SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠BPY=3*x°, the measure of ⌒PAC is 2*x+15, the measure of arc PCY is 3*x-3, P is the center of ⊙P. Find the measure of ⌒PAB.
问题：如图所示，∠BPY=3*x°，弧PAC的角度为2*x+15，弧PCY的角度为3*x-3，⊙P的圆心为P。求弧PAB的角度。
Construction (CDL):
Shape(PY,PYB,BP)
Shape(PB,PBX,XP)
Shape(PX,PXA,AP)
Shape(PA,PAC,CP)
Shape(PC,PCY,YP)
Collinear(BPA)
Cocircular(P,BXACY)
Conditions (CDL):
Equal(MeasureOfAngle(BPY),3*x)
Equal(MeasureOfArc(PAC),2*x+15)
Equal(MeasureOfArc(PCY),3*x-3)
IsCentreOfCircle(P,P)
Goal: Value(MeasureOfArc(PAB))
Answer: 180
Theorem sequence:
flat_angle(1,BPA)
arc_property_center_angle(1,PAB,P)
Problem: As shown in the diagram, the area of triangle CEA is 52, AE=b, CD=b+5, CD is the altitude of △ CEA, ED is perpendicular to CD. Find the length of line CD.
问题：如图所示，三角形CEA的面积为52，AE=b，CD=b+5，CD是三角形CEA的高，ED垂直于CD。求直线CD的长度。
Construction (CDL):
Shape(CE,ED,DC)
Shape(CD,DA,AC)
Collinear(EDA)
Conditions (CDL):
Equal(AreaOfTriangle(CEA),52)
Equal(LengthOfLine(AE),b)
Equal(LengthOfLine(CD),b+5)
IsAltitudeOfTriangle(CD,CEA)
PerpendicularBetweenLine(ED,CD)
Goal: Value(LengthOfLine(CD))
Answer: 13
Theorem sequence:
triangle_area_formula_common(1,CEA)
Problem: As shown in the diagram, AD=6, BD=w+4, CD=24, AB is perpendicular to CB, CD is perpendicular to BD. Find the value of w.
问题：如图所示，AD=6，BD=w+4，CD=24，AB垂直于CB，CD垂直于BD。求w的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(BD),w+4)
Equal(LengthOfLine(CD),24)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(CD,BD)
Goal: Value(w)
Answer: 8
Theorem sequence:
adjacent_complementary_angle(1,CDB,BDA)
triangle_property_angle_sum(1,ABC)
triangle_property_angle_sum(1,CDB)
similar_triangle_judgment_aa(1,BDA,CDB)
similar_triangle_property_line_ratio(1,BDA,CDB)
similar_triangle_property_line_ratio(1,ABD,BCD)
Problem: As shown in the diagram, CA=10, CI=15, ∠ACI=45°, IC and BA are opposite sides of the parallelogram IBAC, BE is perpendicular to IE. Find the area of IBAC.
问题：如图所示，CA=10，CI=15，∠ACI=45°，IC和BA是▱IBAC的一组对边，BE⊥IE。求IBAC的面积。
Construction (CDL):
Shape(IB,BE,EI)
Shape(IE,EA,AC,CI)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(CA),10)
Equal(LengthOfLine(CI),15)
Equal(MeasureOfAngle(ACI),45)
Parallelogram(IBAC)
PerpendicularBetweenLine(BE,IE)
Goal: Value(AreaOfQuadrilateral(IBAC))
Answer: 75*sqrt(2)
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BACI)
parallelogram_property_opposite_line_equal(1,IBAC)
parallelogram_property_opposite_line_equal(1,BACI)
sine_theorem(1,IBE)
altitude_of_quadrilateral_judgment_left_vertex(1,IE,IBAC)
parallelogram_area_formula_common(1,IBAC)
Problem: As shown in the diagram, ∠MLN=31°, ∠QPN=22°, NM is perpendicular to LM. Find the measure of ∠PNQ.
问题：如图所示，∠MLN=31°，∠QPN=22°，NM垂直于LM。求∠PNQ的大小。
Construction (CDL):
Shape(LN,NM,ML)
Shape(NQ,QP,PN)
Collinear(LNP)
Collinear(MNQ)
Conditions (CDL):
Equal(MeasureOfAngle(MLN),31)
Equal(MeasureOfAngle(QPN),22)
PerpendicularBetweenLine(NM,LM)
Goal: Value(MeasureOfAngle(PNQ))
Answer: 59
Theorem sequence:
triangle_property_angle_sum(1,LNM)
vertical_angle(1,LNM,PNQ)
Problem: As shown in the diagram, AD=21, CD=7, BD⊥CD, CB is perpendicular to AB. Find the length of line BD.
