SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=5, AC=7, BC=6, DE=3, △ABC is similar to △DEF.. Find the perimeter of triangle DEF.
问题：如图所示，AB=5，AC=7，BC=6，DE=3，三角形ABC与三角形DEF是相似三角形。求三角形DEF的周长。
Construction (CDL):
Shape(AB,BC,CA)
Shape(DE,EF,FD)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AC),7)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(DE),3)
SimilarBetweenTriangle(ABC,DEF)
Goal: Value(PerimeterOfTriangle(DEF))
Answer: 54/5
Theorem sequence:
similar_triangle_property_line_ratio(1,CAB,FDE)
similar_triangle_property_line_ratio(1,BCA,EFD)
similar_triangle_property_line_ratio(1,ABC,DEF)
triangle_perimeter_formula(1,DEF)
Problem: As shown in the diagram, JK=2*b+3, JM=3*a, LK=21, LM=45, ∠MLR=30°, ∠RLK=70°, JMLK is a parallelogram. Find the measure of ∠KJM.
问题：如图所示，JK=2*b+3，JM=3*a，LK=21，LM=45，∠MLR=30°，∠RLK=70°，JMLK是▱。求∠KJM的大小。
Construction (CDL):
Shape(JM,MR,RJ)
Shape(RM,ML,LR)
Shape(RL,LK,KR)
Shape(JR,RK,KJ)
Collinear(MRK)
Collinear(JRL)
Conditions (CDL):
Equal(LengthOfLine(JK),2*b+3)
Equal(LengthOfLine(JM),3*a)
Equal(LengthOfLine(LK),21)
Equal(LengthOfLine(LM),45)
Equal(MeasureOfAngle(MLR),30)
Equal(MeasureOfAngle(RLK),70)
Parallelogram(JMLK)
Goal: Value(MeasureOfAngle(KJM))
Answer: 100
Theorem sequence:
angle_addition(1,MLR,RLK)
parallelogram_property_opposite_angle_equal(1,JMLK)
Problem: As shown in the diagram, CA=10, CM=6, AC⊥BC, BM⊥CM. Find the perimeter of △ACB.
问题：如图所示，CA=10，CM=6，AC⊥BC，BM⊥CM。求△ACB的周长。
Construction (CDL):
Shape(AC,CM,MA)
Shape(MC,CB,BM)
Collinear(AMB)
Conditions (CDL):
Equal(LengthOfLine(CA),10)
Equal(LengthOfLine(CM),6)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BM,CM)
Goal: Value(PerimeterOfTriangle(ACB))
Answer: 30
Theorem sequence:
adjacent_complementary_angle(1,BMC,CMA)
right_triangle_judgment_angle(1,CMA)
right_triangle_property_pythagorean(1,CMA)
line_addition(1,AM,MB)
right_triangle_judgment_angle(1,BMC)
right_triangle_property_pythagorean(1,BMC)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
triangle_perimeter_formula(1,ACB)
Problem: As shown in the diagram, AC=6, AF=5, BE=GE, BG=6, FD=12, FH=6, GA=6, GC=CA, GE=GC, GH⊥FH, HF is perpendicular to AF, quadrilateral BDFA is a rectangle. Find the sum of the area of triangle EBG, the area of triangle CGA and the area of quadrilateral BDFA.
问题：如图所示，AC=6，AF=5，BE=GE，BG=6，FD=12，FH=6，GA=6，GC=CA，GE=GC，GH⊥FH，HF⊥AF，BDFA是矩形。求△EBG的面积、△CGA的面积与四边形BDFA的面积之和。
Construction (CDL):
Shape(EB,BG,GE)
Shape(CG,GA,AC)
Shape(BD,DH,HG,GB)
Shape(GH,HF,FA,AG)
Collinear(BGA)
Collinear(DHF)
Conditions (CDL):
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(AF),5)
Equal(LengthOfLine(BE),LengthOfLine(GE))
Equal(LengthOfLine(BG),6)
Equal(LengthOfLine(FD),12)
Equal(LengthOfLine(FH),6)
Equal(LengthOfLine(GA),6)
Equal(LengthOfLine(GC),LengthOfLine(CA))
Equal(LengthOfLine(GE),LengthOfLine(GC))
PerpendicularBetweenLine(GH,FH)
PerpendicularBetweenLine(HF,AF)
Rectangle(BDFA)
Goal: Value(Add(AreaOfTriangle(EBG),AreaOfTriangle(CGA),AreaOfQuadrilateral(BDFA)))
Answer: 18*sqrt(3)+60
Theorem sequence:
parallelogram_area_formula_sine(1,DFAB)
isosceles_triangle_judgment_line_equal(1,EBG)
isosceles_triangle_judgment_line_equal(1,BGE)
equilateral_triangle_judgment_isosceles_and_isosceles(1,EBG)
equilateral_triangle_property_angle(1,EBG)
triangle_area_formula_sine(1,EBG)
isosceles_triangle_judgment_line_equal(1,CGA)
isosceles_triangle_judgment_line_equal(1,GAC)
equilateral_triangle_judgment_isosceles_and_isosceles(1,CGA)
equilateral_triangle_property_angle(1,CGA)
triangle_area_formula_sine(1,CGA)
Problem: As shown in the diagram, the area of quadrilateral BCDA is 100, AB=x+15, BE=x, quadrilateral BCDA is a parallelogram, CE is perpendicular to BE. Find the length of line CD.
