweenLine(FR,DS)
Goal: Value(y)
Answer: 8
Theorem sequence:
parallel_property_collinear_extend(3,FR,DS,G)
parallel_property_collinear_extend(3,SD,RG,A)
parallel_property_alternate_interior_angle(1,GR,DA)
Problem: As shown in the diagram, AB=BC, BC=5*sqrt(2), the center of ⊙D is D, AC is the diameter of circle D. Find the area of the ⊙D minus the area of △ABC.
问题：如图所示，AB=BC，BC=5*sqrt(2)，D是圆D的圆心，圆D的直径为AC。求⊙D的面积减去三角形ABC的面积。
Construction (CDL):
Shape(AD,DC,DCA)
Shape(AB,BC,CD,DA)
Shape(BA,DAB)
Shape(CB,DBC)
Collinear(ADC)
Cocircular(D,ABC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(BC))
Equal(LengthOfLine(BC),5*sqrt(2))
IsCentreOfCircle(D,D)
IsDiameterOfCircle(AC,D)
Goal: Value(Sub(AreaOfCircle(D),AreaOfTriangle(ABC)))
Answer: -25+25*pi
Theorem sequence:
diameter_of_circle_property_right_angle(1,ABC,D)
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,ABC)
triangle_area_formula_sine(1,BCA)
diameter_of_circle_property_length_equal(1,AC,D)
circle_property_length_of_radius_and_diameter(1,D)
circle_area_formula(1,D)
Problem: As shown in the diagram, AE=30, BE=20, CE=20, DE=30, AE is perpendicular to BE. Find the area of quadrilateral ACDB.
问题：如图所示，AE=30，BE=20，CE=20，DE=30，AE垂直于BE。求ACDB的面积。
Construction (CDL):
Shape(AE,EB,BA)
Shape(AC,CE,EA)
Shape(CD,DE,EC)
Shape(BE,ED,DB)
Collinear(BEC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AE),30)
Equal(LengthOfLine(BE),20)
Equal(LengthOfLine(CE),20)
Equal(LengthOfLine(DE),30)
PerpendicularBetweenLine(AE,BE)
Goal: Value(AreaOfQuadrilateral(ACDB))
Answer: 1200
Theorem sequence:
adjacent_complementary_angle(1,CEA,AEB)
adjacent_complementary_angle(1,AEB,BED)
adjacent_complementary_angle(1,DEC,CEA)
mirror_congruent_triangle_judgment_sas(1,EDB,ECD)
mirror_congruent_triangle_judgment_sas(1,EBA,EAC)
mirror_congruent_triangle_property_line_equal(1,EDB,ECD)
mirror_congruent_triangle_property_line_equal(1,EBA,EAC)
kite_judgment_equal_and_equal(1,DBAC)
line_addition(1,AE,ED)
line_addition(1,BE,EC)
kite_area_formula_diagonal(1,ACDB)
Problem: As shown in the diagram, the measure of ⌒FAC is 120, the measure of arc FED is 100. Find the measure of ∠DBE.
问题：如图所示，弧FAC的角度为120，弧FED的角度为100。求∠DBE的大小。
Construction (CDL):
Shape(BC,FCE,EB)
Shape(BE,FED,DB)
Shape(BD,DA,AB)
Shape(AD,FDA)
Shape(BA,FAC,CB)
Collinear(EBFA)
Collinear(CBD)
Cocircular(F,EDAC)
Conditions (CDL):
Equal(MeasureOfArc(FAC),120)
Equal(MeasureOfArc(FED),100)
Goal: Value(MeasureOfAngle(DBE))
Answer: 110
Theorem sequence:
arc_property_circumference_angle_external(1,FED,A)
arc_property_circumference_angle_external(1,FAC,D)
triangle_property_angle_sum(1,ABD)
adjacent_complementary_angle(1,ABD,DBE)
Problem: As shown in the diagram, ∠HEL=55°, ∠IHF=110°, IC is parallel to HF. Find the measure of ∠IAE.