问题：如图所示，AD=21，CD=7，BD⊥CD，CB⊥AB。求直线BD的长度。
Construction (CDL):
Shape(AD,DB,BA)
Shape(BD,DC,CB)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AD),21)
Equal(LengthOfLine(CD),7)
PerpendicularBetweenLine(BD,CD)
PerpendicularBetweenLine(CB,AB)
Goal: Value(LengthOfLine(BD))
Answer: 7*sqrt(3)
Theorem sequence:
line_addition(1,AD,DC)
mirror_similar_triangle_judgment_aa(1,ACB,BDC)
mirror_similar_triangle_property_line_ratio(1,ACB,BDC)
mirror_similar_triangle_property_line_ratio(1,BAC,DCB)
right_triangle_judgment_angle(1,BDC)
right_triangle_property_pythagorean(1,BDC)
Problem: As shown in the diagram, ∠AED=3*y+36°, ∠CAE=25*x+20°, ∠DCA=9*y-12°, ∠EDC=12*x+72°, DE and CA are opposite sides of the ▱ DCAE. Find the value of y.
问题：如图所示，∠AED=3*y+36°，∠CAE=25*x+20°，∠DCA=9*y-12°，∠EDC=12*x+72°，DCAE是平行四边形。求y的值。
Construction (CDL):
Shape(DC,CA,AE,ED)
Conditions (CDL):
Equal(MeasureOfAngle(AED),3*y+36)
Equal(MeasureOfAngle(CAE),25*x+20)
Equal(MeasureOfAngle(DCA),9*y-12)
Equal(MeasureOfAngle(EDC),12*x+72)
Parallelogram(DCAE)
Goal: Value(y)
Answer: 8
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,CAED)
Problem: As shown in the diagram, ∠DOB=∠EOD, ∠EOD=∠FOE, ∠FOE=∠AOF, O is the center of circle O, CO is perpendicular to AO. Find the measure of arc OCF.
问题：如图所示，∠DOB=∠EOD，∠EOD=∠FOE，∠FOE=∠AOF，O是圆O的圆心，CO⊥AO。求弧OCF的角度。
Construction (CDL):
Shape(OB,OBD,DO)
Shape(OD,ODE,EO)
Shape(OE,OEF,FO)
Shape(OF,OFA,AO)
Shape(OA,OAC,OC)
Shape(OC,OCB,BO)
Collinear(AOB)
Collinear(COE)
Cocircular(O,BDEFAC)
Conditions (CDL):
Equal(MeasureOfAngle(DOB),MeasureOfAngle(EOD))
Equal(MeasureOfAngle(EOD),MeasureOfAngle(FOE))
Equal(MeasureOfAngle(FOE),MeasureOfAngle(AOF))
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CO,AO)
Goal: Value(MeasureOfArc(O,CF))
Answer: 225
Theorem sequence:
vertical_angle(1,EOB,COA)
angle_addition(1,EOD,DOB)
flat_angle(1,EOC)
angle_addition(1,FOE,EOC)
arc_property_center_angle(1,OCF,O)
Problem: As shown in the diagram, AB=57, KL=21, A is the midpoint of segment KJ, B bisects segment LM, KJ and ML are the non-parallel sides (legs) of trapezoid KJML. Find the length of line JM.
问题：如图所示，AB=57，KL=21，A是线段KJ的中点，B是线段LM的中点，四边形KJML是梯形。求直线JM的长度。
Construction (CDL):
Shape(KA,AB,BL,LK)
Shape(AJ,JM,MB,BA)
Collinear(KAJ)
Collinear(LBM)
Conditions (CDL):
Equal(LengthOfLine(AB),57)
Equal(LengthOfLine(KL),21)
IsMidpointOfLine(A,KJ)
IsMidpointOfLine(B,LM)
Trapezoid(KJML)
Goal: Value(LengthOfLine(JM))
Answer: 93
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,AB,KJML)
midsegment_of_quadrilateral_property_length(1,AB,KJML)
Problem: As shown in the diagram, AZ=y, QZ=z, RQ=12, RS=10, RZ=x, ∠SPA=45°, ∠ZQR=30°, SR is parallel to AZ, PA⊥SA, RZ⊥QZ. Find the value of x.
