SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=z, AC=x, AD=y, BD=6, CD=4, AD⊥BD, BA is perpendicular to CA. Find the value of z.
问题：如图所示，AB=z，AC=x，AD=y，BD=6，CD=4，AD垂直于BD，BA垂直于CA。求z的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),z)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BD),6)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(BA,CA)
Goal: Value(z)
Answer: 2*sqrt(15)
Theorem sequence:
adjacent_complementary_angle(1,CDA,ADB)
angle_addition(1,BAD,DAC)
triangle_property_angle_sum(1,BAD)
similar_triangle_judgment_aa(1,CDA,ADB)
similar_triangle_property_line_ratio(1,CDA,ADB)
similar_triangle_property_line_ratio(1,ACD,BAD)
right_triangle_judgment_angle(1,ADB)
right_triangle_property_pythagorean(1,ADB)
Problem: As shown in the diagram, GJ=GH, ∠JFG=28°, △FGH is a right △. Find the measure of ∠JGH.
问题：如图所示，GJ=GH，∠JFG=28°，△FGH为直角△。求∠JGH的大小。
Construction (CDL):
Shape(JF,FG,GJ)
Shape(HJ,JG,GH)
Collinear(FJH)
Conditions (CDL):
Equal(LengthOfLine(GJ),LengthOfLine(GH))
Equal(MeasureOfAngle(JFG),28)
RightTriangle(FGH)
Goal: Value(MeasureOfAngle(JGH))
Answer: 56
Theorem sequence:
triangle_property_angle_sum(1,FGH)
isosceles_triangle_judgment_line_equal(1,GHJ)
isosceles_triangle_property_angle_equal(1,GHJ)
triangle_property_angle_sum(1,HJG)
Problem: As shown in the diagram, the area of CEBA is 72, the area of quadrilateral GIHF is 50, EC=12, GI=x, quadrilateral ACEB is similar to quadrilateral FGIH. Find the value of x.
问题：如图所示，CEBA的面积为72，GIHF的面积为50，EC=12，GI=x，四边形ACEB与四边形FGIH相似。求x的值。
Construction (CDL):
Shape(BA,AC,CE,EB)
Shape(FG,GI,IH,HF)
Conditions (CDL):
Equal(AreaOfQuadrilateral(CEBA),72)
Equal(AreaOfQuadrilateral(GIHF),50)
Equal(LengthOfLine(EC),12)
Equal(LengthOfLine(GI),x)
SimilarBetweenQuadrilateral(ACEB,FGIH)
Goal: Value(x)
Answer: 10
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,CEBA,GIHF)
similar_quadrilateral_property_line_ratio(1,CEBA,GIHF)
Problem: As shown in the diagram, the measure of ⌒CAD is 196, the tangent to circle C is BD. Find the measure of ∠DPA.
问题：如图所示，⌒CAD的角度为196，圆O的切线为BD。求∠DPA的大小。
Construction (CDL):
Shape(DA,CAD)
Shape(AD,DP,PA)
Shape(PD,CDP)
Shape(AP,CPA)
Shape(PD,DB)
Shape(OD,DA)
Collinear(BDO)
Cocircular(C,ADP)
Conditions (CDL):
Equal(MeasureOfArc(CAD),196)
IsTangentOfCircle(BD,C)
Goal: Value(MeasureOfAngle(DPA))
Answer: 98
Theorem sequence:
arc_property_circumference_angle_external(1,CAD,P)
Problem: As shown in the diagram, BC=15, BE=5, DC=13, CBAD is a parallelogram, BE⊥CE. Find the perimeter of quadrilateral CBAD.
问题：如图所示，BC=15，BE=5，DC=13，BC和AD是▱CBAD的一组对边，BE垂直于CE。求CBAD的周长。
Construction (CDL):
Shape(CB,BE,EC)
Shape(CE,EA,AD,DC)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(BE),5)
Equal(LengthOfLine(DC),13)
Parallelogram(CBAD)
PerpendicularBetweenLine(BE,CE)
Goal: Value(PerimeterOfQuadrilateral(CBAD))
Answer: 56
Theorem sequence:
parallelogram_property_opposite_line_equal(1,CBAD)
parallelogram_property_opposite_line_equal(1,DCBA)
quadrilateral_perimeter_formula(1,CBAD)
Problem: As shown in the diagram, LA=6, LB=3*sqrt(2), LP=PB, LP=PT, PT=PB, LP⊥BP. Find the perimeter of quadrilateral ATBL.
