SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, KJ=6, LK=12, RQ=4, RS=8, ∠JLK=30°, ∠JLK=∠QSR, ∠KJL=∠RQS, ∠LKJ=∠SRQ, ∠LKJ=y°, ∠QSR=x°, ∠RQS=80°, △KJL is similar to △RQS.. Find the value of y.
问题：如图所示，KJ=6，LK=12，RQ=4，RS=8，∠JLK=30°，∠JLK=∠QSR，∠KJL=∠RQS，∠LKJ=∠SRQ，∠LKJ=y°，∠QSR=x°，∠RQS=80°，△KJL与△RQS是相似三角形。求y的值。
Construction (CDL):
Shape(KJ,JL,LK)
Shape(RQ,QS,SR)
Conditions (CDL):
Equal(LengthOfLine(KJ),6)
Equal(LengthOfLine(LK),12)
Equal(LengthOfLine(RQ),4)
Equal(LengthOfLine(RS),8)
Equal(MeasureOfAngle(JLK),30)
Equal(MeasureOfAngle(JLK),MeasureOfAngle(QSR))
Equal(MeasureOfAngle(KJL),MeasureOfAngle(RQS))
Equal(MeasureOfAngle(LKJ),MeasureOfAngle(SRQ))
Equal(MeasureOfAngle(LKJ),y)
Equal(MeasureOfAngle(QSR),x)
Equal(MeasureOfAngle(RQS),80)
SimilarBetweenTriangle(KJL,RQS)
Goal: Value(y)
Answer: 70
Theorem sequence:
triangle_property_angle_sum(1,KJL)
Problem: As shown in the diagram, QP=14, P is the centroid of △QSU. Find the length of line QT.
问题：如图所示，QP=14，三角形QSU的重心为O。求直线QT的长度。
Construction (CDL):
Shape(QR,RP,PQ)
Shape(PR,RS,SP)
Shape(PS,ST,TP)
Shape(PT,TU,UP)
Shape(PU,UV,VP)
Shape(PV,VQ,QP)
Collinear(QRS)
Collinear(QPT)
Collinear(QVU)
Collinear(SPV)
Collinear(STU)
Collinear(UPR)
Conditions (CDL):
Equal(LengthOfLine(QP),14)
IsCentroidOfTriangle(P,QSU)
Goal: Value(LengthOfLine(QT))
Answer: 21
Theorem sequence:
centroid_of_triangle_property_line_ratio(1,P,QSU,T)
line_addition(1,QP,PT)
Problem: As shown in the diagram, AB=x, AD=3*sqrt(2), BC=y, BD=sqrt(2), CD=z, AB⊥CB, CD is perpendicular to BD. Find the value of y.
问题：如图所示，AB=x，AD=3*sqrt(2)，BC=y，BD=sqrt(2)，CD=z，AB⊥CB，CD⊥BD。求y的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),3*sqrt(2))
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),sqrt(2))
Equal(LengthOfLine(CD),z)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(CD,BD)
Goal: Value(y)
Answer: 2*sqrt(5)/3
Theorem sequence:
line_addition(1,AD,DC)
adjacent_complementary_angle(1,CDB,BDA)
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,ABC)
right_triangle_judgment_angle(1,BDA)
right_triangle_property_pythagorean(1,CDB)
right_triangle_property_pythagorean(1,ABC)
right_triangle_property_pythagorean(1,BDA)
Problem: As shown in the diagram, UY=32, UZ=x+6, WY=40, WZ=3*x-6, ∠UWZ=∠UYW, ZU is perpendicular to WU. Find the length of line WZ.
问题：如图所示，UY=32，UZ=x+6，WY=40，WZ=3*x-6，∠UWZ=∠UYW，ZU垂直于WU。求直线WZ的长度。
Construction (CDL):
Shape(WZ,ZU,UW)
Shape(WU,UY,YW)
Collinear(ZUY)
Conditions (CDL):
Equal(LengthOfLine(UY),32)
Equal(LengthOfLine(UZ),x+6)
Equal(LengthOfLine(WY),40)
Equal(LengthOfLine(WZ),3*x-6)
Equal(MeasureOfAngle(UWZ),MeasureOfAngle(UYW))
PerpendicularBetweenLine(ZU,WU)
Goal: Value(LengthOfLine(WZ))
Answer: 30
Theorem sequence:
adjacent_complementary_angle(1,ZUW,WUY)
right_triangle_judgment_angle(1,ZUW)
right_triangle_judgment_angle(1,WUY)
right_triangle_property_pythagorean(1,ZUW)
right_triangle_property_pythagorean(1,WUY)
Problem: As shown in the diagram, AB=YB, AB=x, YA=8, ∠BYA=∠YAB, ∠BYA=y°, AB is perpendicular to YB. Find the value of y.