问题：如图所示，∠HEL=55°，∠IHF=110°，IC平行于HF。求∠IAE的大小。
Construction (CDL):
Shape(DI,IC)
Shape(CI,IH)
Shape(IH,HF)
Shape(FH,HB)
Shape(AI,ID)
Shape(JA,AI)
Shape(IA,AE,EH,HI)
Shape(BH,HE)
Shape(HE,EL)
Shape(KA,AJ)
Shape(EA,AK)
Shape(GE,EA)
Shape(LE,EG)
Collinear(DIHB)
Collinear(JAEL)
Collinear(CIAK)
Collinear(FHEG)
Conditions (CDL):
Equal(MeasureOfAngle(HEL),55)
Equal(MeasureOfAngle(IHF),110)
ParallelBetweenLine(IC,HF)
Goal: Value(MeasureOfAngle(IAE))
Answer: 55
Theorem sequence:
parallel_property_collinear_extend(1,IC,HF,A)
parallel_property_collinear_extend(2,FH,CA,E)
parallel_property_corresponding_angle(2,AC,EF,L)
Problem: As shown in the diagram, ∠DOB=∠EOD, ∠EOD=∠FOE, ∠FOE=∠AOF, the center of ⊙O is O, CO is perpendicular to AO. Find the measure of arc OCB.
问题：如图所示，∠DOB=∠EOD，∠EOD=∠FOE，∠FOE=∠AOF，⊙O的圆心为O，CO⊥AO。求弧OCB的角度。
Construction (CDL):
Shape(AO,OF,OFA)
Shape(FO,OE,OEF)
Shape(EO,OD,ODE)
Shape(DO,OB,OBD)
Shape(BO,OC,OCB)
Shape(CO,OA,OAC)
Collinear(AOB)
Collinear(EOC)
Cocircular(O,ACBDEF)
Conditions (CDL):
Equal(MeasureOfAngle(DOB),MeasureOfAngle(EOD))
Equal(MeasureOfAngle(EOD),MeasureOfAngle(FOE))
Equal(MeasureOfAngle(FOE),MeasureOfAngle(AOF))
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CO,AO)
Goal: Value(MeasureOfArc(OCB))
Answer: 90
Theorem sequence:
arc_property_center_angle(1,OCB,O)
adjacent_complementary_angle(1,BOC,COA)
Problem: As shown in the diagram, QR=20, TC=12, ∠CTQ=120°, ∠QRB=45°, A is the midpoint of segment TQ, B is the midpoint of segment CR, TQ and RC are the non-parallel sides (legs) of trapezoid TQRC. Find the measure of ∠AQR.
问题：如图所示，QR=20，TC=12，∠CTQ=120°，∠QRB=45°，A平分线段TQ，B平分线段CR，TQ和RC是梯形TQRC的腰。求∠AQR的大小。
Construction (CDL):
Shape(TA,AB,BC,CT)
Shape(AQ,QR,RB,BA)
Collinear(TAQ)
Collinear(CBR)
Conditions (CDL):
Equal(LengthOfLine(QR),20)
Equal(LengthOfLine(TC),12)
Equal(MeasureOfAngle(CTQ),120)
Equal(MeasureOfAngle(QRB),45)
IsMidpointOfLine(A,TQ)
IsMidpointOfLine(B,CR)
Trapezoid(TQRC)
Goal: Value(MeasureOfAngle(AQR))
Answer: 60
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,TC,QR)
Problem: As shown in the diagram, AD=8, DB=x, DE=5, DF=6. Find the value of x.
问题：如图所示，AD=8，DB=x，DE=5，DF=6。求x的值。
Construction (CDL):
Shape(FD,DE,CEF)
Shape(ED,DA,CAE)
Shape(BD,DF,CFB)
Shape(AD,DB,CBA)
Collinear(EDB)
Collinear(FDA)
Cocircular(C,FBAE)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(DB),x)
Equal(LengthOfLine(DE),5)
Equal(LengthOfLine(DF),6)
Goal: Value(x)
Answer: 48/5
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,BDE,ADF,C)
Problem: As shown in the diagram, EA=8, ∠ADE=30°, ∠DEA=45°, DG⊥AG. Find the length of line GD.
问题：如图所示，EA=8，∠ADE=30°，∠DEA=45°，DG垂直于AG。求直线GD的长度。
Construction (CDL):
Shape(EA,AG,GE)
Shape(GA,AD,DG)
Collinear(EGD)
Conditions (CDL):
Equal(LengthOfLine(EA),8)
Equal(MeasureOfAngle(ADE),30)
Equal(MeasureOfAngle(DEA),45)
PerpendicularBetweenLine(DG,AG)
Goal: Value(LengthOfLine(GD))
Answer: 4*sqrt(6)
Theorem sequence:
adjacent_complementary_angle(1,DGA,AGE)
triangle_property_angle_sum(1,DGA)
sine_theorem(1,DGA)
sine_theorem(1,ADG)
sine_theorem(1,AGE)
Problem: As shown in the diagram, the area of quadrilateral DCBA is 32, DC=8, EF=5, DCBA is similar to EFGH. Find the area of EFGH.