问题：如图所示，AZ=y，QZ=z，RQ=12，RS=10，RZ=x，∠SPA=45°，∠ZQR=30°，SR∥AZ，PA垂直于SA，RZ⊥QZ。求x的值。
Construction (CDL):
Shape(PA,AS,SP)
Shape(SA,AZ,ZR,RS)
Shape(RZ,ZQ,QR)
Collinear(PAZQ)
Conditions (CDL):
Equal(LengthOfLine(AZ),y)
Equal(LengthOfLine(QZ),z)
Equal(LengthOfLine(RQ),12)
Equal(LengthOfLine(RS),10)
Equal(LengthOfLine(RZ),x)
Equal(MeasureOfAngle(SPA),45)
Equal(MeasureOfAngle(ZQR),30)
ParallelBetweenLine(SR,AZ)
PerpendicularBetweenLine(PA,SA)
PerpendicularBetweenLine(RZ,QZ)
Goal: Value(x)
Answer: 6
Theorem sequence:
sine_theorem(1,RZQ)
Problem: As shown in the diagram, AB=15, AC=3, BC⊥AC. Find the measure of ∠CAB.
问题：如图所示，AB=15，AC=3，BC⊥AC。求∠CAB的大小。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AC),3)
PerpendicularBetweenLine(BC,AC)
Goal: Value(MeasureOfAngle(CAB))
Answer: 180*acos(1/5)/pi
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,ABC)
Problem: As shown in the diagram, DE=8, EH=8, quadrilateral KDEH is a ▱, DE is perpendicular to HE, EH is perpendicular to KH, HK⊥DK, KD⊥ED. Find the perimeter of quadrilateral KDEH.
问题：如图所示，DE=8，EH=8，KH和DE是▱KDEH的一组对边，DE⊥HE，EH⊥KH，HK⊥DK，KD⊥ED。求KDEH的周长。
Construction (CDL):
Shape(KD,DE,EH,HK)
Conditions (CDL):
Equal(LengthOfLine(DE),8)
Equal(LengthOfLine(EH),8)
Parallelogram(KDEH)
PerpendicularBetweenLine(DE,HE)
PerpendicularBetweenLine(EH,KH)
PerpendicularBetweenLine(HK,DK)
PerpendicularBetweenLine(KD,ED)
Goal: Value(PerimeterOfQuadrilateral(KDEH))
Answer: 32
Theorem sequence:
parallelogram_property_opposite_line_equal(1,KDEH)
parallelogram_property_opposite_line_equal(1,DEHK)
quadrilateral_perimeter_formula(1,KDEH)
Problem: As shown in the diagram, AG=Mul(LengthOfLine(HA),2), BH=Mul(LengthOfLine(EB),2), CE=Mul(LengthOfLine(CF),2), DF=Mul(LengthOfLine(DG),2), EH=9, BADC is a square, quadrilateral EHGF is a square. Find the ratio of the area of quadrilateral BADC to the area of quadrilateral EHGF.
问题：如图所示，AG=Mul(LengthOfLine(HA),2)，BH=Mul(LengthOfLine(EB),2)，CE=Mul(LengthOfLine(CF),2)，DF=Mul(LengthOfLine(DG),2)，EH=9，BADC是正方形，EHGF是正方形。求BADC的面积与四边形EHGF的面积之比。
Construction (CDL):
Shape(CE,EB,BC)
Shape(BH,HA,AB)
Shape(AG,GD,DA)
Shape(CD,DF,FC)
Shape(BA,AD,DC,CB)
Collinear(EBH)
Collinear(HAG)
Collinear(GDF)
Collinear(FCE)
Conditions (CDL):
Equal(LengthOfLine(AG),Mul(LengthOfLine(HA),2))
Equal(LengthOfLine(BH),Mul(LengthOfLine(EB),2))
Equal(LengthOfLine(CE),Mul(LengthOfLine(CF),2))
Equal(LengthOfLine(DF),Mul(LengthOfLine(DG),2))
Equal(LengthOfLine(EH),9)
Square(BADC)
Square(EHGF)
Goal: Value(Div(AreaOfQuadrilateral(BADC),AreaOfQuadrilateral(EHGF)))
Answer: 5/9
Theorem sequence:
line_addition(1,EB,BH)
line_addition(1,HA,AG)
right_triangle_judgment_angle(1,BHA)
right_triangle_property_pythagorean(1,BHA)
parallelogram_area_formula_sine(1,BADC)
parallelogram_area_formula_sine(1,EHGF)
Problem: As shown in the diagram, ∠ABC=x°, ∠BCD=2*x+18°, ∠CDA=x°, ∠DAB=4*x-26°. Find the measure of ∠CDA.
