SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AC=b, BA=c, BC=a, ∠ABC=60°, ∠CAB=30°, a=4, BC is perpendicular to AC. Find the value of b.
问题：如图所示，AC=b，BA=c，BC=a，∠ABC=60°，∠CAB=30°，a=4，BC⊥AC。求b的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(BA),c)
Equal(LengthOfLine(BC),a)
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),30)
Equal(a,4)
PerpendicularBetweenLine(BC,AC)
Goal: Value(b)
Answer: 4*sqrt(3)
Theorem sequence:
sine_theorem(1,BCA)
sine_theorem(1,ABC)
Problem: As shown in the diagram, JM=6, KJ=KL, LM is perpendicular to KM. Find the length of line JL.
问题：如图所示，JM=6，KJ=KL，LM垂直于KM。求直线JL的长度。
Construction (CDL):
Shape(JK,KM,MJ)
Shape(MK,KL,LM)
Collinear(JML)
Conditions (CDL):
Equal(LengthOfLine(JM),6)
Equal(LengthOfLine(KJ),LengthOfLine(KL))
PerpendicularBetweenLine(LM,KM)
Goal: Value(LengthOfLine(JL))
Answer: 12
Theorem sequence:
perpendicular_bisector_judgment_distance_equal(1,KM,LJ)
line_addition(1,JM,ML)
Problem: As shown in the diagram, AB=26, AD=10, BC=16, DC=8, AD⊥CD. Find the perimeter of triangle ACB.
问题：如图所示，AB=26，AD=10，BC=16，DC=8，AD垂直于CD。求△ACB的周长。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AC,CB,BA)
Collinear(DCB)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BC),16)
Equal(LengthOfLine(DC),8)
PerpendicularBetweenLine(AD,CD)
Goal: Value(PerimeterOfTriangle(ACB))
Answer: 2*sqrt(41)+42
Theorem sequence:
right_triangle_judgment_angle(1,ADC)
right_triangle_property_pythagorean(1,ADC)
triangle_perimeter_formula(1,ACB)
Problem: As shown in the diagram, AB=x, AC=19, AD=15, BC=z, BD=y, CA is perpendicular to BA, DB is perpendicular to CB. Find the value of y.
问题：如图所示，AB=x，AC=19，AD=15，BC=z，BD=y，CA⊥BA，DB⊥CB。求y的值。
Construction (CDL):
Shape(DB,BA,AD)
Shape(AB,BC,CA)
Collinear(DAC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),19)
Equal(LengthOfLine(AD),15)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BD),y)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(DB,CB)
Goal: Value(y)
Answer: sqrt(510)
Theorem sequence:
adjacent_complementary_angle(1,CAB,BAD)
right_triangle_judgment_angle(1,DBC)
right_triangle_judgment_angle(1,CAB)
right_triangle_judgment_angle(1,BAD)
line_addition(1,DA,AC)
right_triangle_property_pythagorean(1,DBC)
right_triangle_property_pythagorean(1,CAB)
right_triangle_property_pythagorean(1,BAD)
Problem: As shown in the diagram, AB=11, AC=x, BC=9, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=11，AC=x，BC=9，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),11)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),9)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 2*sqrt(10)
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, TU=14, the center of circle Q is Q, the diameter of ⊙Q is TU. Find the radius of ⊙Q.
问题：如图所示，TU=14，圆Q的圆心为Q，TU是圆Q的直径。求圆Q的半径。
Construction (CDL):
Shape(QT,QTU,UQ)
Shape(QV,QVT,TQ)
Shape(QU,QUV,VQ)
Collinear(TQU)
Cocircular(Q,TUV)
Conditions (CDL):
Equal(LengthOfLine(TU),14)
IsCentreOfCircle(Q,Q)
IsDiameterOfCircle(TU,Q)
Goal: Value(RadiusOfCircle(Q))
Answer: 7
Theorem sequence:
diameter_of_circle_property_length_equal(1,TU,Q)
circle_property_length_of_radius_and_diameter(1,Q)
Problem: As shown in the diagram, ∠VWY=67°, ∠XWZ=∠WZX, ∠XZY=65°, ∠ZYW=58°. Find the measure of ∠ZXW.
