SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠BGC=100°, ∠DCF=75°. Find the measure of ∠GAC.
问题：如图所示，∠BGC=100°，∠DCF=75°。求∠GAC的大小。
Construction (CDL):
Shape(GA,AC,CG)
Collinear(EGCF)
Collinear(AGB)
Collinear(ACD)
Shape(BG,GC)
Shape(DC,CF)
Conditions (CDL):
Equal(MeasureOfAngle(BGC),100)
Equal(MeasureOfAngle(DCF),75)
Goal: Value(MeasureOfAngle(GAC))
Answer: 25
Theorem sequence:
flat_angle(1,BGA)
angle_addition(1,BGC,CGA)
vertical_angle(1,ACE,DCF)
triangle_property_angle_sum(1,GAC)
Problem: As shown in the diagram, ∠EFA=63°, F is the center of circle F, DF⊥EF. Find the measure of arc FAD.
问题：如图所示，∠EFA=63°，F是⊙F的圆心，DF垂直于EF。求⌒FAD的角度。
Construction (CDL):
Shape(FCB,BF,FC)
Shape(FBA,AF,FB)
Shape(FAE,EF,FA)
Shape(FED,DF,FE)
Shape(FDC,CF,FD)
Cocircular(F,CBAED)
Conditions (CDL):
Equal(MeasureOfAngle(EFA),63)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(DF,EF)
Goal: Value(MeasureOfArc(FAD))
Answer: 153
Theorem sequence:
angle_addition(1,DFE,EFA)
arc_property_center_angle(1,FAD,F)
Problem: As shown in the diagram, AE=2*x+1, CD=DB, EB=3*x-5, CA⊥EA, DE is perpendicular to BE. Find the value of x.
问题：如图所示，AE=2*x+1，CD=DB，EB=3*x-5，CA垂直于EA，DE垂直于BE。求x的值。
Construction (CDL):
Shape(BD,DE,EB)
Shape(DC,CA,AE,ED)
Collinear(CDB)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AE),2*x+1)
Equal(LengthOfLine(CD),LengthOfLine(DB))
Equal(LengthOfLine(EB),3*x-5)
PerpendicularBetweenLine(CA,EA)
PerpendicularBetweenLine(DE,BE)
Goal: Value(x)
Answer: 6
Theorem sequence:
parallel_judgment_corresponding_angle(2,AC,ED,B)
parallel_property_corresponding_angle(1,DE,CA,B)
similar_triangle_judgment_aa(1,EBD,ABC)
line_addition(1,BD,DC)
line_addition(1,BE,EA)
similar_triangle_property_line_ratio(1,EBD,ABC)
similar_triangle_property_line_ratio(1,DEB,CAB)
Problem: As shown in the diagram, AB=BY, AB=x, AY=8, ∠YAB=y°, AB⊥YB. Find the value of x.
问题：如图所示，AB=BY，AB=x，AY=8，∠YAB=y°，AB⊥YB。求x的值。
Construction (CDL):
Shape(AB,BY,YA)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(BY))
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AY),8)
Equal(MeasureOfAngle(YAB),y)
PerpendicularBetweenLine(AB,YB)
Goal: Value(x)
Answer: 4*sqrt(2)
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,BYA)
isosceles_triangle_property_angle_equal(1,BYA)
triangle_property_angle_sum(1,ABY)
sine_theorem(1,ABY)
Problem: As shown in the diagram, JH=5, JH=y+2, LV=2*x+6, ML=20-5*x, ML=LV, MN=4, QO=3. Find the value of x.
问题：如图所示，JH=5，JH=y+2，LV=2*x+6，ML=20-5*x，ML=LV，MN=4，QO=3。求x的值。
Construction (CDL):
Shape(ML,LQ,QB,BM)
Shape(QL,LV,VU,UQ)
Collinear(NMLVS)
Collinear(ABQUY)
Collinear(GMBP)
Collinear(CLQO)
Collinear(JVUH)
Conditions (CDL):
Equal(LengthOfLine(JH),5)
Equal(LengthOfLine(JH),y+2)
Equal(LengthOfLine(LV),2*x+6)
Equal(LengthOfLine(ML),20-5*x)
Equal(LengthOfLine(ML),LengthOfLine(LV))
Equal(LengthOfLine(MN),4)
Equal(LengthOfLine(QO),3)
Goal: Value(x)
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, ∠GJI=∠LON, ∠GJI=y+30°, ∠HGJ=87°, ∠IHG=98°, ∠IHG=∠NML, ∠JIH=∠ONM, ∠LON=60°, ∠MLO=x-4°, HGJI is similar to MLON. Find the value of y.
