SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AE=12, BE=12, CE=17, DE=17, DE⊥AE. Find the area of ADBC.
问题：如图所示，AE=12，BE=12，CE=17，DE=17，DE垂直于AE。求四边形ADBC的面积。
Construction (CDL):
Shape(AD,DE,EA)
Shape(AE,EC,CA)
Shape(DB,BE,ED)
Shape(EB,BC,CE)
Collinear(DEC)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AE),12)
Equal(LengthOfLine(BE),12)
Equal(LengthOfLine(CE),17)
Equal(LengthOfLine(DE),17)
PerpendicularBetweenLine(DE,AE)
Goal: Value(AreaOfQuadrilateral(ADBC))
Answer: 408
Theorem sequence:
perpendicular_bisector_judgment_per_and_mid(1,AE,DC)
perpendicular_bisector_property_distance_equal(1,AE,DC)
vertical_angle(1,DEA,CEB)
perpendicular_bisector_judgment_per_and_mid(1,BE,CD)
perpendicular_bisector_property_distance_equal(1,BE,CD)
kite_judgment_equal_and_equal(1,ADBC)
line_addition(1,DE,EC)
line_addition(1,AE,EB)
kite_area_formula_diagonal(1,ADBC)
Problem: As shown in the diagram, JL=y+4/5, LC=2*y-11/5, MI=12-3*x, TM=10-2*x, TM=MI, JT∥LM, LM∥CI. Find the value of x.
问题：如图所示，JL=y+4/5，LC=2*y-11/5，MI=12-3*x，TM=10-2*x，TM=MI，JT平行于LM，LM平行于CI。求x的值。
Construction (CDL):
Shape(JL,LM,MT,TJ)
Shape(LC,CI,IM,ML)
Collinear(JLC)
Collinear(TMI)
Conditions (CDL):
Equal(LengthOfLine(JL),y+4/5)
Equal(LengthOfLine(LC),2*y-11/5)
Equal(LengthOfLine(MI),12-3*x)
Equal(LengthOfLine(TM),10-2*x)
Equal(LengthOfLine(TM),LengthOfLine(MI))
ParallelBetweenLine(JT,LM)
ParallelBetweenLine(LM,CI)
Goal: Value(x)
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, AB=32, AD=DC, BC=2*x-3, EF=12, EH=HG, FG=x-5, ∠BCD=∠FGH, ∠DAB=∠HEF. Find the value of x.
问题：如图所示，AB=32，AD=DC，BC=2*x-3，EF=12，EH=HG，FG=x-5，∠BCD=∠FGH，∠DAB=∠HEF。求x的值。
Construction (CDL):
Shape(EF,FH,HE)
Shape(HF,FG,GH)
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(EHG)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),32)
Equal(LengthOfLine(AD),LengthOfLine(DC))
Equal(LengthOfLine(BC),2*x-3)
Equal(LengthOfLine(EF),12)
Equal(LengthOfLine(EH),LengthOfLine(HG))
Equal(LengthOfLine(FG),x-5)
Equal(MeasureOfAngle(BCD),MeasureOfAngle(FGH))
Equal(MeasureOfAngle(DAB),MeasureOfAngle(HEF))
Goal: Value(x)
Answer: 31/2
Theorem sequence:
similar_triangle_judgment_aa(1,FGE,BCA)
similar_triangle_property_line_ratio(1,EFG,ABC)
similar_triangle_property_line_ratio(1,GEF,CAB)
Problem: As shown in the diagram, ∠ABC=110°. Find the measure of ∠ABC.
问题：如图所示，∠ABC=110°。求∠ABC的大小。
Construction (CDL):
Shape(AB,BC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),110)
Goal: Value(MeasureOfAngle(ABC))
Answer: 110
Theorem sequence:
Problem: As shown in the diagram, NP=2*x-30, NR=2*x+10, NMRQ is a rectangle. Find the length of line MP.