问题：如图所示，DCBA的面积为32，DC=8，EF=5，DCBA相似于EFGH。求EFGH的面积。
Construction (CDL):
Shape(DC,CB,BA,AD)
Shape(EF,FG,GH,HE)
Conditions (CDL):
Equal(AreaOfQuadrilateral(DCBA),32)
Equal(LengthOfLine(DC),8)
Equal(LengthOfLine(EF),5)
SimilarBetweenQuadrilateral(DCBA,EFGH)
Goal: Value(AreaOfQuadrilateral(EFGH))
Answer: 25/2
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,DCBA,EFGH)
similar_quadrilateral_property_area_square_ratio(1,DCBA,EFGH)
Problem: As shown in the diagram, CB=4*w-7, CD=11, ED=3*z+10, RS=2*w+13, RU=12, UT=z+16, ∠CBE=2*x+9°, ∠EDC=2*y-31°, ∠STU=y+11°, ∠URS=49°, BEDC is mirror congruent to RSTU. Find the value of w.
问题：如图所示，CB=4*w-7，CD=11，ED=3*z+10，RS=2*w+13，RU=12，UT=z+16，∠CBE=2*x+9°，∠EDC=2*y-31°，∠STU=y+11°，∠URS=49°，四边形BEDC镜像全等于四边形RSTU。求w的值。
Construction (CDL):
Shape(BE,ED,DC,CB)
Shape(UR,RS,ST,TU)
Conditions (CDL):
Equal(LengthOfLine(CB),4*w-7)
Equal(LengthOfLine(CD),11)
Equal(LengthOfLine(ED),3*z+10)
Equal(LengthOfLine(RS),2*w+13)
Equal(LengthOfLine(RU),12)
Equal(LengthOfLine(UT),z+16)
Equal(MeasureOfAngle(CBE),2*x+9)
Equal(MeasureOfAngle(EDC),2*y-31)
Equal(MeasureOfAngle(STU),y+11)
Equal(MeasureOfAngle(URS),49)
MirrorCongruentBetweenQuadrilateral(BEDC,RSTU)
Goal: Value(w)
Answer: 10
Theorem sequence:
mirror_congruent_quadrilateral_property_line_equal(1,CBED,STUR)
Problem: As shown in the diagram, BA=c, BC=a, CA=b, a=8, b=15, c=17, BC is perpendicular to AC. Find the value of sin(CA).
问题：如图所示，BA=c，BC=a，CA=b，a=8，b=15，c=17，BC垂直于AC。求sin(CA)的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(BA),c)
Equal(LengthOfLine(BC),a)
Equal(LengthOfLine(CA),b)
Equal(a,8)
Equal(b,15)
Equal(c,17)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Sin(MeasureOfAngle(CAB)))
Answer: 8/17
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, BD=5, DF=4, EB=DF, EF=DB, EF and BD are opposite sides of the parallelogram EBDF, EB⊥DB. Find the perimeter of quadrilateral EBDF.
问题：如图所示，BD=5，DF=4，EB=DF，EF=DB，四边形EBDF是平行四边形，EB垂直于DB。求EBDF的周长。
Construction (CDL):
Shape(EB,BD,DF,FE)
Conditions (CDL):
Equal(LengthOfLine(BD),5)
Equal(LengthOfLine(DF),4)
Equal(LengthOfLine(EB),LengthOfLine(DF))
Equal(LengthOfLine(EF),LengthOfLine(DB))
Parallelogram(EBDF)
PerpendicularBetweenLine(EB,DB)
Goal: Value(PerimeterOfQuadrilateral(EBDF))
Answer: 18
Theorem sequence:
quadrilateral_perimeter_formula(1,EBDF)
Problem: As shown in the diagram, AB=x+2, BC=8, DC=y-3, EH=10, FE=15, HG=5, ∠ABC=∠FEH, ∠BCD=∠EHG, ∠CDA=∠HGF, ∠DAB=∠GFE, ABCD is similar to FEHG. Find the value of x.