问题：如图所示，∠VWY=67°，∠XWZ=∠WZX，∠XZY=65°，∠ZYW=58°。求∠ZXW的大小。
Construction (CDL):
Shape(VW,WX,XV)
Shape(XW,WZ,ZX)
Shape(XZ,ZY,YX)
Collinear(VXZ)
Collinear(WXY)
Conditions (CDL):
Equal(MeasureOfAngle(VWY),67)
Equal(MeasureOfAngle(XWZ),MeasureOfAngle(WZX))
Equal(MeasureOfAngle(XZY),65)
Equal(MeasureOfAngle(ZYW),58)
Goal: Value(MeasureOfAngle(ZXW))
Answer: 123
Theorem sequence:
triangle_property_angle_sum(1,YXZ)
adjacent_complementary_angle(1,YXZ,ZXW)
Problem: As shown in the diagram, ∠CAY=2*x°, ∠YAF=x°. Find the value of x.
问题：如图所示，∠CAY=2*x°，∠YAF=x°。求x的值。
Construction (CDL):
Shape(YA,AF)
Shape(FA,AD)
Shape(DA,AC)
Shape(CA,AY)
Collinear(YAD)
Collinear(CAF)
Conditions (CDL):
Equal(MeasureOfAngle(CAY),2*x)
Equal(MeasureOfAngle(YAF),x)
Goal: Value(x)
Answer: 60
Theorem sequence:
adjacent_complementary_angle(1,CAY,YAF)
Problem: As shown in the diagram, ∠AED=3*x°, ∠EBN=2*x+45°, ET is parallel to BN. Find the value of x.
问题：如图所示，∠AED=3*x°，∠EBN=2*x+45°，ET平行于BN。求x的值。
Construction (CDL):
Shape(DE,ET)
Shape(TE,EB)
Shape(EB,BN)
Shape(NB,BF)
Shape(FB,BC)
Shape(CB,BE)
Shape(BE,EA)
Shape(AE,ED)
Collinear(DEBF)
Collinear(TEA)
Collinear(NBC)
Conditions (CDL):
Equal(MeasureOfAngle(AED),3*x)
Equal(MeasureOfAngle(EBN),2*x+45)
ParallelBetweenLine(ET,BN)
Goal: Value(x)
Answer: 27
Theorem sequence:
vertical_angle(1,AED,TEB)
parallel_property_ipsilateral_internal_angle(1,ET,BN)
Problem: As shown in the diagram, AC=12, ∠DBC=45°, the center of ⊙D is D, AB is the diameter of circle D, the diameter of circle D is BA. Find the area of the circle D minus the area of triangle CAB.
问题：如图所示，AC=12，∠DBC=45°，D是⊙D的圆心，AB是⊙D的直径，⊙D的直径为BA。求⊙D的面积减去三角形CAB的面积。
Construction (CDL):
Shape(DCA,AC)
Shape(DA,DAB,BD)
Shape(CA,AD,DB,BC)
Shape(DBC,CB)
Collinear(ADB)
Cocircular(D,ABC)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(MeasureOfAngle(DBC),45)
IsCentreOfCircle(D,D)
IsDiameterOfCircle(AB,D)
IsDiameterOfCircle(BA,D)
Goal: Value(Sub(AreaOfCircle(D),AreaOfTriangle(CAB)))
Answer: -72+72*pi
Theorem sequence:
diameter_of_circle_property_right_angle(1,BCA,D)
triangle_property_angle_sum(1,CAB)
sine_theorem(1,ABC)
sine_theorem(1,BCA)
diameter_of_circle_property_length_equal(1,AB,D)
circle_property_length_of_radius_and_diameter(1,D)
triangle_area_formula_sine(1,CAB)
circle_area_formula(1,D)
Problem: As shown in the diagram, the measure of arc OAB is 143, the measure of arc ODC is 75. Find the measure of ∠BEA.
问题：如图所示，弧OAB的角度为143，弧ODC的角度为75。求∠BEA的大小。
Construction (CDL):
Shape(OAB,BE,EA)
Shape(OBD,DB)
Shape(ODC,CE,ED)
Shape(OCA,AE,EC)
Shape(EB,BD,DE)
Collinear(AED)
Collinear(CEB)
Cocircular(O,ABDC)
Conditions (CDL):
Equal(MeasureOfArc(OAB),143)
Equal(MeasureOfArc(ODC),75)
Goal: Value(MeasureOfAngle(BEA))
Answer: 109
Theorem sequence:
arc_property_circumference_angle_external(1,OAB,D)
arc_property_circumference_angle_external(1,ODC,B)
triangle_property_angle_sum(1,EBD)
adjacent_complementary_angle(1,DEB,BEA)
Problem: As shown in the diagram, AZ=4, ∠XYA=∠AYZ, AX⊥YX, YZ⊥AZ. Find the length of line XA.