问题：如图所示，∠ABC=x°，∠BCD=2*x+18°，∠CDA=x°，∠DAB=4*x-26°。求∠CDA的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),x)
Equal(MeasureOfAngle(BCD),2*x+18)
Equal(MeasureOfAngle(CDA),x)
Equal(MeasureOfAngle(DAB),4*x-26)
Goal: Value(MeasureOfAngle(CDA))
Answer: 46
Theorem sequence:
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, RT=13, XY=7, ∠ZXY=124°, XZ is the midsegment of triangle SRT, YX is the midsegment of △ RTS. Find the measure of ∠RYX.
问题：如图所示，RT=13，XY=7，∠ZXY=124°，XZ是△SRT的中位线，YX是△RTS的中位线。求∠RYX的大小。
Construction (CDL):
Shape(SX,XZ,ZS)
Shape(XR,RY,YX)
Shape(XY,YT,TZ,ZX)
Collinear(RXS)
Collinear(RYT)
Collinear(SZT)
Conditions (CDL):
Equal(LengthOfLine(RT),13)
Equal(LengthOfLine(XY),7)
Equal(MeasureOfAngle(ZXY),124)
IsMidsegmentOfTriangle(XZ,SRT)
IsMidsegmentOfTriangle(YX,RTS)
Goal: Value(MeasureOfAngle(RYX))
Answer: 124
Theorem sequence:
midsegment_of_triangle_property_parallel(1,XZ,SRT)
parallel_property_collinear_extend(3,TR,ZX,Y)
parallel_property_ipsilateral_internal_angle(1,XZ,YT)
flat_angle(1,RYT)
angle_addition(1,RYX,XYT)
Problem: As shown in the diagram, ∠FBH=38°, ∠GKM=62°, NO is parallel to MH. Find the measure of ∠CKH.
问题：如图所示，∠FBH=38°，∠GKM=62°，NO∥MH。求∠CKH的大小。
Construction (CDL):
Shape(CK,KB,BC)
Collinear(NCO)
Collinear(MKBH)
Collinear(GKCI)
Collinear(JCBF)
Conditions (CDL):
Equal(MeasureOfAngle(FBH),38)
Equal(MeasureOfAngle(GKM),62)
ParallelBetweenLine(NO,MH)
Goal: Value(MeasureOfAngle(CKH))
Answer: 62
Theorem sequence:
vertical_angle(1,CKH,GKM)
Problem: As shown in the diagram, RS=5, SP=3, ∠PQR=128°, PS and QR are opposite sides of the ▱ PQRS. Find the length of line QR.
问题：如图所示，RS=5，SP=3，∠PQR=128°，PQRS是▱。求直线QR的长度。
Construction (CDL):
Shape(QR,RS,SP,PQ)
Conditions (CDL):
Equal(LengthOfLine(RS),5)
Equal(LengthOfLine(SP),3)
Equal(MeasureOfAngle(PQR),128)
Parallelogram(PQRS)
Goal: Value(LengthOfLine(QR))
Answer: 3
Theorem sequence:
parallelogram_property_opposite_line_equal(1,QRSP)
Problem: As shown in the diagram, AB=x, AC=8, BC=y, ∠BCA=60°, CA is perpendicular to BA. Find the value of y.
问题：如图所示，AB=x，AC=8，BC=y，∠BCA=60°，CA垂直于BA。求y的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 16
Theorem sequence:
triangle_property_angle_sum(1,CAB)
sine_theorem(1,CAB)
Problem: As shown in the diagram, the diameter of circle P is 9, ∠RPS=130°, ∠TPQ=112°, P is the center of circle P. Find the length of arc PQT.
问题：如图所示，圆P的直径为9，∠RPS=130°，∠TPQ=112°，⊙P的圆心为P。求弧PQT的长度。
Construction (CDL):
Shape(PR,PRQ,QP)
Shape(PQ,PQT,TP)
Shape(PT,PTS,SP)
Shape(PS,PSR,RP)
Cocircular(P,RQTS)
Conditions (CDL):
Equal(DiameterOfCircle(P),9)
Equal(MeasureOfAngle(RPS),130)
Equal(MeasureOfAngle(TPQ),112)
IsCentreOfCircle(P,P)
Goal: Value(LengthOfArc(P,QT))
Answer: 14*pi/5
Theorem sequence:
arc_property_center_angle(1,PQT,P)
circle_property_length_of_radius_and_diameter(1,P)
arc_length_formula(1,PQT)
Problem: As shown in the diagram, AB=x, AC=AD, AD=46, ∠ACB=73°, AB is perpendicular to DB. Find the value of x.