问题：如图所示，AB=YB，AB=x，YA=8，∠BYA=∠YAB，∠BYA=y°，AB⊥YB。求y的值。
Construction (CDL):
Shape(AB,BY,YA)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(YB))
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(YA),8)
Equal(MeasureOfAngle(BYA),MeasureOfAngle(YAB))
Equal(MeasureOfAngle(BYA),y)
PerpendicularBetweenLine(AB,YB)
Goal: Value(y)
Answer: 45
Theorem sequence:
triangle_property_angle_sum(1,ABY)
Problem: As shown in the diagram, AE=x, BA=x, CB=12, EC=6, A is the center of ⊙A, CB is the tangent to circle A. Find the value of x.
问题：如图所示，AE=x，BA=x，CB=12，EC=6，A是⊙A的圆心，圆O的切线为CB。求x的值。
Construction (CDL):
Shape(BA,AE,ABE)
Shape(AEB,BA,AE)
Shape(EC,CB,AEB)
Collinear(AEC)
Cocircular(A,EB)
Conditions (CDL):
Equal(LengthOfLine(AE),x)
Equal(LengthOfLine(BA),x)
Equal(LengthOfLine(CB),12)
Equal(LengthOfLine(EC),6)
IsCentreOfCircle(A,A)
IsTangentOfCircle(CB,A)
Goal: Value(x)
Answer: 9
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CB,A,A)
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,CBA)
line_addition(1,AE,EC)
Problem: As shown in the diagram, SQ=SN, ∠QSR=7*x+9°, ∠RSM=3*x+11°, MS is the altitude of △ MQN. Find the value of x.
问题：如图所示，SQ=SN，∠QSR=7*x+9°，∠RSM=3*x+11°，MS是三角形MQN的高。求x的值。
Construction (CDL):
Shape(QS,SR,RQ)
Shape(RS,SM,MR)
Shape(MS,SN,NM)
Collinear(QRM)
Collinear(QSN)
Conditions (CDL):
Equal(LengthOfLine(SQ),LengthOfLine(SN))
Equal(MeasureOfAngle(QSR),7*x+9)
Equal(MeasureOfAngle(RSM),3*x+11)
IsAltitudeOfTriangle(MS,MQN)
Goal: Value(x)
Answer: 7
Theorem sequence:
angle_addition(1,QSR,RSM)
Problem: As shown in the diagram, BD=3, DA=1, AD and CB are opposite sides of the ▱ DACB, DA is perpendicular to CA. Find the area of quadrilateral DACB.
问题：如图所示，BD=3，DA=1，AD和CB是▱DACB的一组对边，DA垂直于CA。求DACB的面积。
Construction (CDL):
Shape(DA,AC,CB,BD)
Conditions (CDL):
Equal(LengthOfLine(BD),3)
Equal(LengthOfLine(DA),1)
Parallelogram(DACB)
PerpendicularBetweenLine(DA,CA)
Goal: Value(AreaOfQuadrilateral(DACB))
Answer: 3
Theorem sequence:
rectangle_judgment_right_angle(1,DACB)
parallelogram_area_formula_sine(1,BDAC)
Problem: As shown in the diagram, ∠CBA=36°, ∠CDE=57°, ∠DBE=42°, ∠ECD=28°. Find the measure of ∠EDB.
问题：如图所示，∠CBA=36°，∠CDE=57°，∠DBE=42°，∠ECD=28°。求∠EDB的大小。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BE,ED,DB)
Shape(EC,CD,DE)
Collinear(AED)
Collinear(CEB)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),36)
Equal(MeasureOfAngle(CDE),57)
Equal(MeasureOfAngle(DBE),42)
Equal(MeasureOfAngle(ECD),28)
Goal: Value(MeasureOfAngle(EDB))
Answer: 53
Theorem sequence:
triangle_property_angle_sum(1,BCD)
angle_addition(1,CDE,EDB)
Problem: As shown in the diagram, AB=16, AC=12, BC=x, BA⊥CA. Find the value of x.
问题：如图所示，AB=16，AC=12，BC=x，BA⊥CA。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 20
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, CA=CD, ∠BCA=3*x+13°, ∠ECB=2*x+7°, EC is the altitude of triangle EDA. Find the measure of ∠ECB.