问题：如图所示，AZ=4，∠XYA=∠AYZ，AX⊥YX，YZ⊥AZ。求直线XA的长度。
Construction (CDL):
Shape(XY,YA,AX)
Shape(AY,YZ,ZA)
Conditions (CDL):
Equal(LengthOfLine(AZ),4)
Equal(MeasureOfAngle(XYA),MeasureOfAngle(AYZ))
PerpendicularBetweenLine(AX,YX)
PerpendicularBetweenLine(YZ,AZ)
Goal: Value(LengthOfLine(XA))
Answer: 4
Theorem sequence:
mirror_congruent_triangle_judgment_aas(3,AXY,AYZ)
mirror_congruent_triangle_property_line_equal(1,YAX,YZA)
Problem: As shown in the diagram, DB=x, DF=5, FB=15, the center of circle D is D, the tangent to ⊙D is BF. Find the value of x.
问题：如图所示，DB=x，DF=5，FB=15，⊙D的圆心为D，BF是圆O的切线。求x的值。
Construction (CDL):
Shape(BF,DAF,AB)
Shape(DA,DAF,FD)
Shape(DF,DFA,AD)
Collinear(DAB)
Cocircular(D,FA)
Conditions (CDL):
Equal(LengthOfLine(DB),x)
Equal(LengthOfLine(DF),5)
Equal(LengthOfLine(FB),15)
IsCentreOfCircle(D,D)
IsTangentOfCircle(BF,D)
Goal: Value(x)
Answer: 5*sqrt(10)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BF,D,D)
right_triangle_judgment_angle(1,BFD)
right_triangle_property_pythagorean(1,BFD)
Problem: As shown in the diagram, the area of triangle CDE is 13, AB=8, CE=6, triangle CDE is similar to triangle ATB.. Find the area of △ATB.
问题：如图所示，三角形CDE的面积为13，AB=8，CE=6，三角形CDE与三角形ATB是相似三角形。求三角形ATB的面积。
Construction (CDL):
Shape(CD,DE,EC)
Shape(AT,TB,BA)
Conditions (CDL):
Equal(AreaOfTriangle(CDE),13)
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(CE),6)
SimilarBetweenTriangle(CDE,ATB)
Goal: Value(AreaOfTriangle(ATB))
Answer: 208/9
Theorem sequence:
similar_triangle_property_line_ratio(1,DEC,TBA)
similar_triangle_property_area_square_ratio(1,DEC,TBA)
Problem: As shown in the diagram, RW=14, RZ=18, UZ=5, the centroid of △RST is Z. Find the length of line SZ.
问题：如图所示，RW=14，RZ=18，UZ=5，Z是△RST的重心。求直线SZ的长度。
Construction (CDL):
Shape(RW,WZ,ZR)
Shape(ZW,WS,SZ)
Shape(ZS,SV,VZ)
Shape(ZV,VT,TZ)
Shape(ZT,TU,UZ)
Shape(ZU,UR,RZ)
Collinear(RWS)
Collinear(SVT)
Collinear(TUR)
Collinear(WZT)
Collinear(SZU)
Collinear(RZV)
Conditions (CDL):
Equal(LengthOfLine(RW),14)
Equal(LengthOfLine(RZ),18)
Equal(LengthOfLine(UZ),5)
IsCentroidOfTriangle(Z,RST)
Goal: Value(LengthOfLine(SZ))
Answer: 10
Theorem sequence:
centroid_of_triangle_property_line_ratio(1,Z,STR,U)
Problem: As shown in the diagram, the measure of ⌒AJH is 114, GB⊥HB, JH is perpendicular to GH. Find the measure of ∠HGB.
