SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠AEC=x°, ∠BED=x°, ∠CEB=x°, ∠DEA=x°. Find the value of x.
问题：如图所示，∠AEC=x°，∠BED=x°，∠CEB=x°，∠DEA=x°。求x的值。
Construction (CDL):
Shape(DE,EA)
Shape(AE,EC)
Shape(CE,EB)
Shape(BE,ED)
Collinear(AEB)
Collinear(CED)
Conditions (CDL):
Equal(MeasureOfAngle(AEC),x)
Equal(MeasureOfAngle(BED),x)
Equal(MeasureOfAngle(CEB),x)
Equal(MeasureOfAngle(DEA),x)
Goal: Value(x)
Answer: 90
Theorem sequence:
adjacent_complementary_angle(1,AEC,CEB)
Problem: As shown in the diagram, ∠SQP=51°, ∠SRQ=33°, RQ⊥PQ. Find the measure of ∠QSR.
问题：如图所示，∠SQP=51°，∠SRQ=33°，RQ⊥PQ。求∠QSR的大小。
Construction (CDL):
Shape(RQ,QS,SR)
Shape(QP,PS,SQ)
Collinear(RSP)
Conditions (CDL):
Equal(MeasureOfAngle(SQP),51)
Equal(MeasureOfAngle(SRQ),33)
PerpendicularBetweenLine(RQ,PQ)
Goal: Value(MeasureOfAngle(QSR))
Answer: 108
Theorem sequence:
angle_addition(1,RQS,SQP)
triangle_property_angle_sum(1,RQS)
Problem: As shown in the diagram, AD=AC, BC=40, BD=22, ∠BCA=y°, ∠DBA=x°, AD is perpendicular to BD. Find the value of x.
问题：如图所示，AD=AC，BC=40，BD=22，∠BCA=y°，∠DBA=x°，AD⊥BD。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Shape(BA,AD,DB)
Collinear(CAD)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(AC))
Equal(LengthOfLine(BC),40)
Equal(LengthOfLine(BD),22)
Equal(MeasureOfAngle(BCA),y)
Equal(MeasureOfAngle(DBA),x)
PerpendicularBetweenLine(AD,BD)
Goal: Value(x)
Answer: 180*asin(3*sqrt(23653)/763)/pi
Theorem sequence:
line_addition(1,CA,AD)
right_triangle_judgment_angle(1,CDB)
right_triangle_property_pythagorean(1,CDB)
right_triangle_judgment_angle(1,ADB)
right_triangle_property_pythagorean(1,ADB)
sine_theorem(1,ADB)
Problem: As shown in the diagram, AE=8, AG=x, GD=y, ∠ADE=30°, ∠DEA=45°, DG⊥AG. Find the value of x.
问题：如图所示，AE=8，AG=x，GD=y，∠ADE=30°，∠DEA=45°，DG垂直于AG。求x的值。
Construction (CDL):
Shape(GE,EA,AG)
Shape(GA,AD,DG)
Collinear(EGD)
Conditions (CDL):
Equal(LengthOfLine(AE),8)
Equal(LengthOfLine(AG),x)
Equal(LengthOfLine(GD),y)
Equal(MeasureOfAngle(ADE),30)
Equal(MeasureOfAngle(DEA),45)
PerpendicularBetweenLine(DG,AG)
Goal: Value(x)
Answer: 4*sqrt(2)
Theorem sequence:
adjacent_complementary_angle(1,DGA,AGE)
triangle_property_angle_sum(1,EAG)
right_triangle_judgment_angle(1,AGE)
isosceles_triangle_judgment_angle_equal(1,GEA)
right_triangle_property_pythagorean(1,AGE)
Problem: As shown in the diagram, AR=RC, AT=16, CT=TB, MC=7, RM=4, SA=SB. Find the length of line MB.