问题：如图所示，NP=2*x-30，NR=2*x+10，NMRQ是长方形。求直线MP的长度。
Construction (CDL):
Shape(NM,MP,PN)
Shape(PM,MR,RP)
Shape(NP,PQ,QN)
Shape(QP,PR,RQ)
Collinear(NPR)
Collinear(MPQ)
Conditions (CDL):
Equal(LengthOfLine(NP),2*x-30)
Equal(LengthOfLine(NR),2*x+10)
Rectangle(NMRQ)
Goal: Value(LengthOfLine(MP))
Answer: 40
Theorem sequence:
parallelogram_property_diagonal_bisection(1,NMRQ,P)
line_addition(1,NP,PR)
parallelogram_property_diagonal_bisection(1,MRQN,P)
rectangle_property_diagonal_equal(1,NMRQ)
line_addition(1,MP,PQ)
Problem: As shown in the diagram, the area of BCAW is 500, BW=30, DG=15, quadrilateral BCAW is similar to quadrilateral DEFG. Find the area of DEFG.
问题：如图所示，BCAW的面积为500，BW=30，DG=15，BCAW与DEFG相似。求四边形DEFG的面积。
Construction (CDL):
Shape(BC,CA,AW,WB)
Shape(DE,EF,FG,GD)
Conditions (CDL):
Equal(AreaOfQuadrilateral(BCAW),500)
Equal(LengthOfLine(BW),30)
Equal(LengthOfLine(DG),15)
SimilarBetweenQuadrilateral(BCAW,DEFG)
Goal: Value(AreaOfQuadrilateral(DEFG))
Answer: 125
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,WBCA,GDEF)
similar_quadrilateral_property_area_square_ratio(1,BCAW,DEFG)
Problem: As shown in the diagram, SR=TS, ∠TAS=93°, the measure of arc ARS is x, the center of ⊙A is A. Find the value of x.
问题：如图所示，SR=TS，∠TAS=93°，弧ARS的角度为x，⊙A的圆心为A。求x的值。
Construction (CDL):
Shape(ATR,RS,SA,AT)
Shape(ARS,SR)
Shape(AS,ST,TA)
Shape(AST,TS)
Cocircular(A,RST)
Conditions (CDL):
Equal(LengthOfLine(SR),LengthOfLine(TS))
Equal(MeasureOfAngle(TAS),93)
Equal(MeasureOfArc(ARS),x)
IsCentreOfCircle(A,A)
Goal: Value(x)
Answer: 93
Theorem sequence:
congruent_arc_judgment_chord_equal(1,ARS,AST)
congruent_arc_property_measure_equal(1,ARS,AST)
arc_property_center_angle(1,AST,A)
Problem: As shown in the diagram, ∠DGH=64°, IL is parallel to BN, WJ∥AK. Find the measure of ∠JGD.
问题：如图所示，∠DGH=64°，IL∥BN，WJ∥AK。求∠JGD的大小。
Construction (CDL):
Shape(AB,BC)
Shape(CB,BD)
Shape(BD,DE)
Shape(ED,DF)
Shape(FD,DG)
Shape(DG,GH)
Shape(HG,GI)
Shape(IG,GJ)
Shape(GJ,JK)
Shape(KJ,JL)
Shape(LJ,JB)
Shape(JB,BA)
Shape(BJ,JG,GD,DB)
Collinear(ABDF)
Collinear(LJGH)
Collinear(CBJK)
Collinear(EDGI)
Conditions (CDL):
Equal(MeasureOfAngle(DGH),64)
ParallelBetweenLine(IL,BN)
ParallelBetweenLine(WJ,AK)
Goal: Value(MeasureOfAngle(JGD))
Answer: 116
Theorem sequence:
adjacent_complementary_angle(1,JGD,DGH)
Problem: As shown in the diagram, BA=3, BD=x-1, CE=x+2, EF=8, ∠GAB=∠EFG, AB is perpendicular to CB, DE is perpendicular to FE. Find the length of line BD.
问题：如图所示，BA=3，BD=x-1，CE=x+2，EF=8，∠GAB=∠EFG，AB⊥CB，DE垂直于FE。求直线BD的长度。
Construction (CDL):
Shape(AB,BC,CG,GA)
Shape(GC,CD,DG)
Shape(GD,DE,EF,FG)
Collinear(AGD)
Collinear(CGF)
Collinear(BCDE)
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(BD))
Answer: 9/5
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, WX=a+12, WZ=4*a-15, XY=YZ, WY is the perpendicular bisector of ZX. Find the length of line WX.