问题：如图所示，⌒AJH的角度为114，GB⊥HB，JH⊥GH。求∠HGB的大小。
Construction (CDL):
Shape(AGF,FB,BG)
Shape(AB,BF,AFJ,JA)
Shape(BA,AJ,JH,HB)
Shape(AJH,HJ)
Shape(GB,BH,HG)
Shape(AHG,GH)
Collinear(FBH)
Collinear(GBAJ)
Cocircular(A,GFJH)
Conditions (CDL):
Equal(MeasureOfArc(AJH),114)
PerpendicularBetweenLine(GB,HB)
PerpendicularBetweenLine(JH,GH)
Goal: Value(MeasureOfAngle(HGB))
Answer: 57
Theorem sequence:
arc_property_circumference_angle_external(1,AJH,G)
Problem: As shown in the diagram, ∠QRS=x°, ∠RST=2*x+7°, ∠STQ=x°, ∠TQR=2*x+5°. Find the measure of ∠STQ.
问题：如图所示，∠QRS=x°，∠RST=2*x+7°，∠STQ=x°，∠TQR=2*x+5°。求∠STQ的大小。
Construction (CDL):
Shape(QR,RS,ST,TQ)
Conditions (CDL):
Equal(MeasureOfAngle(QRS),x)
Equal(MeasureOfAngle(RST),2*x+7)
Equal(MeasureOfAngle(STQ),x)
Equal(MeasureOfAngle(TQR),2*x+5)
Goal: Value(MeasureOfAngle(STQ))
Answer: 58
Theorem sequence:
quadrilateral_property_angle_sum(1,QRST)
Problem: As shown in the diagram, AC=4, CD=3, E is the center of ⊙E, DA is the diameter of circle E, DC⊥AC. Find the circumference of the ⊙E.
问题：如图所示，AC=4，CD=3，E是⊙E的圆心，DA是⊙E的直径，DC垂直于AC。求⊙E的周长。
Construction (CDL):
Shape(EDC,CD)
Shape(ECA,AC)
Shape(DC,CA,AE,ED)
Shape(DE,EA,EAD)
Collinear(DEA)
Cocircular(E,DCA)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(CD),3)
IsCentreOfCircle(E,E)
IsDiameterOfCircle(DA,E)
PerpendicularBetweenLine(DC,AC)
Goal: Value(PerimeterOfCircle(E))
Answer: 5*pi
Theorem sequence:
right_triangle_judgment_angle(1,DCA)
right_triangle_property_pythagorean(1,DCA)
diameter_of_circle_property_length_equal(1,DA,E)
circle_property_length_of_radius_and_diameter(1,E)
circle_perimeter_formula(1,E)
Problem: As shown in the diagram, AD=12, FE=18, D is the center of circle D, the tangent to circle D is EA. Find the length of line AE.
问题：如图所示，AD=12，FE=18，D是圆D的圆心，圆O的切线为EA。求直线AE的长度。
Construction (CDL):
Shape(DAB,BA)
Shape(DBJ,JK,KB)
Shape(BK,KM,MA,AB)
Shape(AM,MF,DFA)
Shape(DFA,FE,EA)
Shape(DJC,CK,KJ)
Shape(MK,KC,CD,DM)
Shape(MD,DF,FM)
Shape(DC,DCH,HD)
Shape(DH,DHF,FD)
Shape(DHF,HG,GF)
Collinear(JKMF)
Collinear(BKC)
Collinear(AMDC)
Collinear(DFE)
Cocircular(D,ABJCHF)
Conditions (CDL):
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(FE),18)
IsCentreOfCircle(D,D)
IsTangentOfCircle(EA,D)
Goal: Value(LengthOfLine(AE))
Answer: 6*sqrt(21)
Theorem sequence:
radius_of_circle_property_length_equal(1,DA,D)
radius_of_circle_property_length_equal(1,DF,D)
line_addition(1,DF,FE)
tangent_of_circle_property_perpendicular(2,EA,D,D)
right_triangle_judgment_angle(1,EAD)
right_triangle_property_pythagorean(1,EAD)
Problem: As shown in the diagram, BA=15, BC=20, CA=25, CB⊥AB. Find the value of tan(BA).
问题：如图所示，BA=15，BC=20，CA=25，CB垂直于AB。求tan(BA)的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(BA),15)
Equal(LengthOfLine(BC),20)
Equal(LengthOfLine(CA),25)
PerpendicularBetweenLine(CB,AB)
Goal: Value(Tan(MeasureOfAngle(BAC)))
Answer: 4/3
Theorem sequence:
cosine_theorem(1,ACB)
Problem: As shown in the diagram, the area of MQPN is 360, NE=18, QP=26, NE⊥PE, MN and QP are the parallel sides of trapezoid MQPN. Find the length of line MN.