问题：如图所示，∠GJI=∠LON，∠GJI=y+30°，∠HGJ=87°，∠IHG=98°，∠IHG=∠NML，∠JIH=∠ONM，∠LON=60°，∠MLO=x-4°，HGJI与MLON相似。求y的值。
Construction (CDL):
Shape(IH,HG,GJ,JI)
Shape(NM,ML,LO,ON)
Conditions (CDL):
Equal(MeasureOfAngle(GJI),MeasureOfAngle(LON))
Equal(MeasureOfAngle(GJI),y+30)
Equal(MeasureOfAngle(HGJ),87)
Equal(MeasureOfAngle(IHG),98)
Equal(MeasureOfAngle(IHG),MeasureOfAngle(NML))
Equal(MeasureOfAngle(JIH),MeasureOfAngle(ONM))
Equal(MeasureOfAngle(LON),60)
Equal(MeasureOfAngle(MLO),x-4)
SimilarBetweenQuadrilateral(HGJI,MLON)
Goal: Value(y)
Answer: 30
Theorem sequence:
Problem: As shown in the diagram, ∠ACE=∠EAC, ∠DBG=136°, ∠DEB=47°, ∠EFA=63°, ∠FED=69°. Find the measure of ∠FAE.
问题：如图所示，∠ACE=∠EAC，∠DBG=136°，∠DEB=47°，∠EFA=63°，∠FED=69°。求∠FAE的大小。
Construction (CDL):
Shape(FA,AE,EF)
Shape(AC,CE,EA)
Shape(DE,EB,BD)
Collinear(AEBG)
Collinear(FEC)
Shape(FE,ED)
Shape(FE,EB)
Shape(DB,BG)
Conditions (CDL):
Equal(MeasureOfAngle(ACE),MeasureOfAngle(EAC))
Equal(MeasureOfAngle(DBG),136)
Equal(MeasureOfAngle(DEB),47)
Equal(MeasureOfAngle(EFA),63)
Equal(MeasureOfAngle(FED),69)
Goal: Value(MeasureOfAngle(FAE))
Answer: 53
Theorem sequence:
flat_angle(1,AEB)
angle_addition(1,FED,DEB)
angle_addition(1,AEF,FEB)
triangle_property_angle_sum(1,FAE)
Problem: As shown in the diagram, RC=x, ST=20, TC=12, the center of ⊙R is R, TS is the tangent to ⊙R. Find the value of x.
问题：如图所示，RC=x，ST=20，TC=12，R是⊙R的圆心，TS是⊙O的切线。求x的值。
Construction (CDL):
Shape(RSC,CR,RS)
Shape(RCS,SR,RC)
Shape(RCS,CT,TS)
Collinear(RCT)
Cocircular(R,SC)
Conditions (CDL):
Equal(LengthOfLine(RC),x)
Equal(LengthOfLine(ST),20)
Equal(LengthOfLine(TC),12)
IsCentreOfCircle(R,R)
IsTangentOfCircle(TS,R)
Goal: Value(x)
Answer: 32/3
Theorem sequence:
tangent_of_circle_property_perpendicular(2,TS,R,R)
right_triangle_judgment_angle(1,TSR)
line_addition(1,RC,CT)
radius_of_circle_property_length_equal(1,RS,R)
radius_of_circle_property_length_equal(1,RC,R)
right_triangle_property_pythagorean(1,TSR)
Problem: As shown in the diagram, AB=x, AC=11, BC=9, ∠BCA=28°. Find the value of x.
问题：如图所示，AB=x，AC=11，BC=9，∠BCA=28°。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),11)
Equal(LengthOfLine(BC),9)
Equal(MeasureOfAngle(BCA),28)
Goal: Value(x)
Answer: sqrt(202-198*cos(7*pi/45))
Theorem sequence:
cosine_theorem(1,CAB)
Problem: As shown in the diagram, AC=5*y, AY=2*x-5, CB=3*x-18, YB=2*y+12, AC and YB are opposite sides of the parallelogram AYBC. Find the value of y.
问题：如图所示，AC=5*y，AY=2*x-5，CB=3*x-18，YB=2*y+12，AC和YB是平行四边形AYBC的一组对边。求y的值。
Construction (CDL):
Shape(AY,YB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),5*y)
Equal(LengthOfLine(AY),2*x-5)
Equal(LengthOfLine(CB),3*x-18)
Equal(LengthOfLine(YB),2*y+12)
Parallelogram(AYBC)
Goal: Value(y)
Answer: 4
Theorem sequence:
parallelogram_property_opposite_line_equal(1,YBCA)
Problem: As shown in the diagram, ∠BAD=65°, ∠DAE=110°, ∠EAC=x°, CA is perpendicular to BA. Find the value of x.
