SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠BMN=56°, ∠GLC=3*y-11°, ∠HND=4*x°, AD∥BE, BE is parallel to CF. Find the value of y.
问题：如图所示，∠BMN=56°，∠GLC=3*y-11°，∠HND=4*x°，AD平行于BE，BE∥CF。求y的值。
Construction (CDL):
Shape(HN,ND)
Shape(DN,NM)
Shape(NM,ME)
Shape(EM,ML)
Shape(ML,LF)
Shape(FL,LG)
Shape(AN,NH)
Shape(MN,NA)
Shape(BM,MN)
Shape(LM,MB)
Shape(CL,LM)
Shape(GL,LC)
Collinear(HNMLG)
Collinear(AND)
Collinear(BME)
Collinear(CLF)
Conditions (CDL):
Equal(MeasureOfAngle(BMN),56)
Equal(MeasureOfAngle(GLC),3*y-11)
Equal(MeasureOfAngle(HND),4*x)
ParallelBetweenLine(AD,BE)
ParallelBetweenLine(BE,CF)
Goal: Value(y)
Answer: 45
Theorem sequence:
parallel_property_collinear_extend(3,FC,EB,L)
parallel_property_collinear_extend(3,BE,CL,M)
parallel_property_corresponding_angle(2,LC,MB,N)
adjacent_complementary_angle(1,GLC,CLM)
Problem: As shown in the diagram, the area of ACEB is 72, the area of MNPQ is 50, EC=12, NP=x, quadrilateral ACEB is similar to quadrilateral MNPQ. Find The ratio of similar quadrilaterals ACEB and MNPQ.
问题：如图所示，四边形ACEB的面积为72，四边形MNPQ的面积为50，EC=12，NP=x，ACEB相似于MNPQ。求四边形ACEB和四边形MNPQ的相似比。
Construction (CDL):
Shape(AC,CE,EB,BA)
Shape(MN,NP,PQ,QM)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ACEB),72)
Equal(AreaOfQuadrilateral(MNPQ),50)
Equal(LengthOfLine(EC),12)
Equal(LengthOfLine(NP),x)
SimilarBetweenQuadrilateral(ACEB,MNPQ)
Goal: Value(RatioOfSimilarQuadrilateral(ACEB,MNPQ))
Answer: 6/5
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,ACEB,MNPQ)
Problem: As shown in the diagram, JL=4*x-8, KJ=JL, KJ=LK, KJ=x+7, LK=JL. Find the length of line JL.
问题：如图所示，JL=4*x-8，KJ=JL，KJ=LK，KJ=x+7，LK=JL。求直线JL的长度。
Construction (CDL):
Shape(KJ,JL,LK)
Conditions (CDL):
Equal(LengthOfLine(JL),4*x-8)
Equal(LengthOfLine(KJ),LengthOfLine(JL))
Equal(LengthOfLine(KJ),LengthOfLine(LK))
Equal(LengthOfLine(KJ),x+7)
Equal(LengthOfLine(LK),LengthOfLine(JL))
Goal: Value(LengthOfLine(JL))
Answer: 12
Theorem sequence:
Problem: As shown in the diagram, ∠DIB=x°, the measure of arc EAB is 143, the measure of ⌒EDC is 75. Find the value of x.
问题：如图所示，∠DIB=x°，弧EAB的角度为143，⌒EDC的角度为75。求x的值。
Construction (CDL):
Shape(ECA,AC)
Shape(AI,IC,CA)
Shape(EAB,BI,IA)
Shape(ID,EDC,CI)
Shape(EBD,DI,IB)
Collinear(AID)
Collinear(CIB)
Cocircular(E,ABDC)
Conditions (CDL):
Equal(MeasureOfAngle(DIB),x)
Equal(MeasureOfArc(EAB),143)
Equal(MeasureOfArc(EDC),75)
Goal: Value(x)
Answer: 71
Theorem sequence:
arc_property_circumference_angle_external(1,EAB,C)
arc_property_circumference_angle_external(1,EDC,A)
triangle_property_angle_sum(1,AIC)
vertical_angle(1,AIC,DIB)
Problem: As shown in the diagram, the area of triangle ABC is 40, the area of triangle DEF is 30, CA=5, FD=x, △DEF is similar to △ABC.. Find the ratio of similar △DEF and △ABC.