问题：如图所示，CA=CD，∠BCA=3*x+13°，∠ECB=2*x+7°，EC是三角形EDA的高。求∠ECB的大小。
Construction (CDL):
Shape(AB,BC,CA)
Shape(CB,BE,EC)
Shape(CE,ED,DC)
Collinear(ABE)
Collinear(ACD)
Conditions (CDL):
Equal(LengthOfLine(CA),LengthOfLine(CD))
Equal(MeasureOfAngle(BCA),3*x+13)
Equal(MeasureOfAngle(ECB),2*x+7)
IsAltitudeOfTriangle(EC,EDA)
Goal: Value(MeasureOfAngle(ECB))
Answer: 35
Theorem sequence:
angle_addition(1,ECB,BCA)
Problem: As shown in the diagram, AB=15, AC=25, CB=20, CB⊥AB. Find the value of cos(AC).
问题：如图所示，AB=15，AC=25，CB=20，CB垂直于AB。求cos(AC)的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AC),25)
Equal(LengthOfLine(CB),20)
PerpendicularBetweenLine(CB,AB)
Goal: Value(Cos(MeasureOfAngle(ACB)))
Answer: 4/5
Theorem sequence:
cosine_theorem(1,CBA)
Problem: As shown in the diagram, ∠POM=43°, HP is parallel to NO, PO is parallel to HN. Find the measure of ∠FHG.
问题：如图所示，∠POM=43°，HP平行于NO，PO平行于HN。求∠FHG的大小。
Construction (CDL):
Shape(OP,PH,HN,NO)
Shape(PO,OM)
Shape(GH,HP)
Shape(FH,HG)
Collinear(KPOC)
Collinear(GHNA)
Collinear(BPHF)
Collinear(MONQ)
Conditions (CDL):
Equal(MeasureOfAngle(POM),43)
ParallelBetweenLine(HP,NO)
ParallelBetweenLine(PO,HN)
Goal: Value(MeasureOfAngle(FHG))
Answer: 137
Theorem sequence:
parallel_property_collinear_extend(1,PO,HN,K)
parallel_property_collinear_extend(1,ON,PH,M)
parallel_property_collinear_extend(2,NH,PK,G)
parallel_property_collinear_extend(2,HP,OM,B)
parallel_property_corresponding_angle(1,PB,OM,K)
parallel_property_corresponding_angle(2,HG,PK,B)
flat_angle(1,FHP)
angle_addition(1,FHG,GHP)
Problem: As shown in the diagram, CX=4*x+12, EC=12*x-26, ∠BCX=57°, ∠CBE=33°, ∠ECB=57°, ∠XBC=33°. Find the value of x.
问题：如图所示，CX=4*x+12，EC=12*x-26，∠BCX=57°，∠CBE=33°，∠ECB=57°，∠XBC=33°。求x的值。
Construction (CDL):
Shape(XB,BC,CX)
Shape(CB,BE,EC)
Conditions (CDL):
Equal(LengthOfLine(CX),4*x+12)
Equal(LengthOfLine(EC),12*x-26)
Equal(MeasureOfAngle(BCX),57)
Equal(MeasureOfAngle(CBE),33)
Equal(MeasureOfAngle(ECB),57)
Equal(MeasureOfAngle(XBC),33)
Goal: Value(x)
Answer: 19/4
Theorem sequence:
mirror_congruent_triangle_judgment_aas(2,XBC,ECB)
mirror_congruent_triangle_property_line_equal(1,BCX,BEC)
Problem: As shown in the diagram, AF=5, AH=7, BF=x, EH=7, C is the center of circle C, the center of ⊙D is D, the tangent to ⊙C is AG, the tangent to circle D is AG. Find the value of x.
问题：如图所示，AF=5，AH=7，BF=x，EH=7，⊙C的圆心为C，⊙D的圆心为D，AG是圆O的切线，AG是⊙O的切线。求x的值。
Construction (CDL):
Shape(DBF,FB)
Shape(BF,DFG,DGB)
Shape(FA,AG,DFG)
Shape(GA,AH,CGH)
Shape(CHE,EH)
Shape(HE,CEG,CGH)
Collinear(BFA)
Collinear(EHA)
Cocircular(D,BFG)
Cocircular(C,GHE)
Conditions (CDL):
Equal(LengthOfLine(AF),5)
Equal(LengthOfLine(AH),7)
Equal(LengthOfLine(BF),x)
Equal(LengthOfLine(EH),7)
IsCentreOfCircle(C,C)
IsCentreOfCircle(D,D)
IsTangentOfCircle(AG,C)
IsTangentOfCircle(AG,D)
Goal: Value(x)
Answer: 73/5
Theorem sequence:
line_addition(1,BF,FA)
line_addition(1,EH,HA)
circle_property_circular_power_tangent_and_segment_line(1,AG,AFB,D)
circle_property_circular_power_tangent_and_segment_line(1,AG,AHE,C)
Problem: As shown in the diagram, BE=ED, the measure of arc NED is 120. Find the measure of ⌒NBE.
