SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, the measure of ⌒DAE is 110, the measure of ⌒DBC is 150, D is the center of circle D. Find the measure of ∠AJC.
问题：如图所示，弧DAE的角度为110，⌒DBC的角度为150，圆D的圆心为D。求∠AJC的大小。
Construction (CDL):
Shape(AJ,JC,CA)
Shape(BJ,JE,JEB)
Shape(CJ,JB,DBC)
Shape(AC,DCA)
Shape(EJ,JA,DAE)
Collinear(AJB)
Collinear(EJC)
Cocircular(D,EBCA)
Conditions (CDL):
Equal(MeasureOfArc(DAE),110)
Equal(MeasureOfArc(DBC),150)
IsCentreOfCircle(D,D)
Goal: Value(MeasureOfAngle(AJC))
Answer: 50
Theorem sequence:
arc_property_circumference_angle_external(1,DAE,C)
arc_property_circumference_angle_external(1,DBC,A)
triangle_property_angle_sum(1,AJC)
Problem: As shown in the diagram, KL=17, KM=x, LA=10, MA=x, the tangent to ⊙M is LK, LK⊥MK. Find the value of x.
问题：如图所示，KL=17，KM=x，LA=10，MA=x，LK是⊙O的切线，LK垂直于MK。求x的值。
Construction (CDL):
Shape(KM,MA,MAK)
Shape(AL,LK,MAK)
Shape(AM,MK,MKA)
Collinear(MAL)
Cocircular(M,AK)
Conditions (CDL):
Equal(LengthOfLine(KL),17)
Equal(LengthOfLine(KM),x)
Equal(LengthOfLine(LA),10)
Equal(LengthOfLine(MA),x)
IsTangentOfCircle(LK,M)
PerpendicularBetweenLine(LK,MK)
Goal: Value(x)
Answer: 189/20
Theorem sequence:
line_addition(1,MA,AL)
right_triangle_judgment_angle(1,LKM)
right_triangle_property_pythagorean(1,LKM)
Problem: As shown in the diagram, ∠GFJ=10*x-4°, ∠JEH=8*x+6°, C is the center of ⊙C. Find the value of x.
问题：如图所示，∠GFJ=10*x-4°，∠JEH=8*x+6°，圆C的圆心为C。求x的值。
Construction (CDL):
Shape(C)
Shape(EH,HJ,JE)
Shape(FJ,JG,GF)
Shape(HE,CEH)
Shape(FG,CGF)
Shape(GJ,JH,CHG)
Shape(EJ,JF,CFE)
Collinear(EJG)
Collinear(FJH)
Cocircular(C,HGFE)
Conditions (CDL):
Equal(MeasureOfAngle(GFJ),10*x-4)
Equal(MeasureOfAngle(JEH),8*x+6)
IsCentreOfCircle(C,C)
Goal: Value(x)
Answer: 5
Theorem sequence:
arc_property_circumference_angle_external(1,CHG,E)
arc_property_circumference_angle_external(1,CHG,F)
Problem: As shown in the diagram, ∠NMP=56°, ∠PNM=45°. Find the measure of ∠QPM.
问题：如图所示，∠NMP=56°，∠PNM=45°。求∠QPM的大小。
Construction (CDL):
Shape(MP,PN,NM)
Shape(QP,PM)
Collinear(QPN)
Conditions (CDL):
Equal(MeasureOfAngle(NMP),56)
Equal(MeasureOfAngle(PNM),45)
Goal: Value(MeasureOfAngle(QPM))
Answer: 101
Theorem sequence:
triangle_property_angle_sum(1,MPN)
adjacent_complementary_angle(1,QPM,MPN)
Problem: As shown in the diagram, ∠CBA=90°, ∠IAG=157°, BJ is parallel to AD, CF is parallel to BJ. Find the measure of ∠ECH.
问题：如图所示，∠CBA=90°，∠IAG=157°，BJ∥AD，CF∥BJ。求∠ECH的大小。
Construction (CDL):
Shape(HC,CF)
Shape(EC,CH)
Shape(BC,CE)
Shape(BC,CF)
Shape(CB,BJ)
Shape(JB,BA)
Shape(BA,AD)
Shape(DA,AI)
Shape(GA,AB)
Shape(IA,AG)
Collinear(HCB)
Collinear(DAG)
Collinear(BAI)
Collinear(ECF)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),90)
Equal(MeasureOfAngle(IAG),157)
ParallelBetweenLine(BJ,AD)
ParallelBetweenLine(CF,BJ)
Goal: Value(MeasureOfAngle(ECH))
Answer: 113
Theorem sequence:
adjacent_complementary_angle(1,DAI,IAG)
parallel_property_corresponding_angle(2,BJ,AD,I)
angle_addition(1,CBJ,JBA)
parallel_property_ipsilateral_internal_angle(1,CF,BJ)
vertical_angle(1,ECH,FCB)
Problem: As shown in the diagram, ∠GKH=122°, K is the center of circle K. Find the measure of ⌒KGJ.