问题：如图所示，AR=RC，AT=16，CT=TB，MC=7，RM=4，SA=SB。求直线MB的长度。
Construction (CDL):
Shape(AR,RM,MA)
Shape(MR,RC,CM)
Shape(MC,CT,TM)
Shape(MT,TB,BM)
Shape(MB,BS,SM)
Shape(MS,SA,AM)
Collinear(AMT)
Collinear(ARC)
Collinear(ASB)
Collinear(CTB)
Collinear(CMS)
Collinear(BMR)
Conditions (CDL):
Equal(LengthOfLine(AR),LengthOfLine(RC))
Equal(LengthOfLine(AT),16)
Equal(LengthOfLine(CT),LengthOfLine(TB))
Equal(LengthOfLine(MC),7)
Equal(LengthOfLine(RM),4)
Equal(LengthOfLine(SA),LengthOfLine(SB))
Goal: Value(LengthOfLine(MB))
Answer: 8
Theorem sequence:
median_of_triangle_judgment(1,AT,ACB)
median_of_triangle_judgment(1,BR,BAC)
centroid_of_triangle_judgment_intersection(1,M,CBA,T,R)
centroid_of_triangle_property_line_ratio(1,M,BAC,R)
Problem: As shown in the diagram, ∠ADB=120°, ∠BCA=3*y+40°, ∠DBC=x°, ∠ECB=3*x-70°, DA∥BC. Find the value of y.
问题：如图所示，∠ADB=120°，∠BCA=3*y+40°，∠DBC=x°，∠ECB=3*x-70°，DA平行于BC。求y的值。
Construction (CDL):
Shape(EC,CB)
Shape(AD,DB,BC,CA)
Collinear(ACE)
Conditions (CDL):
Equal(MeasureOfAngle(ADB),120)
Equal(MeasureOfAngle(BCA),3*y+40)
Equal(MeasureOfAngle(DBC),x)
Equal(MeasureOfAngle(ECB),3*x-70)
ParallelBetweenLine(DA,BC)
Goal: Value(y)
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,ECB,BCA)
parallel_property_ipsilateral_internal_angle(1,DA,BC)
Problem: As shown in the diagram, AB=20, AC=x, BC=y, ∠ABC=60°, BC is perpendicular to AC. Find the value of y.
问题：如图所示，AB=20，AC=x，BC=y，∠ABC=60°，BC垂直于AC。求y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(ABC),60)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 10
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=5*x+8, AC=7*x, BC=6*x+5, the perimeter of triangle CBA is 139. Find the length of line BA.
问题：如图所示，AB=5*x+8，AC=7*x，BC=6*x+5，三角形CBA的周长为139。求直线BA的长度。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),5*x+8)
Equal(LengthOfLine(AC),7*x)
Equal(LengthOfLine(BC),6*x+5)
Equal(PerimeterOfTriangle(CBA),139)
Goal: Value(LengthOfLine(BA))
Answer: 43
Theorem sequence:
triangle_perimeter_formula(1,CBA)
Problem: As shown in the diagram, RQ=18, RS=37, SQ=23. Find the measure of ∠QRS.
问题：如图所示，RQ=18，RS=37，SQ=23。求∠QRS的大小。
Construction (CDL):
Shape(QR,RS,SQ)
Conditions (CDL):
Equal(LengthOfLine(RQ),18)
Equal(LengthOfLine(RS),37)
Equal(LengthOfLine(SQ),23)
Goal: Value(MeasureOfAngle(QRS))
Answer: 180*acos(97/111)/pi
Theorem sequence:
cosine_theorem(1,RSQ)
Problem: As shown in the diagram, ∠BAE=58°, ∠CBE=26°, ∠EBA=47°, AB∥EC. Find the measure of ∠BEC.
问题：如图所示，∠BAE=58°，∠CBE=26°，∠EBA=47°，AB∥EC。求∠BEC的大小。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BE,EC,CB)
Shape(ED,DC,CB)
Collinear(AED)
Collinear(BCD)
Conditions (CDL):
Equal(MeasureOfAngle(BAE),58)
Equal(MeasureOfAngle(CBE),26)
Equal(MeasureOfAngle(EBA),47)
ParallelBetweenLine(AB,EC)
Goal: Value(MeasureOfAngle(BEC))
Answer: 47
Theorem sequence:
parallel_property_alternate_interior_angle(2,AB,EC)
Problem: As shown in the diagram, GH=x+9, JH=5*x-2, FJHG is a rhombus. Find the value of x.
问题：如图所示，GH=x+9，JH=5*x-2，四边形FJHG是菱形。求x的值。
Construction (CDL):
Shape(GF,FK,KG)
Shape(GK,KH,HG)
Shape(FJ,JK,KF)
Shape(KJ,JH,HK)
Collinear(FKH)
Collinear(JKG)
Conditions (CDL):
Equal(LengthOfLine(GH),x+9)
Equal(LengthOfLine(JH),5*x-2)
Rhombus(FJHG)
Goal: Value(x)
Answer: 11/4
Theorem sequence:
Problem: As shown in the diagram, AB=x, BC=10, ∠CBA=30°, BA⊥CA. Find the value of x.