问题：如图所示，三角形ABC的面积为40，△DEF的面积为30，CA=5，FD=x，三角形DEF相似于三角形ABC。求三角形DEF和三角形ABC的相似比。
Construction (CDL):
Shape(AB,BC,CA)
Shape(DE,EF,FD)
Conditions (CDL):
Equal(AreaOfTriangle(ABC),40)
Equal(AreaOfTriangle(DEF),30)
Equal(LengthOfLine(CA),5)
Equal(LengthOfLine(FD),x)
SimilarBetweenTriangle(DEF,ABC)
Goal: Value(RatioOfSimilarTriangle(DEF,ABC))
Answer: sqrt(3)/2
Theorem sequence:
similar_triangle_property_area_square_ratio(1,DEF,ABC)
Problem: As shown in the diagram, AB=y, AC=15, BC=x, BD=z, CD=19, AB is perpendicular to DB, DC is perpendicular to BC. Find the value of z.
问题：如图所示，AB=y，AC=15，BC=x，BD=z，CD=19，AB⊥DB，DC垂直于BC。求z的值。
Construction (CDL):
Shape(AB,BC,CA)
Shape(CB,BD,DC)
Collinear(ACD)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),15)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(BD),z)
Equal(LengthOfLine(CD),19)
PerpendicularBetweenLine(AB,DB)
PerpendicularBetweenLine(DC,BC)
Goal: Value(z)
Answer: sqrt(646)
Theorem sequence:
line_addition(1,AC,CD)
adjacent_complementary_angle(1,DCB,BCA)
right_triangle_judgment_angle(1,BCA)
right_triangle_judgment_angle(1,DCB)
right_triangle_judgment_angle(1,ABD)
right_triangle_property_pythagorean(1,BCA)
right_triangle_property_pythagorean(1,DCB)
right_triangle_property_pythagorean(1,ABD)
Problem: As shown in the diagram, ∠PRQ=115°, R is the center of circle R, NR⊥PR. Find the measure of arc RPM.
问题：如图所示，∠PRQ=115°，圆R的圆心为R，NR垂直于PR。求弧RPM的角度。
Construction (CDL):
Shape(RPN,NR,RP)
Shape(RN,RNM,MR)
Shape(PR,RQ,RQP)
Shape(RMQ,QR,RM)
Collinear(MRP)
Cocircular(R,PNMQ)
Conditions (CDL):
Equal(MeasureOfAngle(PRQ),115)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(NR,PR)
Goal: Value(MeasureOfArc(RPM))
Answer: 180
Theorem sequence:
flat_angle(1,MRP)
arc_property_center_angle(1,RPM,R)
Problem: As shown in the diagram, the area of GFEH is 525, the area of quadrilateral XBAC is 1575, AB=18, EF=x, quadrilateral GFEH is similar to quadrilateral XBAC. Find The ratio of similar quadrilaterals GFEH and XBAC.
问题：如图所示，GFEH的面积为525，四边形XBAC的面积为1575，AB=18，EF=x，GFEH与XBAC相似。求GFEH和XBAC的相似比。
Construction (CDL):
Shape(GF,FE,EH,HG)
Shape(XB,BA,AC,CX)
Conditions (CDL):
Equal(AreaOfQuadrilateral(GFEH),525)
Equal(AreaOfQuadrilateral(XBAC),1575)
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(EF),x)
SimilarBetweenQuadrilateral(GFEH,XBAC)
Goal: Value(RatioOfSimilarQuadrilateral(GFEH,XBAC))
Answer: sqrt(3)/3
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,GFEH,XBAC)
Problem: As shown in the diagram, BC=18, DC=12, ∠BAD=115°, quadrilateral ADCB is a parallelogram. Find the length of line AB.
问题：如图所示，BC=18，DC=12，∠BAD=115°，ADCB是▱。求直线AB的长度。
Construction (CDL):
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(BC),18)
Equal(LengthOfLine(DC),12)
Equal(MeasureOfAngle(BAD),115)
Parallelogram(ADCB)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DCBA)
Problem: As shown in the diagram, ∠BAD=∠DBA, ∠DFB=63°, ∠ECG=136°, ∠EDC=47°, ∠FDE=69°. Find the measure of ∠BAD.