问题：如图所示，四边形MQPN的面积为360，NE=18，QP=26，NE垂直于PE，MQPN是梯形。求直线MN的长度。
Construction (CDL):
Shape(MQ,QE,EN,NM)
Shape(NE,EP,PN)
Collinear(QEP)
Conditions (CDL):
Equal(AreaOfQuadrilateral(MQPN),360)
Equal(LengthOfLine(NE),18)
Equal(LengthOfLine(QP),26)
PerpendicularBetweenLine(NE,PE)
Trapezoid(MQPN)
Goal: Value(LengthOfLine(MN))
Answer: 14
Theorem sequence:
adjacent_complementary_angle(1,QEN,NEP)
altitude_of_quadrilateral_judgment_right_vertex(2,NE,MQPN)
trapezoid_area_formula(1,MQPN)
Problem: As shown in the diagram, MJ=2*x-5, NJ=x-2, NM=3*x-9, △MJN is an isosceles △. Find the length of line JN.
问题：如图所示，MJ=2*x-5，NJ=x-2，NM=3*x-9，三角形MJN为等腰三角形。求直线JN的长度。
Construction (CDL):
Shape(MJ,JN,NM)
Conditions (CDL):
Equal(LengthOfLine(MJ),2*x-5)
Equal(LengthOfLine(NJ),x-2)
Equal(LengthOfLine(NM),3*x-9)
IsoscelesTriangle(MJN)
Goal: Value(LengthOfLine(JN))
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, QV=8, Q is the center of circle Q. Find the diameter of circle Q.
问题：如图所示，QV=8，⊙Q的圆心为Q。求圆Q的直径。
Construction (CDL):
Shape(QT,QTU,UQ)
Shape(QVT,TQ,QV)
Shape(QUV,VQ,QU)
Collinear(TQU)
Cocircular(Q,TUV)
Conditions (CDL):
Equal(LengthOfLine(QV),8)
IsCentreOfCircle(Q,Q)
Goal: Value(DiameterOfCircle(Q))
Answer: 16
Theorem sequence:
radius_of_circle_property_length_equal(1,QV,Q)
circle_property_length_of_radius_and_diameter(1,Q)
Problem: As shown in the diagram, ∠TZY=30*x°, ∠TZY=∠ZYW, ∠WTZ=20*x°, ∠WTZ=∠YWT, quadrilateral TZYW is a isosceles trapezoid. Find the measure of ∠TZY.
问题：如图所示，∠TZY=30*x°，∠TZY=∠ZYW，∠WTZ=20*x°，∠WTZ=∠YWT，四边形TZYW是等腰梯形。求∠TZY的大小。
Construction (CDL):
Shape(TZ,ZY,YW,WT)
Conditions (CDL):
Equal(MeasureOfAngle(TZY),30*x)
Equal(MeasureOfAngle(TZY),MeasureOfAngle(ZYW))
Equal(MeasureOfAngle(WTZ),20*x)
Equal(MeasureOfAngle(WTZ),MeasureOfAngle(YWT))
IsoscelesTrapezoid(TZYW)
Goal: Value(MeasureOfAngle(TZY))
Answer: 108
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,TW,ZY)
Problem: As shown in the diagram, BA=6, DA=9, DC=6.86, ∠BAF=32°, ∠CBF=40.1°, ∠FAD=20°, quadrilateral DCBA is a ▱. Find the length of line BC.
问题：如图所示，BA=6，DA=9，DC=6.86，∠BAF=32°，∠CBF=40.1°，∠FAD=20°，CD和BA是▱DCBA的一组对边。求直线BC的长度。
Construction (CDL):
Shape(DC,CF,FD)
Shape(FC,CB,BF)
Shape(FB,BA,AF)
Shape(DF,FA,AD)
Collinear(DFB)
Collinear(CFA)
Conditions (CDL):
Equal(LengthOfLine(BA),6)
Equal(LengthOfLine(DA),9)
Equal(LengthOfLine(DC),6.86)
Equal(MeasureOfAngle(BAF),32)
Equal(MeasureOfAngle(CBF),40.1)
Equal(MeasureOfAngle(FAD),20)
Parallelogram(DCBA)
Goal: Value(LengthOfLine(BC))
Answer: 9
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ADCB)
Problem: As shown in the diagram, BC=15, BE=6, DA=10, DE=3*x-2, EC=12, ∠ADE=∠CBE. Find the length of line AE.