问题：如图所示，AB=x，BC=10，∠CBA=30°，BA垂直于CA。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),10)
Equal(MeasureOfAngle(CBA),30)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 5*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,BAC)
sine_theorem(1,BAC)
Problem: As shown in the diagram, BC=6, A is the center of ⊙A, the diameter of ⊙A is BC. Find the circumference of the circle A.
问题：如图所示，BC=6，⊙A的圆心为A，BC是⊙A的直径。求圆A的周长。
Construction (CDL):
Shape(ABC,CA,AB)
Shape(ACB,BA,AC)
Collinear(BAC)
Cocircular(A,BC)
Conditions (CDL):
Equal(LengthOfLine(BC),6)
IsCentreOfCircle(A,A)
IsDiameterOfCircle(BC,A)
Goal: Value(PerimeterOfCircle(A))
Answer: 6*pi
Theorem sequence:
diameter_of_circle_property_length_equal(1,BC,A)
circle_property_length_of_radius_and_diameter(1,A)
circle_perimeter_formula(1,A)
Problem: As shown in the diagram, ∠BGE=52°, ∠GFE=30°, the measure of ⌒OBA is 108, the measure of ⌒OFE is 118. Find the measure of arc OCF.
问题：如图所示，∠BGE=52°，∠GFE=30°，弧OBA的角度为108，⌒OFE的角度为118。求⌒OCF的角度。
Construction (CDL):
Shape(OAC,AD,DC)
Shape(OCF,FC)
Shape(OAC,CG,GB,BA)
Shape(GF,FE,EG)
Shape(OFE,EF)
Shape(OEB,BG,GE)
Shape(OBA,AB)
Shape(CF,FG,GC)
Collinear(DAB)
Collinear(DCGE)
Collinear(FGB)
Cocircular(O,ACFEB)
Conditions (CDL):
Equal(MeasureOfAngle(BGE),52)
Equal(MeasureOfAngle(GFE),30)
Equal(MeasureOfArc(OBA),108)
Equal(MeasureOfArc(OFE),118)
Goal: Value(MeasureOfArc(OCF))
Answer: 44
Theorem sequence:
adjacent_complementary_angle(1,BGE,EGF)
triangle_property_angle_sum(1,GFE)
arc_property_circumference_angle_external(1,OCF,E)
Problem: As shown in the diagram, AB=x, AC=74, BC=24, CB is perpendicular to AB. Find the value of x.
问题：如图所示，AB=x，AC=74，BC=24，CB⊥AB。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),74)
Equal(LengthOfLine(BC),24)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 70
Theorem sequence:
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, ∠BFC=72°, ∠CFD=4*a-1°, ∠DFA=2*a-5°. Find the measure of ∠CFD.
问题：如图所示，∠BFC=72°，∠CFD=4*a-1°，∠DFA=2*a-5°。求∠CFD的大小。
Construction (CDL):
Shape(FBA,AF,FB)
Shape(FAD,DF,FA)
Shape(FDC,CF,FD)
Shape(FCB,BF,FC)
Collinear(AFC)
Cocircular(F,BADC)
Conditions (CDL):
Equal(MeasureOfAngle(BFC),72)
Equal(MeasureOfAngle(CFD),4*a-1)
Equal(MeasureOfAngle(DFA),2*a-5)
Goal: Value(MeasureOfAngle(CFD))
Answer: 123
Theorem sequence:
adjacent_complementary_angle(1,CFD,DFA)
Problem: As shown in the diagram, ∠JMK=55°, ∠JMK=∠PMR, ∠MRP=70°, quadrilateral JMLK is a rectangle, MRPL is a rhombus. Find the measure of ∠PLK.
问题：如图所示，∠JMK=55°，∠JMK=∠PMR，∠MRP=70°，JMLK是矩形，四边形MRPL是菱形。求∠PLK的大小。
Construction (CDL):
Shape(JM,MK,KJ)
Shape(KM,ML,LK)
Shape(MR,RP,PM)
Shape(MP,PL,LM)
Conditions (CDL):
Equal(MeasureOfAngle(JMK),55)
Equal(MeasureOfAngle(JMK),MeasureOfAngle(PMR))
Equal(MeasureOfAngle(MRP),70)
Rectangle(JMLK)
Rhombus(MRPL)
Goal: Value(MeasureOfAngle(PLK))
Answer: 160
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,RPLM)
angle_addition(1,PLM,MLK)
Problem: As shown in the diagram, AB=61, AC=29, BC=50, ∠ACB=x°. Find the value of x.