问题：如图所示，∠GKH=122°，K是⊙K的圆心。求⌒KGJ的角度。
Construction (CDL):
Shape(HK,KJ,KJH)
Shape(GK,KH,KHG)
Shape(JK,KG,KGL,KLJ)
Collinear(GKJ)
Cocircular(K,LJHG)
Conditions (CDL):
Equal(MeasureOfAngle(GKH),122)
IsCentreOfCircle(K,K)
Goal: Value(MeasureOfArc(KGJ))
Answer: 180
Theorem sequence:
flat_angle(1,JKG)
arc_property_center_angle(1,KGJ,K)
Problem: As shown in the diagram, ∠AQT=32°, ∠ASR=5*x+4°, ∠QTA=6*x-2°. Find the measure of ∠ASR.
问题：如图所示，∠AQT=32°，∠ASR=5*x+4°，∠QTA=6*x-2°。求∠ASR的大小。
Construction (CDL):
Shape(C)
Shape(RA,AS,SR)
Shape(AQ,QT,TA)
Shape(RS,CSR)
Shape(TQ,CQT)
Shape(QA,AR,CRQ)
Shape(SA,AT,CTS)
Collinear(QAS)
Collinear(RAT)
Cocircular(C,QTSR)
Conditions (CDL):
Equal(MeasureOfAngle(AQT),32)
Equal(MeasureOfAngle(ASR),5*x+4)
Equal(MeasureOfAngle(QTA),6*x-2)
Goal: Value(MeasureOfAngle(ASR))
Answer: 34
Theorem sequence:
arc_property_circumference_angle_external(1,CRQ,T)
arc_property_circumference_angle_external(1,CRQ,S)
Problem: As shown in the diagram, XZ=8*x-4, YX=2*x+3, ZY=4*x-1, ZY=YX, ∠YXZ and ∠XZY are the base angles of the isosceles triangle YXZ. Find the length of line YX.
问题：如图所示，XZ=8*x-4，YX=2*x+3，ZY=4*x-1，ZY=YX，∠YXZ和∠XZY是等腰△YXZ的底角。求直线YX的长度。
Construction (CDL):
Shape(YX,XZ,ZY)
Conditions (CDL):
Equal(LengthOfLine(XZ),8*x-4)
Equal(LengthOfLine(YX),2*x+3)
Equal(LengthOfLine(ZY),4*x-1)
Equal(LengthOfLine(ZY),LengthOfLine(YX))
IsoscelesTriangle(YXZ)
Goal: Value(LengthOfLine(YX))
Answer: 7
Theorem sequence:
Problem: As shown in the diagram, PT=27/2, SP=PR, SP=x, UQ=3, UQ=QV, WQ=9, ∠RTS=∠VWU, ∠TSR=∠UVW. Find the value of x.
问题：如图所示，PT=27/2，SP=PR，SP=x，UQ=3，UQ=QV，WQ=9，∠RTS=∠VWU，∠TSR=∠UVW。求x的值。
Construction (CDL):
Shape(SP,PT,TS)
Shape(PR,RT,TP)
Shape(UQ,QW,WU)
Shape(QV,VW,WQ)
Collinear(SPR)
Collinear(UQV)
Conditions (CDL):
Equal(LengthOfLine(PT),27/2)
Equal(LengthOfLine(SP),LengthOfLine(PR))
Equal(LengthOfLine(SP),x)
Equal(LengthOfLine(UQ),3)
Equal(LengthOfLine(UQ),LengthOfLine(QV))
Equal(LengthOfLine(WQ),9)
Equal(MeasureOfAngle(RTS),MeasureOfAngle(VWU))
Equal(MeasureOfAngle(TSR),MeasureOfAngle(UVW))
Goal: Value(x)
Answer: 9/2
Theorem sequence:
line_addition(1,SP,PR)
line_addition(1,UQ,QV)
mirror_similar_triangle_judgment_aa(1,RTS,UVW)
mirror_similar_triangle_property_line_ratio(1,RTS,UVW)
mirror_similar_triangle_property_line_ratio(1,TSR,WUV)
mirror_similar_triangle_judgment_sas(1,SPT,VWQ)
mirror_similar_triangle_property_line_ratio(1,SPT,VWQ)
mirror_similar_triangle_property_line_ratio(1,TSP,WQV)
Problem: As shown in the diagram, AF=2*x, AG=y, BD=x, BF=24, CG=12.25, DC=9, DC is the tangent to ⊙E. Find the value of x.