问题：如图所示，WX=a+12，WZ=4*a-15，XY=YZ，WY垂直平分AB。求直线WX的长度。
Construction (CDL):
Shape(WZ,ZY,YW)
Shape(WY,YX,XW)
Collinear(ZYX)
Conditions (CDL):
Equal(LengthOfLine(WX),a+12)
Equal(LengthOfLine(WZ),4*a-15)
Equal(LengthOfLine(XY),LengthOfLine(YZ))
IsPerpendicularBisectorOfLine(WY,ZX)
Goal: Value(LengthOfLine(WX))
Answer: 21
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,WY,ZX)
Problem: As shown in the diagram, BC=y, BD=12, CD=12, ∠CBD=x°, BD⊥CD. Find the value of x.
问题：如图所示，BC=y，BD=12，CD=12，∠CBD=x°，BD⊥CD。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Shape(CB,BD,DC)
Conditions (CDL):
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),12)
Equal(LengthOfLine(CD),12)
Equal(MeasureOfAngle(CBD),x)
PerpendicularBetweenLine(BD,CD)
Goal: Value(x)
Answer: 45
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DCB)
isosceles_triangle_property_angle_equal(1,DCB)
triangle_property_angle_sum(1,DCB)
Problem: As shown in the diagram, ∠DHB=38°, ∠FDA=52°, HB is perpendicular to DB. Find the measure of ∠HDF.
问题：如图所示，∠DHB=38°，∠FDA=52°，HB垂直于DB。求∠HDF的大小。
Construction (CDL):
Shape(CH,HI)
Shape(IH,HD)
Shape(HD,DF)
Shape(FD,DA)
Shape(AD,DB)
Shape(DB,BE)
Shape(EB,BG)
Shape(GB,BH)
Shape(BH,HC)
Shape(HB,BD,DH)
Collinear(CHDA)
Collinear(IHBE)
Collinear(GBDF)
Conditions (CDL):
Equal(MeasureOfAngle(DHB),38)
Equal(MeasureOfAngle(FDA),52)
PerpendicularBetweenLine(HB,DB)
Goal: Value(MeasureOfAngle(HDF))
Answer: 128
Theorem sequence:
adjacent_complementary_angle(1,HDF,FDA)
Problem: As shown in the diagram, the measure of arc DBC is 170, D is the center of circle D. Find the measure of ∠CAB.
问题：如图所示，⌒DBC的角度为170，D是圆D的圆心。求∠CAB的大小。
Construction (CDL):
Shape(DCA,AC)
Shape(DC,CA,AD)
Shape(DA,AB,BD)
Shape(DAB,BA)
Shape(DB,DBC,CD)
Cocircular(D,CAB)
Conditions (CDL):
Equal(MeasureOfArc(DBC),170)
IsCentreOfCircle(D,D)
Goal: Value(MeasureOfAngle(CAB))
Answer: 85
Theorem sequence:
arc_property_center_angle(1,DBC,D)
round_angle(1,CDB,BDC)
radius_of_circle_property_length_equal(1,DC,D)
radius_of_circle_property_length_equal(1,DA,D)
radius_of_circle_property_length_equal(1,DB,D)
isosceles_triangle_judgment_line_equal(1,DCA)
isosceles_triangle_property_angle_equal(1,DCA)
isosceles_triangle_judgment_line_equal(1,DAB)
isosceles_triangle_property_angle_equal(1,DAB)
angle_addition(1,CAD,DAB)
angle_addition(1,BDA,ADC)
triangle_property_angle_sum(1,DCA)
triangle_property_angle_sum(1,DAB)
Problem: As shown in the diagram, ∠PNO=56°, the measure of arc BMN is 70. Find the measure of arc BOP.
问题：如图所示，∠PNO=56°，弧BMN的角度为70。求弧BOP的角度。
Construction (CDL):
Shape(BPM,MP)
Shape(BMN,NP,PM)
Shape(PN,NO,BOP)
Shape(BNO,ON)
Cocircular(B,PMNO)
Conditions (CDL):
Equal(MeasureOfAngle(PNO),56)
Equal(MeasureOfArc(BMN),70)
Goal: Value(MeasureOfArc(BOP))
Answer: 112
Theorem sequence:
arc_property_circumference_angle_external(1,BOP,N)
Problem: As shown in the diagram, ∠ACB=3*y+36°, ∠BDA=9*y-12°, ∠CBD=12*x+72°, ∠DAC=25*x+20°, BC and DA are opposite sides of the ▱ BDAC. Find the value of x.