问题：如图所示，四边形BCDA的面积为100，AB=x+15，BE=x，四边形BCDA是平行四边形，CE垂直于BE。求直线CD的长度。
Construction (CDL):
Shape(BC,CE,EB)
Shape(BE,ED,DA,AB)
Collinear(CED)
Conditions (CDL):
Equal(AreaOfQuadrilateral(BCDA),100)
Equal(LengthOfLine(AB),x+15)
Equal(LengthOfLine(BE),x)
Parallelogram(BCDA)
PerpendicularBetweenLine(CE,BE)
Goal: Value(LengthOfLine(CD))
Answer: 20
Theorem sequence:
altitude_of_quadrilateral_judgment_left_vertex(1,BE,BCDA)
parallelogram_area_formula_common(1,BCDA)
parallelogram_property_opposite_line_equal(1,CDAB)
Problem: As shown in the diagram, ∠ABC=42°, ∠ADC=100°, ∠DAB=x°. Find the value of x.
问题：如图所示，∠ABC=42°，∠ADC=100°，∠DAB=x°。求x的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC)
Collinear(BDC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),42)
Equal(MeasureOfAngle(ADC),100)
Equal(MeasureOfAngle(DAB),x)
Goal: Value(x)
Answer: 58
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADC)
triangle_property_angle_sum(1,ABD)
Problem: As shown in the diagram, AB=3/5, AC=x, BC=1/5, AB⊥CB. Find the value of x.
问题：如图所示，AB=3/5，AC=x，BC=1/5，AB⊥CB。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),3/5)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),1/5)
PerpendicularBetweenLine(AB,CB)
Goal: Value(x)
Answer: sqrt(10)/5
Theorem sequence:
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,ABC)
Problem: As shown in the diagram, AB=x, AC=8, BC=6, AC⊥BC. Find the value of x.
问题：如图所示，AB=x，AC=8，BC=6，AC垂直于BC。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),6)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 10
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, ∠AIL=4*x+6°, ∠CME=z°, ∠GLP=142°, ∠ILG=2*y+8°, IA is parallel to LG, MN∥IA. Find the value of x.
问题：如图所示，∠AIL=4*x+6°，∠CME=z°，∠GLP=142°，∠ILG=2*y+8°，IA平行于LG，MN∥IA。求x的值。
Construction (CDL):
Shape(CM,ME)
Shape(EM,MN)
Shape(Im,MC)
Shape(NM,MI)
Shape(FI,IM)
Shape(MI,IA)
Shape(LI,IF)
Shape(AI,IL)
Shape(QL,LI)
Shape(PL,LQ)
Shape(IL,LG)
Shape(GL,LP)
Collinear(QLG)
Collinear(FIA)
Collinear(CMN)
Collinear(EMILP)
Conditions (CDL):
Equal(MeasureOfAngle(AIL),4*x+6)
Equal(MeasureOfAngle(CME),z)
Equal(MeasureOfAngle(GLP),142)
Equal(MeasureOfAngle(ILG),2*y+8)
ParallelBetweenLine(IA,LG)
ParallelBetweenLine(MN,IA)
Goal: Value(x)
Answer: 34
Theorem sequence:
parallel_property_corresponding_angle(2,IA,LG,P)
Problem: As shown in the diagram, the measure of arc FCB is 146, F is the center of circle F, AB is the tangent to ⊙F. Find the measure of ∠ABC.