问题：如图所示，AF=2*x，AG=y，BD=x，BF=24，CG=12.25，DC=9，圆O的切线为DC。求x的值。
Construction (CDL):
Shape(GA,AF,EGF)
Shape(CD,DB,EBC)
Shape(BF,EFB)
Shape(GC,ECG)
Shape(FB,EBC,CG,EGF)
Collinear(DBFA)
Collinear(CGA)
Cocircular(E,CGFB)
Conditions (CDL):
Equal(LengthOfLine(AF),2*x)
Equal(LengthOfLine(AG),y)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(BF),24)
Equal(LengthOfLine(CG),12.25)
Equal(LengthOfLine(DC),9)
IsTangentOfCircle(DC,E)
Goal: Value(x)
Answer: 3
Theorem sequence:
line_addition(1,DB,BF)
circle_property_circular_power_tangent_and_segment_line(1,DC,DBF,E)
Problem: As shown in the diagram, AE=24, DB=23, DE=7, quadrilateral ADBC is a ▱, AE⊥DE. Find the area of quadrilateral ADBC.
问题：如图所示，AE=24，DB=23，DE=7，ADBC是平行四边形，AE⊥DE。求ADBC的面积。
Construction (CDL):
Shape(AE,ED,DA)
Shape(AD,DB,BC,CA)
Collinear(EDB)
Conditions (CDL):
Equal(LengthOfLine(AE),24)
Equal(LengthOfLine(DB),23)
Equal(LengthOfLine(DE),7)
Parallelogram(ADBC)
PerpendicularBetweenLine(AE,DE)
Goal: Value(AreaOfQuadrilateral(ADBC))
Answer: 552
Theorem sequence:
altitude_of_quadrilateral_judgment_left_vertex(3,AE,ADBC)
parallelogram_area_formula_common(1,ADBC)
Problem: As shown in the diagram, AC=8, BC=15, the center of circle D is D, AB is the diameter of circle D. Find the circumference of the circle D.
问题：如图所示，AC=8，BC=15，D是⊙D的圆心，⊙D的直径为AB。求⊙D的周长。
Construction (CDL):
Shape(CA,DAC)
Shape(BC,DCB)
Shape(AD,DB,DBA)
Shape(AC,CB,BD,DA)
Collinear(ADB)
Cocircular(D,ACB)
Conditions (CDL):
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),15)
IsCentreOfCircle(D,D)
IsDiameterOfCircle(AB,D)
Goal: Value(PerimeterOfCircle(D))
Answer: 17*pi
Theorem sequence:
diameter_of_circle_property_right_angle(1,ACB,D)
right_triangle_judgment_angle(1,ACB)
line_addition(1,AD,DB)
right_triangle_property_pythagorean(1,ACB)
radius_of_circle_property_length_equal(1,DA,D)
radius_of_circle_property_length_equal(1,DB,D)
circle_perimeter_formula(1,D)
Problem: As shown in the diagram, AB=y+11, AC=x+7, BD=2*x+3, CD=3*y-5, quadrilateral DBAC is a ▱. Find the value of x.
问题：如图所示，AB=y+11，AC=x+7，BD=2*x+3，CD=3*y-5，DBAC是平行四边形。求x的值。
Construction (CDL):
Shape(DB,BA,AC,CD)
Conditions (CDL):
Equal(LengthOfLine(AB),y+11)
Equal(LengthOfLine(AC),x+7)
Equal(LengthOfLine(BD),2*x+3)
Equal(LengthOfLine(CD),3*y-5)
Parallelogram(DBAC)
Goal: Value(x)
Answer: 4
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DBAC)
Problem: As shown in the diagram, NP=4*x-2, PL=3*x+6, ∠LMP=25°, ∠PMN=25°, MN⊥PN, PL⊥ML. Find the length of line NP.
问题：如图所示，NP=4*x-2，PL=3*x+6，∠LMP=25°，∠PMN=25°，MN垂直于PN，PL垂直于ML。求直线NP的长度。
Construction (CDL):
Shape(LM,MP,PL)
Shape(PM,MN,NP)
Conditions (CDL):
Equal(LengthOfLine(NP),4*x-2)
Equal(LengthOfLine(PL),3*x+6)
Equal(MeasureOfAngle(LMP),25)
Equal(MeasureOfAngle(PMN),25)
PerpendicularBetweenLine(MN,PN)
PerpendicularBetweenLine(PL,ML)
Goal: Value(LengthOfLine(NP))
Answer: 30
Theorem sequence:
mirror_congruent_triangle_judgment_aas(1,PMN,PLM)
mirror_congruent_triangle_property_line_equal(1,MNP,MPL)
Problem: As shown in the diagram, KJ=2, LK=4, LP=2, MP=3, PQ=6, RS=6, HS∥KQ, JR∥HS, LP is parallel to KQ. Find the length of line ML.