问题：如图所示，AB=x，AC=AD，AD=46，∠ACB=73°，AB垂直于DB。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Shape(AB,BD,DA)
Collinear(CBD)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),LengthOfLine(AD))
Equal(LengthOfLine(AD),46)
Equal(MeasureOfAngle(ACB),73)
PerpendicularBetweenLine(AB,DB)
Goal: Value(x)
Answer: 46*sin(73*pi/180)
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ACD)
isosceles_triangle_property_angle_equal(1,ACD)
sine_theorem(1,ABD)
Problem: As shown in the diagram, RQ=12, RS=x, TS=8, Q is the center of circle Q, SR is the tangent to ⊙Q. Find the value of x.
问题：如图所示，RQ=12，RS=x，TS=8，Q是⊙Q的圆心，SR是⊙O的切线。求x的值。
Construction (CDL):
Shape(QR,RS,SQ)
Shape(QR,QRT,TQ)
Shape(QT,QTR,RQ)
Collinear(ARS)
Collinear(QTS)
Cocircular(Q,TR)
Conditions (CDL):
Equal(LengthOfLine(RQ),12)
Equal(LengthOfLine(RS),x)
Equal(LengthOfLine(TS),8)
IsCentreOfCircle(Q,Q)
IsTangentOfCircle(SR,Q)
Goal: Value(x)
Answer: 16
Theorem sequence:
tangent_of_circle_property_perpendicular(1,SR,Q,Q)
radius_of_circle_property_length_equal(1,QR,Q)
radius_of_circle_property_length_equal(1,QT,Q)
line_addition(1,QT,TS)
right_triangle_judgment_angle(1,QRS)
right_triangle_property_pythagorean(1,QRS)
Problem: As shown in the diagram, RQ=5*x, RQ=SQ, RS=6*x-5, RS=RQ, SQ=3*x+10. Find the length of line QR.
问题：如图所示，RQ=5*x，RQ=SQ，RS=6*x-5，RS=RQ，SQ=3*x+10。求直线QR的长度。
Construction (CDL):
Shape(RQ,QS,SR)
Conditions (CDL):
Equal(LengthOfLine(RQ),5*x)
Equal(LengthOfLine(RQ),LengthOfLine(SQ))
Equal(LengthOfLine(RS),6*x-5)
Equal(LengthOfLine(RS),LengthOfLine(RQ))
Equal(LengthOfLine(SQ),3*x+10)
Goal: Value(LengthOfLine(QR))
Answer: 25
Theorem sequence:
Problem: As shown in the diagram, ∠DCB=64°, ADCB is a rhombus. Find the measure of ∠BAC.
问题：如图所示，∠DCB=64°，ADCB是菱形。求∠BAC的大小。
Construction (CDL):
Shape(AD,DO,OA)
Shape(DC,CO,OD)
Shape(OC,CB,BO)
Shape(OB,BA,AO)
Collinear(AOC)
Collinear(DOB)
Conditions (CDL):
Equal(MeasureOfAngle(DCB),64)
Rhombus(ADCB)
Goal: Value(MeasureOfAngle(BAC))
Answer: 32
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,ADCB)
kite_property_diagonal_perpendicular_bisection(1,ADCB,O)
perpendicular_bisector_property_bisector(1,AO,DB)
angle_addition(1,BAC,CAD)
Problem: As shown in the diagram, JM=JK, ML=LK, ∠KJM=59°, ∠MLK=67°. Find the measure of ∠LKJ.
问题：如图所示，JM=JK，ML=LK，∠KJM=59°，∠MLK=67°。求∠LKJ的大小。
Construction (CDL):
Shape(JM,MK,KJ)
Shape(ML,LK,KM)
Conditions (CDL):
Equal(LengthOfLine(JM),LengthOfLine(JK))
Equal(LengthOfLine(ML),LengthOfLine(LK))
Equal(MeasureOfAngle(KJM),59)
Equal(MeasureOfAngle(MLK),67)
Goal: Value(MeasureOfAngle(LKJ))
Answer: 117
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,LKM)
isosceles_triangle_property_angle_equal(1,LKM)
isosceles_triangle_judgment_line_equal(1,JMK)
isosceles_triangle_property_angle_equal(1,JMK)
triangle_property_angle_sum(1,JMK)
triangle_property_angle_sum(1,MLK)
angle_addition(1,LKM,MKJ)
Problem: As shown in the diagram, ∠PON=124°, MPON is a rhombus. Find the measure of ∠POM.