问题：如图所示，LA=6，LB=3*sqrt(2)，LP=PB，LP=PT，PT=PB，LP⊥BP。求四边形ATBL的周长。
Construction (CDL):
Shape(LA,AP,PL)
Shape(LP,PB,BL)
Shape(AT,TP,PA)
Shape(BP,PT,TB)
Collinear(APB)
Collinear(LPT)
Conditions (CDL):
Equal(LengthOfLine(LA),6)
Equal(LengthOfLine(LB),3*sqrt(2))
Equal(LengthOfLine(LP),LengthOfLine(PB))
Equal(LengthOfLine(LP),LengthOfLine(PT))
Equal(LengthOfLine(PT),LengthOfLine(PB))
PerpendicularBetweenLine(LP,BP)
Goal: Value(PerimeterOfQuadrilateral(ATBL))
Answer: 6*sqrt(2)+12
Theorem sequence:
adjacent_complementary_angle(1,APL,LPB)
adjacent_complementary_angle(1,LPB,BPT)
adjacent_complementary_angle(1,TPA,APL)
mirror_congruent_triangle_judgment_sas(1,PBL,PTB)
mirror_congruent_triangle_judgment_sas(1,PLA,PAT)
mirror_congruent_triangle_property_line_equal(1,PBL,PTB)
mirror_congruent_triangle_property_line_equal(1,PLA,PAT)
quadrilateral_perimeter_formula(1,ATBL)
Problem: As shown in the diagram, ∠ADB=64°, AC⊥DC, BA is perpendicular to CA, BF⊥DF, CD is perpendicular to BD. Find the measure of ∠BAF.
问题：如图所示，∠ADB=64°，AC⊥DC，BA⊥CA，BF⊥DF，CD垂直于BD。求∠BAF的大小。
Construction (CDL):
Shape(BA,AF,FB)
Shape(AC,CD,DF,FA)
Shape(BF,FD,DB)
Collinear(AFD)
Conditions (CDL):
Equal(MeasureOfAngle(ADB),64)
PerpendicularBetweenLine(AC,DC)
PerpendicularBetweenLine(BA,CA)
PerpendicularBetweenLine(BF,DF)
PerpendicularBetweenLine(CD,BD)
Goal: Value(MeasureOfAngle(BAF))
Answer: 26
Theorem sequence:
parallel_judgment_per_per(1,AB,CD)
parallel_judgment_per_per(1,CA,DB)
parallelogram_judgment_parallel_and_parallel(1,ACDB)
rectangle_judgment_right_angle(1,ACDB)
triangle_property_angle_sum(1,ADB)
Problem: As shown in the diagram, RS=4, RT=x, UT=9, VS=5. Find the value of x.
问题：如图所示，RS=4，RT=x，UT=9，VS=5。求x的值。
Construction (CDL):
Shape(TB,BU,BUT)
Shape(TR,RS,BST)
Shape(VS,BSV)
Shape(UB,BT,BTS,SV,BVU)
Collinear(RTBU)
Collinear(RSV)
Cocircular(B,TSVU)
Conditions (CDL):
Equal(LengthOfLine(RS),4)
Equal(LengthOfLine(RT),x)
Equal(LengthOfLine(UT),9)
Equal(LengthOfLine(VS),5)
Goal: Value(x)
Answer: 3
Theorem sequence:
line_addition(1,RT,TU)
line_addition(1,RS,SV)
circle_property_circular_power_segment_and_segment_line(1,RTU,RSV,B)
Problem: As shown in the diagram, AB=6, AD=4, BC=x+3, CD=x, ∠CBD=∠DBA. Find the value of x.
问题：如图所示，AB=6，AD=4，BC=x+3，CD=x，∠CBD=∠DBA。求x的值。
Construction (CDL):
Shape(DC,CB,BD)
Shape(AD,DB,BA)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(BC),x+3)
Equal(LengthOfLine(CD),x)
Equal(MeasureOfAngle(CBD),MeasureOfAngle(DBA))
Goal: Value(x)
Answer: 6
Theorem sequence:
bisector_of_angle_judgment_angle_equal(1,BD,CBA)
bisector_of_angle_property_line_ratio(1,BD,CBA)
Problem: As shown in the diagram, CD=30, ED=10, ∠CED=60°, quadrilateral DEBC is a parallelogram. Find the perimeter of DEBC.
问题：如图所示，CD=30，ED=10，∠CED=60°，ED和BC是平行四边形DEBC的一组对边。求四边形DEBC的周长。
Construction (CDL):
Shape(DE,EF,FD)
Shape(DF,FB,BC,CD)
Collinear(EFB)
Conditions (CDL):
Equal(LengthOfLine(CD),30)
Equal(LengthOfLine(ED),10)
Equal(MeasureOfAngle(CED),60)
Parallelogram(DEBC)
Goal: Value(PerimeterOfQuadrilateral(DEBC))
Answer: 80
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DEBC)
parallelogram_property_opposite_line_equal(1,CDEB)
quadrilateral_perimeter_formula(1,DEBC)
Problem: As shown in the diagram, BC=14, BD=x, FB=20, FC=11, ∠DCF=∠BCD. Find the value of x.