问题：如图所示，∠ACB=3*y+36°，∠BDA=9*y-12°，∠CBD=12*x+72°，∠DAC=25*x+20°，四边形BDAC是平行四边形。求x的值。
Construction (CDL):
Shape(BD,DA,AC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),3*y+36)
Equal(MeasureOfAngle(BDA),9*y-12)
Equal(MeasureOfAngle(CBD),12*x+72)
Equal(MeasureOfAngle(DAC),25*x+20)
Parallelogram(BDAC)
Goal: Value(x)
Answer: 4
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BDAC)
Problem: As shown in the diagram, AD=11, BC=25, BACD is a kite. Find the area of BACD.
问题：如图所示，AD=11，BC=25，四边形BACD是风筝形。求BACD的面积。
Construction (CDL):
Shape(BA,AO,OB)
Shape(BO,OD,DB)
Shape(AC,CO,OA)
Shape(OC,CD,DO)
Collinear(COB)
Collinear(AOD)
Conditions (CDL):
Equal(LengthOfLine(AD),11)
Equal(LengthOfLine(BC),25)
Kite(BACD)
Goal: Value(AreaOfQuadrilateral(BACD))
Answer: 275/2
Theorem sequence:
kite_area_formula_diagonal(1,BACD)
Problem: As shown in the diagram, ∠CBD=55°, ∠FBG=35°, B is the center of ⊙B. Find the measure of arc BFA.
问题：如图所示，∠CBD=55°，∠FBG=35°，圆B的圆心为B。求⌒BFA的角度。
Construction (CDL):
Shape(BCA,AB,BC)
Shape(BAG,GB,BA)
Shape(BGF,FB,BG)
Shape(BFD,DB,BF)
Shape(BDC,CB,BD)
Collinear(ABD)
Collinear(CBG)
Cocircular(B,CAGFD)
Conditions (CDL):
Equal(MeasureOfAngle(CBD),55)
Equal(MeasureOfAngle(FBG),35)
IsCentreOfCircle(B,B)
Goal: Value(MeasureOfArc(BFA))
Answer: 270
Theorem sequence:
vertical_angle(1,CBD,GBA)
angle_addition(1,FBG,GBA)
round_angle(1,FBA,ABF)
arc_property_center_angle(1,BFA,B)
Problem: As shown in the diagram, PQ=25*sqrt(3), RQ=25, PQ⊥RQ. Find the measure of ∠QRP.
问题：如图所示，PQ=25*sqrt(3)，RQ=25，PQ垂直于RQ。求∠QRP的大小。
Construction (CDL):
Shape(PQ,QR,RP)
Conditions (CDL):
Equal(LengthOfLine(PQ),25*sqrt(3))
Equal(LengthOfLine(RQ),25)
PerpendicularBetweenLine(PQ,RQ)
Goal: Value(MeasureOfAngle(QRP))
Answer: 60
Theorem sequence:
right_triangle_judgment_angle(1,PQR)
right_triangle_property_pythagorean(1,PQR)
cosine_theorem(1,RPQ)
Problem: As shown in the diagram, AB=z, AN=x, AY=5, YB=14, YN=y, BA⊥YA, YN is perpendicular to AN. Find the value of y.
问题：如图所示，AB=z，AN=x，AY=5，YB=14，YN=y，BA垂直于YA，YN垂直于AN。求y的值。
Construction (CDL):
Shape(AY,YN,NA)
Shape(AN,NB,BA)
Collinear(YNB)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AN),x)
Equal(LengthOfLine(AY),5)
Equal(LengthOfLine(YB),14)
Equal(LengthOfLine(YN),y)
PerpendicularBetweenLine(BA,YA)
PerpendicularBetweenLine(YN,AN)
Goal: Value(y)
Answer: 25/14
Theorem sequence:
adjacent_complementary_angle(1,YNA,ANB)
line_addition(1,YN,NB)
right_triangle_judgment_angle(1,YNA)
right_triangle_judgment_angle(1,ANB)
right_triangle_judgment_angle(1,BAY)
right_triangle_property_pythagorean(1,YNA)
right_triangle_property_pythagorean(1,ANB)
right_triangle_property_pythagorean(1,BAY)
Problem: As shown in the diagram, KL=6, ∠JKL=148°, the center of ⊙K is K. Find the area of the sector KLJ.