问题：如图所示，∠BAD=∠DBA，∠DFB=63°，∠ECG=136°，∠EDC=47°，∠FDE=69°。求∠BAD的大小。
Construction (CDL):
Shape(FB,BD,DF)
Shape(BA,AD,DB)
Shape(ED,DC,CE)
Shape(FD,DE)
Shape(EC,CG)
Collinear(FDA)
Collinear(BDCG)
Conditions (CDL):
Equal(MeasureOfAngle(BAD),MeasureOfAngle(DBA))
Equal(MeasureOfAngle(DFB),63)
Equal(MeasureOfAngle(ECG),136)
Equal(MeasureOfAngle(EDC),47)
Equal(MeasureOfAngle(FDE),69)
Goal: Value(MeasureOfAngle(BAD))
Answer: 32
Theorem sequence:
angle_addition(1,FDE,EDC)
vertical_angle(1,FDC,ADB)
triangle_property_angle_sum(1,DBA)
Problem: As shown in the diagram, JK=16, KL=12, ST=3, ∠JKL=87°, ∠LJK=46°, ∠RST=87°, ∠TRS=46°. Find the length of line SR.
问题：如图所示，JK=16，KL=12，ST=3，∠JKL=87°，∠LJK=46°，∠RST=87°，∠TRS=46°。求直线SR的长度。
Construction (CDL):
Shape(JK,KL,LJ)
Shape(TR,RS,ST)
Conditions (CDL):
Equal(LengthOfLine(JK),16)
Equal(LengthOfLine(KL),12)
Equal(LengthOfLine(ST),3)
Equal(MeasureOfAngle(JKL),87)
Equal(MeasureOfAngle(LJK),46)
Equal(MeasureOfAngle(RST),87)
Equal(MeasureOfAngle(TRS),46)
Goal: Value(LengthOfLine(SR))
Answer: 4
Theorem sequence:
similar_triangle_judgment_aa(1,LJK,TRS)
similar_triangle_property_line_ratio(1,LJK,TRS)
similar_triangle_property_line_ratio(1,JKL,RST)
Problem: As shown in the diagram, ∠AGC=60°, DG is perpendicular to AG. Find the measure of ∠EGA.
问题：如图所示，∠AGC=60°，DG⊥AG。求∠EGA的大小。
Construction (CDL):
Shape(GCA,AG,GC)
Shape(GA,GAD,DG)
Shape(GD,GDE,EG)
Shape(GE,GEB,BG)
Shape(CG,GB,GBC)
Collinear(CGE)
Collinear(AGB)
Cocircular(G,CADEB)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),60)
PerpendicularBetweenLine(DG,AG)
Goal: Value(MeasureOfAngle(EGA))
Answer: 120
Theorem sequence:
adjacent_complementary_angle(1,EGA,AGC)
Problem: As shown in the diagram, BC=103, ∠ABC=43°, ∠CAB=51°. Find the perimeter of triangle ABC.
问题：如图所示，BC=103，∠ABC=43°，∠CAB=51°。求△ABC的周长。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(BC),103)
Equal(MeasureOfAngle(ABC),43)
Equal(MeasureOfAngle(CAB),51)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: (-103*sqrt(3*sqrt(5)+15)-103*sqrt(10)-103*sqrt(sqrt(5)+5)-103*sqrt(2)+103*sqrt(6)+103*sqrt(30)+103*sqrt(5*sqrt(5)+25)+103*sqrt(15*sqrt(5)+75)+1648*sin(43*pi/180)+1648*cos(pi/45))/(-sqrt(3)*sqrt(sqrt(5)+5)-sqrt(10)-sqrt(sqrt(5)+5)-sqrt(2)+sqrt(6)+sqrt(30)+sqrt(5)*sqrt(sqrt(5)+5)+sqrt(15)*sqrt(sqrt(5)+5))
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,BCA)
sine_theorem(1,CAB)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, ∠AGC=85°, ∠CGE=x°, ∠DGA=40°, EG is perpendicular to DG. Find the value of x.