问题：如图所示，∠PON=124°，MPON是菱形。求∠POM的大小。
Construction (CDL):
Shape(MP,PR,RM)
Shape(PO,OR,RP)
Shape(RO,ON,NR)
Shape(RN,NM,MR)
Collinear(MRO)
Collinear(PRN)
Conditions (CDL):
Equal(MeasureOfAngle(PON),124)
Rhombus(MPON)
Goal: Value(MeasureOfAngle(POM))
Answer: 62
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,ONMP,R)
perpendicular_bisector_property_bisector(1,OR,NP)
angle_addition(1,POR,RON)
Problem: As shown in the diagram, AB=4*z, AT=2*x-5, AU=3, DA=6, EA=y, FA=5, S bisects segment ED, T is the midpoint of segment EF, U bisects segment DF. Find the value of y.
问题：如图所示，AB=4*z，AT=2*x-5，AU=3，DA=6，EA=y，FA=5，S是线段ED的中点，T平分线段EF，U是线段DF的中点。求y的值。
Construction (CDL):
Shape(SA,AE,ES)
Shape(SD,DA,AS)
Shape(AD,DU,UA)
Shape(AU,UF,FA)
Shape(AF,FT,TA)
Shape(AT,TE,EA)
Collinear(DSE)
Collinear(DUF)
Collinear(ETF)
Collinear(SAF)
Collinear(DAT)
Collinear(EAU)
Conditions (CDL):
Equal(LengthOfLine(AB),4*z)
Equal(LengthOfLine(AT),2*x-5)
Equal(LengthOfLine(AU),3)
Equal(LengthOfLine(DA),6)
Equal(LengthOfLine(EA),y)
Equal(LengthOfLine(FA),5)
IsMidpointOfLine(S,ED)
IsMidpointOfLine(T,EF)
IsMidpointOfLine(U,DF)
Goal: Value(y)
Answer: 6
Theorem sequence:
median_of_triangle_judgment(1,DT,DFE)
median_of_triangle_judgment(1,FS,FED)
centroid_of_triangle_judgment_intersection(1,A,EDF,S,T)
centroid_of_triangle_property_line_ratio(1,A,EDF,U)
Problem: As shown in the diagram, AB=y, AC=x, BC=3, ∠CAB=45°, BC⊥AC. Find the value of y.
问题：如图所示，AB=y，AC=x，BC=3，∠CAB=45°，BC垂直于AC。求y的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),3)
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 3*sqrt(2)
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, ∠GHI=∠LMN, ∠GJI=y+30°, ∠HGJ=87°, ∠HIJ=∠MNO, ∠IHG=98°, ∠IJG=∠NOL, ∠LON=60°, ∠MLO=x-4°, quadrilateral HGJI is similar to quadrilateral MLON. Find the value of x.
问题：如图所示，∠GHI=∠LMN，∠GJI=y+30°，∠HGJ=87°，∠HIJ=∠MNO，∠IHG=98°，∠IJG=∠NOL，∠LON=60°，∠MLO=x-4°，HGJI与MLON相似。求x的值。
Construction (CDL):
Shape(HG,GJ,JI,IH)
Shape(ML,LO,ON,NM)
Conditions (CDL):
Equal(MeasureOfAngle(GHI),MeasureOfAngle(LMN))
Equal(MeasureOfAngle(GJI),y+30)
Equal(MeasureOfAngle(HGJ),87)
Equal(MeasureOfAngle(HIJ),MeasureOfAngle(MNO))
Equal(MeasureOfAngle(IHG),98)
Equal(MeasureOfAngle(IJG),MeasureOfAngle(NOL))
Equal(MeasureOfAngle(LON),60)
Equal(MeasureOfAngle(MLO),x-4)
SimilarBetweenQuadrilateral(HGJI,MLON)
Goal: Value(x)
Answer: 91
Theorem sequence:
similar_quadrilateral_property_angle_equal(1,GJIH,LONM)
Problem: As shown in the diagram, AB=x, AC=20, BC=20, ∠CAB=45°, BC⊥AC. Find the value of x.
问题：如图所示，AB=x，AC=20，BC=20，∠CAB=45°，BC垂直于AC。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BC),20)
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 20*sqrt(2)
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, BA=5*x, DC=3*x+6, the measure of ⌒OBA is 127, the measure of arc OCD is 127. Find the value of x.