问题：如图所示，BC=14，BD=x，FB=20，FC=11，∠DCF=∠BCD。求x的值。
Construction (CDL):
Shape(CF,FD,DC)
Shape(CD,DB,BC)
Collinear(FDB)
Conditions (CDL):
Equal(LengthOfLine(BC),14)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(FB),20)
Equal(LengthOfLine(FC),11)
Equal(MeasureOfAngle(DCF),MeasureOfAngle(BCD))
Goal: Value(x)
Answer: 56/5
Theorem sequence:
line_addition(1,FD,DB)
bisector_of_angle_judgment_angle_equal(1,CD,BCF)
bisector_of_angle_property_line_ratio(1,CD,BCF)
Problem: As shown in the diagram, ∠KLM=57°, ∠LKJ=28°, ∠MKL=71°. Find the measure of ∠KJL.
问题：如图所示，∠KLM=57°，∠LKJ=28°，∠MKL=71°。求∠KJL的大小。
Construction (CDL):
Shape(JL,LK,KJ)
Shape(KL,LM,MK)
Collinear(JLM)
Conditions (CDL):
Equal(MeasureOfAngle(KLM),57)
Equal(MeasureOfAngle(LKJ),28)
Equal(MeasureOfAngle(MKL),71)
Goal: Value(MeasureOfAngle(KJL))
Answer: 29
Theorem sequence:
flat_angle(1,JLM)
angle_addition(1,JLK,KLM)
triangle_property_angle_sum(1,JLK)
Problem: As shown in the diagram, PD=10, PQ=10, QE=20, CQ⊥PQ, PD is perpendicular to BD. Find the length of line PE.
问题：如图所示，PD=10，PQ=10，QE=20，CQ⊥PQ，PD⊥BD。求直线PE的长度。
Construction (CDL):
Shape(PQ,QE,EP)
Shape(CQ,QP,PD,DB,PBC)
Shape(AD,DP,PE,PEA)
Shape(GD,DA,PAG)
Shape(BD,DG,PGB)
Shape(FQ,QC,PCF)
Shape(EQ,QF,PFE)
Collinear(GDP)
Collinear(PQF)
Collinear(BDA)
Collinear(CQE)
Cocircular(P,AGBCFE)
Conditions (CDL):
Equal(LengthOfLine(PD),10)
Equal(LengthOfLine(PQ),10)
Equal(LengthOfLine(QE),20)
PerpendicularBetweenLine(CQ,PQ)
PerpendicularBetweenLine(PD,BD)
Goal: Value(LengthOfLine(PE))
Answer: 10*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,CQP,PQE)
right_triangle_judgment_angle(1,PQE)
right_triangle_property_pythagorean(1,PQE)
Problem: As shown in the diagram, ∠EAD=42°. Find the measure of ∠CAB.
问题：如图所示，∠EAD=42°。求∠CAB的大小。
Construction (CDL):
Shape(CA,AB,ABC)
Shape(DA,AC,ACD)
Shape(EA,AD,ADE)
Shape(BA,AE,AEB)
Collinear(BAD)
Collinear(CAE)
Cocircular(A,BCDE)
Conditions (CDL):
Equal(MeasureOfAngle(EAD),42)
Goal: Value(MeasureOfAngle(CAB))
Answer: 42
Theorem sequence:
vertical_angle(1,CAB,EAD)
Problem: As shown in the diagram, AC=15, BM=8, CM and CB are one pair of adjacent sides of the kite AMCB. Find the area of quadrilateral AMCB.
问题：如图所示，AC=15，BM=8，CM和CB是风筝形AMCB的一组临边。求四边形AMCB的面积。
Construction (CDL):
Shape(CB,BO,OC)
Shape(CO,OM,MC)
Shape(BA,AO,OB)
Shape(OA,AM,MO)
Collinear(BOM)
Collinear(AOC)
Conditions (CDL):
Equal(LengthOfLine(AC),15)
Equal(LengthOfLine(BM),8)
Kite(AMCB)
Goal: Value(AreaOfQuadrilateral(AMCB))
Answer: 60
Theorem sequence:
kite_area_formula_diagonal(1,AMCB)
Problem: As shown in the diagram, RW=SR, ST=2*x+2, TW=4*x-4, RT⊥WT. Find the length of line SW.
问题：如图所示，RW=SR，ST=2*x+2，TW=4*x-4，RT垂直于WT。求直线SW的长度。
Construction (CDL):
Shape(WR,RT,TW)
Shape(RS,ST,TR)
Collinear(WTS)
Conditions (CDL):
Equal(LengthOfLine(RW),LengthOfLine(SR))
Equal(LengthOfLine(ST),2*x+2)
Equal(LengthOfLine(TW),4*x-4)
PerpendicularBetweenLine(RT,WT)
Goal: Value(LengthOfLine(SW))
Answer: 16
Theorem sequence:
adjacent_complementary_angle(1,STR,RTW)
isosceles_triangle_judgment_line_equal(1,RSW)
isosceles_triangle_property_angle_equal(1,RSW)
mirror_congruent_triangle_judgment_aas(1,RTW,RST)
mirror_congruent_triangle_property_line_equal(1,RTW,RST)
line_addition(1,WT,TS)
Problem: As shown in the diagram, AB=15, AC=25, CB=20, CB⊥AB. Find the value of sin(BA).