问题：如图所示，∠AGC=85°，∠CGE=x°，∠DGA=40°，EG⊥DG。求x的值。
Construction (CDL):
Shape(GAD,DG,GA)
Shape(GDE,EG,GD)
Shape(GEC,CG,GE)
Shape(GCA,AG,GC)
Cocircular(G,ADEC)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),85)
Equal(MeasureOfAngle(CGE),x)
Equal(MeasureOfAngle(DGA),40)
PerpendicularBetweenLine(EG,DG)
Goal: Value(x)
Answer: 145
Theorem sequence:
angle_addition(1,DGA,AGC)
angle_addition(1,EGD,DGC)
round_angle(1,CGE,EGC)
Problem: As shown in the diagram, AV=12, BV=12, ∠AVB=v°, ∠VBA=52°. Find the measure of ∠AVB.
问题：如图所示，AV=12，BV=12，∠AVB=v°，∠VBA=52°。求∠AVB的大小。
Construction (CDL):
Shape(AV,VB,BA)
Conditions (CDL):
Equal(LengthOfLine(AV),12)
Equal(LengthOfLine(BV),12)
Equal(MeasureOfAngle(AVB),v)
Equal(MeasureOfAngle(VBA),52)
Goal: Value(MeasureOfAngle(AVB))
Answer: 76
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,VBA)
isosceles_triangle_property_angle_equal(1,VBA)
triangle_property_angle_sum(1,AVB)
Problem: As shown in the diagram, CB=4, ∠CBO=45°, CBOA is a ▱, BO is perpendicular to CO. Find the area of quadrilateral CBOA.
问题：如图所示，CB=4，∠CBO=45°，BC和OA是▱CBOA的一组对边，BO垂直于CO。求四边形CBOA的面积。
Construction (CDL):
Shape(CB,BO,OC)
Shape(CO,OA,AC)
Conditions (CDL):
Equal(LengthOfLine(CB),4)
Equal(MeasureOfAngle(CBO),45)
Parallelogram(CBOA)
PerpendicularBetweenLine(BO,CO)
Goal: Value(AreaOfQuadrilateral(CBOA))
Answer: 8
Theorem sequence:
triangle_property_angle_sum(1,CBO)
sine_theorem(1,BOC)
parallelogram_area_formula_sine(1,CBOA)
Problem: As shown in the diagram, AC=17, DC=15, ∠ABC=30°, CD⊥BD. Find the area of triangle ABC.
问题：如图所示，AC=17，DC=15，∠ABC=30°，CD⊥BD。求△ABC的面积。
Construction (CDL):
Shape(AD,DC,CA)
Shape(DB,BC,CD)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),17)
Equal(LengthOfLine(DC),15)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(CD,BD)
Goal: Value(AreaOfTriangle(ABC))
Answer: 60+225*sqrt(3)/2
Theorem sequence:
sine_theorem(1,CDB)
sine_theorem(1,CAB)
sine_theorem(1,ABC)
cosine_theorem(1,ABC)
triangle_area_formula_sine(1,CAB)
Problem: As shown in the diagram, AD=15, DE=10, ∠DEC=60°, ED and CA are opposite sides of the parallelogram DECA. Find the area of DECA.
问题：如图所示，AD=15，DE=10，∠DEC=60°，DECA是平行四边形。求四边形DECA的面积。
Construction (CDL):
Shape(DE,EC,CA,AD)
Conditions (CDL):
Equal(LengthOfLine(AD),15)
Equal(LengthOfLine(DE),10)
Equal(MeasureOfAngle(DEC),60)
Parallelogram(DECA)
Goal: Value(AreaOfQuadrilateral(DECA))
Answer: 75*sqrt(3)
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ECAD)
parallelogram_area_formula_sine(1,DECA)
Problem: As shown in the diagram, BA=7, BC=6, CA=3, EF=9, △DFE is similar to △ACB.. Find the perimeter of triangle DFE.
问题：如图所示，BA=7，BC=6，CA=3，EF=9，三角形DFE与三角形ACB是相似三角形。求三角形DFE的周长。
Construction (CDL):
Shape(AC,CB,BA)
Shape(DF,FE,ED)
Conditions (CDL):
Equal(LengthOfLine(BA),7)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(CA),3)
Equal(LengthOfLine(EF),9)
SimilarBetweenTriangle(DFE,ACB)
Goal: Value(PerimeterOfTriangle(DFE))
Answer: 24
Theorem sequence:
similar_triangle_property_line_ratio(1,DFE,ACB)
similar_triangle_property_line_ratio(1,FED,CBA)
similar_triangle_property_line_ratio(1,EDF,BAC)
triangle_perimeter_formula(1,DFE)
Problem: As shown in the diagram, ∠ABC=70°, the measure of ⌒EAC is x, the measure of arc ECA is y, E is the center of ⊙E, BA is the tangent to circle E, BC is the tangent to ⊙E. Find the value of y.