问题：如图所示，KJ=2，LK=4，LP=2，MP=3，PQ=6，RS=6，HS∥KQ，JR∥HS，LP∥KQ。求直线ML的长度。
Construction (CDL):
Shape(LP,PM,ML)
Shape(KQ,QP,PL,LK)
Shape(JR,RQ,Qk,KJ)
Shape(HS,SR,RJ,JH)
Collinear(HJKLM)
Collinear(SRQPM)
Conditions (CDL):
Equal(LengthOfLine(KJ),2)
Equal(LengthOfLine(LK),4)
Equal(LengthOfLine(LP),2)
Equal(LengthOfLine(MP),3)
Equal(LengthOfLine(PQ),6)
Equal(LengthOfLine(RS),6)
ParallelBetweenLine(HS,KQ)
ParallelBetweenLine(JR,HS)
ParallelBetweenLine(LP,KQ)
Goal: Value(LengthOfLine(ML))
Answer: 2
Theorem sequence:
parallel_property_corresponding_angle(1,LP,KQ,M)
similar_triangle_judgment_aa(1,PML,QMK)
line_addition(1,QP,PM)
line_addition(1,KL,LM)
similar_triangle_property_line_ratio(1,LPM,KQM)
similar_triangle_property_line_ratio(1,PML,QMK)
Problem: As shown in the diagram, ∠AGB=101°, GC∥AE. Find the measure of ∠GAE.
问题：如图所示，∠AGB=101°，GC∥AE。求∠GAE的大小。
Construction (CDL):
Shape(BG,GD)
Shape(DG,GC)
Shape(AG,GB)
Shape(CG,GA)
Shape(FA,AG)
Shape(GA,AE)
Shape(HA,AF)
Shape(EA,AH)
Collinear(BGC)
Collinear(FAE)
Collinear(DCAH)
Conditions (CDL):
Equal(MeasureOfAngle(AGB),101)
ParallelBetweenLine(GC,AE)
Goal: Value(MeasureOfAngle(GAE))
Answer: 101
Theorem sequence:
adjacent_complementary_angle(1,CGA,AGB)
parallel_property_ipsilateral_internal_angle(1,GC,AE)
Problem: As shown in the diagram, PQ=17, QR=15, RP=8, QR is perpendicular to PR. Find the value of sin(RP).
问题：如图所示，PQ=17，QR=15，RP=8，QR垂直于PR。求sin(RP)的值。
Construction (CDL):
Shape(QR,RP,PQ)
Conditions (CDL):
Equal(LengthOfLine(PQ),17)
Equal(LengthOfLine(QR),15)
Equal(LengthOfLine(RP),8)
PerpendicularBetweenLine(QR,PR)
Goal: Value(Sin(MeasureOfAngle(RPQ)))
Answer: 15/17
Theorem sequence:
sine_theorem(1,QRP)
Problem: As shown in the diagram, LM=2*y+6, MP=4, PL=5*x-3, ∠LMP=60°, ∠MPL=60°. Find the value of x.
问题：如图所示，LM=2*y+6，MP=4，PL=5*x-3，∠LMP=60°，∠MPL=60°。求x的值。
Construction (CDL):
Shape(MP,PL,LM)
Conditions (CDL):
Equal(LengthOfLine(LM),2*y+6)
Equal(LengthOfLine(MP),4)
Equal(LengthOfLine(PL),5*x-3)
Equal(MeasureOfAngle(LMP),60)
Equal(MeasureOfAngle(MPL),60)
Goal: Value(x)
Answer: 7/5
Theorem sequence:
triangle_property_angle_sum(1,MPL)
sine_theorem(1,LMP)
sine_theorem(1,MPL)
Problem: As shown in the diagram, ∠CBE=55°, ∠DAE=57°, ∠EDA=47°. Find the measure of ∠BEC.
问题：如图所示，∠CBE=55°，∠DAE=57°，∠EDA=47°。求∠BEC的大小。
Construction (CDL):
Shape(DA,AE,ED)
Shape(BE,EC,CB)
Collinear(DEC)
Collinear(AEB)
Conditions (CDL):
Equal(MeasureOfAngle(CBE),55)
Equal(MeasureOfAngle(DAE),57)
Equal(MeasureOfAngle(EDA),47)
Goal: Value(MeasureOfAngle(BEC))
Answer: 76
Theorem sequence:
triangle_property_angle_sum(1,DAE)
vertical_angle(1,AED,BEC)
Problem: As shown in the diagram, ∠EHB=72°, ∠GCF=x-12°, CA is parallel to HE. Find the value of x.