问题：如图所示，BC=15，BE=6，DA=10，DE=3*x-2，EC=12，∠ADE=∠CBE。求直线AE的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(BE,EC,CB)
Collinear(AEC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(BE),6)
Equal(LengthOfLine(DA),10)
Equal(LengthOfLine(DE),3*x-2)
Equal(LengthOfLine(EC),12)
Equal(MeasureOfAngle(ADE),MeasureOfAngle(CBE))
Goal: Value(LengthOfLine(AE))
Answer: 8
Theorem sequence:
vertical_angle(1,DEA,BEC)
similar_triangle_judgment_aa(1,ADE,CBE)
similar_triangle_property_line_ratio(1,EAD,ECB)
similar_triangle_property_line_ratio(1,DEA,BEC)
Problem: As shown in the diagram, QR=4, SA=5, ST=x, Q is the center of ⊙Q, SR is the tangent to circle Q, ST is the tangent to ⊙Q. Find the value of x.
问题：如图所示，QR=4，SA=5，ST=x，⊙Q的圆心为Q，⊙O的切线为SR，ST是⊙O的切线。求x的值。
Construction (CDL):
Shape(SR,QAR,AS)
Shape(RQ,QA,QAR)
Shape(QT,QTA,AQ)
Shape(QTA,TS,SA)
Shape(QR,QRT,TQ)
Collinear(QAS)
Cocircular(Q,TAR)
Conditions (CDL):
Equal(LengthOfLine(QR),4)
Equal(LengthOfLine(SA),5)
Equal(LengthOfLine(ST),x)
IsCentreOfCircle(Q,Q)
IsTangentOfCircle(SR,Q)
IsTangentOfCircle(ST,Q)
Goal: Value(x)
Answer: sqrt(65)
Theorem sequence:
radius_of_circle_property_length_equal(1,QR,Q)
radius_of_circle_property_length_equal(1,QA,Q)
radius_of_circle_property_length_equal(1,QT,Q)
line_addition(1,QA,AS)
tangent_of_circle_property_perpendicular(1,ST,Q,Q)
right_triangle_judgment_angle(1,QTS)
right_triangle_property_pythagorean(1,QTS)
Problem: As shown in the diagram, the area of BLAN is 72, the area of quadrilateral EFGH is 50, BN=6, EH=x, EFGH is similar to BLAN. Find the value of x.
问题：如图所示，四边形BLAN的面积为72，EFGH的面积为50，BN=6，EH=x，四边形EFGH相似于四边形BLAN。求x的值。
Construction (CDL):
Shape(EF,FG,GH,HE)
Shape(BL,LA,AN,NB)
Conditions (CDL):
Equal(AreaOfQuadrilateral(BLAN),72)
Equal(AreaOfQuadrilateral(EFGH),50)
Equal(LengthOfLine(BN),6)
Equal(LengthOfLine(EH),x)
SimilarBetweenQuadrilateral(EFGH,BLAN)
Goal: Value(x)
Answer: 5
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,EFGH,BLAN)
similar_quadrilateral_property_line_ratio(1,HEFG,NBLA)
Problem: As shown in the diagram, HQ=y, JQ=2*z, KQ=7, LQ=3, QM=4, QP=2*x-6, J is the midpoint of segment HK, L bisects segment KM, P bisects segment MH. Find the value of z.
问题：如图所示，HQ=y，JQ=2*z，KQ=7，LQ=3，QM=4，QP=2*x-6，J是线段HK的中点，L是线段KM的中点，P是线段MH的中点。求z的值。
Construction (CDL):
Shape(HJ,JQ,QH)
Shape(JK,KQ,QJ)
Shape(QK,KL,LQ)
Shape(QL,LM,MQ)
Shape(QM,MP,PQ)
Shape(QP,PH,HQ)
Collinear(HJK)
Collinear(KLM)
Collinear(MPH)
Collinear(HQL)
Collinear(JQM)
Collinear(KQP)
Conditions (CDL):
Equal(LengthOfLine(HQ),y)
Equal(LengthOfLine(JQ),2*z)
Equal(LengthOfLine(KQ),7)
Equal(LengthOfLine(LQ),3)
Equal(LengthOfLine(QM),4)
Equal(LengthOfLine(QP),2*x-6)
IsMidpointOfLine(J,HK)
IsMidpointOfLine(L,KM)
IsMidpointOfLine(P,MH)
Goal: Value(z)
Answer: 1
Theorem sequence:
median_of_triangle_judgment(1,KP,KMH)
median_of_triangle_judgment(1,HL,HKM)
centroid_of_triangle_judgment_intersection(1,Q,MHK,P,L)
centroid_of_triangle_property_line_ratio(1,Q,MHK,J)
Problem: As shown in the diagram, AF=24, AH=25, BH=11, ∠DBH=30°, ∠GCH=28°, H is the incenter of △ABC, BG⊥HG, HD is perpendicular to BD, HF⊥AF. Find the measure of ∠CAH.