问题：如图所示，∠BAD=65°，∠DAE=110°，∠EAC=x°，CA⊥BA。求x的值。
Construction (CDL):
Shape(ABC,CA,AB)
Shape(ADB,BA,AD)
Shape(ACE,EA,AC)
Shape(AED,DA,AE)
Cocircular(A,BCED)
Conditions (CDL):
Equal(MeasureOfAngle(BAD),65)
Equal(MeasureOfAngle(DAE),110)
Equal(MeasureOfAngle(EAC),x)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 95
Theorem sequence:
angle_addition(1,CAB,BAD)
angle_addition(1,CAD,DAE)
round_angle(1,EAC,CAE)
Problem: As shown in the diagram, ML=3, PJ=x, HP is the midsegment of △ MLK, JH is the midsegment of triangle LKM, PJ is the midsegment of △ KML. Find the value of x.
问题：如图所示，ML=3，PJ=x，HP是三角形MLK的中位线，JH是三角形LKM的中位线，PJ是三角形KML的中位线。求x的值。
Construction (CDL):
Shape(KP,PJ,JK)
Shape(JP,PH,HJ)
Shape(PM,MH,HP)
Shape(JH,HL,LJ)
Collinear(KPM)
Collinear(MHL)
Collinear(KJL)
Conditions (CDL):
Equal(LengthOfLine(ML),3)
Equal(LengthOfLine(PJ),x)
IsMidsegmentOfTriangle(HP,MLK)
IsMidsegmentOfTriangle(JH,LKM)
IsMidsegmentOfTriangle(PJ,KML)
Goal: Value(x)
Answer: 3/2
Theorem sequence:
midsegment_of_triangle_property_parallel(1,PJ,KML)
parallel_property_corresponding_angle(1,PJ,ML,K)
similar_triangle_judgment_aa(1,JKP,LKM)
line_addition(1,KP,PM)
similar_triangle_property_line_ratio(1,KPJ,KML)
similar_triangle_property_line_ratio(1,JKP,LKM)
Problem: As shown in the diagram, QP=24, RT=16, ST=9, PQ∥TR. Find the length of line PS.
问题：如图所示，QP=24，RT=16，ST=9，PQ平行于TR。求直线PS的长度。
Construction (CDL):
Shape(RQ,QP,PT,TR)
Shape(SR,RT,TS)
Collinear(PTS)
Collinear(QRS)
Conditions (CDL):
Equal(LengthOfLine(QP),24)
Equal(LengthOfLine(RT),16)
Equal(LengthOfLine(ST),9)
ParallelBetweenLine(PQ,TR)
Goal: Value(LengthOfLine(PS))
Answer: 27/2
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,PQ,TR)
parallel_property_ipsilateral_internal_angle(1,RT,QP)
flat_angle(1,PTS)
flat_angle(1,SRQ)
angle_addition(1,PTR,RTS)
angle_addition(1,SRT,TRQ)
similar_triangle_judgment_aa(1,SRT,SQP)
similar_triangle_property_line_ratio(1,RTS,QPS)
similar_triangle_property_line_ratio(1,SRT,SQP)
Problem: As shown in the diagram, AB=10, the measure of arc OBA is 60, the radius of ⊙O is 10, O is the center of ⊙O, BX⊥OX. Find the length of line AX.
问题：如图所示，AB=10，弧OBA的角度为60，⊙O的半径为10，O是⊙O的圆心，BX垂直于OX。求直线AX的长度。
Construction (CDL):
Shape(OBY,YX,XB)
Shape(OYA,AX,XY)
Shape(OAB,BX,XA)
Shape(BX,XO)
Shape(OX,XA)
Collinear(AXB)
Collinear(YXO)
Cocircular(O,BYA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(MeasureOfArc(OBA),60)
Equal(RadiusOfCircle(O),10)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(BX,OX)
Goal: Value(LengthOfLine(AX))
Answer: 5
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,O,OX,BA)
line_addition(1,AX,XB)
Problem: As shown in the diagram, PM=20, QR=6*x, TS=2*x, M bisects segment RS, P bisects segment QT, quadrilateral QTSR is a trapezoid. Find the value of x.