问题：如图所示，AB=15，AC=25，CB=20，CB⊥AB。求sin(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AC),25)
Equal(LengthOfLine(CB),20)
PerpendicularBetweenLine(CB,AB)
Goal: Value(Sin(MeasureOfAngle(BAC)))
Answer: 4/5
Theorem sequence:
sine_theorem(1,CBA)
Problem: As shown in the diagram, AB=30, CD=30, the measure of ⌒XCZ is 40, the center of ⊙X is X, AM is perpendicular to YM, DN⊥ZN. Find the measure of ⌒XZD.
问题：如图所示，AB=30，CD=30，⌒XCZ的角度为40，圆X的圆心为X，AM⊥YM，DN垂直于ZN。求⌒XZD的角度。
Construction (CDL):
Shape(AM,MY,XYA)
Shape(YM,MB,XBY)
Shape(CX,XM,MA,XAC)
Shape(BM,MX,XD,XDB)
Shape(XC,CN,NX)
Shape(XN,ND,DX)
Shape(ZN,NC,XCZ)
Shape(DN,NZ,XZD)
Collinear(YMX)
Collinear(XNZ)
Collinear(AMB)
Collinear(CND)
Cocircular(X,CZDBYA)
Conditions (CDL):
Equal(LengthOfLine(AB),30)
Equal(LengthOfLine(CD),30)
Equal(MeasureOfArc(XCZ),40)
IsCentreOfCircle(X,X)
PerpendicularBetweenLine(AM,YM)
PerpendicularBetweenLine(DN,ZN)
Goal: Value(MeasureOfArc(XZD))
Answer: 40
Theorem sequence:
arc_property_center_angle(1,XCZ,X)
radius_of_circle_property_length_equal(1,XC,X)
radius_of_circle_property_length_equal(1,XD,X)
isosceles_triangle_judgment_line_equal(1,XCD)
isosceles_triangle_property_angle_equal(1,XCD)
adjacent_complementary_angle(1,XND,DNZ)
adjacent_complementary_angle(1,CNX,XND)
mirror_congruent_triangle_judgment_aas(1,XCN,XND)
mirror_congruent_triangle_property_angle_equal(1,XCN,XND)
arc_property_center_angle(1,XZD,X)
Problem: As shown in the diagram, ∠EHI=∠IGF, ∠GFI=3*x+23°, ∠IEH=6*x+2°. Find the value of x.
问题：如图所示，∠EHI=∠IGF，∠GFI=3*x+23°，∠IEH=6*x+2°。求x的值。
Construction (CDL):
Shape(HE,OEH)
Shape(GI,IH,OHG)
Shape(FG,OGF)
Shape(EI,IF,OFE)
Shape(EH,HI,IE)
Shape(IG,GF,FI)
Collinear(EIG)
Collinear(HIF)
Cocircular(O,EHGF)
Conditions (CDL):
Equal(MeasureOfAngle(EHI),MeasureOfAngle(IGF))
Equal(MeasureOfAngle(GFI),3*x+23)
Equal(MeasureOfAngle(IEH),6*x+2)
Goal: Value(x)
Answer: 7
Theorem sequence:
vertical_angle(1,HIE,FIG)
mirror_similar_triangle_judgment_aa(1,EHI,FIG)
mirror_similar_triangle_property_angle_equal(1,EHI,FIG)
Problem: As shown in the diagram, FB=7-x, GF=5, SA=2, ST=x, FB is the altitude of triangle FGB, SA is the altitude of triangle STA, GB⊥FB, TA is perpendicular to SA, △SRT is similar to △FEG.. Find the length of line FB.