问题：如图所示，∠EHB=72°，∠GCF=x-12°，CA平行于HE。求x的值。
Construction (CDL):
Shape(FC,CA)
Shape(AC,CH)
Shape(GC,CF)
Shape(HC,CG)
Shape(CH,HE)
Shape(EH,HB)
Shape(DH,HC)
Shape(BH,HD)
Collinear(ACG)
Collinear(EHD)
Collinear(FCHB)
Conditions (CDL):
Equal(MeasureOfAngle(EHB),72)
Equal(MeasureOfAngle(GCF),x-12)
ParallelBetweenLine(CA,HE)
Goal: Value(x)
Answer: 84
Theorem sequence:
parallel_property_corresponding_angle(2,CA,HE,B)
vertical_angle(1,GCF,ACH)
Problem: As shown in the diagram, AB=x, AD=y, BC=12, BD=8, CA=z, BD is perpendicular to AD, CA⊥BA. Find the value of x.
问题：如图所示，AB=x，AD=y，BC=12，BD=8，CA=z，BD垂直于AD，CA垂直于BA。求x的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(AB,BD,DA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BC),12)
Equal(LengthOfLine(BD),8)
Equal(LengthOfLine(CA),z)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 4*sqrt(6)
Theorem sequence:
line_addition(1,CD,DB)
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,CAB)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,ADC)
right_triangle_property_pythagorean(1,CAB)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,ADC)
Problem: As shown in the diagram, GK=6, HG=9, KJ=4, KH is parallel to JI. Find the length of line GI.
问题：如图所示，GK=6，HG=9，KJ=4，KH平行于JI。求直线GI的长度。
Construction (CDL):
Shape(GK,KH,HG)
Shape(KJ,JI,IH,HK)
Collinear(GKJ)
Collinear(IHG)
Conditions (CDL):
Equal(LengthOfLine(GK),6)
Equal(LengthOfLine(HG),9)
Equal(LengthOfLine(KJ),4)
ParallelBetweenLine(KH,JI)
Goal: Value(LengthOfLine(GI))
Answer: 15
Theorem sequence:
parallel_property_corresponding_angle(1,KH,JI,G)
similar_triangle_judgment_aa(1,HGK,IGJ)
line_addition(1,GK,KJ)
similar_triangle_property_line_ratio(1,HGK,IGJ)
similar_triangle_property_line_ratio(1,KHG,JIG)
Problem: As shown in the diagram, BC=6, ∠ABC=240°, the center of circle B is B. Find the area of the sector BAC minus the area of △BAC.
问题：如图所示，BC=6，∠ABC=240°，B是圆B的圆心。求扇形BAC的面积减去△BAC的面积。
Construction (CDL):
Shape(AC,CB,BA)
Shape(AC,BCA)
Shape(AB,BC,BCA)
Cocircular(B,CA)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
Equal(MeasureOfAngle(ABC),240)
IsCentreOfCircle(B,B)
Goal: Value(Sub(AreaOfSector(BAC),AreaOfTriangle(BAC)))
Answer: -9*sqrt(3)+12*pi
Theorem sequence:
round_angle(1,CBA,ABC)
radius_of_circle_property_length_equal(1,BC,B)
radius_of_circle_property_length_equal(1,BA,B)
triangle_area_formula_sine(1,BAC)
arc_property_center_angle(1,BAC,B)
sector_area_formula(1,BAC)
Problem: As shown in the diagram, DE=10*x, DF=17/2*x+3, EF=7/2*x+4, FD=DE, △DEF is an isosceles △. Find the length of line EF.
问题：如图所示，DE=10*x，DF=17/2*x+3，EF=7/2*x+4，FD=DE，∠DEF和∠EFD是等腰△DEF的底角。求直线EF的长度。
Construction (CDL):
Shape(DE,EF,FD)
Conditions (CDL):
Equal(LengthOfLine(DE),10*x)
Equal(LengthOfLine(DF),17/2*x+3)
Equal(LengthOfLine(EF),7/2*x+4)
Equal(LengthOfLine(FD),LengthOfLine(DE))
IsoscelesTriangle(DEF)
Goal: Value(LengthOfLine(EF))
Answer: 11
Theorem sequence:
Problem: As shown in the diagram, AC=30, BA=41, CN=27, NB=19, AC⊥NC. Find the area of △BAN.
问题：如图所示，AC=30，BA=41，CN=27，NB=19，AC⊥NC。求三角形BAN的面积。
Construction (CDL):
Shape(BA,AN,NB)
Shape(NA,AC,CN)
Collinear(BNC)
Conditions (CDL):
Equal(LengthOfLine(AC),30)
Equal(LengthOfLine(BA),41)
Equal(LengthOfLine(CN),27)
Equal(LengthOfLine(NB),19)
PerpendicularBetweenLine(AC,NC)
Goal: Value(AreaOfTriangle(BAN))
Answer: 285
Theorem sequence:
altitude_of_triangle_judgment(2,AC,ANB)
triangle_area_formula_common(1,ANB)
Problem: As shown in the diagram, CS=15, DS=9, RD=10, the center of circle R is R, the center of circle S is S. Find the length of line RC.