问题：如图所示，PM=20，QR=6*x，TS=2*x，M平分线段RS，P平分线段QT，QT和SR是梯形QTSR的腰。求x的值。
Construction (CDL):
Shape(QP,PM,MR,RQ)
Shape(PT,TS,SM,MP)
Collinear(QPT)
Collinear(RMS)
Conditions (CDL):
Equal(LengthOfLine(PM),20)
Equal(LengthOfLine(QR),6*x)
Equal(LengthOfLine(TS),2*x)
IsMidpointOfLine(M,RS)
IsMidpointOfLine(P,QT)
Trapezoid(QTSR)
Goal: Value(x)
Answer: 5
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,PM,QTSR)
midsegment_of_quadrilateral_property_length(1,PM,QTSR)
Problem: As shown in the diagram, ∠GCH=2*x°, ∠HCD=6*x+28°, the center of circle C is C, FC⊥GC. Find the measure of ⌒CFE.
问题：如图所示，∠GCH=2*x°，∠HCD=6*x+28°，C是圆C的圆心，FC⊥GC。求⌒CFE的角度。
Construction (CDL):
Shape(CHG,GC,CH)
Shape(CGF,FC,CG)
Shape(CFE,EC,CF)
Shape(CED,DC,CE)
Shape(CDH,HC,CD)
Collinear(HCE)
Collinear(GCD)
Cocircular(C,HGFED)
Conditions (CDL):
Equal(MeasureOfAngle(GCH),2*x)
Equal(MeasureOfAngle(HCD),6*x+28)
IsCentreOfCircle(C,C)
PerpendicularBetweenLine(FC,GC)
Goal: Value(MeasureOfArc(CFE))
Answer: 52
Theorem sequence:
flat_angle(1,GCD)
angle_addition(1,GCH,HCD)
vertical_angle(1,GCH,DCE)
adjacent_complementary_angle(1,DCF,FCG)
angle_addition(1,DCE,ECF)
arc_property_center_angle(1,CFE,C)
Problem: As shown in the diagram, RT=x, ST=7, ∠RTS=30°, ∠TSR=120°. Find the value of x.
问题：如图所示，RT=x，ST=7，∠RTS=30°，∠TSR=120°。求x的值。
Construction (CDL):
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(RT),x)
Equal(LengthOfLine(ST),7)
Equal(MeasureOfAngle(RTS),30)
Equal(MeasureOfAngle(TSR),120)
Goal: Value(x)
Answer: 7*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,SRT)
sine_theorem(1,SRT)
sine_theorem(1,RTS)
Problem: As shown in the diagram, the length of arc SRT is 8, ∠TSR=70°, S is the center of ⊙S. Find the circumference of the circle S.
问题：如图所示，弧SRT的长度为8，∠TSR=70°，S是圆S的圆心。求圆S的周长。
Construction (CDL):
Shape(STR,RS,ST)
Shape(SRT,TS,SR)
Cocircular(S,TR)
Conditions (CDL):
Equal(LengthOfArc(SRT),8)
Equal(MeasureOfAngle(TSR),70)
IsCentreOfCircle(S,S)
Goal: Value(PerimeterOfCircle(S))
Answer: 288/7
Theorem sequence:
arc_property_center_angle(1,SRT,S)
arc_length_formula(1,SRT)
circle_perimeter_formula(1,S)
Problem: As shown in the diagram, ∠ABD=45°, ∠BCD=70°, ∠DAB=40°. Find the measure of ∠DBC.
问题：如图所示，∠ABD=45°，∠BCD=70°，∠DAB=40°。求∠DBC的大小。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Collinear(CDA)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),45)
Equal(MeasureOfAngle(BCD),70)
Equal(MeasureOfAngle(DAB),40)
Goal: Value(MeasureOfAngle(DBC))
Answer: 25
Theorem sequence:
triangle_property_angle_sum(1,CAB)
angle_addition(1,ABD,DBC)
Problem: As shown in the diagram, NQ=2*x+3, QK=5*x-9, KJ⊥NJ, NM is perpendicular to KM, quadrilateral JNMK is a rectangle. Find the length of line JQ.
问题：如图所示，NQ=2*x+3，QK=5*x-9，KJ垂直于NJ，NM⊥KM，四边形JNMK是矩形。求直线JQ的长度。
Construction (CDL):
Shape(JN,NQ,QJ)
Shape(JQ,QK,KJ)
Shape(KQ,QM,MK)
Shape(QN,NM,MQ)
Collinear(JQM)
Collinear(NQK)
Conditions (CDL):
Equal(LengthOfLine(NQ),2*x+3)
Equal(LengthOfLine(QK),5*x-9)
PerpendicularBetweenLine(KJ,NJ)
PerpendicularBetweenLine(NM,KM)
Rectangle(JNMK)
Goal: Value(LengthOfLine(JQ))
Answer: 11
Theorem sequence:
parallelogram_property_diagonal_bisection(1,JNMK,Q)
parallelogram_property_diagonal_bisection(1,NMKJ,Q)
rectangle_property_diagonal_equal(1,JNMK)
line_addition(1,JQ,QM)
line_addition(1,NQ,QK)
Problem: As shown in the diagram, AB=5*x-11, AD=3*x+5, BC=15, CD=15, AC is perpendicular to DC. Find the length of line AB.