问题：如图所示，AF=24，AH=25，BH=11，∠DBH=30°，∠GCH=28°，三角形ABC的内切圆圆心为H，BG垂直于HG，HD⊥BD，HF⊥AF。求∠CAH的大小。
Construction (CDL):
Shape(AD,DH,HA)
Shape(HD,DB,BH)
Shape(HB,BG,GH)
Shape(HG,GC,CH)
Shape(HC,CF,FH)
Shape(HF,FA,AH)
Collinear(ADB)
Collinear(BGC)
Collinear(CFA)
Conditions (CDL):
Equal(LengthOfLine(AF),24)
Equal(LengthOfLine(AH),25)
Equal(LengthOfLine(BH),11)
Equal(MeasureOfAngle(DBH),30)
Equal(MeasureOfAngle(GCH),28)
IsIncenterOfTriangle(H,ABC)
PerpendicularBetweenLine(BG,HG)
PerpendicularBetweenLine(HD,BD)
PerpendicularBetweenLine(HF,AF)
Goal: Value(MeasureOfAngle(CAH))
Answer: 32
Theorem sequence:
angle_addition(1,DBH,HBG)
angle_addition(1,GCH,HCF)
angle_addition(1,FAH,HAD)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, AB=2*x+2, AC=y+10, BD=2*y+5, CD=x+5, CD and AB are opposite sides of the ▱ CABD. Find the value of x.
问题：如图所示，AB=2*x+2，AC=y+10，BD=2*y+5，CD=x+5，四边形CABD是平行四边形。求x的值。
Construction (CDL):
Shape(CA,AB,BD,DC)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x+2)
Equal(LengthOfLine(AC),y+10)
Equal(LengthOfLine(BD),2*y+5)
Equal(LengthOfLine(CD),x+5)
Parallelogram(CABD)
Goal: Value(x)
Answer: 3
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABDC)
Problem: As shown in the diagram, AB=c, AC=b, CB=a, a=8, b=15, c=17, BC is perpendicular to AC. Find the value of tan(AB).
问题：如图所示，AB=c，AC=b，CB=a，a=8，b=15，c=17，BC垂直于AC。求tan(AB)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),c)
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(CB),a)
Equal(a,8)
Equal(b,15)
Equal(c,17)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Tan(MeasureOfAngle(ABC)))
Answer: 15/8
Theorem sequence:
cosine_theorem(1,BCA)
Problem: As shown in the diagram, AB=2*x+6, AD=4*y+6, BC=6*y-8, DC=3*x-2, BC and AD are opposite sides of the ▱ CBAD. Find the value of x.
问题：如图所示，AB=2*x+6，AD=4*y+6，BC=6*y-8，DC=3*x-2，BC和AD是平行四边形CBAD的一组对边。求x的值。
Construction (CDL):
Shape(CB,BA,AD,DC)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x+6)
Equal(LengthOfLine(AD),4*y+6)
Equal(LengthOfLine(BC),6*y-8)
Equal(LengthOfLine(DC),3*x-2)
Parallelogram(CBAD)
Goal: Value(x)
Answer: 8
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BADC)
Problem: As shown in the diagram, AB=16, AC=12, BA is perpendicular to CA. Find the perimeter of triangle CBA.
问题：如图所示，AB=16，AC=12，BA垂直于CA。求三角形CBA的周长。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
Equal(LengthOfLine(AC),12)
PerpendicularBetweenLine(BA,CA)
Goal: Value(PerimeterOfTriangle(CBA))
Answer: 48
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
triangle_perimeter_formula(1,CBA)
Problem: As shown in the diagram, AC=10, AD=x, BC=z, BD=y, CD=4, BD is perpendicular to AD, CA⊥BA. Find the value of z.