问题：如图所示，∠ABC=70°，弧EAC的角度为x，⌒ECA的角度为y，⊙E的圆心为E，⊙O的切线为BA，圆O的切线为BC。求y的值。
Construction (CDL):
Shape(AB,BC,EAC)
Shape(EAC,CE,EA)
Shape(ECD,EDA,AE,EC)
Cocircular(E,ACD)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),70)
Equal(MeasureOfArc(EAC),x)
Equal(MeasureOfArc(ECA),y)
IsCentreOfCircle(E,E)
IsTangentOfCircle(BA,E)
IsTangentOfCircle(BC,E)
Goal: Value(y)
Answer: 250
Theorem sequence:
tangent_of_circle_property_perpendicular(1,BA,E,E)
tangent_of_circle_property_perpendicular(2,BC,E,E)
quadrilateral_property_angle_sum(1,ABCE)
arc_property_center_angle(1,ECA,E)
round_angle(1,CEA,AEC)
Problem: As shown in the diagram, DR=28, KD=6*x+23, P bisects segment JK, Q is the midpoint of segment KL, R bisects segment JL. Find the value of x.
问题：如图所示，DR=28，KD=6*x+23，P平分线段JK，Q平分线段KL，R是线段JL的中点。求x的值。
Construction (CDL):
Shape(LQ,QD,DL)
Shape(QK,KD,DQ)
Shape(DK,KP,PD)
Shape(DP,PJ,JD)
Shape(DJ,JR,RD)
Shape(LD,DR,RL)
Collinear(LDP)
Collinear(QDJ)
Collinear(KDR)
Collinear(LQK)
Collinear(KPJ)
Collinear(JRL)
Conditions (CDL):
Equal(LengthOfLine(DR),28)
Equal(LengthOfLine(KD),6*x+23)
IsMidpointOfLine(P,JK)
IsMidpointOfLine(Q,KL)
IsMidpointOfLine(R,JL)
Goal: Value(x)
Answer: 11/2
Theorem sequence:
median_of_triangle_judgment(1,KR,KJL)
median_of_triangle_judgment(1,JQ,JLK)
centroid_of_triangle_judgment_intersection(1,D,LKJ,Q,R)
centroid_of_triangle_property_line_ratio(1,D,KJL,R)
Problem: As shown in the diagram, LM=10, NM=10, PL=2*x-4, PN=x+5, NM is perpendicular to PM. Find the length of line NP.
问题：如图所示，LM=10，NM=10，PL=2*x-4，PN=x+5，NM垂直于PM。求直线NP的长度。
Construction (CDL):
Shape(LP,PM,ML)
Shape(MP,PN,NM)
Collinear(LMN)
Conditions (CDL):
Equal(LengthOfLine(LM),10)
Equal(LengthOfLine(NM),10)
Equal(LengthOfLine(PL),2*x-4)
Equal(LengthOfLine(PN),x+5)
PerpendicularBetweenLine(NM,PM)
Goal: Value(LengthOfLine(NP))
Answer: 14
Theorem sequence:
perpendicular_bisector_judgment_per_and_mid(1,PM,NL)
perpendicular_bisector_property_distance_equal(1,PM,NL)
Problem: As shown in the diagram, MJ=13, MX=5, RS=x, M is the center of circle M, MX⊥RX. Find the value of x.