问题：如图所示，弧FCB的角度为146，⊙F的圆心为F，AB是⊙O的切线。求∠ABC的大小。
Construction (CDL):
Shape(CB,BF,FC)
Shape(CF,FB,FBC)
Shape(BC,FCB)
Shape(AB,BC)
Shape(FB,BD)
Collinear(ABD)
Cocircular(F,BC)
Conditions (CDL):
Equal(MeasureOfArc(FCB),146)
IsCentreOfCircle(F,F)
IsTangentOfCircle(AB,F)
Goal: Value(MeasureOfAngle(ABC))
Answer: 73
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AB,F,F)
arc_property_center_angle(1,FCB,F)
radius_of_circle_property_length_equal(1,FB,F)
radius_of_circle_property_length_equal(1,FC,F)
isosceles_triangle_judgment_line_equal(1,FCB)
isosceles_triangle_property_angle_equal(1,FCB)
triangle_property_angle_sum(1,FCB)
angle_addition(1,ABC,CBF)
Problem: As shown in the diagram, AB=25, AC=15, BC=19, ∠ACB=x°. Find the value of x.
问题：如图所示，AB=25，AC=15，BC=19，∠ACB=x°。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),25)
Equal(LengthOfLine(AC),15)
Equal(LengthOfLine(BC),19)
Equal(MeasureOfAngle(ACB),x)
Goal: Value(x)
Answer: 180*acos(-13/190)/pi
Theorem sequence:
cosine_theorem(1,CBA)
Problem: As shown in the diagram, AC=y, AD=10, BC=AB, CD=AD, CD=BC, ∠ACB=x°, CB⊥AB. Find the value of y.
问题：如图所示，AC=y，AD=10，BC=AB，CD=AD，CD=BC，∠ACB=x°，CB垂直于AB。求y的值。
Construction (CDL):
Shape(CB,BA,AC)
Shape(CA,AD,DC)
Conditions (CDL):
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BC),LengthOfLine(AB))
Equal(LengthOfLine(CD),LengthOfLine(AD))
Equal(LengthOfLine(CD),LengthOfLine(BC))
Equal(MeasureOfAngle(ACB),x)
PerpendicularBetweenLine(CB,AB)
Goal: Value(y)
Answer: 10*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, JK=x, JL=5, KL=4, the center of ⊙L is L, JK is the tangent to ⊙L. Find the value of x.
问题：如图所示，JK=x，JL=5，KL=4，⊙L的圆心为L，圆O的切线为JK。求x的值。
Construction (CDL):
Shape(CL,LK,LKC)
Shape(KL,LC,LCK)
Shape(KJ,JC,LKC)
Collinear(JCL)
Cocircular(L,CK)
Conditions (CDL):
Equal(LengthOfLine(JK),x)
Equal(LengthOfLine(JL),5)
Equal(LengthOfLine(KL),4)
IsCentreOfCircle(L,L)
IsTangentOfCircle(JK,L)
Goal: Value(x)
Answer: 3
Theorem sequence:
tangent_of_circle_property_perpendicular(1,JK,L,L)
right_triangle_judgment_angle(1,LKJ)
right_triangle_property_pythagorean(1,LKJ)
Problem: As shown in the diagram, AB=19, AC=x, BC=y, ∠CBA=45°, AC is perpendicular to BC. Find the value of y.
问题：如图所示，AB=19，AC=x，BC=y，∠CBA=45°，AC垂直于BC。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),19)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(y)
Answer: 19*sqrt(2)/2
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,BAC)
Problem: As shown in the diagram, ∠KJP=2*x+4°, ∠PLK=7*x+5°, quadrilateral JMLK is a rectangle. Find the value of x.
问题：如图所示，∠KJP=2*x+4°，∠PLK=7*x+5°，JMLK是矩形。求x的值。
Construction (CDL):
Shape(JM,MP,PJ)
Shape(PM,ML,LP)
Shape(KP,PL,LK)
Shape(JP,PK,KJ)
Collinear(JPL)
Collinear(MPK)
Conditions (CDL):
Equal(MeasureOfAngle(KJP),2*x+4)
Equal(MeasureOfAngle(PLK),7*x+5)
Rectangle(JMLK)
Goal: Value(x)
Answer: 9
Theorem sequence:
triangle_property_angle_sum(1,KJL)
Problem: As shown in the diagram, ∠FDG=28°, ∠FGH and ∠GHF are the base angles of the isosceles △ FGH, triangleGFD is an isosceles triangle. Find the measure of ∠HFG.