问题：如图所示，BE=ED，⌒NED的角度为120。求弧NBE的角度。
Construction (CDL):
Shape(NBE,EB)
Shape(NED,DE)
Shape(BE,ED,NDB)
Cocircular(N,BED)
Conditions (CDL):
Equal(LengthOfLine(BE),LengthOfLine(ED))
Equal(MeasureOfArc(NED),120)
Goal: Value(MeasureOfArc(NBE))
Answer: 120
Theorem sequence:
congruent_arc_judgment_chord_equal(1,NBE,NED)
congruent_arc_property_measure_equal(1,NBE,NED)
Problem: As shown in the diagram, Add(AreaOfQuadrilateral(ECGB)=AreaOfQuadrilateral(GDAF)), AD=20, BF=11, CG=6, DC=9, EC=9, FA=3, ECGB is a rectangle, quadrilateral GDAF is a rectangle. Find the value of x.
问题：如图所示，Add(AreaOfQuadrilateral(ECGB)=AreaOfQuadrilateral(GDAF))，AD=20，BF=11，CG=6，DC=9，EC=9，FA=3，ECGB是长方形，四边形GDAF是矩形。求x的值。
Construction (CDL):
Shape(EC,CG,GB,BE)
Shape(BG,GD,DA,AF,FB)
Collinear(GBF)
Collinear(CGD)
Conditions (CDL):
Equal(Add(AreaOfQuadrilateral(ECGB),AreaOfQuadrilateral(GDAF)),x)
Equal(LengthOfLine(AD),20)
Equal(LengthOfLine(BF),11)
Equal(LengthOfLine(CG),6)
Equal(LengthOfLine(DC),9)
Equal(LengthOfLine(EC),9)
Equal(LengthOfLine(FA),3)
Rectangle(ECGB)
Rectangle(GDAF)
Goal: Value(x)
Answer: 114
Theorem sequence:
line_addition(1,CG,GD)
parallelogram_area_formula_sine(1,ECGB)
parallelogram_area_formula_sine(1,GDAF)
Problem: As shown in the diagram, AB=10, AC=x, DC=x+4, the center of circle E is E, the tangent to ⊙E is AB. Find the value of x.
问题：如图所示，AB=10，AC=x，DC=x+4，E是⊙E的圆心，AB是⊙O的切线。求x的值。
Construction (CDL):
Shape(CA,AB,ECB)
Shape(EDC,CD)
Shape(DC,ECB,EBD)
Collinear(ACD)
Cocircular(E,CBD)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(DC),x+4)
IsCentreOfCircle(E,E)
IsTangentOfCircle(AB,E)
Goal: Value(x)
Answer: -1+sqrt(51)
Theorem sequence:
line_addition(1,AC,CD)
circle_property_circular_power_tangent_and_segment_line(1,AB,ACD,E)
Problem: As shown in the diagram, FG=18, HG=x-4, HJ=x-5, JK=15, GJ is parallel to FK. Find the value of x.
问题：如图所示，FG=18，HG=x-4，HJ=x-5，JK=15，GJ平行于FK。求x的值。
Construction (CDL):
Shape(HG,GJ,JH)
Shape(GF,FK,KJ,JG)
Collinear(HGF)
Collinear(HJK)
Conditions (CDL):
Equal(LengthOfLine(FG),18)
Equal(LengthOfLine(HG),x-4)
Equal(LengthOfLine(HJ),x-5)
Equal(LengthOfLine(JK),15)
ParallelBetweenLine(GJ,FK)
Goal: Value(x)
Answer: 10
Theorem sequence:
parallel_property_corresponding_angle(1,GJ,FK,H)
similar_triangle_judgment_aa(1,JHG,KHF)
line_addition(1,HG,GF)
line_addition(1,HJ,JK)
similar_triangle_property_line_ratio(1,JHG,KHF)
similar_triangle_property_line_ratio(1,GJH,FKH)
Problem: As shown in the diagram, AB=3, AC=1, quadrilateral ACDB is a rectangle. Find the perimeter of ACDB.