问题：如图所示，FB=7-x，GF=5，SA=2，ST=x，FB是三角形FGB的高，SA是△STA的高，GB⊥FB，TA⊥SA，三角形SRT与三角形FEG是相似三角形。求直线FB的长度。
Construction (CDL):
Shape(SR,RT,TS)
Shape(ST,TA,AS)
Shape(FE,EG,GF)
Shape(FG,GB,BF)
Collinear(RTA)
Collinear(EGB)
Conditions (CDL):
Equal(LengthOfLine(FB),7-x)
Equal(LengthOfLine(GF),5)
Equal(LengthOfLine(SA),2)
Equal(LengthOfLine(ST),x)
IsAltitudeOfTriangle(FB,FGB)
IsAltitudeOfTriangle(SA,STA)
PerpendicularBetweenLine(GB,FB)
PerpendicularBetweenLine(TA,SA)
SimilarBetweenTriangle(SRT,FEG)
Goal: Value(LengthOfLine(FB))
Answer: 2
Theorem sequence:
similar_triangle_property_angle_equal(1,TSR,GFE)
adjacent_complementary_angle(1,RTS,STA)
adjacent_complementary_angle(1,EGF,FGB)
similar_triangle_judgment_aa(1,STA,FGB)
similar_triangle_property_line_ratio(1,TAS,GBF)
similar_triangle_property_line_ratio(1,AST,BFG)
Problem: As shown in the diagram, ∠ANB=∠CND, ∠BNC=30°. Find the measure of ∠ANB.
问题：如图所示，∠ANB=∠CND，∠BNC=30°。求∠ANB的大小。
Construction (CDL):
Shape(AN,NB,NBA)
Shape(ED,DN,NA,NAE)
Shape(DE,NED)
Shape(CN,ND,NDC)
Shape(BC,NCB)
Shape(BN,NC,CB)
Collinear(AND)
Cocircular(N,AEDCB)
Conditions (CDL):
Equal(MeasureOfAngle(ANB),MeasureOfAngle(CND))
Equal(MeasureOfAngle(BNC),30)
Goal: Value(MeasureOfAngle(ANB))
Answer: 75
Theorem sequence:
angle_addition(1,ANB,BNC)
adjacent_complementary_angle(1,ANC,CND)
Problem: As shown in the diagram, DB=12, ∠DBC=18°, CD is perpendicular to BD. Find the area of △BCD.
问题：如图所示，DB=12，∠DBC=18°，CD⊥BD。求△BCD的面积。
Construction (CDL):
Shape(BC,CD,DB)
Conditions (CDL):
Equal(LengthOfLine(DB),12)
Equal(MeasureOfAngle(DBC),18)
PerpendicularBetweenLine(CD,BD)
Goal: Value(AreaOfTriangle(BCD))
Answer: 36*(-sqrt(2)+sqrt(10))/sqrt(sqrt(5)+5)
Theorem sequence:
triangle_property_angle_sum(1,BCD)
sine_theorem(1,BCD)
triangle_area_formula_sine(1,BCD)
Problem: As shown in the diagram, ∠EGB=85°, ∠GAS=2*x-17°, GR∥AS. Find the value of x.
问题：如图所示，∠EGB=85°，∠GAS=2*x-17°，GR∥AS。求x的值。
Construction (CDL):
Shape(EG,GB)
Shape(BG,GR)
Shape(AG,GE)
Shape(RG,GA)
Shape(DA,AG)
Shape(SA,AG)
Shape(HA,AD)
Shape(SA,AH)
Collinear(BGAH)
Collinear(EGR)
Collinear(DAS)
Conditions (CDL):
Equal(MeasureOfAngle(EGB),85)
Equal(MeasureOfAngle(GAS),2*x-17)
ParallelBetweenLine(GR,AS)
Goal: Value(x)
Answer: 56
Theorem sequence:
adjacent_complementary_angle(1,EGB,BGR)
parallel_property_corresponding_angle(1,GR,AS,B)
Problem: As shown in the diagram, ∠HUN=123°, KA∥UQ, UN is parallel to KG. Find the measure of ∠GKF.
问题：如图所示，∠HUN=123°，KA平行于UQ，UN平行于KG。求∠GKF的大小。
Construction (CDL):
Shape(GK,KF)
Shape(UK,KG)
Shape(NU,UK)
Shape(HU,UN)
Shape(AK,KF)
Shape(AK,KU)
Shape(KU,UQ)
Shape(QU,UH)
Shape(KA,AJ)
Shape(JA,AM)
Shape(QA,AK)
Shape(AM,AQ)
Shape(UQ,QA)
Shape(EQ,QA)
Shape(IQ,QU)
Shape(EQ,QI)
Collinear(GKAM)
Collinear(NUQE)
Collinear(FKUH)
Collinear(JAQI)
Conditions (CDL):
Equal(MeasureOfAngle(HUN),123)
ParallelBetweenLine(KA,UQ)
ParallelBetweenLine(UN,KG)
Goal: Value(MeasureOfAngle(GKF))
Answer: 57
Theorem sequence:
parallel_property_corresponding_angle(1,UN,KG,H)
adjacent_complementary_angle(1,UKG,GKF)
Problem: As shown in the diagram, CA=42, DM=MS, the center of circle M is M, AD⊥MD, MS⊥TS. Find the length of line SC.