问题：如图所示，∠FDG=28°，∠FGH和∠GHF是等腰△FGH的底角，∠GFD和∠FDG是等腰△GFD的底角。求∠HFG的大小。
Construction (CDL):
Shape(FD,DG,GF)
Shape(FG,GH,HF)
Collinear(DGH)
Conditions (CDL):
Equal(MeasureOfAngle(FDG),28)
IsoscelesTriangle(FGH)
IsoscelesTriangle(GFD)
Goal: Value(MeasureOfAngle(HFG))
Answer: 68
Theorem sequence:
isosceles_triangle_property_angle_equal(1,GFD)
triangle_property_angle_sum(1,FDG)
flat_angle(1,DGH)
angle_addition(1,DGF,FGH)
isosceles_triangle_property_angle_equal(1,FGH)
triangle_property_angle_sum(1,FGH)
Problem: As shown in the diagram, AB=21, CA=16, CB=15. Find the value of cos(CB).
问题：如图所示，AB=21，CA=16，CB=15。求cos(CB)的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),21)
Equal(LengthOfLine(CA),16)
Equal(LengthOfLine(CB),15)
Goal: Value(Cos(MeasureOfAngle(CBA)))
Answer: 41/63
Theorem sequence:
cosine_theorem(1,BAC)
Problem: As shown in the diagram, AD=x-3, AF=y+1, DB=12, FB=AF, LM=8, LO=10, LO=NO, MN=16, ∠AFB=∠NOL, ∠DAF=∠MNO, BD⊥AD, FB is perpendicular to DB, LM is perpendicular to NM, OL⊥ML, quadrilateral BDAF is similar to quadrilateral LMNO. Find the value of x.
问题：如图所示，AD=x-3，AF=y+1，DB=12，FB=AF，LM=8，LO=10，LO=NO，MN=16，∠AFB=∠NOL，∠DAF=∠MNO，BD垂直于AD，FB⊥DB，LM垂直于NM，OL垂直于ML，BDAF相似于LMNO。求x的值。
Construction (CDL):
Shape(BD,DA,AF,FB)
Shape(LM,MN,NO,OL)
Conditions (CDL):
Equal(LengthOfLine(AD),x-3)
Equal(LengthOfLine(AF),y+1)
Equal(LengthOfLine(DB),12)
Equal(LengthOfLine(FB),LengthOfLine(AF))
Equal(LengthOfLine(LM),8)
Equal(LengthOfLine(LO),10)
Equal(LengthOfLine(LO),LengthOfLine(NO))
Equal(LengthOfLine(MN),16)
Equal(MeasureOfAngle(AFB),MeasureOfAngle(NOL))
Equal(MeasureOfAngle(DAF),MeasureOfAngle(MNO))
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(FB,DB)
PerpendicularBetweenLine(LM,NM)
PerpendicularBetweenLine(OL,ML)
SimilarBetweenQuadrilateral(BDAF,LMNO)
Goal: Value(x)
Answer: 27
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,BDAF,LMNO)
similar_quadrilateral_property_line_ratio(1,DAFB,MNOL)
Problem: As shown in the diagram, PS=3, SR=5, ∠PQR=128°, PQRS is a ▱. Find the measure of ∠QRS.
问题：如图所示，PS=3，SR=5，∠PQR=128°，QP和RS是平行四边形PQRS的一组对边。求∠QRS的大小。
Construction (CDL):
Shape(QR,RS,SP,PQ)
Conditions (CDL):
Equal(LengthOfLine(PS),3)
Equal(LengthOfLine(SR),5)
Equal(MeasureOfAngle(PQR),128)
Parallelogram(PQRS)
Goal: Value(MeasureOfAngle(QRS))
Answer: 52
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,QP,RS)
Problem: As shown in the diagram, NM=2*x+1, PQ=3*x-7, the measure of arc JNM is equal to the measure of ⌒JQP. Find the length of line PQ.
问题：如图所示，NM=2*x+1，PQ=3*x-7，⌒JNM的角度与弧JQP的角度相等。求直线PQ的长度。
Construction (CDL):
Shape(QP,JPN,NM,JMQ)
Shape(MN,JNM)
Shape(PQ,JQP)
Cocircular(J,NMQP)
Conditions (CDL):
Equal(LengthOfLine(NM),2*x+1)
Equal(LengthOfLine(PQ),3*x-7)
Equal(MeasureOfArc(JNM),MeasureOfArc(JQP))
Goal: Value(LengthOfLine(PQ))
Answer: 17
Theorem sequence:
congruent_arc_judgment_measure_equal(1,JNM,JQP)
congruent_arc_property_chord_equal(1,JNM,JQP)
Problem: As shown in the diagram, AE=23, BD=23, IC=8, IF=10, IG=10, AE∥IC, IC∥DB, DG is perpendicular to IG, EF is perpendicular to IF. Find the sum of the area of quadrilateral CEAI and the area of quadrilateral IDBC.