问题：如图所示，AB=61，AC=29，BC=50，∠ACB=x°。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),61)
Equal(LengthOfLine(AC),29)
Equal(LengthOfLine(BC),50)
Equal(MeasureOfAngle(ACB),x)
Goal: Value(x)
Answer: 180*acos(-19/145)/pi
Theorem sequence:
cosine_theorem(1,CBA)
Problem: As shown in the diagram, BE=9, Q is the centroid of triangle BAC. Find the length of line BQ.
问题：如图所示，BE=9，Q是△BAC的重心。求直线BQ的长度。
Construction (CDL):
Shape(BF,FQ,QB)
Shape(FA,AQ,QF)
Shape(AE,EQ,QA)
Shape(EC,CQ,QE)
Shape(CD,DQ,QC)
Shape(QD,DB,BQ)
Collinear(BFA)
Collinear(BQE)
Collinear(BDC)
Collinear(AQD)
Collinear(AEC)
Collinear(CQF)
Conditions (CDL):
Equal(LengthOfLine(BE),9)
IsCentroidOfTriangle(Q,BAC)
Goal: Value(LengthOfLine(BQ))
Answer: 6
Theorem sequence:
line_addition(1,BQ,QE)
centroid_of_triangle_property_line_ratio(1,Q,BAC,E)
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 radius of circle D.
问题：如图所示，AC=8，BC=15，D是⊙D的圆心，AB是圆D的直径。求⊙D的半径。
Construction (CDL):
Shape(DBA,AD,DB)
Shape(DAC,CA)
Shape(DCB,BC)
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(RadiusOfCircle(D))
Answer: 17/2
Theorem sequence:
diameter_of_circle_property_right_angle(1,ACB,D)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
diameter_of_circle_property_length_equal(1,AB,D)
circle_property_length_of_radius_and_diameter(1,D)
Problem: As shown in the diagram, AB=x, AC=z, AD=y, BC=12, BD=8, AD is perpendicular to CD, CA is perpendicular to BA. Find the value of y.
问题：如图所示，AB=x，AC=z，AD=y，BC=12，BD=8，AD垂直于CD，CA垂直于BA。求y的值。
Construction (CDL):
Shape(DC,CA,AD)
Shape(DA,AB,BD)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),z)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BC),12)
Equal(LengthOfLine(BD),8)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 4*sqrt(2)
Theorem sequence:
line_addition(1,CD,DB)
right_triangle_judgment_angle(1,ADC)
mirror_similar_triangle_judgment_aa(1,ADC,BCA)
mirror_similar_triangle_property_line_ratio(1,DCA,ABC)
mirror_similar_triangle_property_line_ratio(1,ADC,BCA)
right_triangle_property_pythagorean(1,ADC)
Problem: As shown in the diagram, WZ=15, XY=YZ, WY perpendicular bisects XZ, XY⊥WY. Find the length of line WX.
问题：如图所示，WZ=15，XY=YZ，WY是XZ的垂直平分线，XY⊥WY。求直线WX的长度。
Construction (CDL):
Shape(WY,YX,XW)
Shape(YW,WZ,ZY)
Collinear(DWYT)
Collinear(XYZ)
Conditions (CDL):
Equal(LengthOfLine(WZ),15)
Equal(LengthOfLine(XY),LengthOfLine(YZ))
IsPerpendicularBisectorOfLine(WY,XZ)
PerpendicularBetweenLine(XY,WY)
Goal: Value(LengthOfLine(WX))
Answer: 15
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,WY,XZ)
Problem: As shown in the diagram, GR=7, PG=z+4, PT=3*y-1, TQ=2*x, TR=8, TF perpendicular bisects RQ, TG perpendicular bisects PR, TH perpendicular bisects QP. Find the value of z.