问题：如图所示，CS=15，DS=9，RD=10，圆R的圆心为R，圆S的圆心为S。求直线RC的长度。
Construction (CDL):
Shape(DS,SB,SBA,RDA)
Shape(BS,SD,RED,SEB)
Shape(CD,RDA,SCA)
Shape(DC,SCE,RED)
Shape(RC,SCA,RAE,SCE)
Collinear(RCDSB)
Cocircular(R,EDA)
Cocircular(S,ACEB)
Conditions (CDL):
Equal(LengthOfLine(CS),15)
Equal(LengthOfLine(DS),9)
Equal(LengthOfLine(RD),10)
IsCentreOfCircle(R,R)
IsCentreOfCircle(S,S)
Goal: Value(LengthOfLine(RC))
Answer: 4
Theorem sequence:
line_addition(1,RC,CD)
line_addition(1,CD,DS)
Problem: As shown in the diagram, JK=JL, JL=5*x-6, KJ=3*x+6, KL=KJ, LK=4*x. Find the value of x.
问题：如图所示，JK=JL，JL=5*x-6，KJ=3*x+6，KL=KJ，LK=4*x。求x的值。
Construction (CDL):
Shape(KJ,JL,LK)
Conditions (CDL):
Equal(LengthOfLine(JK),LengthOfLine(JL))
Equal(LengthOfLine(JL),5*x-6)
Equal(LengthOfLine(KJ),3*x+6)
Equal(LengthOfLine(KL),LengthOfLine(KJ))
Equal(LengthOfLine(LK),4*x)
Goal: Value(x)
Answer: 6
Theorem sequence:
Problem: As shown in the diagram, ST=24, TV=18, UW=21, VS=12, triangle WRU is similar to triangle VST.. Find the perimeter of △WRU.
问题：如图所示，ST=24，TV=18，UW=21，VS=12，△WRU相似于△VST。求三角形WRU的周长。
Construction (CDL):
Shape(WR,RU,UW)
Shape(VS,ST,TV)
Collinear(R,S,T,U)
Conditions (CDL):
Equal(LengthOfLine(ST),24)
Equal(LengthOfLine(TV),18)
Equal(LengthOfLine(UW),21)
Equal(LengthOfLine(VS),12)
SimilarBetweenTriangle(WRU,VST)
Goal: Value(PerimeterOfTriangle(WRU))
Answer: 63
Theorem sequence:
similar_triangle_property_line_ratio(1,WRU,VST)
similar_triangle_property_line_ratio(1,RUW,STV)
similar_triangle_property_line_ratio(1,UWR,TVS)
triangle_perimeter_formula(1,WRU)
Problem: As shown in the diagram, PT=6, ST=8, TR=4, PT is parallel to QR. Find the length of line QR.
问题：如图所示，PT=6，ST=8，TR=4，PT平行于QR。求直线QR的长度。
Construction (CDL):
Shape(SP,PT,TS)
Shape(PQ,QR,RT,TP)
Collinear(SPQ)
Collinear(STR)
Conditions (CDL):
Equal(LengthOfLine(PT),6)
Equal(LengthOfLine(ST),8)
Equal(LengthOfLine(TR),4)
ParallelBetweenLine(PT,QR)
Goal: Value(LengthOfLine(QR))
Answer: 9
Theorem sequence:
line_addition(1,RT,TS)
parallel_property_corresponding_angle(1,PT,QR,S)
similar_triangle_judgment_aa(1,TSP,RSQ)
similar_triangle_property_line_ratio(1,PTS,QRS)
similar_triangle_property_line_ratio(1,SPT,SQR)
Problem: As shown in the diagram, CB=x, CD=16, DB=8, ∠DCB=y°, CB⊥DB. Find the value of x.
问题：如图所示，CB=x，CD=16，DB=8，∠DCB=y°，CB垂直于DB。求x的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(CB,BD,DC)
Conditions (CDL):
Equal(LengthOfLine(CB),x)
Equal(LengthOfLine(CD),16)
Equal(LengthOfLine(DB),8)
Equal(MeasureOfAngle(DCB),y)
PerpendicularBetweenLine(CB,DB)
Goal: Value(x)
Answer: 8*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,CBD)
right_triangle_property_pythagorean(1,CBD)
Problem: As shown in the diagram, AB=3, AC=5, AE=y, BE=x, CD=3.5, DE=3, CD∥BE. Find the length of line AD.