问题：如图所示，MJ=13，MX=5，RS=x，M是圆M的圆心，MX垂直于RX。求x的值。
Construction (CDL):
Shape(MSR,RX,XS)
Shape(XR,RM,MX)
Shape(MR,MRJ,JM)
Shape(XM,MJ,MJS,SX)
Collinear(RXS)
Cocircular(M,RJS)
Conditions (CDL):
Equal(LengthOfLine(MJ),13)
Equal(LengthOfLine(MX),5)
Equal(LengthOfLine(RS),x)
IsCentreOfCircle(M,M)
PerpendicularBetweenLine(MX,RX)
Goal: Value(x)
Answer: 24
Theorem sequence:
radius_of_circle_property_length_equal(1,MR,M)
radius_of_circle_property_length_equal(1,MJ,M)
right_triangle_judgment_angle(1,MXR)
right_triangle_property_pythagorean(1,MXR)
adjacent_complementary_angle(1,SXM,MXR)
circle_property_chord_perpendicular_bisect_chord(1,M,MX,SR)
line_addition(1,RX,XS)
Problem: As shown in the diagram, PR=6*x-4, PS=11*y-12, QP=8*y-6, TP=4*x+2, QRST is a parallelogram. Find the value of y.
问题：如图所示，PR=6*x-4，PS=11*y-12，QP=8*y-6，TP=4*x+2，QT和RS是▱QRST的一组对边。求y的值。
Construction (CDL):
Shape(QR,RP,PQ)
Shape(PR,RS,SP)
Shape(PS,ST,TP)
Shape(QP,PT,TQ)
Collinear(QPS)
Collinear(TPR)
Conditions (CDL):
Equal(LengthOfLine(PR),6*x-4)
Equal(LengthOfLine(PS),11*y-12)
Equal(LengthOfLine(QP),8*y-6)
Equal(LengthOfLine(TP),4*x+2)
Parallelogram(QRST)
Goal: Value(y)
Answer: 2
Theorem sequence:
parallelogram_property_diagonal_bisection(1,QRST,P)
Problem: As shown in the diagram, BD=2, triangle CBA is an equilateral triangle, D is the center of circle D, E bisects segment CA. Find the area of the circle D minus the area of triangle CBA.
问题：如图所示，BD=2，△CBA为等边△，圆D的圆心为D，E是线段CA的中点。求⊙D的面积减去三角形CBA的面积。
Construction (CDL):
Shape(DBA,AB)
Shape(BA,AD,DB)
Shape(AE,ED,DA)
Shape(DE,EC,CD)
Shape(BD,DC,CB)
Shape(DAC,CA)
Shape(BC,DCB)
Collinear(BDE)
Collinear(AEC)
Cocircular(D,CBA)
Conditions (CDL):
Equal(LengthOfLine(BD),2)
EquilateralTriangle(CBA)
IsCentreOfCircle(D,D)
IsMidpointOfLine(E,CA)
Goal: Value(Sub(AreaOfCircle(D),AreaOfTriangle(CBA)))
Answer: -3*sqrt(3)+4*pi
Theorem sequence:
line_addition(1,BD,DE)
line_addition(1,AE,EC)
radius_of_circle_property_length_equal(1,DB,D)
radius_of_circle_property_length_equal(1,DA,D)
equilateral_triangle_property_angle(1,CBA)
median_of_triangle_judgment(1,BE,BAC)
isosceles_triangle_property_line_coincidence(2,BAC,E)
right_triangle_judgment_angle(1,AEB)
right_triangle_judgment_angle(1,AED)
right_triangle_property_pythagorean(1,AEB)
right_triangle_property_pythagorean(1,AED)
triangle_area_formula_sine(1,CBA)
circle_area_formula(1,D)
Problem: As shown in the diagram, DG=9, DK=16, FK=12, K is the centroid of triangle CDF. Find the length of line KH.
问题：如图所示，DG=9，DK=16，FK=12，K是三角形CDF的重心。求直线KH的长度。
Construction (CDL):
Shape(CG,GK,KC)
Shape(GD,DK,KG)
Shape(KD,DJ,JK)
Shape(KJ,JF,FK)
Shape(KF,FH,HK)
Shape(KH,HC,CK)
Collinear(CGD)
Collinear(DJF)
Collinear(FHC)
Collinear(CKJ)
Collinear(GKF)
Collinear(DKH)
Conditions (CDL):
Equal(LengthOfLine(DG),9)
Equal(LengthOfLine(DK),16)
Equal(LengthOfLine(FK),12)
IsCentroidOfTriangle(K,CDF)
Goal: Value(LengthOfLine(KH))
Answer: 8
Theorem sequence:
centroid_of_triangle_property_line_ratio(1,K,DFC,H)
Problem: As shown in the diagram, AB=14, AC=x, BC=x, ∠CAB=45°, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=14，AC=x，BC=x，∠CAB=45°，BC⊥AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),14)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 7*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠ADG=36°, ∠AGF=104°, ∠EFC=40°, AB is perpendicular to CB, GB⊥CB. Find the measure of ∠BFG.