问题：如图所示，KL=6，∠JKL=148°，⊙K的圆心为K。求扇形KLJ的面积。
Construction (CDL):
Shape(KL,KLJ,JK)
Shape(KJ,KJL,LK)
Cocircular(K,JL)
Conditions (CDL):
Equal(LengthOfLine(KL),6)
Equal(MeasureOfAngle(JKL),148)
IsCentreOfCircle(K,K)
Goal: Value(AreaOfSector(KLJ))
Answer: 74*pi/5
Theorem sequence:
radius_of_circle_property_length_equal(1,KL,K)
arc_property_center_angle(1,KLJ,K)
sector_area_formula(1,KLJ)
Problem: As shown in the diagram, ∠KJL=2*x+27°, ∠KLA=100°, ∠LKJ=2*x-11°. Find the measure of ∠LKJ.
问题：如图所示，∠KJL=2*x+27°，∠KLA=100°，∠LKJ=2*x-11°。求∠LKJ的大小。
Construction (CDL):
Shape(KJ,JL,LK)
Shape(KL,LA)
Collinear(JLA)
Conditions (CDL):
Equal(MeasureOfAngle(KJL),2*x+27)
Equal(MeasureOfAngle(KLA),100)
Equal(MeasureOfAngle(LKJ),2*x-11)
Goal: Value(MeasureOfAngle(LKJ))
Answer: 31
Theorem sequence:
adjacent_complementary_angle(1,JKL,KLA)
triangle_property_angle_sum(1,KJL)
Problem: As shown in the diagram, AB=c, CA=b, CB=a, ∠ABC=60°, ∠CAB=30°, b=18, BC is perpendicular to AC. Find the value of c.
问题：如图所示，AB=c，CA=b，CB=a，∠ABC=60°，∠CAB=30°，b=18，BC垂直于AC。求c的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),c)
Equal(LengthOfLine(CA),b)
Equal(LengthOfLine(CB),a)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),30)
Equal(b,18)
PerpendicularBetweenLine(BC,AC)
Goal: Value(c)
Answer: 12*sqrt(3)
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, AB=x, AC=y, BC=6, ∠BAC=30°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=x，AC=y，BC=6，∠BAC=30°，AC⊥BC。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(BC),6)
Equal(MeasureOfAngle(BAC),30)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 12
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, AD=15, PF=6, the centroid of triangle ACE is P. Find the length of line PC.
问题：如图所示，AD=15，PF=6，三角形ACE的重心为O。求直线PC的长度。
Construction (CDL):
Shape(AB,BP,PA)
Shape(PB,BC,CP)
Shape(PC,CD,DP)
Shape(AP,PF,FA)
Shape(PD,DE,EP)
Shape(FP,PE,EF)
Collinear(ABC)
Collinear(CDE)
Collinear(EFA)
Collinear(BPE)
Collinear(CPF)
Collinear(APD)
Conditions (CDL):
Equal(LengthOfLine(AD),15)
Equal(LengthOfLine(PF),6)
IsCentroidOfTriangle(P,ACE)
Goal: Value(LengthOfLine(PC))
Answer: 12
Theorem sequence:
centroid_of_triangle_property_line_ratio(1,P,CEA,F)
Problem: As shown in the diagram, the area of ABCD is 846, the area of quadrilateral EFGH is 376, CB=x, FG=24, ABCD is similar to EFGH. Find The ratio of similar quadrilaterals ABCD and EFGH.
问题：如图所示，ABCD的面积为846，EFGH的面积为376，CB=x，FG=24，ABCD相似于EFGH。求ABCD和EFGH的相似比。
Construction (CDL):
Shape(DA,AB,BC,CD)
Shape(HE,EF,FG,GH)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ABCD),846)
Equal(AreaOfQuadrilateral(EFGH),376)
Equal(LengthOfLine(CB),x)
Equal(LengthOfLine(FG),24)
SimilarBetweenQuadrilateral(ABCD,EFGH)
Goal: Value(RatioOfSimilarQuadrilateral(ABCD,EFGH))
Answer: 3/2
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,ABCD,EFGH)
Problem: As shown in the diagram, AB=15*x+9, AC=7*x, BC=11*x+5, the perimeter of triangle CBA is 320. Find the length of line CB.
问题：如图所示，AB=15*x+9，AC=7*x，BC=11*x+5，三角形CBA的周长为320。求直线CB的长度。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),15*x+9)
Equal(LengthOfLine(AC),7*x)
Equal(LengthOfLine(BC),11*x+5)
Equal(PerimeterOfTriangle(CBA),320)
Goal: Value(LengthOfLine(CB))
Answer: 107
Theorem sequence:
triangle_perimeter_formula(1,CBA)
Problem: As shown in the diagram, ∠PRQ=115°, the center of circle R is R, NR is perpendicular to PR. Find the measure of ⌒RMQ.