问题：如图所示，GR=7，PG=z+4，PT=3*y-1，TQ=2*x，TR=8，TF是RQ的垂直平分线，TG是PR的垂直平分线，TH是QP的垂直平分线。求z的值。
Construction (CDL):
Shape(QH,HT,TQ)
Shape(HP,PT,TH)
Shape(TP,PG,GT)
Shape(TG,GR,RT)
Shape(TR,RF,FT)
Shape(TF,FQ,QT)
Collinear(QHP)
Collinear(PGR)
Collinear(RFQ)
Conditions (CDL):
Equal(LengthOfLine(GR),7)
Equal(LengthOfLine(PG),z+4)
Equal(LengthOfLine(PT),3*y-1)
Equal(LengthOfLine(TQ),2*x)
Equal(LengthOfLine(TR),8)
IsPerpendicularBisectorOfLine(TF,RQ)
IsPerpendicularBisectorOfLine(TG,PR)
IsPerpendicularBisectorOfLine(TH,QP)
Goal: Value(z)
Answer: 3
Theorem sequence:
Problem: As shown in the diagram, AB=12, AC=16, ED=5, EB is parallel to DC. Find the length of line AE.
问题：如图所示，AB=12，AC=16，ED=5，EB∥DC。求直线AE的长度。
Construction (CDL):
Shape(BA,AE,EB)
Shape(CB,BE,ED,DC)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),16)
Equal(LengthOfLine(ED),5)
ParallelBetweenLine(EB,DC)
Goal: Value(LengthOfLine(AE))
Answer: 15
Theorem sequence:
line_addition(1,AE,ED)
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
similar_triangle_property_line_ratio(1,BAE,CAD)
similar_triangle_property_line_ratio(1,EBA,DCA)
Problem: As shown in the diagram, ∠LCR=3*x+5°, ∠MCN=60°, ∠MCR=x-1°. Find the measure of ∠NCL.
问题：如图所示，∠LCR=3*x+5°，∠MCN=60°，∠MCR=x-1°。求∠NCL的大小。
Construction (CDL):
Shape(CMR,RC,CM)
Shape(CRL,LC,CR)
Shape(CLN,NC,CL)
Shape(CNM,MC,CN)
Collinear(MCL)
Cocircular(C,RLNM)
Conditions (CDL):
Equal(MeasureOfAngle(LCR),3*x+5)
Equal(MeasureOfAngle(MCN),60)
Equal(MeasureOfAngle(MCR),x-1)
Goal: Value(MeasureOfAngle(NCL))
Answer: 120
Theorem sequence:
adjacent_complementary_angle(1,MCN,NCL)
Problem: As shown in the diagram, FG=16, HF=12, EH is the altitude of triangle EGF, EH is perpendicular to FH, FE⊥GE. Find the length of line EH.
问题：如图所示，FG=16，HF=12，EH是△EGF的高，EH⊥FH，FE⊥GE。求直线EH的长度。
Construction (CDL):
Shape(HE,EG,GH)
Shape(EH,HF,FE)
Collinear(GHF)
Conditions (CDL):
Equal(LengthOfLine(FG),16)
Equal(LengthOfLine(HF),12)
IsAltitudeOfTriangle(EH,EGF)
PerpendicularBetweenLine(EH,FH)
PerpendicularBetweenLine(FE,GE)
Goal: Value(LengthOfLine(EH))
Answer: 4*sqrt(3)
Theorem sequence:
line_addition(1,GH,HF)
right_triangle_judgment_angle(1,GHE)
mirror_similar_triangle_judgment_aa(1,EGH,FEG)
mirror_similar_triangle_property_line_ratio(1,HEG,EGF)
mirror_similar_triangle_property_line_ratio(1,EGH,FEG)
right_triangle_property_pythagorean(1,GHE)
Problem: As shown in the diagram, AB=3, AC=AB, BC=4*y-5, ∠BAC=2*x°, ∠CBA=2*x°. Find the value of x.
问题：如图所示，AB=3，AC=AB，BC=4*y-5，∠BAC=2*x°，∠CBA=2*x°。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),LengthOfLine(AB))
Equal(LengthOfLine(BC),4*y-5)
Equal(MeasureOfAngle(BAC),2*x)
Equal(MeasureOfAngle(CBA),2*x)
Goal: Value(x)
Answer: 30
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ACB)
isosceles_triangle_property_angle_equal(1,ACB)
triangle_property_angle_sum(1,ACB)
Problem: As shown in the diagram, MP=10, ∠MPQ=8*x°, ∠QLM=2*x*x-10°, LMPQ is a rhombus. Find the perimeter of quadrilateral LMPQ.