问题：如图所示，AB=5*x-11，AD=3*x+5，BC=15，CD=15，AC垂直于DC。求直线AB的长度。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD,DA)
Collinear(BCD)
Conditions (CDL):
Equal(LengthOfLine(AB),5*x-11)
Equal(LengthOfLine(AD),3*x+5)
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(CD),15)
PerpendicularBetweenLine(AC,DC)
Goal: Value(LengthOfLine(AB))
Answer: 29
Theorem sequence:
adjacent_complementary_angle(1,BCA,ACD)
perpendicular_bisector_judgment_per_and_mid(1,AC,BD)
perpendicular_bisector_property_distance_equal(1,AC,BD)
Problem: As shown in the diagram, BE=x, DE=x+10, EA=x+1, EC=x+3. Find the value of x.
问题：如图所示，BE=x，DE=x+10，EA=x+1，EC=x+3。求x的值。
Construction (CDL):
Shape(FBA,AE,EB)
Shape(FAD,DF,FE,EA)
Shape(FDC,CE,EF,FD)
Shape(FCB,BE,EC)
Collinear(BEFD)
Collinear(AEC)
Cocircular(F,CBAD)
Conditions (CDL):
Equal(LengthOfLine(BE),x)
Equal(LengthOfLine(DE),x+10)
Equal(LengthOfLine(EA),x+1)
Equal(LengthOfLine(EC),x+3)
Goal: Value(x)
Answer: 1/2
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,CEA,BED,F)
Problem: As shown in the diagram, KJ=6, KL=12, RQ=4, RS=8, ∠JLK=30°, ∠JLK=∠QSR, ∠KJL=∠RQS, ∠LKJ=∠SRQ, ∠LKJ=y°, ∠QSR=x°, ∠RQS=80°. Find the value of x.
问题：如图所示，KJ=6，KL=12，RQ=4，RS=8，∠JLK=30°，∠JLK=∠QSR，∠KJL=∠RQS，∠LKJ=∠SRQ，∠LKJ=y°，∠QSR=x°，∠RQS=80°。求x的值。
Construction (CDL):
Shape(KJ,JL,LK)
Shape(RQ,QS,SR)
Conditions (CDL):
Equal(LengthOfLine(KJ),6)
Equal(LengthOfLine(KL),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)
Goal: Value(x)
Answer: 30
Theorem sequence:
Problem: As shown in the diagram, AB=x, BC=5, DE=x-4, FE=3, ∠ABC=∠FED, ∠CAB=∠EDF. Find the length of line DE.
问题：如图所示，AB=x，BC=5，DE=x-4，FE=3，∠ABC=∠FED，∠CAB=∠EDF。求直线DE的长度。
Construction (CDL):
Shape(AB,BC,CA)
Shape(DF,FE,ED)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(BC),5)
Equal(LengthOfLine(DE),x-4)
Equal(LengthOfLine(FE),3)
Equal(MeasureOfAngle(ABC),MeasureOfAngle(FED))
Equal(MeasureOfAngle(CAB),MeasureOfAngle(EDF))
Goal: Value(LengthOfLine(DE))
Answer: 6
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,CAB,FED)
mirror_similar_triangle_property_line_ratio(1,CAB,FED)
mirror_similar_triangle_property_line_ratio(1,ABC,DFE)
Problem: As shown in the diagram, AC=x, BA=7, BC=19, BA⊥CA. Find the value of x.
问题：如图所示，AC=x，BA=7，BC=19，BA垂直于CA。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BA),7)
Equal(LengthOfLine(BC),19)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 2*sqrt(78)
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, RP=y+4, RS=27, TP=2*y-5, TQ=5*x, ∠PQT=95°, ∠RQP=33°, ∠TSP=3*z°, QR and TS are opposite sides of the ▱ QTSR. Find the value of x.