问题：如图所示，∠PRQ=115°，⊙R的圆心为R，NR垂直于PR。求⌒RMQ的角度。
Construction (CDL):
Shape(RPN,NR,RP)
Shape(RNM,MR,RN)
Shape(RMQ,QR,RM)
Shape(RQP,PR,RQ)
Collinear(PRM)
Cocircular(R,PNMQ)
Conditions (CDL):
Equal(MeasureOfAngle(PRQ),115)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(NR,PR)
Goal: Value(MeasureOfArc(RMQ))
Answer: 65
Theorem sequence:
adjacent_complementary_angle(1,PRQ,QRM)
arc_property_center_angle(1,RMQ,R)
Problem: As shown in the diagram, AB=12, AC=y, BC=x, ∠ACB=60°, ∠BAC=30°, CB is perpendicular to AB. Find the value of y.
问题：如图所示，AB=12，AC=y，BC=x，∠ACB=60°，∠BAC=30°，CB垂直于AB。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ACB),60)
Equal(MeasureOfAngle(BAC),30)
PerpendicularBetweenLine(CB,AB)
Goal: Value(y)
Answer: 8*sqrt(3)
Theorem sequence:
sine_theorem(1,ACB)
Problem: As shown in the diagram, BO=15, BO=CA, CE=7, CO=13, CO=BA, OE is perpendicular to AE. Find the perimeter of OCAB.
问题：如图所示，BO=15，BO=CA，CE=7，CO=13，CO=BA，OE垂直于AE。求四边形OCAB的周长。
Construction (CDL):
Shape(OC,CE,EO)
Shape(OE,EA,AB,BO)
Collinear(CEA)
Conditions (CDL):
Equal(LengthOfLine(BO),15)
Equal(LengthOfLine(BO),LengthOfLine(CA))
Equal(LengthOfLine(CE),7)
Equal(LengthOfLine(CO),13)
Equal(LengthOfLine(CO),LengthOfLine(BA))
PerpendicularBetweenLine(OE,AE)
Goal: Value(PerimeterOfQuadrilateral(OCAB))
Answer: 56
Theorem sequence:
quadrilateral_perimeter_formula(1,OCAB)
Problem: As shown in the diagram, ∠ACD=50°, ∠CDE=78°, ∠FGA=120°, ∠GFB=56°. Find the measure of ∠EAG.
问题：如图所示，∠ACD=50°，∠CDE=78°，∠FGA=120°，∠GFB=56°。求∠EAG的大小。
Construction (CDL):
Shape(CD,DA,AC)
Shape(EA,AG,GE)
Shape(GF,FB,BG)
Shape(CA,AE)
Shape(EG,GB)
Shape(GA,AD)
Shape(FG,GA)
Collinear(CAGB)
Collinear(EAD)
Collinear(EGF)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),50)
Equal(MeasureOfAngle(CDE),78)
Equal(MeasureOfAngle(FGA),120)
Equal(MeasureOfAngle(GFB),56)
Goal: Value(MeasureOfAngle(EAG))
Answer: 52
Theorem sequence:
vertical_angle(1,EAG,DAC)
triangle_property_angle_sum(1,CDA)
Problem: As shown in the diagram, RQ=5, VS=11, S is the midpoint of segment RT, V bisects segment QU, QR and UT are the parallel sides of trapezoid QUTR. Find the length of line UT.
问题：如图所示，RQ=5，VS=11，S平分线段RT，V平分线段QU，QU和TR是梯形QUTR的腰。求直线UT的长度。
Construction (CDL):
Shape(QV,VS,SR,RQ)
Shape(VU,UT,TS,SV)
Collinear(QVU)
Collinear(RST)
Conditions (CDL):
Equal(LengthOfLine(RQ),5)
Equal(LengthOfLine(VS),11)
IsMidpointOfLine(S,RT)
IsMidpointOfLine(V,QU)
Trapezoid(QUTR)
Goal: Value(LengthOfLine(UT))
Answer: 17
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,VS,QUTR)
midsegment_of_quadrilateral_property_length(1,VS,QUTR)
Problem: As shown in the diagram, FK=3*x-1, JK=6*y-2, KG=4*y+3, KH=2*x+3, FG and JH are opposite sides of the ▱ FJHG. Find the value of x.