问题：如图所示，CA=42，DM=MS，M是⊙M的圆心，AD垂直于MD，MS⊥TS。求直线SC的长度。
Construction (CDL):
Shape(CD,DA,MAC)
Shape(TS,SC,MCT)
Shape(AM,MS,ST,MTA)
Shape(AD,DM,MA)
Shape(DC,CM,MD)
Shape(MC,CS,SM)
Collinear(ADC)
Collinear(CST)
Cocircular(M,CTA)
Conditions (CDL):
Equal(LengthOfLine(CA),42)
Equal(LengthOfLine(DM),LengthOfLine(MS))
IsCentreOfCircle(M,M)
PerpendicularBetweenLine(AD,MD)
PerpendicularBetweenLine(MS,TS)
Goal: Value(LengthOfLine(SC))
Answer: 21
Theorem sequence:
line_addition(1,AD,DC)
radius_of_circle_property_length_equal(1,MA,M)
radius_of_circle_property_length_equal(1,MC,M)
isosceles_triangle_judgment_line_equal(1,MAC)
isosceles_triangle_property_angle_equal(1,MAC)
adjacent_complementary_angle(1,ADM,MDC)
adjacent_complementary_angle(1,CSM,MST)
mirror_congruent_triangle_judgment_aas(1,MDC,MAD)
mirror_congruent_triangle_property_line_equal(1,MDC,MAD)
mirror_congruent_triangle_judgment_hl(1,CSM,CMD)
mirror_congruent_triangle_property_line_equal(1,MCS,MDC)
Problem: As shown in the diagram, ∠FLE=35°, ∠IEL=28°, ∠LAG=25°, ∠LGI=51°, EF⊥LF, LE⊥DE, LI is perpendicular to EI. Find the measure of ∠GLA.
问题：如图所示，∠FLE=35°，∠IEL=28°，∠LAG=25°，∠LGI=51°，EF垂直于LF，LE垂直于DE，LI⊥EI。求∠GLA的大小。
Construction (CDL):
Shape(LA,AG,GL)
Shape(LG,GI,IL)
Shape(LI,IE,EL)
Shape(LE,EF,FL)
Shape(FE,ED,DF)
Collinear(AGIE)
Collinear(LFD)
Conditions (CDL):
Equal(MeasureOfAngle(FLE),35)
Equal(MeasureOfAngle(IEL),28)
Equal(MeasureOfAngle(LAG),25)
Equal(MeasureOfAngle(LGI),51)
PerpendicularBetweenLine(EF,LF)
PerpendicularBetweenLine(LE,DE)
PerpendicularBetweenLine(LI,EI)
Goal: Value(MeasureOfAngle(GLA))
Answer: 26
Theorem sequence:
flat_angle(1,AGI)
angle_addition(1,AGL,LGI)
triangle_property_angle_sum(1,AGL)
Problem: As shown in the diagram, ∠UZY=2*x+24°, ∠VZU=4*x°, ∠WZX=∠XZY. Find the measure of ∠WZV.
问题：如图所示，∠UZY=2*x+24°，∠VZU=4*x°，∠WZX=∠XZY。求∠WZV的大小。
Construction (CDL):
Shape(WZ,ZV,ZVW)
Shape(XZ,ZW,ZWX)
Shape(YZ,ZX,ZXY)
Shape(UZ,ZY,ZYU)
Shape(VZ,Zu,ZUV)
Collinear(VZY)
Collinear(WZU)
Cocircular(Z,VWXYU)
Conditions (CDL):
Equal(MeasureOfAngle(UZY),2*x+24)
Equal(MeasureOfAngle(VZU),4*x)
Equal(MeasureOfAngle(WZX),MeasureOfAngle(XZY))
Goal: Value(MeasureOfAngle(WZV))
Answer: 76
Theorem sequence:
adjacent_complementary_angle(1,VZU,UZY)
vertical_angle(1,WZV,UZY)
Problem: As shown in the diagram, JH=x, KM=22, JH is the midsegment of triangle LKM. Find the value of x.
问题：如图所示，JH=x，KM=22，JH是三角形LKM的中位线。求x的值。
Construction (CDL):
Shape(LJ,JH,HL)
Shape(JK,KB,BJ)
Shape(JB,BM,MH,HJ)
Collinear(KJL)
Collinear(LHM)
Collinear(KBM)
Conditions (CDL):
Equal(LengthOfLine(JH),x)
Equal(LengthOfLine(KM),22)
IsMidsegmentOfTriangle(JH,LKM)
Goal: Value(x)
Answer: 11
Theorem sequence:
line_addition(1,KJ,JL)
midsegment_of_triangle_property_parallel(1,JH,LKM)
parallel_property_corresponding_angle(1,JH,KM,L)
similar_triangle_judgment_aa(1,HLJ,MLK)
similar_triangle_property_line_ratio(1,HLJ,MLK)
similar_triangle_property_line_ratio(1,LJH,LKM)
Problem: As shown in the diagram, ∠LWX=53°, GN is parallel to IK, WL∥XE. Find the measure of ∠WXE.