问题：如图所示，AB=3，AC=1，ACDB是矩形。求四边形ACDB的周长。
Construction (CDL):
Shape(BA,AC,CD,DB)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),1)
Rectangle(ACDB)
Goal: Value(PerimeterOfQuadrilateral(ACDB))
Answer: 8
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ACDB)
parallelogram_property_opposite_line_equal(1,CDBA)
quadrilateral_perimeter_formula(1,BACD)
Problem: As shown in the diagram, AB=x, AD=3*sqrt(3), BD=9, CD=y, ∠ABC=30°, ∠BCA=60°, AD is perpendicular to CD, CA⊥BA. Find the value of y.
问题：如图所示，AB=x，AD=3*sqrt(3)，BD=9，CD=y，∠ABC=30°，∠BCA=60°，AD垂直于CD，CA⊥BA。求y的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(DA,AB,BD)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),3*sqrt(3))
Equal(LengthOfLine(BD),9)
Equal(LengthOfLine(CD),y)
Equal(MeasureOfAngle(ABC),30)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 3
Theorem sequence:
triangle_property_angle_sum(1,ADC)
sine_theorem(1,ADC)
sine_theorem(1,CAD)
Problem: As shown in the diagram, AB=x, AC=4, BC=8, CA⊥BA. Find the value of x.
问题：如图所示，AB=x，AC=4，BC=8，CA垂直于BA。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BC),8)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 4*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, JK=17/2, KL=4, KM=5, KL⊥ML. Find the area of △JMK.
问题：如图所示，JK=17/2，KL=4，KM=5，KL垂直于ML。求△JMK的面积。
Construction (CDL):
Shape(JL,LK,KJ)
Shape(KL,LM,MK)
Collinear(JLM)
Conditions (CDL):
Equal(LengthOfLine(JK),17/2)
Equal(LengthOfLine(KL),4)
Equal(LengthOfLine(KM),5)
PerpendicularBetweenLine(KL,ML)
Goal: Value(AreaOfTriangle(JMK))
Answer: 21
Theorem sequence:
adjacent_complementary_angle(1,JLK,KLM)
right_triangle_judgment_angle(1,KLM)
right_triangle_judgment_angle(1,JLK)
right_triangle_property_pythagorean(1,JLK)
right_triangle_property_pythagorean(1,KLM)
line_addition(1,JL,LM)
altitude_of_triangle_judgment(1,KL,KJM)
triangle_area_formula_common(1,KJM)
Problem: As shown in the diagram, BC=12, △BAC is an equilateral △. Find the area of △BAC.
问题：如图所示，BC=12，三角形BAC为等边三角形。求三角形BAC的面积。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(BC),12)
EquilateralTriangle(BAC)
Goal: Value(AreaOfTriangle(BAC))
Answer: 36*sqrt(3)
Theorem sequence:
cosine_theorem(1,BAC)
triangle_area_formula_sine(1,BAC)
Problem: As shown in the diagram, ∠ADB=9*y-2°, ∠CEA=135°, ∠DAE=5*y+5°, ∠EDA=4*z+9°. Find the value of y.
问题：如图所示，∠ADB=9*y-2°，∠CEA=135°，∠DAE=5*y+5°，∠EDA=4*z+9°。求y的值。
Construction (CDL):
Shape(AE,ED,DA)
Shape(CE,EA)
Shape(AD,DB)
Collinear(CEDB)
Conditions (CDL):
Equal(MeasureOfAngle(ADB),9*y-2)
Equal(MeasureOfAngle(CEA),135)
Equal(MeasureOfAngle(DAE),5*y+5)
Equal(MeasureOfAngle(EDA),4*z+9)
Goal: Value(y)
Answer: 13
Theorem sequence:
flat_angle(1,CED)
angle_addition(1,CEA,AED)
triangle_property_angle_sum(1,AED)
flat_angle(1,EDB)
angle_addition(1,EDA,ADB)
Problem: As shown in the diagram, AD=62, AO=54, EC=27, x=Add(AreaOfTriangle(CBO), BE is perpendicular to CE, BDAO is a rectangle. Find the value of x.