问题：如图所示，AB=x+2，BC=8，DC=y-3，EH=10，FE=15，HG=5，∠ABC=∠FEH，∠BCD=∠EHG，∠CDA=∠HGF，∠DAB=∠GFE，四边形ABCD与四边形FEHG相似。求x的值。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(FE,EH,HG,GF)
Conditions (CDL):
Equal(LengthOfLine(AB),x+2)
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(DC),y-3)
Equal(LengthOfLine(EH),10)
Equal(LengthOfLine(FE),15)
Equal(LengthOfLine(HG),5)
Equal(MeasureOfAngle(ABC),MeasureOfAngle(FEH))
Equal(MeasureOfAngle(BCD),MeasureOfAngle(EHG))
Equal(MeasureOfAngle(CDA),MeasureOfAngle(HGF))
Equal(MeasureOfAngle(DAB),MeasureOfAngle(GFE))
SimilarBetweenQuadrilateral(ABCD,FEHG)
Goal: Value(x)
Answer: 10
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,ABCD,FEHG)
similar_quadrilateral_property_line_ratio(1,BCDA,EHGF)
Problem: As shown in the diagram, AC=24, AD=36, BD=BC, ∠BAC=y°, ∠CDA=x°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AC=24，AD=36，BD=BC，∠BAC=y°，∠CDA=x°，AC⊥BC。求x的值。
Construction (CDL):
Shape(DA,AB,BD)
Shape(BA,AC,CB)
Collinear(DBC)
Conditions (CDL):
Equal(LengthOfLine(AC),24)
Equal(LengthOfLine(AD),36)
Equal(LengthOfLine(BD),LengthOfLine(BC))
Equal(MeasureOfAngle(BAC),y)
Equal(MeasureOfAngle(CDA),x)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 180*acos(sqrt(5)/3)/pi
Theorem sequence:
line_addition(1,DB,BC)
right_triangle_judgment_angle(1,ACB)
right_triangle_judgment_angle(1,ACD)
right_triangle_property_pythagorean(1,ACB)
right_triangle_property_pythagorean(1,ACD)
cosine_theorem(1,DAB)
Problem: As shown in the diagram, ∠ADG=36°, ∠AGF=104°, ∠EFC=40°, AB⊥CB, GB⊥CB. Find the measure of ∠CFB.
问题：如图所示，∠ADG=36°，∠AGF=104°，∠EFC=40°，AB垂直于CB，GB垂直于CB。求∠CFB的大小。
Construction (CDL):
Shape(AD,DG,GA)
Shape(GB,BF,FG)
Shape(FB,BC,CF)
Shape(EF,FC)
Shape(AG,GF)
Collinear(AGB)
Collinear(BFE)
Collinear(DGFC)
Conditions (CDL):
Equal(MeasureOfAngle(ADG),36)
Equal(MeasureOfAngle(AGF),104)
Equal(MeasureOfAngle(EFC),40)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(GB,CB)
Goal: Value(MeasureOfAngle(CFB))
Answer: 140
Theorem sequence:
adjacent_complementary_angle(1,EFC,CFB)
Problem: As shown in the diagram, AG=5, CE=5, CF=6, CG=2, DB=8, EB=AD, FM=1, GF=6, EA is parallel to BD, FH⊥EH. Find the area of EBDA minus the area of △CFG.
问题：如图所示，AG=5，CE=5，CF=6，CG=2，DB=8，EB=AD，FM=1，GF=6，EA平行于BD，FH垂直于EH。求EBDA的面积减去三角形CFG的面积。
Construction (CDL):
Shape(EB,BM,MF,FC,CE)
Shape(CF,FH,HC)
Shape(GH,HF,FG)
Shape(GF,FM,MD,DA,AG)
Collinear(ECHGA)
Collinear(BMD)
Collinear(HFM)
Conditions (CDL):
Equal(LengthOfLine(AG),5)
Equal(LengthOfLine(CE),5)
Equal(LengthOfLine(CF),6)
Equal(LengthOfLine(CG),2)
Equal(LengthOfLine(DB),8)
Equal(LengthOfLine(EB),LengthOfLine(AD))
Equal(LengthOfLine(FM),1)
Equal(LengthOfLine(GF),6)
ParallelBetweenLine(EA,BD)
PerpendicularBetweenLine(FH,EH)
Goal: Value(Sub(AreaOfQuadrilateral(EBDA),AreaOfTriangle(CFG)))
Answer: 10+9*sqrt(35)
Theorem sequence:
adjacent_complementary_angle(1,GHF,FHE)
trapezoid_judgment_parallel(1,EBDA)
altitude_of_triangle_judgment(1,FH,FGC)
triangle_area_formula_common(1,FGC)
line_addition(1,EC,CH)
line_addition(1,HG,GA)
line_addition(1,EH,HA)
altitude_of_quadrilateral_judgment(2,MH,DAEB)
trapezoid_area_formula(1,DAEB)
isosceles_triangle_judgment_line_equal(1,FGC)
isosceles_triangle_property_line_coincidence(1,FGC,H)
right_triangle_judgment_angle(1,FHC)
line_addition(1,CH,HG)
line_addition(1,HF,FM)
right_triangle_property_pythagorean(1,FHC)