问题：如图所示，RP=y+4，RS=27，TP=2*y-5，TQ=5*x，∠PQT=95°，∠RQP=33°，∠TSP=3*z°，QTSR是平行四边形。求x的值。
Construction (CDL):
Shape(QT,TP,PQ)
Shape(PT,TS,SP)
Shape(QP,PR,RQ)
Shape(RP,PS,SR)
Collinear(QPS)
Collinear(TPR)
Conditions (CDL):
Equal(LengthOfLine(RP),y+4)
Equal(LengthOfLine(RS),27)
Equal(LengthOfLine(TP),2*y-5)
Equal(LengthOfLine(TQ),5*x)
Equal(MeasureOfAngle(PQT),95)
Equal(MeasureOfAngle(RQP),33)
Equal(MeasureOfAngle(TSP),3*z)
Parallelogram(QTSR)
Goal: Value(x)
Answer: 27/5
Theorem sequence:
parallelogram_property_opposite_line_equal(1,QTSR)
Problem: As shown in the diagram, EN=3, △DCB is an equilateral △, N is the center of ⊙N, the centroid of △DCB is N, the incenter of triangle DCB is N, BG is the tangent to ⊙N, CF is the tangent to circle N, the tangent to circle N is DE. Find the area of triangle DCB minus the area of the ⊙N.
问题：如图所示，EN=3，三角形DCB为等边三角形，N是圆N的圆心，三角形DCB的重心为O，N是△DCB的内心，BG是圆O的切线，CF是圆O的切线，⊙O的切线为DE。求三角形DCB的面积减去圆N的面积。
Construction (CDL):
Shape(NQG,NGE,EN,NQ)
Shape(NEF,NFQ,QN,NE)
Shape(NGE,GD,DE)
Shape(NEF,EC,CF)
Shape(NFQ,FB,BQ)
Shape(NQG,QB,BG)
Collinear(ENQB)
Collinear(DGB)
Collinear(CFB)
Collinear(DEC)
Cocircular(N,GEFQ)
Conditions (CDL):
Equal(LengthOfLine(EN),3)
EquilateralTriangle(DCB)
IsCentreOfCircle(N,N)
IsCentroidOfTriangle(N,DCB)
IsIncenterOfTriangle(N,DCB)
IsTangentOfCircle(BG,N)
IsTangentOfCircle(CF,N)
IsTangentOfCircle(DE,N)
Goal: Value(Sub(AreaOfTriangle(DCB),AreaOfCircle(N)))
Answer: -9*pi+27*sqrt(3)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,DE,N,N)
equilateral_triangle_property_angle(1,DCB)
centroid_of_triangle_property_line_ratio(1,N,BDC,E)
line_addition(1,EN,NB)
sine_theorem(1,BDE)
triangle_area_formula_sine(1,DCB)
radius_of_circle_property_length_equal(1,NE,N)
circle_area_formula(1,N)
Problem: As shown in the diagram, AH=y, DA=8, DH=x, HF=6, ∠DAH=30°, ∠FBC=45°, DC∥HF, AH⊥DH, CF is perpendicular to BF. Find the value of y.
问题：如图所示，AH=y，DA=8，DH=x，HF=6，∠DAH=30°，∠FBC=45°，DC∥HF，AH⊥DH，CF垂直于BF。求y的值。
Construction (CDL):
Shape(DA,AH,HD)
Shape(DH,HF,FC,CD)
Shape(CF,FB,BC)
Collinear(AHFB)
Conditions (CDL):
Equal(LengthOfLine(AH),y)
Equal(LengthOfLine(DA),8)
Equal(LengthOfLine(DH),x)
Equal(LengthOfLine(HF),6)
Equal(MeasureOfAngle(DAH),30)
Equal(MeasureOfAngle(FBC),45)
ParallelBetweenLine(DC,HF)
PerpendicularBetweenLine(AH,DH)
PerpendicularBetweenLine(CF,BF)
Goal: Value(y)
Answer: 4*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,AHD)
sine_theorem(1,AHD)
Problem: As shown in the diagram, BG=10, EG=10, the radius of ⊙G is 26, G is the center of circle G, CB⊥GB, FE is perpendicular to GE. Find the length of line DE.
问题：如图所示，BG=10，EG=10，圆G的半径为26，G是⊙G的圆心，CB⊥GB，FE⊥GE。求直线DE的长度。
Construction (CDL):
Shape(GAF,FG,GA)
Shape(GF,FE,EG)
Shape(AG,GB,BA)
Shape(GFD,DE,EF)
Shape(GCA,AB,BC)
Shape(GE,ED,GDC,CB,BG)
Collinear(FED)
Collinear(ABC)
Cocircular(G,AFDC)
Conditions (CDL):
Equal(LengthOfLine(BG),10)
Equal(LengthOfLine(EG),10)
Equal(RadiusOfCircle(G),26)
IsCentreOfCircle(G,G)
PerpendicularBetweenLine(CB,GB)
PerpendicularBetweenLine(FE,GE)
Goal: Value(LengthOfLine(DE))
Answer: 24
Theorem sequence:
radius_of_circle_property_length_equal(1,GF,G)
right_triangle_judgment_angle(1,FEG)
right_triangle_property_pythagorean(1,FEG)
circle_property_chord_perpendicular_bisect_chord(1,G,GE,FD)
Problem: As shown in the diagram, RA=3, TA=8, UB=x+2, WB=3*x-6, TA is the median of △ TRS, WB is the median of △ WUV, triangle RST is similar to triangle UVW.. Find the length of line UB.