问题：如图所示，AB=3，AC=5，AE=y，BE=x，CD=3.5，DE=3，CD∥BE。求直线AD的长度。
Construction (CDL):
Shape(AE,EB,BA)
Shape(ED,DC,CB,BE)
Collinear(AED)
Collinear(ABC)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(AE),y)
Equal(LengthOfLine(BE),x)
Equal(LengthOfLine(CD),3.5)
Equal(LengthOfLine(DE),3)
ParallelBetweenLine(CD,BE)
Goal: Value(LengthOfLine(AD))
Answer: 15/2
Theorem sequence:
line_addition(1,AE,ED)
parallel_property_corresponding_angle(2,CD,BE,A)
similar_triangle_judgment_aa(1,EBA,DCA)
similar_triangle_property_line_ratio(1,EBA,DCA)
similar_triangle_property_line_ratio(1,BAE,CAD)
Problem: As shown in the diagram, ∠ACB=2*x°, ∠BAC=90°, ∠CBA=x°, CA is perpendicular to BA. Find the value of x.
问题：如图所示，∠ACB=2*x°，∠BAC=90°，∠CBA=x°，CA⊥BA。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),2*x)
Equal(MeasureOfAngle(BAC),90)
Equal(MeasureOfAngle(CBA),x)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 30
Theorem sequence:
triangle_property_angle_sum(1,BAC)
Problem: As shown in the diagram, KJ=2, LK=4, LM=2, LP=2, MP=3, PQ=6, RQ=3, RS=6, HS∥KQ, JR∥HS, LP is parallel to KQ. Find the length of line JH.
问题：如图所示，KJ=2，LK=4，LM=2，LP=2，MP=3，PQ=6，RQ=3，RS=6，HS平行于KQ，JR平行于HS，LP∥KQ。求直线JH的长度。
Construction (CDL):
Shape(LP,PM,ML)
Shape(KQ,QP,PL,LK)
Shape(JR,RQ,QK,KJ)
Shape(HS,SR,RJ,JH)
Collinear(HJKLM)
Collinear(SRQPM)
Conditions (CDL):
Equal(LengthOfLine(KJ),2)
Equal(LengthOfLine(LK),4)
Equal(LengthOfLine(LM),2)
Equal(LengthOfLine(LP),2)
Equal(LengthOfLine(MP),3)
Equal(LengthOfLine(PQ),6)
Equal(LengthOfLine(RQ),3)
Equal(LengthOfLine(RS),6)
ParallelBetweenLine(HS,KQ)
ParallelBetweenLine(JR,HS)
ParallelBetweenLine(LP,KQ)
Goal: Value(LengthOfLine(JH))
Answer: 4
Theorem sequence:
line_addition(1,HJ,JM)
line_addition(1,JK,KL)
line_addition(1,JL,LM)
line_addition(1,SR,RQ)
line_addition(1,SQ,QP)
line_addition(1,SP,PM)
line_addition(1,RQ,QP)
line_addition(1,RP,PM)
parallel_property_corresponding_angle(1,JR,HS,M)
similar_triangle_judgment_aa(1,RMJ,SMH)
similar_triangle_property_line_ratio(1,RMJ,SMH)
similar_triangle_property_line_ratio(1,JRM,HSM)
Problem: As shown in the diagram, AC=x-3, BA=x, CB=9, AC⊥BC. Find the value of x.
问题：如图所示，AC=x-3，BA=x，CB=9，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),x-3)
Equal(LengthOfLine(BA),x)
Equal(LengthOfLine(CB),9)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 15
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, ∠BEC=38°, the measure of ⌒DCA is 160. Find the measure of ∠AEC.
问题：如图所示，∠BEC=38°，弧DCA的角度为160。求∠AEC的大小。
Construction (CDL):
Shape(EA,DAE)
Shape(AE,EB,DBA)
Shape(BE,EC,DCB)
Shape(CE,DEC)
Cocircular(D,ECBA)
Conditions (CDL):
Equal(MeasureOfAngle(BEC),38)
Equal(MeasureOfArc(DCA),160)
Goal: Value(MeasureOfAngle(AEC))
Answer: 80
Theorem sequence:
arc_property_circumference_angle_external(1,DCA,E)
Problem: As shown in the diagram, ∠ABD=115°, ∠BDE=x°, ∠CED=x+24°, ∠DEA=2*y-56°, BA∥DE. Find the value of x.
问题：如图所示，∠ABD=115°，∠BDE=x°，∠CED=x+24°，∠DEA=2*y-56°，BA平行于DE。求x的值。
Construction (CDL):
Shape(CE,ED)
Shape(EA,AB,BD,DE)
Collinear(CEA)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),115)
Equal(MeasureOfAngle(BDE),x)
Equal(MeasureOfAngle(CED),x+24)
Equal(MeasureOfAngle(DEA),2*y-56)
ParallelBetweenLine(BA,DE)
Goal: Value(x)
Answer: 65
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,BA,DE)
Problem: As shown in the diagram, BA=x+3, BC=2*x-8, DB=3, EB=5, EA∥CD. Find the length of line BC.