问题：如图所示，AE=23，BD=23，IC=8，IF=10，IG=10，AE平行于IC，IC∥DB，DG垂直于IG，EF⊥IF。求四边形CEAI的面积与四边形IDBC的面积之和。
Construction (CDL):
Shape(AI,IF,FA)
Shape(FI,IC,CE,EF)
Shape(ID,DG,GI)
Shape(IG,GB,BC,CI)
Collinear(AFE)
Collinear(DGB)
Collinear(FIG)
Conditions (CDL):
Equal(LengthOfLine(AE),23)
Equal(LengthOfLine(BD),23)
Equal(LengthOfLine(IC),8)
Equal(LengthOfLine(IF),10)
Equal(LengthOfLine(IG),10)
ParallelBetweenLine(AE,IC)
ParallelBetweenLine(IC,DB)
PerpendicularBetweenLine(DG,IG)
PerpendicularBetweenLine(EF,IF)
Goal: Value(Add(AreaOfQuadrilateral(CEAI),AreaOfQuadrilateral(IDBC)))
Answer: 310
Theorem sequence:
trapezoid_judgment_parallel(1,CEAI)
trapezoid_judgment_parallel(1,IDBC)
altitude_of_quadrilateral_judgment_right_vertex(2,IF,CEAI)
altitude_of_quadrilateral_judgment_left_vertex(2,IG,IDBC)
trapezoid_area_formula(1,CEAI)
trapezoid_area_formula(1,IDBC)
Problem: As shown in the diagram, AE=19, CB=44, ∠ACE=30°, DB and AC are opposite sides of the parallelogram BDAC, DE⊥CE. Find the perimeter of quadrilateral BDAC.
问题：如图所示，AE=19，CB=44，∠ACE=30°，四边形BDAC是▱，DE⊥CE。求BDAC的周长。
Construction (CDL):
Shape(BD,DE,EC,CB)
Shape(CE,EA,AC)
Collinear(DEA)
Conditions (CDL):
Equal(LengthOfLine(AE),19)
Equal(LengthOfLine(CB),44)
Equal(MeasureOfAngle(ACE),30)
Parallelogram(BDAC)
PerpendicularBetweenLine(DE,CE)
Goal: Value(PerimeterOfQuadrilateral(BDAC))
Answer: 164
Theorem sequence:
adjacent_complementary_angle(1,DEC,CEA)
sine_theorem(1,ACE)
parallelogram_property_opposite_line_equal(1,BDAC)
parallelogram_property_opposite_line_equal(1,DACB)
quadrilateral_perimeter_formula(1,BDAC)
Problem: As shown in the diagram, ∠ANE=30°, ∠CEB=110°, ∠ECA=∠AFN, ∠ENJ=130°. Find the measure of ∠NEA.
问题：如图所示，∠ANE=30°，∠CEB=110°，∠ECA=∠AFN，∠ENJ=130°。求∠NEA的大小。
Construction (CDL):
Shape(CA,AE,EC)
Shape(EA,AN,NE)
Shape(AF,FN,NA)
Shape(GA,AC)
Shape(DA,AG)
Shape(FA,AD)
Shape(CE,EB)
Shape(BE,EN)
Shape(EN,NJ)
Shape(JN,NM)
Shape(MN,NF)
Collinear(DAEB)
Collinear(GANM)
Collinear(FNJ)
Collinear(CAF)
Collinear(CEN)
Conditions (CDL):
Equal(MeasureOfAngle(ANE),30)
Equal(MeasureOfAngle(CEB),110)
Equal(MeasureOfAngle(ECA),MeasureOfAngle(AFN))
Equal(MeasureOfAngle(ENJ),130)
Goal: Value(MeasureOfAngle(NEA))
Answer: 110
Theorem sequence:
vertical_angle(1,NEA,CEB)
Problem: As shown in the diagram, ∠HEL=105°, EL is parallel to GC. Find the measure of ∠CGP.