问题：如图所示，MP=10，∠MPQ=8*x°，∠QLM=2*x*x-10°，LMPQ是菱形。求四边形LMPQ的周长。
Construction (CDL):
Shape(QL,LR,RQ)
Shape(RL,LM,MR)
Shape(QR,RP,PQ)
Shape(RM,MP,PR)
Collinear(LRP)
Collinear(QRM)
Conditions (CDL):
Equal(LengthOfLine(MP),10)
Equal(MeasureOfAngle(MPQ),8*x)
Equal(MeasureOfAngle(QLM),2*x*x-10)
Rhombus(LMPQ)
Goal: Value(PerimeterOfQuadrilateral(LMPQ))
Answer: 40
Theorem sequence:
quadrilateral_perimeter_formula(1,QLMP)
Problem: As shown in the diagram, AD=22, AE=23, CB=37, AE is perpendicular to CE, DBCA is a trapezoid. Find the area of DBCA.
问题：如图所示，AD=22，AE=23，CB=37，AE⊥CE，DBCA是梯形。求四边形DBCA的面积。
Construction (CDL):
Shape(DB,BE,EA,AD)
Shape(AE,EC,CA)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AD),22)
Equal(LengthOfLine(AE),23)
Equal(LengthOfLine(CB),37)
PerpendicularBetweenLine(AE,CE)
Trapezoid(DBCA)
Goal: Value(AreaOfQuadrilateral(DBCA))
Answer: 1357/2
Theorem sequence:
adjacent_complementary_angle(1,BEA,AEC)
altitude_of_quadrilateral_judgment_right_vertex(2,AE,DBCA)
trapezoid_area_formula(1,DBCA)
Problem: As shown in the diagram, ML=3*x+1, MN=4*x-2, ∠LPM=5*y°, △MLN is an equilateral △, MP is the median of triangle MLN. Find the length of line LM.
问题：如图所示，ML=3*x+1，MN=4*x-2，∠LPM=5*y°，三角形MLN为等边三角形，MP是三角形MLN的中线。求直线LM的长度。
Construction (CDL):
Shape(ML,LN,NM)
Shape(MP,PN,NM)
Collinear(LPN)
Conditions (CDL):
Equal(LengthOfLine(ML),3*x+1)
Equal(LengthOfLine(MN),4*x-2)
Equal(MeasureOfAngle(LPM),5*y)
EquilateralTriangle(MLN)
IsMedianOfTriangle(MP,MLN)
Goal: Value(LengthOfLine(LM))
Answer: 10
Theorem sequence:
Problem: As shown in the diagram, AB=9*x-2, AD=6*x+2, BC=10, CD=8, ∠DBC=∠ABD. Find the value of x.
问题：如图所示，AB=9*x-2，AD=6*x+2，BC=10，CD=8，∠DBC=∠ABD。求x的值。
Construction (CDL):
Shape(DA,AB,BD)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),9*x-2)
Equal(LengthOfLine(AD),6*x+2)
Equal(LengthOfLine(BC),10)
Equal(LengthOfLine(CD),8)
Equal(MeasureOfAngle(DBC),MeasureOfAngle(ABD))
Goal: Value(x)
Answer: 3
Theorem sequence:
bisector_of_angle_judgment_angle_equal(1,BD,ABC)
bisector_of_angle_property_line_ratio(1,BD,ABC)
Problem: As shown in the diagram, ∠NMQ=10*x°, ∠PNM=20*x°, ∠PNM=∠MQP, ∠QPN=∠NMQ, quadrilateral MQPN is a parallelogram. Find the measure of ∠NMQ.
问题：如图所示，∠NMQ=10*x°，∠PNM=20*x°，∠PNM=∠MQP，∠QPN=∠NMQ，MQPN是▱。求∠NMQ的大小。
Construction (CDL):
Shape(MQ,QP,PN,NM)
Conditions (CDL):
Equal(MeasureOfAngle(NMQ),10*x)
Equal(MeasureOfAngle(PNM),20*x)
Equal(MeasureOfAngle(PNM),MeasureOfAngle(MQP))
Equal(MeasureOfAngle(QPN),MeasureOfAngle(NMQ))
Parallelogram(MQPN)
Goal: Value(MeasureOfAngle(NMQ))
Answer: 60
Theorem sequence:
quadrilateral_property_angle_sum(1,MQPN)
Problem: As shown in the diagram, ∠AHD=2*x+23°, ∠EAH=7*x-5°, AE∥HD. Find the value of x.