问题：如图所示，RA=3，TA=8，UB=x+2，WB=3*x-6，TA是三角形TRS的中线，WB是三角形WUV的中线，三角形RST相似于三角形UVW。求直线UB的长度。
Construction (CDL):
Shape(TR,RA,AT)
Shape(TA,AS,ST)
Shape(WU,UB,BW)
Shape(WB,BV,VW)
Collinear(RAS)
Collinear(UBV)
Conditions (CDL):
Equal(LengthOfLine(RA),3)
Equal(LengthOfLine(TA),8)
Equal(LengthOfLine(UB),x+2)
Equal(LengthOfLine(WB),3*x-6)
IsMedianOfTriangle(TA,TRS)
IsMedianOfTriangle(WB,WUV)
SimilarBetweenTriangle(RST,UVW)
Goal: Value(LengthOfLine(UB))
Answer: 36
Theorem sequence:
similar_triangle_property_angle_equal(1,RST,UVW)
similar_triangle_property_line_ratio(1,TRS,WUV)
similar_triangle_property_line_ratio(1,STR,VWU)
line_addition(1,RA,AS)
line_addition(1,UB,BV)
similar_triangle_judgment_sas(1,RAT,UBW)
similar_triangle_property_line_ratio(1,RAT,UBW)
similar_triangle_property_line_ratio(1,TRA,WUB)
Problem: As shown in the diagram, AE=x+7, BE=4, EC=9, EF=x. Find the value of x.
问题：如图所示，AE=x+7，BE=4，EC=9，EF=x。求x的值。
Construction (CDL):
Shape(DAC,CE,EA)
Shape(DCB,BE,EC)
Shape(DBF,FE,EB)
Shape(DFA,AE,EF)
Collinear(AEB)
Collinear(CEF)
Cocircular(D,ACBF)
Conditions (CDL):
Equal(LengthOfLine(AE),x+7)
Equal(LengthOfLine(BE),4)
Equal(LengthOfLine(EC),9)
Equal(LengthOfLine(EF),x)
Goal: Value(x)
Answer: 28/5
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,AEB,CEF,D)
Problem: As shown in the diagram, TR=44, VA=21, M is the midpoint of segment TV, N is the midpoint of segment RA, TV and AR are the non-parallel sides (legs) of trapezoid TVAR. Find the length of line MN.
问题：如图所示，TR=44，VA=21，M平分线段TV，N平分线段RA，四边形TVAR是梯形。求直线MN的长度。
Construction (CDL):
Shape(TM,MN,NR,RT)
Shape(MV,VA,AN,NM)
Collinear(TMV)
Collinear(RNA)
Conditions (CDL):
Equal(LengthOfLine(TR),44)
Equal(LengthOfLine(VA),21)
IsMidpointOfLine(M,TV)
IsMidpointOfLine(N,RA)
Trapezoid(TVAR)
Goal: Value(LengthOfLine(MN))
Answer: 65/2
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,MN,TVAR)
midsegment_of_quadrilateral_property_length(1,MN,TVAR)
Problem: As shown in the diagram, AC=5*sqrt(26), AD=25, CB=sqrt(26), CD=5, DB=1, AD is perpendicular to CD, BC is perpendicular to AC. Find the ratio of the length of line CD to the length of line DB.
问题：如图所示，AC=5*sqrt(26)，AD=25，CB=sqrt(26)，CD=5，DB=1，AD垂直于CD，BC⊥AC。求直线CD的长度与直线DB的长度之比。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),5*sqrt(26))
Equal(LengthOfLine(AD),25)
Equal(LengthOfLine(CB),sqrt(26))
Equal(LengthOfLine(CD),5)
Equal(LengthOfLine(DB),1)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Div(LengthOfLine(CD),LengthOfLine(DB)))
Answer: 5
Theorem sequence:
Problem: As shown in the diagram, ∠AGB=30°, the center of circle G is G, CG is perpendicular to DG. Find the measure of ⌒GBA.