问题：如图所示，∠HEL=105°，EL平行于GC。求∠CGP的大小。
Construction (CDL):
Shape(HE,EL)
Shape(JE,EH)
Shape(LE,EG)
Shape(GE,EL)
Shape(DJ,JE)
Shape(EJ,JA)
Shape(KJ,JD)
Shape(AJ,JK)
Shape(EG,GC)
Shape(CG,GP)
Shape(AG,GE)
Shape(PG,GA)
Shape(JA,AG)
Shape(IA,AJ)
Shape(GA,AF)
Shape(FA,AI)
Collinear(HEGP)
Collinear(DJAF)
Collinear(KJEL)
Collinear(IAGC)
Conditions (CDL):
Equal(MeasureOfAngle(HEL),105)
ParallelBetweenLine(EL,GC)
Goal: Value(MeasureOfAngle(CGP))
Answer: 75
Theorem sequence:
adjacent_complementary_angle(1,HEL,LEG)
parallel_property_corresponding_angle(2,EL,GC,P)
Problem: As shown in the diagram, AB=x, AC=z, AP=y, BC=12, BP=8, AP⊥CP, CA⊥BA. Find the value of z.
问题：如图所示，AB=x，AC=z，AP=y，BC=12，BP=8，AP垂直于CP，CA⊥BA。求z的值。
Construction (CDL):
Shape(CA,AP,PC)
Shape(PA,AB,BP)
Collinear(CPB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),z)
Equal(LengthOfLine(AP),y)
Equal(LengthOfLine(BC),12)
Equal(LengthOfLine(BP),8)
PerpendicularBetweenLine(AP,CP)
PerpendicularBetweenLine(CA,BA)
Goal: Value(z)
Answer: 4*sqrt(3)
Theorem sequence:
line_addition(1,CP,PB)
angle_addition(1,CAP,PAB)
triangle_property_angle_sum(1,CAP)
adjacent_complementary_angle(1,BPA,APC)
similar_triangle_judgment_aa(1,APC,BPA)
similar_triangle_property_line_ratio(1,APC,BPA)
similar_triangle_property_line_ratio(1,CAP,ABP)
right_triangle_judgment_angle(1,APC)
right_triangle_property_pythagorean(1,APC)
Problem: As shown in the diagram, triangle NGD is congruent to triangle LDG, ∠DNG=36°, ∠LDG=x°, DG⊥LG. Find the value of x.
问题：如图所示，三角形NGD与三角形LDG是全等三角形，∠DNG=36°，∠LDG=x°，DG⊥LG。求x的值。
Construction (CDL):
Shape(DN,NG,GD)
Shape(DG,GL,LD)
Conditions (CDL):
CongruentBetweenTriangle(NGD,LDG)
Equal(MeasureOfAngle(DNG),36)
Equal(MeasureOfAngle(LDG),x)
PerpendicularBetweenLine(DG,LG)
Goal: Value(x)
Answer: 54
Theorem sequence:
congruent_triangle_property_angle_equal(1,NGD,LDG)
triangle_property_angle_sum(1,DGL)
Problem: As shown in the diagram, ∠AXW=82°, ∠YXA=33°, ZW and YX are opposite sides of the ▱ WZYX. Find the measure of ∠XWZ.
问题：如图所示，∠AXW=82°，∠YXA=33°，WZYX是▱。求∠XWZ的大小。
Construction (CDL):
Shape(XW,WA,AX)
Shape(AW,WZ,ZA)
Shape(AZ,ZY,YA)
Shape(XA,AY,YX)
Collinear(XAZ)
Collinear(WAY)
Conditions (CDL):
Equal(MeasureOfAngle(AXW),82)
Equal(MeasureOfAngle(YXA),33)
Parallelogram(WZYX)
Goal: Value(MeasureOfAngle(XWZ))
Answer: 65
Theorem sequence:
angle_addition(1,YXA,AXW)
parallel_property_ipsilateral_internal_angle(1,XY,WZ)
Problem: As shown in the diagram, RT=3*x+3, SR=4*x-4, SR=RT, ST=12+x. Find the length of line RT.
问题：如图所示，RT=3*x+3，SR=4*x-4，SR=RT，ST=12+x。求直线RT的长度。
Construction (CDL):
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(RT),3*x+3)
Equal(LengthOfLine(SR),4*x-4)
Equal(LengthOfLine(SR),LengthOfLine(RT))
Equal(LengthOfLine(ST),12+x)
Goal: Value(LengthOfLine(RT))
Answer: 24
Theorem sequence:
Problem: As shown in the diagram, AH=x+2, BE=3*x, DB=x, DH=8. Find the value of x.