问题：如图所示，∠AHD=2*x+23°，∠EAH=7*x-5°，AE∥HD。求x的值。
Construction (CDL):
Shape(GH,HC)
Shape(AH,HD)
Shape(EA,AH)
Shape(BA,AF)
Collinear(GHAF)
Collinear(CHD)
Collinear(BAE)
Conditions (CDL):
Equal(MeasureOfAngle(AHD),2*x+23)
Equal(MeasureOfAngle(EAH),7*x-5)
ParallelBetweenLine(AE,HD)
Goal: Value(x)
Answer: 18
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AE,HD)
Problem: As shown in the diagram, MN=3*x-4, NQ=15.4, PN=2*y+5, PQ=11.1, RM=17.9, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, quadrilateral MRPN is a parallelogram. Find the measure of ∠NMR.
问题：如图所示，MN=3*x-4，NQ=15.4，PN=2*y+5，PQ=11.1，RM=17.9，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，MN和RP是▱MRPN的一组对边。求∠NMR的大小。
Construction (CDL):
Shape(QM,MR,RQ)
Shape(QR,RP,PQ)
Shape(QP,PN,NQ)
Shape(QN,NM,MQ)
Collinear(MQP)
Collinear(RQN)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(NQ),15.4)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),11.1)
Equal(LengthOfLine(RM),17.9)
Equal(LengthOfLine(RP),20)
Equal(LengthOfLine(RQ),3*z-3)
Equal(MeasureOfAngle(MRQ),38)
Equal(MeasureOfAngle(NQP),83)
Equal(MeasureOfAngle(QNM),33)
Parallelogram(MRPN)
Goal: Value(MeasureOfAngle(NMR))
Answer: 109
Theorem sequence:
triangle_property_angle_sum(1,MRN)
Problem: As shown in the diagram, the measure of ⌒DCA is 122. Find the measure of ∠ABC.
问题：如图所示，弧DCA的角度为122。求∠ABC的大小。
Construction (CDL):
Shape(DAB,BA)
Shape(DBC,CB)
Shape(DCA,AB,BC)
Cocircular(D,ABC)
Conditions (CDL):
Equal(MeasureOfArc(DCA),122)
Goal: Value(MeasureOfAngle(ABC))
Answer: 61
Theorem sequence:
arc_property_circumference_angle_external(1,DCA,B)
Problem: As shown in the diagram, NL=5, QR=35, RS=37, SQ=12, triangle LMN is similar to triangle QRS.. Find the perimeter of △LMN.
问题：如图所示，NL=5，QR=35，RS=37，SQ=12，△LMN与△QRS是相似三角形。求△LMN的周长。
Construction (CDL):
Shape(SQ,QR,RS)
Shape(NL,LM,MN)
Conditions (CDL):
Equal(LengthOfLine(NL),5)
Equal(LengthOfLine(QR),35)
Equal(LengthOfLine(RS),37)
Equal(LengthOfLine(SQ),12)
SimilarBetweenTriangle(LMN,QRS)
Goal: Value(PerimeterOfTriangle(LMN))
Answer: 35
Theorem sequence:
similar_triangle_property_line_ratio(1,LMN,QRS)
similar_triangle_property_line_ratio(1,MNL,RSQ)
similar_triangle_property_line_ratio(1,NLM,SQR)
triangle_perimeter_formula(1,LMN)
Problem: As shown in the diagram, ∠CAB=51°, ∠CEB=25°, ∠EBC=28°, ∠FCB=35°, BF is perpendicular to CF, CB is perpendicular to DB, CG is perpendicular to BG. Find the measure of ∠FBD.
问题：如图所示，∠CAB=51°，∠CEB=25°，∠EBC=28°，∠FCB=35°，BF垂直于CF，CB垂直于DB，CG⊥BG。求∠FBD的大小。
Construction (CDL):
Shape(CE,EA,AC)
Shape(CA,AG,GC)
Shape(CG,GB,BC)
Shape(CB,BF,FC)
Shape(FB,BD,DF)
Collinear(EAGB)
Collinear(CFD)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),51)
Equal(MeasureOfAngle(CEB),25)
Equal(MeasureOfAngle(EBC),28)
Equal(MeasureOfAngle(FCB),35)
PerpendicularBetweenLine(BF,CF)
PerpendicularBetweenLine(CB,DB)
PerpendicularBetweenLine(CG,BG)
Goal: Value(MeasureOfAngle(FBD))
Answer: 35
Theorem sequence:
triangle_property_angle_sum(1,CGB)
angle_addition(1,FCB,BCG)
quadrilateral_property_angle_sum(1,CGBF)
angle_addition(1,GBC,CBF)
angle_addition(1,CBF,FBD)
Problem: As shown in the diagram, RS=3*x, UT=7*x-44, the measure of arc PSR is 155, the measure of ⌒QTU is 205, the radius of circle P is equal to the radius of ⊙Q, the center of circle P is P, the center of ⊙Q is Q. Find the value of x.