问题：如图所示，∠ADG=36°，∠AGF=104°，∠EFC=40°，AB垂直于CB，GB垂直于CB。求∠BFG的大小。
Construction (CDL):
Shape(AD,DG,GA)
Shape(AG,GF)
Shape(FB,BF,FG)
Shape(GF,FE)
Shape(EF,FC)
Shape(FB,BC,CF)
Collinear(DGFC)
Collinear(BFE)
Conditions (CDL):
Equal(MeasureOfAngle(ADG),36)
Equal(MeasureOfAngle(AGF),104)
Equal(MeasureOfAngle(EFC),40)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(GB,CB)
Goal: Value(MeasureOfAngle(BFG))
Answer: 40
Theorem sequence:
vertical_angle(1,BFG,EFC)
Problem: As shown in the diagram, JK=2*b+3, JM=3*a, LK=21, LM=45, ∠MLJ=30°, ∠RLK=70°, JM and KL are opposite sides of the parallelogram JKLM, JK and ML are opposite sides of the parallelogram JMLK. Find the value of a.
问题：如图所示，JK=2*b+3，JM=3*a，LK=21，LM=45，∠MLJ=30°，∠RLK=70°，四边形JKLM是平行四边形，JMLK是▱。求a的值。
Construction (CDL):
Shape(JM,MR,RJ)
Shape(JR,RK,KJ)
Shape(RM,ML,LR)
Shape(KR,RL,LK)
Collinear(JRL)
Collinear(MRK)
Conditions (CDL):
Equal(LengthOfLine(JK),2*b+3)
Equal(LengthOfLine(JM),3*a)
Equal(LengthOfLine(LK),21)
Equal(LengthOfLine(LM),45)
Equal(MeasureOfAngle(MLJ),30)
Equal(MeasureOfAngle(RLK),70)
Parallelogram(JKLM)
Parallelogram(JMLK)
Goal: Value(a)
Answer: 7
Theorem sequence:
parallelogram_property_opposite_line_equal(1,JMLK)
Problem: As shown in the diagram, the measure of ⌒AQS is 103, the measure of arc AUT is 43. Find the measure of ∠SRQ.
问题：如图所示，⌒AQS的角度为103，弧AUT的角度为43。求∠SRQ的大小。
Construction (CDL):
Shape(US,ASU)
Shape(QT,ATQ)
Shape(UR,RT,ATU)
Shape(SU,AUT,TQ,AQS)
Collinear(RUSF)
Collinear(RTQ)
Cocircular(A,SUTQ)
Conditions (CDL):
Equal(MeasureOfArc(AQS),103)
Equal(MeasureOfArc(AUT),43)
Goal: Value(MeasureOfAngle(SRQ))
Answer: 30
Theorem sequence:
circle_property_circular_power_segment_and_segment_angle(1,RUS,RTQ,A)
Problem: As shown in the diagram, AB=y, BD=x, CD=12, ∠BCA=55°, ∠CAB=47°, BD⊥AD. Find the value of x.
问题：如图所示，AB=y，BD=x，CD=12，∠BCA=55°，∠CAB=47°，BD垂直于AD。求x的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA.AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CD),12)
Equal(MeasureOfAngle(BCA),55)
Equal(MeasureOfAngle(CAB),47)
PerpendicularBetweenLine(BD,AD)
Goal: Value(x)
Answer: 12/tan(7*pi/36)
Theorem sequence:
adjacent_complementary_angle(1,CDB,BDA)
triangle_property_angle_sum(1,BCD)
sine_theorem(1,DBC)
Problem: As shown in the diagram, ∠FDC=40°, the measure of ⌒ECA is x**2+12*x, the measure of ⌒EFC is x**2+2*x, DC is the tangent to circle E. Find the value of x.