问题：如图所示，BA=5*x，DC=3*x+6，⌒OBA的角度为127，弧OCD的角度为127。求x的值。
Construction (CDL):
Shape(OB,BA,AO)
Shape(OC,CD,DO)
Cocircular(O,BACD)
Conditions (CDL):
Equal(LengthOfLine(BA),5*x)
Equal(LengthOfLine(DC),3*x+6)
Equal(MeasureOfArc(OBA),127)
Equal(MeasureOfArc(OCD),127)
Goal: Value(x)
Answer: 3
Theorem sequence:
congruent_arc_judgment_measure_equal(1,OBA,OCD)
congruent_arc_property_chord_equal(1,OBA,OCD)
Problem: As shown in the diagram, AB=30, CD=30, the measure of ⌒XCZ is 40, X is the center of circle X, AM is perpendicular to YM, DN⊥ZN. Find the length of line AM.
问题：如图所示，AB=30，CD=30，弧XCZ的角度为40，圆X的圆心为X，AM垂直于YM，DN垂直于ZN。求直线AM的长度。
Construction (CDL):
Shape(XYA,AM,MY)
Shape(XAC,CN,NX,XM,MA)
Shape(XCZ,ZN,NC)
Shape(XZD,DN,NZ)
Shape(XDB,BM,MX,XN,ND)
Shape(XBY,YM,MB)
Collinear(AMB)
Collinear(YMX)
Collinear(CND)
Collinear(XNZ)
Cocircular(X,ACZDBY)
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(AM))
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, LM=x, LN=12, MN=16, the center of circle L is L, the tangent to circle L is MN. Find the value of x.
问题：如图所示，LM=x，LN=12，MN=16，L是⊙L的圆心，MN是⊙O的切线。求x的值。
Construction (CDL):
Shape(NM,ML,LN)
Cocircular(L,N)
Conditions (CDL):
Equal(LengthOfLine(LM),x)
Equal(LengthOfLine(LN),12)
Equal(LengthOfLine(MN),16)
IsCentreOfCircle(L,L)
IsTangentOfCircle(MN,L)
Goal: Value(x)
Answer: 20
Theorem sequence:
tangent_of_circle_property_perpendicular(1,MN,L,L)
right_triangle_judgment_angle(1,LNM)
right_triangle_property_pythagorean(1,LNM)
Problem: As shown in the diagram, AB=8, AC=8, BC=7, BH=6, DE=4, HF=4, △BAC is similar to △EDF.. Find the ratio of similar △BAC and △EDF.
问题：如图所示，AB=8，AC=8，BC=7，BH=6，DE=4，HF=4，△BAC相似于△EDF。求△BAC和△EDF的相似比。
Construction (CDL):
Shape(BA,AC,CB)
Shape(ED,DF,FE)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),7)
Equal(LengthOfLine(BH),6)
Equal(LengthOfLine(DE),4)
Equal(LengthOfLine(HF),4)
SimilarBetweenTriangle(BAC,EDF)
Goal: Value(RatioOfSimilarTriangle(BAC,EDF))
Answer: 2
Theorem sequence:
similar_triangle_property_line_ratio(1,CBA,FED)
Problem: As shown in the diagram, the diameter of ⊙B is 20, BD=5*x, ∠ACE=45°, B is the center of circle B, BD⊥ED. Find the value of x.
问题：如图所示，⊙B的直径为20，BD=5*x，∠ACE=45°，⊙B的圆心为B，BD垂直于ED。求x的值。
Construction (CDL):
Shape(BAC,CB,BA)
Shape(BC,CD,DB)
Shape(BCE,ED,DC)
Shape(BEA,AB,BD,DE)
Collinear(ABC)
Collinear(CDE)
Cocircular(B,ACE)
Conditions (CDL):
Equal(DiameterOfCircle(B),20)
Equal(LengthOfLine(BD),5*x)
Equal(MeasureOfAngle(ACE),45)
IsCentreOfCircle(B,B)
PerpendicularBetweenLine(BD,ED)
Goal: Value(x)
Answer: sqrt(2)
Theorem sequence:
circle_property_length_of_radius_and_diameter(1,B)
adjacent_complementary_angle(1,CDB,BDE)
right_triangle_judgment_angle(1,CDB)
right_triangle_property_pythagorean(1,CDB)
radius_of_circle_property_length_equal(1,BC,B)
triangle_property_angle_sum(1,CDB)
isosceles_triangle_judgment_angle_equal(1,DBC)
Problem: As shown in the diagram, AB=2*x, AC=XC, BC=10, MN=5, MP=x, NP=PQ, ∠BAC=∠MNP, ∠CXB=∠PQM. Find the value of x.