问题：如图所示，RS=3*x，UT=7*x-44，⌒PSR的角度为155，弧QTU的角度为205，圆P的半径与圆Q的半径相等，P是圆P的圆心，⊙Q的圆心为Q。求x的值。
Construction (CDL):
Shape(PSR,RS)
Shape(PRS,SR)
Shape(QUT,TU)
Shape(PS,SR,RP)
Shape(QU,UT,TQ)
Shape(QTU,UQ,QT)
Cocircular(P,RS)
Cocircular(Q,UT)
Conditions (CDL):
Equal(LengthOfLine(RS),3*x)
Equal(LengthOfLine(UT),7*x-44)
Equal(MeasureOfArc(PSR),155)
Equal(MeasureOfArc(QTU),205)
Equal(RadiusOfCircle(P),RadiusOfCircle(Q))
IsCentreOfCircle(P,P)
IsCentreOfCircle(Q,Q)
Goal: Value(x)
Answer: 11
Theorem sequence:
radius_of_circle_property_length_equal(1,PR,P)
radius_of_circle_property_length_equal(1,PS,P)
radius_of_circle_property_length_equal(1,QT,Q)
radius_of_circle_property_length_equal(1,QU,Q)
arc_property_center_angle(1,PSR,P)
arc_property_center_angle(1,QTU,Q)
round_angle(1,TQU,UQT)
congruent_triangle_judgment_sas(1,PSR,QUT)
congruent_triangle_property_line_equal(1,PSR,QUT)
Problem: As shown in the diagram, AB=x, BC=4, ∠CAB=45°, BC⊥AC. Find the value of x.
问题：如图所示，AB=x，BC=4，∠CAB=45°，BC垂直于AC。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),4)
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 4*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,ABC)
right_triangle_judgment_angle(1,BCA)
isosceles_triangle_judgment_angle_equal(1,CAB)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, MN=6, MQ=5, ON=3+3/5, QP=x, QN is parallel to PO. Find the length of line QP.
问题：如图所示，MN=6，MQ=5，ON=3+3/5，QP=x，QN∥PO。求直线QP的长度。
Construction (CDL):
Shape(MQ,QN,NM)
Shape(QP,PO,ON,NQ)
Collinear(MQP)
Collinear(MNO)
Conditions (CDL):
Equal(LengthOfLine(MN),6)
Equal(LengthOfLine(MQ),5)
Equal(LengthOfLine(ON),3+3/5)
Equal(LengthOfLine(QP),x)
ParallelBetweenLine(QN,PO)
Goal: Value(LengthOfLine(QP))
Answer: 3
Theorem sequence:
line_addition(1,MQ,QP)
line_addition(1,MN,NO)
parallel_property_corresponding_angle(1,QN,PO,M)
parallel_property_corresponding_angle(2,OP,NQ,M)
similar_triangle_judgment_aa(1,MQN,MPO)
similar_triangle_property_line_ratio(1,QNM,POM)
similar_triangle_property_line_ratio(1,NMQ,OMP)
Problem: As shown in the diagram, AB=42, BC=38, CA=x, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=42，BC=38，CA=x，BC垂直于AC。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),42)
Equal(LengthOfLine(BC),38)
Equal(LengthOfLine(CA),x)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 8*sqrt(5)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, AB=2*x+3, BC=5*x, ADCB is a rhombus. Find the measure of ∠AEB.
问题：如图所示，AB=2*x+3，BC=5*x，ADCB是菱形。求∠AEB的大小。
Construction (CDL):
Shape(EA,AD,DE)
Shape(ED,DC,CE)
Shape(EC,CB,BE)
Shape(EB,BA,AE)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x+3)
Equal(LengthOfLine(BC),5*x)
Rhombus(ADCB)
Goal: Value(MeasureOfAngle(AEB))
Answer: 90
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,ADCB,E)