问题：如图所示，AC=10，AD=x，BC=z，BD=y，CD=4，BD⊥AD，CA⊥BA。求z的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BD),y)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(z)
Answer: 25
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADC)
mirror_similar_triangle_judgment_aa(1,ADC,BCA)
mirror_similar_triangle_property_line_ratio(1,ADC,BCA)
mirror_similar_triangle_property_line_ratio(1,DCA,ABC)
Problem: As shown in the diagram, PY=3, XZ=18, ZY=XW, XY is parallel to WZ. Find the length of line PW.
问题：如图所示，PY=3，XZ=18，ZY=XW，XY∥WZ。求直线PW的长度。
Construction (CDL):
Shape(XW,WP,PX)
Shape(PW,WZ,ZP)
Shape(XP,PY,YX)
Shape(PZ,ZY,YP)
Collinear(XPZ)
Collinear(WPY)
Conditions (CDL):
Equal(LengthOfLine(PY),3)
Equal(LengthOfLine(XZ),18)
Equal(LengthOfLine(ZY),LengthOfLine(XW))
ParallelBetweenLine(XY,WZ)
Goal: Value(LengthOfLine(PW))
Answer: 15
Theorem sequence:
trapezoid_judgment_parallel(1,XWZY)
isosceles_trapezoid_judgment_line_equal(1,XWZY)
isosceles_trapezoid_property_diagonal_equal(1,XWZY)
line_addition(1,WP,PY)
Problem: As shown in the diagram, ∠ABC=y°, ∠ACD=z°, ∠BCA=48°, ∠CAB=42°, ∠DAC=3*x°, AD is parallel to BC, BA∥CD. Find the value of x.
问题：如图所示，∠ABC=y°，∠ACD=z°，∠BCA=48°，∠CAB=42°，∠DAC=3*x°，AD平行于BC，BA平行于CD。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),y)
Equal(MeasureOfAngle(ACD),z)
Equal(MeasureOfAngle(BCA),48)
Equal(MeasureOfAngle(CAB),42)
Equal(MeasureOfAngle(DAC),3*x)
ParallelBetweenLine(AD,BC)
ParallelBetweenLine(BA,CD)
Goal: Value(x)
Answer: 16
Theorem sequence:
parallel_property_alternate_interior_angle(1,AD,BC)
Problem: As shown in the diagram, GK=HK, HK=KJ, ∠HGK=28°. Find the measure of ∠KJH.
问题：如图所示，GK=HK，HK=KJ，∠HGK=28°。求∠KJH的大小。
Construction (CDL):
Shape(HG,GK,KH)
Shape(HK,KJ,JH)
Collinear(GKJ)
Conditions (CDL):
Equal(LengthOfLine(GK),LengthOfLine(HK))
Equal(LengthOfLine(HK),LengthOfLine(KJ))
Equal(MeasureOfAngle(HGK),28)
Goal: Value(MeasureOfAngle(KJH))
Answer: 62
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,KHG)
isosceles_triangle_property_angle_equal(1,KHG)
triangle_property_angle_sum(1,KHG)
adjacent_complementary_angle(1,GKH,HKJ)
isosceles_triangle_judgment_line_equal(1,KJH)
isosceles_triangle_property_angle_equal(1,KJH)
triangle_property_angle_sum(1,KJH)
Problem: As shown in the diagram, ML=w, NM=2*y+5, NQ=3*x+2, QL=12, RM=4*x-2, RQ=3*y, RM and QN are opposite sides of the parallelogram MRQN. Find the length of line NQ.
问题：如图所示，ML=w，NM=2*y+5，NQ=3*x+2，QL=12，RM=4*x-2，RQ=3*y，RM和QN是平行四边形MRQN的一组对边。求直线NQ的长度。
Construction (CDL):
Shape(NM,ML,LN)
Shape(LM,MR,RL)
Shape(LR,RQ,QL)
Shape(NL,LQ,QN)
Collinear(NLR)
Collinear(MLQ)
Conditions (CDL):
Equal(LengthOfLine(ML),w)
Equal(LengthOfLine(NM),2*y+5)
Equal(LengthOfLine(NQ),3*x+2)
Equal(LengthOfLine(QL),12)
Equal(LengthOfLine(RM),4*x-2)
Equal(LengthOfLine(RQ),3*y)
Parallelogram(MRQN)
Goal: Value(LengthOfLine(NQ))
Answer: 14
Theorem sequence:
parallelogram_property_opposite_line_equal(1,MRQN)