问题：如图所示，AH=x+2，BE=3*x，DB=x，DH=8。求x的值。
Construction (CDL):
Shape(BD,DH,FHB)
Shape(AH,FHA)
Shape(BE,FEB)
Shape(FBH,HA,FAE,EB)
Collinear(DBE)
Collinear(DHA)
Cocircular(F,BHAE)
Conditions (CDL):
Equal(LengthOfLine(AH),x+2)
Equal(LengthOfLine(BE),3*x)
Equal(LengthOfLine(DB),x)
Equal(LengthOfLine(DH),8)
Goal: Value(x)
Answer: 1+sqrt(21)
Theorem sequence:
line_addition(1,DB,BE)
line_addition(1,DH,HA)
circle_property_circular_power_segment_and_segment_line(1,DBE,DHA,F)
Problem: As shown in the diagram, BD=9, BE=3, EF=x, FC=5. Find the value of x.
问题：如图所示，BD=9，BE=3，EF=x，FC=5。求x的值。
Construction (CDL):
Shape(BE,EF,AFB)
Shape(CF,AFC)
Shape(BD,ADB)
Shape(FC,ACD,DB,ABF)
Collinear(EFC)
Collinear(EBD)
Cocircular(A,FCDB)
Conditions (CDL):
Equal(LengthOfLine(BD),9)
Equal(LengthOfLine(BE),3)
Equal(LengthOfLine(EF),x)
Equal(LengthOfLine(FC),5)
Goal: Value(x)
Answer: 4
Theorem sequence:
line_addition(1,EF,FC)
line_addition(1,EB,BD)
circle_property_circular_power_segment_and_segment_line(1,EFC,EBD,A)
Problem: As shown in the diagram, KN=9, LN=16, PM=Mul(LengthOfLine(KP),2), PR∥KL, KN is perpendicular to QN, RM is perpendicular to PM. Find the length of line ML.
问题：如图所示，KN=9，LN=16，PM=Mul(LengthOfLine(KP),2)，PR平行于KL，KN垂直于QN，RM垂直于PM。求直线ML的长度。
Construction (CDL):
Shape(LR,RQ,QN,NL)
Shape(NQ,QP,PK,KN)
Shape(RM,MQ,QR)
Shape(QM,MP,PQ)
Collinear(LNK)
Collinear(RQP)
Collinear(LRM)
Collinear(KPM)
Collinear(NQM)
Conditions (CDL):
Equal(LengthOfLine(KN),9)
Equal(LengthOfLine(LN),16)
Equal(LengthOfLine(PM),Mul(LengthOfLine(KP),2))
ParallelBetweenLine(PR,KL)
PerpendicularBetweenLine(KN,QN)
PerpendicularBetweenLine(RM,PM)
Goal: Value(LengthOfLine(ML))
Answer: 20
Theorem sequence:
line_addition(1,LN,NK)
mirror_similar_triangle_judgment_aa(1,LMK,MKN)
mirror_similar_triangle_property_line_ratio(1,LMK,MKN)
mirror_similar_triangle_property_line_ratio(1,MKL,NMK)
right_triangle_judgment_angle(1,LMK)
right_triangle_property_pythagorean(1,LMK)
Problem: As shown in the diagram, ∠ADC=x°, ∠BAD=x°, ∠CBA=2*x°, ∠DCB=2*x°. Find the measure of ∠DCB.
问题：如图所示，∠ADC=x°，∠BAD=x°，∠CBA=2*x°，∠DCB=2*x°。求∠DCB的大小。
Construction (CDL):
Shape(BA,AD,DC,CB)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),x)
Equal(MeasureOfAngle(BAD),x)
Equal(MeasureOfAngle(CBA),2*x)
Equal(MeasureOfAngle(DCB),2*x)
Goal: Value(MeasureOfAngle(DCB))
Answer: 120
Theorem sequence:
quadrilateral_property_angle_sum(1,BADC)
Problem: As shown in the diagram, ∠WXH=130°, ∠YZI=20°, Mul(LengthOfLine(IJ)=LengthOfLine(YJ)). Find the measure of ∠JHI.