问题：如图所示，FK=3*x-1，JK=6*y-2，KG=4*y+3，KH=2*x+3，四边形FJHG是平行四边形。求x的值。
Construction (CDL):
Shape(FJ,JK,KF)
Shape(KJ,JH,HK)
Shape(KH,HG,GK)
Shape(FK,KG,GF)
Collinear(FKH)
Collinear(JKG)
Conditions (CDL):
Equal(LengthOfLine(FK),3*x-1)
Equal(LengthOfLine(JK),6*y-2)
Equal(LengthOfLine(KG),4*y+3)
Equal(LengthOfLine(KH),2*x+3)
Parallelogram(FJHG)
Goal: Value(x)
Answer: 4
Theorem sequence:
parallelogram_property_diagonal_bisection(1,FJHG,K)
Problem: As shown in the diagram, AC=5, BC=x, ∠ABC=60°, ∠CAB=30°. Find the value of x.
问题：如图所示，AC=5，BC=x，∠ABC=60°，∠CAB=30°。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),30)
Goal: Value(x)
Answer: 5*sqrt(3)/3
Theorem sequence:
sine_theorem(1,CAB)
Problem: As shown in the diagram, ∠FOE=118°, ∠LDA=104°, OD is parallel to FI. Find the measure of ∠OFI.
问题：如图所示，∠FOE=118°，∠LDA=104°，OD∥FI。求∠OFI的大小。
Construction (CDL):
Shape(LD,DA)
Shape(AD,DI)
Shape(DI,IB)
Shape(BI,IC)
Shape(CI,IF)
Shape(IF,FJ)
Shape(JF,FK)
Shape(KF,FO)
Shape(FO,OE)
Shape(EO,OH)
Shape(HO,OD)
Shape(ID,DL)
Shape(DO,OF,FI,ID)
Collinear(LDIC)
Collinear(HOFJ)
Collinear(ADOE)
Collinear(BIFK)
Conditions (CDL):
Equal(MeasureOfAngle(FOE),118)
Equal(MeasureOfAngle(LDA),104)
ParallelBetweenLine(OD,FI)
Goal: Value(MeasureOfAngle(OFI))
Answer: 118
Theorem sequence:
adjacent_complementary_angle(1,DOF,FOE)
parallel_property_ipsilateral_internal_angle(1,OD,FI)
Problem: As shown in the diagram, ∠GOI=3*y+1°, ∠HBI=3*x+11°, ∠OIE=4*x-5°, GE∥OI, IB is parallel to OH, OI∥HB. Find the value of x.
问题：如图所示，∠GOI=3*y+1°，∠HBI=3*x+11°，∠OIE=4*x-5°，GE平行于OI，IB∥OH，OI∥HB。求x的值。
Construction (CDL):
Shape(GO,OI,IE,EG)
Shape(OH,HB,BI,IO)
Collinear(GOH)
Collinear(BIE)
Conditions (CDL):
Equal(MeasureOfAngle(GOI),3*y+1)
Equal(MeasureOfAngle(HBI),3*x+11)
Equal(MeasureOfAngle(OIE),4*x-5)
ParallelBetweenLine(GE,OI)
ParallelBetweenLine(IB,OH)
ParallelBetweenLine(OI,HB)
Goal: Value(x)
Answer: 16
Theorem sequence:
parallel_property_corresponding_angle(2,BH,IO,E)
Problem: As shown in the diagram, AC=5*sqrt(26), AD=25, CB=sqrt(26), CD=5, DB=1, ∠BCD=y°, ∠DCA=x°, AD is perpendicular to CD, BC is perpendicular to AC. Find the value of cos(CA).
问题：如图所示，AC=5*sqrt(26)，AD=25，CB=sqrt(26)，CD=5，DB=1，∠BCD=y°，∠DCA=x°，AD⊥CD，BC垂直于AC。求cos(CA)的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),5*sqrt(26))
Equal(LengthOfLine(AD),25)
Equal(LengthOfLine(CB),sqrt(26))
Equal(LengthOfLine(CD),5)
Equal(LengthOfLine(DB),1)
Equal(MeasureOfAngle(BCD),y)
Equal(MeasureOfAngle(DCA),x)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Cos(MeasureOfAngle(CAD)))
Answer: 5*sqrt(26)/26
Theorem sequence:
cosine_theorem(1,ADC)
Problem: As shown in the diagram, A is the center of ⊙A, DF is the tangent to circle A. Find the measure of ∠AFD.