问题：如图所示，∠AGB=30°，G是圆G的圆心，CG垂直于DG。求弧GBA的角度。
Construction (CDL):
Shape(GBA,AG,GB)
Shape(GCB,BG,GC)
Shape(GDC,CG,GD)
Shape(GFD,DG,GF)
Shape(GAF,FG,GA)
Collinear(AGD)
Collinear(BGF)
Cocircular(G,CBAFD)
Conditions (CDL):
Equal(MeasureOfAngle(AGB),30)
IsCentreOfCircle(G,G)
PerpendicularBetweenLine(CG,DG)
Goal: Value(MeasureOfArc(GBA))
Answer: 30
Theorem sequence:
arc_property_center_angle(1,GBA,G)
Problem: As shown in the diagram, UY=32, WY=40, WZ=3*x-6, ZU=x+6, ∠UWZ=∠UYW, ZU is perpendicular to WU. Find the length of line UZ.
问题：如图所示，UY=32，WY=40，WZ=3*x-6，ZU=x+6，∠UWZ=∠UYW，ZU垂直于WU。求直线UZ的长度。
Construction (CDL):
Shape(WZ,ZU,UW)
Shape(WU,UY,YW)
Collinear(ZUY)
Conditions (CDL):
Equal(LengthOfLine(UY),32)
Equal(LengthOfLine(WY),40)
Equal(LengthOfLine(WZ),3*x-6)
Equal(LengthOfLine(ZU),x+6)
Equal(MeasureOfAngle(UWZ),MeasureOfAngle(UYW))
PerpendicularBetweenLine(ZU,WU)
Goal: Value(LengthOfLine(UZ))
Answer: 18
Theorem sequence:
adjacent_complementary_angle(1,ZUW,WUY)
similar_triangle_judgment_aa(1,ZUW,WUY)
right_triangle_judgment_angle(1,WUY)
right_triangle_property_pythagorean(1,WUY)
similar_triangle_property_line_ratio(1,ZUW,WUY)
similar_triangle_property_line_ratio(1,WZU,YWU)
Problem: As shown in the diagram, NC=10, ND=2, M is the center of ⊙M, NC is the tangent to ⊙M. Find the perimeter of triangle MNC.
问题：如图所示，NC=10，ND=2，圆M的圆心为M，⊙O的切线为NC。求△MNC的周长。
Construction (CDL):
Shape(MCD,DM,MC)
Shape(MDC,CM,MD)
Shape(MDC,DN,NC)
Collinear(NDM)
Cocircular(M,DC)
Conditions (CDL):
Equal(LengthOfLine(NC),10)
Equal(LengthOfLine(ND),2)
IsCentreOfCircle(M,M)
IsTangentOfCircle(NC,M)
Goal: Value(PerimeterOfTriangle(MNC))
Answer: 60
Theorem sequence:
tangent_of_circle_property_perpendicular(2,NC,M,M)
line_addition(1,ND,DM)
radius_of_circle_property_length_equal(1,MD,M)
radius_of_circle_property_length_equal(1,MC,M)
right_triangle_judgment_angle(1,NCM)
right_triangle_property_pythagorean(1,NCM)
triangle_perimeter_formula(1,NCM)
Problem: As shown in the diagram, ∠BAC=124°. Find the measure of ∠CAE.
问题：如图所示，∠BAC=124°。求∠CAE的大小。
Construction (CDL):
Shape(BA,AC)
Shape(DA,AB)
Shape(CA,AE)
Shape(EA,AD)
Collinear(BAE)
Collinear(CAD)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),124)
Goal: Value(MeasureOfAngle(CAE))
Answer: 56
Theorem sequence:
flat_angle(1,BAE)
angle_addition(1,BAC,CAE)
Problem: As shown in the diagram, NM=39, PM=36, PN=15, MP⊥NP. Find the value of cos(NM).
问题：如图所示，NM=39，PM=36，PN=15，MP⊥NP。求cos(NM)的值。
Construction (CDL):
Shape(PN,NM,MP)
Conditions (CDL):
Equal(LengthOfLine(NM),39)
Equal(LengthOfLine(PM),36)
Equal(LengthOfLine(PN),15)
PerpendicularBetweenLine(MP,NP)
Goal: Value(Cos(MeasureOfAngle(NMP)))
Answer: 12/13
Theorem sequence:
cosine_theorem(1,MPN)
Problem: As shown in the diagram, AE=2, BE=x, DE=4, EC=5. Find the value of x.
问题：如图所示，AE=2，BE=x，DE=4，EC=5。求x的值。
Construction (CDL):
Shape(XAD,DE,EA)
Shape(XDB,BX,XE,ED)
Shape(XBC,CE,EX,XB)
Shape(XCA,AE,EC)
Collinear(AEB)
Collinear(DEC)
Cocircular(X,CADB)
Conditions (CDL):
Equal(LengthOfLine(AE),2)
Equal(LengthOfLine(BE),x)
Equal(LengthOfLine(DE),4)
Equal(LengthOfLine(EC),5)
Goal: Value(x)
Answer: 10
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,CED,AEB,X)