问题：如图所示，∠LWX=53°，GN∥IK，WL∥XE。求∠WXE的大小。
Construction (CDL):
Shape(LW,WX)
Shape(EX,XN)
Shape(GW,WL)
Shape(EX,XN)
Shape(YW,WG)
Shape(XW,WY)
Shape(ZX,XW)
Shape(NX,XZ)
Collinear(GWXN)
Collinear(IYZK)
Collinear(LWYM)
Collinear(EXZH)
Conditions (CDL):
Equal(MeasureOfAngle(LWX),53)
ParallelBetweenLine(GN,IK)
ParallelBetweenLine(WL,XE)
Goal: Value(MeasureOfAngle(WXE))
Answer: 127
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,WL,XE)
Problem: As shown in the diagram, PQ=721, PS=756, QR=547, RS=593, ∠QPS=58°. Find the length of line QS.
问题：如图所示，PQ=721，PS=756，QR=547，RS=593，∠QPS=58°。求直线QS的长度。
Construction (CDL):
Shape(PS,SQ,QP)
Shape(QS,SR,RQ)
Conditions (CDL):
Equal(LengthOfLine(PQ),721)
Equal(LengthOfLine(PS),756)
Equal(LengthOfLine(QR),547)
Equal(LengthOfLine(RS),593)
Equal(MeasureOfAngle(QPS),58)
Goal: Value(LengthOfLine(QS))
Answer: 7*sqrt(22273-22248*cos(29*pi/90))
Theorem sequence:
cosine_theorem(1,PSQ)
Problem: As shown in the diagram, the area of triangle BAC is 875, the area of △EDF is 315, BP=x, EQ=21, BP is the altitude of triangle BAC, EQ is the altitude of triangle EDF, △BAC is similar to △EDF.. Find the ratio of the length of line BP to the length of line EQ.
问题：如图所示，△BAC的面积为875，三角形EDF的面积为315，BP=x，EQ=21，BP是△BAC的高，EQ是三角形EDF的高，△BAC相似于△EDF。求直线BP的长度与直线EQ的长度之比。
Construction (CDL):
Shape(BA,AP,PB)
Shape(BP,PC,CB)
Shape(ED,DQ,QE)
Shape(EQ,QF,FE)
Collinear(APC)
Collinear(DQF)
Conditions (CDL):
Equal(AreaOfTriangle(BAC),875)
Equal(AreaOfTriangle(EDF),315)
Equal(LengthOfLine(BP),x)
Equal(LengthOfLine(EQ),21)
IsAltitudeOfTriangle(BP,BAC)
IsAltitudeOfTriangle(EQ,EDF)
SimilarBetweenTriangle(BAC,EDF)
Goal: Value(Div(LengthOfLine(BP),LengthOfLine(EQ)))
Answer: 5/3
Theorem sequence:
similar_triangle_property_angle_equal(1,ACB,DFE)
similar_triangle_judgment_aa(1,BAP,EDQ)
similar_triangle_property_line_ratio(1,CBA,FED)
similar_triangle_property_line_ratio(1,APB,DQE)
similar_triangle_property_line_ratio(1,PBA,QED)
similar_triangle_property_area_square_ratio(1,BAC,EDF)
Problem: As shown in the diagram, QH=y, QJ=2*z, QK=7, QL=3, QM=4, QP=2*x-6, J is the midpoint of segment KH, L is the midpoint of segment MK, P bisects segment HM. Find the value of x.
问题：如图所示，QH=y，QJ=2*z，QK=7，QL=3，QM=4，QP=2*x-6，J是线段KH的中点，L平分线段MK，P是线段HM的中点。求x的值。
Construction (CDL):
Shape(HJ,JQ,QH)
Shape(HQ,QP,PH)
Shape(JK,KQ,QJ)
Shape(QK,KL,LQ)
Shape(QL,LM,MQ)
Shape(QM,MP,PQ)
Collinear(KJH)
Collinear(KLM)
Collinear(HPM)
Collinear(JQM)
Collinear(KQP)
Collinear(HQL)
Conditions (CDL):
Equal(LengthOfLine(QH),y)
Equal(LengthOfLine(QJ),2*z)
Equal(LengthOfLine(QK),7)
Equal(LengthOfLine(QL),3)
Equal(LengthOfLine(QM),4)
Equal(LengthOfLine(QP),2*x-6)
IsMidpointOfLine(J,KH)
IsMidpointOfLine(L,MK)
IsMidpointOfLine(P,HM)
Goal: Value(x)
Answer: 19/4
Theorem sequence:
median_of_triangle_judgment(1,KP,KMH)
median_of_triangle_judgment(1,MJ,MHK)
centroid_of_triangle_judgment_intersection(1,Q,HKM,J,P)
centroid_of_triangle_property_line_ratio(1,Q,KMH,P)
Problem: As shown in the diagram, PM=12, QR=16, TS=4*x, M is the midpoint of segment RS, P is the midpoint of segment QT, QTSR is a trapezoid. Find the value of x.