问题：如图所示，∠FDC=40°，⌒ECA的角度为x**2+12*x，弧EFC的角度为x**2+2*x，⊙O的切线为DC。求x的值。
Construction (CDL):
Shape(FA,EAF)
Shape(FD,DC,ECF)
Shape(AF,EFC,ECA)
Collinear(DFA)
Cocircular(E,AFC)
Conditions (CDL):
Equal(MeasureOfAngle(FDC),40)
Equal(MeasureOfArc(ECA),x**2+12*x)
Equal(MeasureOfArc(EFC),x**2+2*x)
IsTangentOfCircle(DC,E)
Goal: Value(x)
Answer: 8
Theorem sequence:
circle_property_circular_power_tangent_and_segment_angle(2,DC,DFA,E)
Problem: As shown in the diagram, GR=7, PG=z+4, PT=3*y-1, TQ=2*x, TR=8, TF is the perpendicular bisector of RQ, TG is the perpendicular bisector of PR, TH is the perpendicular bisector of QP. Find the value of x.
问题：如图所示，GR=7，PG=z+4，PT=3*y-1，TQ=2*x，TR=8，TF垂直平分AB，TG是PR的垂直平分线，TH垂直平分AB。求x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(x)
Answer: 4
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,TF,RQ)
Problem: As shown in the diagram, ∠GKM=62°, ∠HBF=38°, NO∥MH. Find the measure of ∠KCN.
问题：如图所示，∠GKM=62°，∠HBF=38°，NO平行于MH。求∠KCN的大小。
Construction (CDL):
Shape(NC,CJ)
Shape(JC,CI)
Shape(IC,CO)
Shape(KC,CN)
Shape(OC,CB)
Shape(CK,KB,BC)
Shape(MK,KC)
Shape(CB,BH)
Shape(GK,KM)
Shape(BK,KG)
Shape(HB,BF)
Shape(FB,BK)
Collinear(NCO)
Collinear(MKBH)
Collinear(JCBF)
Collinear(ICKG)
Conditions (CDL):
Equal(MeasureOfAngle(GKM),62)
Equal(MeasureOfAngle(HBF),38)
ParallelBetweenLine(NO,MH)
Goal: Value(MeasureOfAngle(KCN))
Answer: 62
Theorem sequence:
parallel_property_collinear_extend(3,NO,MH,C)
parallel_property_collinear_extend(3,HM,CN,K)
parallel_property_corresponding_angle(1,KM,CN,G)
Problem: As shown in the diagram, XY=2*x+3, XY=ZY, XZ=8*x-4, ZY=4*x-1, YX and YZ are the legs of the isosceles triangle YXZ. Find the length of line XZ.
问题：如图所示，XY=2*x+3，XY=ZY，XZ=8*x-4，ZY=4*x-1，YX和YZ是等腰△YXZ的腰。求直线XZ的长度。
Construction (CDL):
Shape(YX,XZ,ZY)
Conditions (CDL):
Equal(LengthOfLine(XY),2*x+3)
Equal(LengthOfLine(XY),LengthOfLine(ZY))
Equal(LengthOfLine(XZ),8*x-4)
Equal(LengthOfLine(ZY),4*x-1)
IsoscelesTriangle(YXZ)
Goal: Value(LengthOfLine(XZ))
Answer: 12
Theorem sequence:
Problem: As shown in the diagram, VW=3, XV=2*a-2, ZV=5*a/4+1/4, ∠ZYV=53°, quadrilateral XWZY is a rhombus. Find the length of line XZ.
问题：如图所示，VW=3，XV=2*a-2，ZV=5*a/4+1/4，∠ZYV=53°，XWZY是菱形。求直线XZ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*a/4+1/4)
Equal(MeasureOfAngle(ZYV),53)
Rhombus(XWZY)
Goal: Value(LengthOfLine(XZ))
Answer: 8
Theorem sequence:
line_addition(1,XV,VZ)
kite_property_diagonal_perpendicular_bisection(1,YXWZ,V)
Problem: As shown in the diagram, AB=18, AC=x, BC=y, ∠CBA=30°, AC⊥BC. Find the value of x.
问题：如图所示，AB=18，AC=x，BC=y，∠CBA=30°，AC⊥BC。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),18)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(CBA),30)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 9
Theorem sequence:
sine_theorem(1,ACB)