问题：如图所示，A是圆A的圆心，圆O的切线为DF。求∠AFD的大小。
Construction (CDL):
Shape(AF,AFB,BA)
Shape(AB,ABF,FA)
Shape(AF,FD)
Collinear(FABE)
Cocircular(A,FB)
Conditions (CDL):
IsCentreOfCircle(A,A)
IsTangentOfCircle(DF,A)
Goal: Value(MeasureOfAngle(AFD))
Answer: 90
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DF,A,A)
Problem: As shown in the diagram, WX=9*x, WY=6*x+3, WY=YX, XW=WY, YX=4*x+5. Find the length of line XY.
问题：如图所示，WX=9*x，WY=6*x+3，WY=YX，XW=WY，YX=4*x+5。求直线XY的长度。
Construction (CDL):
Shape(XW,WY,YX)
Conditions (CDL):
Equal(LengthOfLine(WX),9*x)
Equal(LengthOfLine(WY),6*x+3)
Equal(LengthOfLine(WY),LengthOfLine(YX))
Equal(LengthOfLine(XW),LengthOfLine(WY))
Equal(LengthOfLine(YX),4*x+5)
Goal: Value(LengthOfLine(XY))
Answer: 9
Theorem sequence:
Problem: As shown in the diagram, ST=11*x-2, TU=8*x+4, UV=15*x, triangle RST is mirror congruent to triangle VTU, RS⊥TS, TU is perpendicular to VU. Find the value of x.
问题：如图所示，ST=11*x-2，TU=8*x+4，UV=15*x，△RST与△VTU是镜像全等三角形，RS⊥TS，TU垂直于VU。求x的值。
Construction (CDL):
Shape(RS,ST,TR)
Shape(TU,UV,VT)
Collinear(STU)
Conditions (CDL):
Equal(LengthOfLine(ST),11*x-2)
Equal(LengthOfLine(TU),8*x+4)
Equal(LengthOfLine(UV),15*x)
MirrorCongruentBetweenTriangle(RST,VTU)
PerpendicularBetweenLine(RS,TS)
PerpendicularBetweenLine(TU,VU)
Goal: Value(x)
Answer: 2
Theorem sequence:
mirror_congruent_triangle_property_line_equal(1,RST,VTU)
Problem: As shown in the diagram, CB=20, EC=24, ∠CAE=37°, CA and BD are opposite sides of the ▱ ACBD, AE⊥CE. Find the area of quadrilateral ACBD.
问题：如图所示，CB=20，EC=24，∠CAE=37°，ACBD是▱，AE垂直于CE。求ACBD的面积。
Construction (CDL):
Shape(EC,CA,AE)
Shape(AC,CB,BD,DA)
Collinear(EAD)
Conditions (CDL):
Equal(LengthOfLine(CB),20)
Equal(LengthOfLine(EC),24)
Equal(MeasureOfAngle(CAE),37)
Parallelogram(ACBD)
PerpendicularBetweenLine(AE,CE)
Goal: Value(AreaOfQuadrilateral(ACBD))
Answer: 480
Theorem sequence:
altitude_of_quadrilateral_judgment_right_vertex(5,CE,BDAC)
parallelogram_property_opposite_line_equal(1,CBDA)
parallelogram_area_formula_common(1,BDAC)
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°, quadrilateral BEDC is mirror congruent to quadrilateral RSTU. Find the value of x.
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°，BEDC与RSTU镜像全等。求x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(x)
Answer: 20
Theorem sequence:
mirror_congruent_quadrilateral_property_angle_equal(1,BEDC,RSTU)
Problem: As shown in the diagram, CB=12, CB=AB, ∠BAC=44°. Find the measure of ∠CBA.
问题：如图所示，CB=12，CB=AB，∠BAC=44°。求∠CBA的大小。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(CB),12)
Equal(LengthOfLine(CB),LengthOfLine(AB))
Equal(MeasureOfAngle(BAC),44)
Goal: Value(MeasureOfAngle(CBA))
Answer: 92
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,BAC)
isosceles_triangle_property_angle_equal(1,BAC)
triangle_property_angle_sum(1,BAC)