问题：如图所示，AB=2*x，AC=XC，BC=10，MN=5，MP=x，NP=PQ，∠BAC=∠MNP，∠CXB=∠PQM。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Shape(BC,CX,XB)
Shape(MN,NP,PM)
Shape(MP,PQ,QM)
Collinear(ACX)
Collinear(NPQ)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x)
Equal(LengthOfLine(AC),LengthOfLine(XC))
Equal(LengthOfLine(BC),10)
Equal(LengthOfLine(MN),5)
Equal(LengthOfLine(MP),x)
Equal(LengthOfLine(NP),LengthOfLine(PQ))
Equal(MeasureOfAngle(BAC),MeasureOfAngle(MNP))
Equal(MeasureOfAngle(CXB),MeasureOfAngle(PQM))
Goal: Value(x)
Answer: 5
Theorem sequence:
similar_triangle_judgment_aa(1,BAX,MNQ)
similar_triangle_property_line_ratio(1,BAX,MNQ)
similar_triangle_property_line_ratio(1,XBA,QMN)
line_addition(1,AC,CX)
line_addition(1,NP,PQ)
similar_triangle_judgment_sas(1,ACB,NPM)
similar_triangle_property_line_ratio(1,CBA,PMN)
similar_triangle_property_line_ratio(1,ACB,NPM)
Problem: As shown in the diagram, ∠AJH=z°, ∠FHG=x°, ∠HGA=y°, the measure of arc AHG is 78, A is the center of circle A, GB⊥HB, JH is perpendicular to GH. Find the value of z.
问题：如图所示，∠AJH=z°，∠FHG=x°，∠HGA=y°，弧AHG的角度为78，⊙A的圆心为A，GB⊥HB，JH垂直于GH。求z的值。
Construction (CDL):
Shape(AGF,FAB,BG)
Shape(AFJ,JA,AB,BF)
Shape(AJ,JH,HA)
Shape(BA,AH,HB)
Shape(GB,BH,HG)
Shape(AJH,HJ)
Shape(AHG,GH)
Collinear(FBH)
Collinear(GBAJ)
Cocircular(A,FJHG)
Conditions (CDL):
Equal(MeasureOfAngle(AJH),z)
Equal(MeasureOfAngle(FHG),x)
Equal(MeasureOfAngle(HGA),y)
Equal(MeasureOfArc(AHG),78)
IsCentreOfCircle(A,A)
PerpendicularBetweenLine(GB,HB)
PerpendicularBetweenLine(JH,GH)
Goal: Value(z)
Answer: 39
Theorem sequence:
radius_of_circle_property_length_equal(1,AH,A)
radius_of_circle_property_length_equal(1,AG,A)
isosceles_triangle_judgment_line_equal(1,AHG)
isosceles_triangle_property_angle_equal(1,AHG)
arc_property_center_angle(1,AHG,A)
triangle_property_angle_sum(1,AHG)
triangle_property_angle_sum(1,GJH)
Problem: As shown in the diagram, LJ=3*y-7, LM=9*x, NJ=y+15, NK=36, quadrilateral KNML is a parallelogram. Find the value of x.
问题：如图所示，LJ=3*y-7，LM=9*x，NJ=y+15，NK=36，四边形KNML是▱。求x的值。
Construction (CDL):
Shape(KN,NJ,JK)
Shape(JN,NM,MJ)
Shape(JM,ML,LJ)
Shape(JL,LK,KJ)
Collinear(KJM)
Collinear(NJL)
Conditions (CDL):
Equal(LengthOfLine(LJ),3*y-7)
Equal(LengthOfLine(LM),9*x)
Equal(LengthOfLine(NJ),y+15)
Equal(LengthOfLine(NK),36)
Parallelogram(KNML)
Goal: Value(x)
Answer: 4
Theorem sequence:
parallelogram_property_opposite_line_equal(1,KNML)
Problem: As shown in the diagram, DF=2*(x+5)-7, EG=3*(x-2), DG⊥GG, quadrilateral DGFE is a rectangle. Find the length of line EG.
问题：如图所示，DF=2*(x+5)-7，EG=3*(x-2)，DG垂直于GG，DGFE是长方形。求直线EG的长度。
Construction (CDL):
Shape(DG,GB,BD)
Shape(BG,GF,FB)
Shape(BF,FE,EB)
Shape(BE,ED,DB)
Collinear(DBF)
Collinear(GBE)
Conditions (CDL):
Equal(LengthOfLine(DF),2*(x+5)-7)
Equal(LengthOfLine(EG),3*(x-2))
PerpendicularBetweenLine(DG,GF)
Rectangle(DGFE)
Goal: Value(LengthOfLine(EG))
Answer: 21
Theorem sequence:
rectangle_property_diagonal_equal(1,DGFE)