问题：如图所示，PM=12，QR=16，TS=4*x，M是线段RS的中点，P平分线段QT，四边形QTSR是梯形。求x的值。
Construction (CDL):
Shape(QP,PM,MR,RQ)
Shape(PT,TS,SM,MP)
Collinear(RMS)
Collinear(QPT)
Conditions (CDL):
Equal(LengthOfLine(PM),12)
Equal(LengthOfLine(QR),16)
Equal(LengthOfLine(TS),4*x)
IsMidpointOfLine(M,RS)
IsMidpointOfLine(P,QT)
Trapezoid(QTSR)
Goal: Value(x)
Answer: 2
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,PM,QTSR)
midsegment_of_quadrilateral_property_length(1,PM,QTSR)
Problem: As shown in the diagram, BC=5, TC=5, the center of circle D is D, the diameter of circle D is AB, quadrilateral BATC is a square. Find the area of quadrilateral BATC minus the area of the sector DAB.
问题：如图所示，BC=5，TC=5，圆D的圆心为D，AB是⊙D的直径，四边形BATC是正方形。求BATC的面积减去扇形DAB的面积。
Construction (CDL):
Shape(AT,TC,CB,DAB)
Shape(BD,DA,DAB)
Shape(AD,DB,DBA)
Collinear(BDA)
Cocircular(D,AB)
Conditions (CDL):
Equal(LengthOfLine(BC),5)
Equal(LengthOfLine(TC),5)
IsCentreOfCircle(D,D)
IsDiameterOfCircle(AB,D)
Square(BATC)
Goal: Value(Sub(AreaOfQuadrilateral(BATC),AreaOfSector(DAB)))
Answer: 25-25*pi/8
Theorem sequence:
diameter_of_circle_property_length_equal(1,AB,D)
circle_property_length_of_radius_and_diameter(1,D)
parallelogram_area_formula_sine(1,ATCB)
flat_angle(1,BDA)
arc_property_center_angle(1,DAB,D)
sector_area_formula(1,DAB)
Problem: As shown in the diagram, ∠LWX=3*a+40°, ∠WXE=2*a+25°, ∠XZK=5*b-26°, WL∥XE, XN is parallel to ZK. Find the value of b.
问题：如图所示，∠LWX=3*a+40°，∠WXE=2*a+25°，∠XZK=5*b-26°，WL平行于XE，XN∥ZK。求b的值。
Construction (CDL):
Shape(LW,WX)
Shape(EX,XN)
Shape(GW,WL)
Shape(EX,XN)
Shape(YW,WG)
Shape(XW,WY)
Shape(ZX,XW)
Shape(NX,XZ)
Shape(IY,YW)
Shape(MY,YI)
Shape(WY,YZ)
Shape(ZY,YM)
Shape(YZ,ZX)
Shape(KZ,ZX)
Shape(HZ,ZY)
Shape(KZ,ZH)
Collinear(GWXN)
Collinear(IYZK)
Collinear(LWYM)
Collinear(EXZH)
Conditions (CDL):
Equal(MeasureOfAngle(LWX),3*a+40)
Equal(MeasureOfAngle(WXE),2*a+25)
Equal(MeasureOfAngle(XZK),5*b-26)
ParallelBetweenLine(WL,XE)
ParallelBetweenLine(XN,ZK)
Goal: Value(b)
Answer: 27
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,WL,XE)
adjacent_complementary_angle(1,WXE,EXN)
parallel_property_corresponding_angle(1,XN,ZK,E)
Problem: As shown in the diagram, AC=12, CE=4, ∠CBE=35°, BCAD is a parallelogram, BE is perpendicular to CE. Find the area of quadrilateral BCAD.
问题：如图所示，AC=12，CE=4，∠CBE=35°，CB和AD是▱BCAD的一组对边，BE垂直于CE。求四边形BCAD的面积。
Construction (CDL):
Shape(BE,EC,CB)
Shape(BC,CA,AD,DB)
Collinear(ECA)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(CE),4)
Equal(MeasureOfAngle(CBE),35)
Parallelogram(BCAD)
PerpendicularBetweenLine(BE,CE)
Goal: Value(AreaOfQuadrilateral(BCAD))
Answer: 48*sqrt(tan(7*pi/36)**2+1)*sin(11*pi/36)/tan(7*pi/36)
Theorem sequence:
triangle_property_angle_sum(1,ECB)
sine_theorem(1,ECB)
right_triangle_judgment_angle(1,BEC)
right_triangle_property_pythagorean(1,BEC)
adjacent_complementary_angle(1,ECB,BCA)
parallelogram_area_formula_sine(1,BCAD)