问题：如图所示，AD=62，AO=54，EC=27，x=Add(AreaOfTriangle(CBO)，BE⊥CE，四边形BDAO是长方形。求x的值。
Construction (CDL):
Shape(BD,DA,AO,OB)
Shape(CB,BE,EC)
Shape(CE,EO,OC)
Collinear(BEO)
Conditions (CDL):
Equal(LengthOfLine(AD),62)
Equal(LengthOfLine(AO),54)
Equal(LengthOfLine(EC),27)
Equal(x,Add(AreaOfTriangle(CBO),AreaOfQuadrilateral(BDAO)))
PerpendicularBetweenLine(BE,CE)
Rectangle(BDAO)
Goal: Value(x)
Answer: 4185
Theorem sequence:
parallelogram_property_opposite_line_equal(1,BDAO)
parallelogram_property_opposite_line_equal(1,DAOB)
altitude_of_triangle_judgment(1,CE,CBO)
triangle_area_formula_common(1,CBO)
parallelogram_area_formula_sine(1,BDAO)
Problem: As shown in the diagram, AC=BA, ∠ABC=3*x+8°, ∠CAB=2*x+20°. Find the value of x.
问题：如图所示，AC=BA，∠ABC=3*x+8°，∠CAB=2*x+20°。求x的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(BA))
Equal(MeasureOfAngle(ABC),3*x+8)
Equal(MeasureOfAngle(CAB),2*x+20)
Goal: Value(x)
Answer: 18
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, the area of ABLT is 147, the area of quadrilateral CEFD is 27, EH=x, GB=14, BG is perpendicular to LG, EH is perpendicular to FH, quadrilateral CEFD is similar to quadrilateral ABLT. Find The ratio of similar quadrilaterals CEFD and ABLT.
问题：如图所示，ABLT的面积为147，CEFD的面积为27，EH=x，GB=14，BG垂直于LG，EH垂直于FH，CEFD与ABLT相似。求四边形CEFD和四边形ABLT的相似比。
Construction (CDL):
Shape(CE,EF,FD,DC)
Shape(EH,HF,FE)
Shape(AB,BL,LT,TA)
Shape(BG,GL,LB)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ABLT),147)
Equal(AreaOfQuadrilateral(CEFD),27)
Equal(LengthOfLine(EH),x)
Equal(LengthOfLine(GB),14)
PerpendicularBetweenLine(BG,LG)
PerpendicularBetweenLine(EH,FH)
SimilarBetweenQuadrilateral(CEFD,ABLT)
Goal: Value(RatioOfSimilarQuadrilateral(CEFD,ABLT))
Answer: 3/7
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,CEFD,ABLT)
Problem: As shown in the diagram, BA=4*y-11, CA=2*y+3, DA=3*y+5, EA=2*x+4, CE and DB are opposite sides of the ▱ CDBE. Find the value of x.
问题：如图所示，BA=4*y-11，CA=2*y+3，DA=3*y+5，EA=2*x+4，DC和BE是▱CDBE的一组对边。求x的值。
Construction (CDL):
Shape(CD,DA,AC)
Shape(AD,DB,BA)
Shape(AB,BE,EA)
Shape(CA,AE,EC)
Collinear(DAE)
Collinear(BAC)
Conditions (CDL):
Equal(LengthOfLine(BA),4*y-11)
Equal(LengthOfLine(CA),2*y+3)
Equal(LengthOfLine(DA),3*y+5)
Equal(LengthOfLine(EA),2*x+4)
Parallelogram(CDBE)
Goal: Value(x)
Answer: 11
Theorem sequence:
parallelogram_property_diagonal_bisection(1,CDBE,A)
parallelogram_property_diagonal_bisection(1,DBEC,A)
Problem: As shown in the diagram, the measure of arc EBG is x, the measure of ⌒EGI is 100, the measure of arc EIJ is 120, the measure of ⌒EJB is 90, the center of circle E is E. Find the value of x.
问题：如图所示，弧EBG的角度为x，⌒EGI的角度为100，弧EIJ的角度为120，弧EJB的角度为90，圆E的圆心为E。求x的值。
Construction (CDL):
Shape(EGI,IG)
Shape(GI,IE,EG)
Shape(EI,EIJ,JE)
Shape(EJ,JB,BE)
Shape(GE,EB,EBG)
Shape(BJ,EJB)
Cocircular(E,IJBG)
Conditions (CDL):
Equal(MeasureOfArc(EBG),x)
Equal(MeasureOfArc(EGI),100)
Equal(MeasureOfArc(EIJ),120)
Equal(MeasureOfArc(EJB),90)
IsCentreOfCircle(E,E)
Goal: Value(x)
Answer: 50
Theorem sequence:
arc_addition_measure(1,EIJ,EJB)
arc_addition_measure(1,EGI,EIB)
round_arc(1,EBG,EGB)
Problem: As shown in the diagram, AB=7, BC=4, ∠CBA=29°, CD⊥AD. Find the area of triangle CBA.
问题：如图所示，AB=7，BC=4，∠CBA=29°，CD垂直于AD。求三角形CBA的面积。
Construction (CDL):
Shape(CB,BD,DC)
Shape(CD,DA,AC)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(AB),7)
Equal(LengthOfLine(BC),4)
Equal(MeasureOfAngle(CBA),29)
PerpendicularBetweenLine(CD,AD)
Goal: Value(AreaOfTriangle(CBA))
Answer: 14*sin(29*pi/180)
Theorem sequence:
adjacent_complementary_angle(1,BDC,CDA)
sine_theorem(1,CBD)
altitude_of_triangle_judgment(1,CD,CBA)
triangle_area_formula_common(1,CBA)
Problem: As shown in the diagram, ∠FEU=8*x+4°, ∠FTV=9*x-11°, DA is parallel to CV. Find the value of x.
问题：如图所示，∠FEU=8*x+4°，∠FTV=9*x-11°，DA∥CV。求x的值。
Construction (CDL):
Shape(FE,EU)
Shape(FT,TV)
Collinear(DEUA)
Collinear(CTV)
Collinear(FRETB)
Conditions (CDL):
Equal(MeasureOfAngle(FEU),8*x+4)
Equal(MeasureOfAngle(FTV),9*x-11)
ParallelBetweenLine(DA,CV)
Goal: Value(x)
Answer: 15
Theorem sequence:
parallel_property_collinear_extend(3,DA,CV,E)
parallel_property_collinear_extend(3,VC,AE,T)
parallel_property_corresponding_angle(1,EA,TV,F)
Problem: As shown in the diagram, AC=40, BD=Mul(LengthOfLine(AC),3/5), BE=3*x, BC and AD are opposite sides of the parallelogram BADC. Find the value of x.
问题：如图所示，AC=40，BD=Mul(LengthOfLine(AC),3/5)，BE=3*x，BADC是平行四边形。求x的值。
Construction (CDL):
Shape(BA,AE,EB)
Shape(EA,AD,DE)
Shape(ED,DC,CE)
Shape(BE,EC,CB)
Collinear(AEC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(AC),40)
Equal(LengthOfLine(BD),Mul(LengthOfLine(AC),3/5))
Equal(LengthOfLine(BE),3*x)
Parallelogram(BADC)
Goal: Value(x)
Answer: 4
Theorem sequence:
parallelogram_property_diagonal_bisection(1,BADC,E)
line_addition(1,BE,ED)
Problem: As shown in the diagram, VW=3, XV=2*a-2, ZV=5/4*a+1, ∠ZYW=53°, quadrilateral XWZY is a rhombus. Find the length of line XW.
问题：如图所示，VW=3，XV=2*a-2，ZV=5/4*a+1，∠ZYW=53°，四边形XWZY是菱形。求直线XW的长度。
Construction (CDL):
Shape(XW,WV,VX)
Shape(YX,XV,VY)
Shape(VW,WZ,ZV)
Shape(YV,VZ,ZY)
Collinear(XVZ)
Collinear(YVW)
Conditions (CDL):
Equal(LengthOfLine(VW),3)
Equal(LengthOfLine(XV),2*a-2)
Equal(LengthOfLine(ZV),5/4*a+1)
Equal(MeasureOfAngle(ZYW),53)
Rhombus(XWZY)
Goal: Value(LengthOfLine(XW))
Answer: 3*sqrt(5)
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,WZYX,V)
right_triangle_judgment_angle(1,WVX)
right_triangle_property_pythagorean(1,WVX)
Problem: As shown in the diagram, AB=30, CD=30, the measure of arc XCZ is 40, the center of circle X is X, AM is perpendicular to YM, DN is perpendicular to ZN. Find the length of line MB.
问题：如图所示，AB=30，CD=30，⌒XCZ的角度为40，X是⊙X的圆心，AM⊥YM，DN⊥ZN。求直线MB的长度。
Construction (CDL):
Shape(AM,MY,XYA)
Shape(YM,MB,XBY)
Shape(XM,MA,XAC,CN,NX)
Shape(NC,XCZ,ZN)
Shape(DN,NZ,XZD)
Shape(BM,MX,XN,ND,XDB)
Collinear(AMB)
Collinear(CND)
Collinear(YMX)
Collinear(XNZ)
Cocircular(X,YACZDB)
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(LengthOfLine(MB))
Answer: 15
Theorem sequence:
adjacent_complementary_angle(1,AMY,YMB)
adjacent_complementary_angle(1,YMB,BMX)
line_addition(1,AM,MB)
circle_property_chord_perpendicular_bisect_chord(1,X,XM,BA)
Problem: As shown in the diagram, AD=x, AE=12, DE=9, DE⊥AE. Find the value of x.
问题：如图所示，AD=x，AE=12，DE=9，DE⊥AE。求x的值。
Construction (CDL):
Shape(DB,BE,ED)
Shape(EB,BC,CE)
Shape(EC,CA,AE)
Shape(EA,AD,DE)
Collinear(DEC)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(AE),12)
Equal(LengthOfLine(DE),9)
PerpendicularBetweenLine(DE,AE)
Goal: Value(x)
Answer: 15
Theorem sequence:
right_triangle_judgment_angle(1,DEA)
right_triangle_property_pythagorean(1,DEA)
Problem: As shown in the diagram, AI=7, BH=3, IB and IH are one pair of adjacent sides of the kite ABIH. Find the area of ABIH.
问题：如图所示，AI=7，BH=3，ABIH是风筝形。求四边形ABIH的面积。
Construction (CDL):
Shape(AB,BH,HA)
Shape(BI,IH,HB)
Collinear(AOI)
Collinear(BOH)
Conditions (CDL):
Equal(LengthOfLine(AI),7)
Equal(LengthOfLine(BH),3)
Kite(ABIH)
Goal: Value(AreaOfQuadrilateral(ABIH))
Answer: 21/2
Theorem sequence:
kite_area_formula_diagonal(1,ABIH)
Problem: As shown in the diagram, QY=6, RQ=4, RT=5, YX=2, RT is the altitude of triangle RYS, XV is the altitude of triangle XWQ, RT is perpendicular to ST, SR is perpendicular to YR, WV⊥XV, triangle SRY is similar to triangle QXW.. Find the length of line QV.
问题：如图所示，QY=6，RQ=4，RT=5，YX=2，RT是△RYS的高，XV是三角形XWQ的高，RT垂直于ST，SR垂直于YR，WV⊥XV，△SRY与△QXW是相似三角形。求直线QV的长度。
Construction (CDL):
Shape(WV,VX,XW)
Shape(XV,VA,AY,YX)
Shape(YA,AQ,QY)
Shape(AT,TR,RQ,QA)
Shape(TS,SR,RT)
Collinear(XYQR)
Collinear(WVAQ)
Collinear(STAY)
Conditions (CDL):
Equal(LengthOfLine(QY),6)
Equal(LengthOfLine(RQ),4)
Equal(LengthOfLine(RT),5)
Equal(LengthOfLine(YX),2)
IsAltitudeOfTriangle(RT,RYS)
IsAltitudeOfTriangle(XV,XWQ)
PerpendicularBetweenLine(RT,ST)
PerpendicularBetweenLine(SR,YR)
PerpendicularBetweenLine(WV,XV)
SimilarBetweenTriangle(SRY,QXW)
Goal: Value(LengthOfLine(QV))
Answer: 4
Theorem sequence:
triangle_property_angle_sum(1,SRY)
triangle_property_angle_sum(1,RYT)
similar_triangle_property_angle_equal(1,SRY,QXW)
similar_triangle_judgment_aa(1,XVQ,YTR)
line_addition(1,YQ,QR)
line_addition(1,XY,YQ)
similar_triangle_property_line_ratio(1,VQX,TRY)
similar_triangle_property_line_ratio(1,XVQ,YTR)
Problem: As shown in the diagram, AD=BC, ∠DCB=101°, AB is parallel to DC. Find the measure of ∠ADC.
问题：如图所示，AD=BC，∠DCB=101°，AB平行于DC。求∠ADC的大小。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(BC))
Equal(MeasureOfAngle(DCB),101)
ParallelBetweenLine(AB,DC)
Goal: Value(MeasureOfAngle(ADC))
Answer: 101
Theorem sequence:
trapezoid_judgment_parallel(1,ADCB)
isosceles_trapezoid_judgment_line_equal(1,ADCB)
isosceles_trapezoid_property_angle_equal(1,ADCB)