问题：如图所示，∠WXH=130°，∠YZI=20°，Mul(LengthOfLine(IJ)=LengthOfLine(YJ))。求∠JHI的大小。
Construction (CDL):
Shape(JH,HI,IJ)
Shape(HX,XY,YI,IH)
Shape(IY,YZ,ZI)
Shape(WX,XH)
Collinear(GHIZ)
Collinear(WXYZ)
Collinear(JHX)
Collinear(JIY)
Conditions (CDL):
Equal(MeasureOfAngle(WXH),130)
Equal(MeasureOfAngle(YZI),20)
Equal(Mul(LengthOfLine(IJ),LengthOfLine(YJ)),Mul(LengthOfLine(XJ),LengthOfLine(HJ)))
Goal: Value(MeasureOfAngle(JHI))
Answer: 70
Theorem sequence:
flat_angle(1,WXY)
angle_addition(1,WXH,HXY)
mirror_similar_triangle_judgment_sas(1,JHI,JXY)
mirror_similar_triangle_property_angle_equal(1,HIJ,YJX)
mirror_similar_triangle_property_angle_equal(1,IJH,XYJ)
vertical_angle(1,GIJ,ZIY)
triangle_property_angle_sum(1,IYZ)
flat_angle(1,XYZ)
angle_addition(1,XYI,IYZ)
Problem: As shown in the diagram, AC=CD, ∠BCA=3*x+13°, ∠ECB=2*x+7°, EC is the altitude of triangle EDA. Find the measure of ∠BCA.
问题：如图所示，AC=CD，∠BCA=3*x+13°，∠ECB=2*x+7°，EC是△EDA的高。求∠BCA的大小。
Construction (CDL):
Shape(AB,BC,CA)
Shape(CB,BE,EC)
Shape(CE,ED,DC)
Collinear(ACD)
Collinear(ABE)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(CD))
Equal(MeasureOfAngle(BCA),3*x+13)
Equal(MeasureOfAngle(ECB),2*x+7)
IsAltitudeOfTriangle(EC,EDA)
Goal: Value(MeasureOfAngle(BCA))
Answer: 55
Theorem sequence:
angle_addition(1,ECB,BCA)
Problem: As shown in the diagram, AP=12, KA=10, AK is perpendicular to LK, KL⊥PL, LP is perpendicular to AP, PA is perpendicular to KA. Find the perimeter of LPAK.
问题：如图所示，AP=12，KA=10，AK垂直于LK，KL⊥PL，LP垂直于AP，PA垂直于KA。求四边形LPAK的周长。
Construction (CDL):
Shape(LP,PA,AK,KL)
Conditions (CDL):
Equal(LengthOfLine(AP),12)
Equal(LengthOfLine(KA),10)
PerpendicularBetweenLine(AK,LK)
PerpendicularBetweenLine(KL,PL)
PerpendicularBetweenLine(LP,AP)
PerpendicularBetweenLine(PA,KA)
Goal: Value(PerimeterOfQuadrilateral(LPAK))
Answer: 44
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,LK,PA)
parallel_judgment_ipsilateral_internal_angle(1,PL,AK)
parallelogram_judgment_parallel_and_parallel(1,LPAK)
parallelogram_property_opposite_line_equal(1,LPAK)
parallelogram_property_opposite_line_equal(1,PAKL)
quadrilateral_perimeter_formula(1,LPAK)
Problem: As shown in the diagram, AB=15, AC=y, BC=x, ∠ABC=30°, BC⊥AC. Find the value of x.
问题：如图所示，AB=15，AC=y，BC=x，∠ABC=30°，BC垂直于AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 15*sqrt(3)/2
Theorem sequence:
triangle_property_angle_sum(1,BCA)
sine_theorem(1,BCA)
Problem: As shown in the diagram, ∠GFC=7*x-1°, FA is parallel to DE, ED is perpendicular to HD. Find the value of x.
问题：如图所示，∠GFC=7*x-1°，FA平行于DE，ED垂直于HD。求x的值。
Construction (CDL):
Shape(CF,FA)
Shape(AF,FD)
Shape(GF,FC)
Shape(DF,FG)
Shape(FD,DE)
Shape(ED,DH)
Shape(BD,DF)
Shape(HD,DB)
Collinear(CFDH)
Collinear(AFG)
Collinear(EDB)
Conditions (CDL):
Equal(MeasureOfAngle(GFC),7*x-1)
ParallelBetweenLine(FA,DE)
PerpendicularBetweenLine(ED,HD)
Goal: Value(x)
Answer: 13
Theorem sequence:
parallel_property_corresponding_angle(2,FA,DE,H)
vertical_angle(1,GFC,AFD)