问题：如图所示，BA=x+3，BC=2*x-8，DB=3，EB=5，EA平行于CD。求直线BC的长度。
Construction (CDL):
Shape(AE,EB,BA)
Shape(CD,DB,BC)
Collinear(ABC)
Collinear(EBD)
Conditions (CDL):
Equal(LengthOfLine(BA),x+3)
Equal(LengthOfLine(BC),2*x-8)
Equal(LengthOfLine(DB),3)
Equal(LengthOfLine(EB),5)
ParallelBetweenLine(EA,CD)
Goal: Value(LengthOfLine(BC))
Answer: 6
Theorem sequence:
parallel_property_alternate_interior_angle(1,EA,CD)
vertical_angle(1,EBA,DBC)
similar_triangle_judgment_aa(1,AEB,CDB)
similar_triangle_property_line_ratio(1,AEB,CDB)
similar_triangle_property_line_ratio(1,EBA,DBC)
Problem: As shown in the diagram, ML=3*x+1, NM=4*x-2, ∠LPM=5*y°, triangle MLN is an equilateral triangle, MP is the altitude of triangle MLN, ML and MN are the legs of the isosceles triangle MLN. Find the value of y.
问题：如图所示，ML=3*x+1，NM=4*x-2，∠LPM=5*y°，三角形MLN为等边三角形，MP是△MLN的高，ML和MN是等腰△MLN的腰。求y的值。
Construction (CDL):
Shape(ML,LP,PM)
Shape(MP,PN,NM)
Collinear(LPN)
Conditions (CDL):
Equal(LengthOfLine(ML),3*x+1)
Equal(LengthOfLine(NM),4*x-2)
Equal(MeasureOfAngle(LPM),5*y)
EquilateralTriangle(MLN)
IsAltitudeOfTriangle(MP,MLN)
IsoscelesTriangle(MLN)
Goal: Value(y)
Answer: 18
Theorem sequence:
Problem: As shown in the diagram, AB=12, BC=x, CA=y, ∠CAB=60°, BC⊥AC. Find the value of y.
问题：如图所示，AB=12，BC=x，CA=y，∠CAB=60°，BC⊥AC。求y的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CA),y)
Equal(MeasureOfAngle(CAB),60)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 6
Theorem sequence:
cosine_theorem(1,CAB)
cosine_theorem(1,ABC)
Problem: As shown in the diagram, ∠SQR=6*x+8°, ∠TQR=4*x-14°, QT bisects ∠SQR. Find the measure of ∠SQT.
问题：如图所示，∠SQR=6*x+8°，∠TQR=4*x-14°，QT是∠SQR的角平分线。求∠SQT的大小。
Construction (CDL):
Shape(PQ,QS)
Shape(SQ,QT)
Shape(TQ,QR)
Collinear(PQR)
Conditions (CDL):
Equal(MeasureOfAngle(SQR),6*x+8)
Equal(MeasureOfAngle(TQR),4*x-14)
IsBisectorOfAngle(QT,SQR)
Goal: Value(MeasureOfAngle(SQT))
Answer: 58
Theorem sequence:
angle_addition(1,SQT,TQR)
Problem: As shown in the diagram, BN=NC, BN=y, EH=20-5*x, EH=HP, HP=2*x+6, NC=3/5*y+2, AJ is parallel to MF, MF∥LK. Find the value of y.
问题：如图所示，BN=NC，BN=y，EH=20-5*x，EH=HP，HP=2*x+6，NC=3/5*y+2，AJ∥MF，MF平行于LK。求y的值。
Construction (CDL):
Shape(AE,EG)
Shape(GE,EB)
Shape(EB,BI)
Shape(IB,BJ)
Shape(HE,EA)
Shape(BE,EH,HN,NB)
Shape(JB,BN)
Shape(MH,HE)
Shape(BN,NF)
Shape(PH,HM)
Shape(NH,HP,PC,CN)
Shape(FN,NC)
Shape(LP,PH)
Shape(NC,CK)
Shape(DP,PL)
Shape(CP,PD)
Shape(OC,CP)
Shape(KC,CO)
Collinear(AEBJ)
Collinear(MHNF)
Collinear(LPCK)
Collinear(GEHPD)
Collinear(IBNCO)
Conditions (CDL):
Equal(LengthOfLine(BN),LengthOfLine(NC))
Equal(LengthOfLine(BN),y)
Equal(LengthOfLine(EH),20-5*x)
Equal(LengthOfLine(EH),LengthOfLine(HP))
Equal(LengthOfLine(HP),2*x+6)
Equal(LengthOfLine(NC),3/5*y+2)
ParallelBetweenLine(AJ,MF)
ParallelBetweenLine(MF,LK)
Goal: Value(y)
Answer: 5
Theorem sequence: