SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, GF=12, HG=6, HJ=8, JK=x-4, GJ is parallel to FK. Find the value of x.
问题：如图所示，GF=12，HG=6，HJ=8，JK=x-4，GJ∥FK。求x的值。
Construction (CDL):
Shape(HG,GJ,JH)
Shape(GF,FK,KJ,JG)
Collinear(HGF)
Collinear(KJH)
Conditions (CDL):
Equal(LengthOfLine(GF),12)
Equal(LengthOfLine(HG),6)
Equal(LengthOfLine(HJ),8)
Equal(LengthOfLine(JK),x-4)
ParallelBetweenLine(GJ,FK)
Goal: Value(x)
Answer: 20
Theorem sequence:
parallel_property_corresponding_angle(1,GJ,FK,H)
similar_triangle_judgment_aa(1,JHG,KHF)
line_addition(1,HG,GF)
line_addition(1,HJ,JK)
similar_triangle_property_line_ratio(1,JHG,KHF)
similar_triangle_property_line_ratio(1,GJH,FKH)
Problem: As shown in the diagram, ML=w, MN=2*y+5, MR=4*x-2, QL=12, QN=3*x+2, QR=3*y, quadrilateral NMRQ is a ▱. Find the value of w.
问题：如图所示，ML=w，MN=2*y+5，MR=4*x-2，QL=12，QN=3*x+2，QR=3*y，NQ和MR是▱NMRQ的一组对边。求w的值。
Construction (CDL):
Shape(NM,ML,LN)
Shape(LM,MR,RL)
Shape(NL,LQ,QN)
Shape(LR,RQ,QL)
Collinear(NLR)
Collinear(MLQ)
Conditions (CDL):
Equal(LengthOfLine(ML),w)
Equal(LengthOfLine(MN),2*y+5)
Equal(LengthOfLine(MR),4*x-2)
Equal(LengthOfLine(QL),12)
Equal(LengthOfLine(QN),3*x+2)
Equal(LengthOfLine(QR),3*y)
Parallelogram(NMRQ)
Goal: Value(w)
Answer: 12
Theorem sequence:
parallelogram_property_diagonal_bisection(1,MRQN,L)
Problem: As shown in the diagram, BA=6, CA=x, CB=x, ∠BAC=45°, ∠CBA=45°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，BA=6，CA=x，CB=x，∠BAC=45°，∠CBA=45°，AC⊥BC。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(BA),6)
Equal(LengthOfLine(CA),x)
Equal(LengthOfLine(CB),x)
Equal(MeasureOfAngle(BAC),45)
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 3*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, ∠YVW=25°, the measure of ⌒AXZ is 110, the measure of arc AYW is x. Find the value of x.
问题：如图所示，∠YVW=25°，⌒AXZ的角度为110，弧AYW的角度为x。求x的值。
Construction (CDL):
Shape(YV,VW,AYW)
Shape(AZY,YZ)
Shape(AYW,WX,AXZ,ZY)
Shape(AWX,XW)
Collinear(VYZ)
Collinear(VWX)
Cocircular(A,ZYWX)
Conditions (CDL):
Equal(MeasureOfAngle(YVW),25)
Equal(MeasureOfArc(AXZ),110)
Equal(MeasureOfArc(AYW),x)
Goal: Value(x)
Answer: 60
Theorem sequence:
circle_property_circular_power_segment_and_segment_angle(1,VYZ,VWX,A)
Problem: As shown in the diagram, KJ=11, ∠JKL=65°, the center of circle K is K. Find the area of the sector KJL.
问题：如图所示，KJ=11，∠JKL=65°，圆K的圆心为K。求扇形KJL的面积。
Construction (CDL):
Shape(KJ,KJL,LK)
Shape(JK,KL,KLJ)
Cocircular(K,JL)
Conditions (CDL):
Equal(LengthOfLine(KJ),11)
Equal(MeasureOfAngle(JKL),65)
IsCentreOfCircle(K,K)
Goal: Value(AreaOfSector(KJL))
Answer: 7139*pi/72
Theorem sequence:
round_angle(1,JKL,LKJ)
radius_of_circle_property_length_equal(1,KJ,K)
arc_property_center_angle(1,KJL,K)
sector_area_formula(1,KJL)
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 is perpendicular to QZ, SPQR is a trapezoid. Find the perimeter of quadrilateral SPQR.
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，四边形SPQR是梯形。求SPQR的周长。
Construction (CDL):
Shape(SP,PA,AS)
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)
Trapezoid(SPQR)
Goal: Value(PerimeterOfQuadrilateral(SPQR))
Answer: 6*sqrt(2)+6*sqrt(3)+38
Theorem sequence:
triangle_property_angle_sum(1,RZQ)
sine_theorem(1,RZQ)
sine_theorem(1,QRZ)
altitude_of_quadrilateral_judgment_left_vertex(2,SA,SPQR)
adjacent_complementary_angle(1,PZR,RZQ)
altitude_of_quadrilateral_judgment_right_vertex(2,RZ,SPQR)
parallel_judgment_corresponding_angle(1,AS,ZR,P)
parallelogram_judgment_parallel_and_parallel(1,SAZR)
parallelogram_property_opposite_line_equal(1,AZRS)
triangle_property_angle_sum(1,SPA)
isosceles_triangle_judgment_angle_equal(1,ASP)
sine_theorem(1,SPA)
line_addition(1,PA,AZ)
line_addition(1,PZ,ZQ)
quadrilateral_perimeter_formula(1,SPQR)
Problem: As shown in the diagram, ∠EYQ=3*y+1°, ∠MAQ=3*x+11°, ∠YQF=4*x-5°, EF∥YQ, QA is parallel to YM, YQ∥MA. Find the value of y.
问题：如图所示，∠EYQ=3*y+1°，∠MAQ=3*x+11°，∠YQF=4*x-5°，EF平行于YQ，QA∥YM，YQ平行于MA。求y的值。
Construction (CDL):
Shape(EY,YQ,QF,FE)
Shape(YM,MA,AQ,QY)
Collinear(EYM)
Collinear(FQA)
Conditions (CDL):
Equal(MeasureOfAngle(EYQ),3*y+1)
Equal(MeasureOfAngle(MAQ),3*x+11)
Equal(MeasureOfAngle(YQF),4*x-5)
ParallelBetweenLine(EF,YQ)
ParallelBetweenLine(QA,YM)
ParallelBetweenLine(YQ,MA)
Goal: Value(y)
Answer: 40
Theorem sequence:
parallel_property_corresponding_angle(2,AM,QY,F)
parallelogram_judgment_parallel_and_parallel(1,YMAQ)
parallelogram_property_opposite_angle_equal(1,YMAQ)
adjacent_complementary_angle(1,EYQ,QYM)
Problem: As shown in the diagram, ∠AXW=82°, ∠YXA=33°, quadrilateral XWZY is a ▱. Find the measure of ∠WZY.
问题：如图所示，∠AXW=82°，∠YXA=33°，WX和ZY是平行四边形XWZY的一组对边。求∠WZY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(XWZY)
Goal: Value(MeasureOfAngle(WZY))
Answer: 115
Theorem sequence:
angle_addition(1,YXA,AXW)
parallelogram_property_opposite_angle_equal(1,XWZY)
Problem: As shown in the diagram, FE=6, FG=3, FH=4, HG=2, triangle DEF is similar to triangle GFH.. Find the perimeter of △DEF.
问题：如图所示，FE=6，FG=3，FH=4，HG=2，△DEF相似于△GFH。求△DEF的周长。
Construction (CDL):
Shape(DE,EF,FD)
Shape(CD,DF,FG,GC)
Shape(GF,FH,HG)
Collinear(CDE)
Collinear(EFH)
Collinear(HGC)
Conditions (CDL):
Equal(LengthOfLine(FE),6)
Equal(LengthOfLine(FG),3)
Equal(LengthOfLine(FH),4)
Equal(LengthOfLine(HG),2)
SimilarBetweenTriangle(DEF,GFH)
Goal: Value(PerimeterOfTriangle(DEF))
Answer: 27/2
Theorem sequence:
similar_triangle_property_line_ratio(1,DEF,GFH)
similar_triangle_property_line_ratio(1,FDE,HGF)
similar_triangle_property_line_ratio(1,EFD,FHG)
triangle_perimeter_formula(1,DEF)
Problem: As shown in the diagram, ∠LWX=3*a+40°, ∠WXE=2*a+25°, ∠XZK=5*b-26°, WL∥XE, XN is parallel to ZK. Find the value of a.
问题：如图所示，∠LWX=3*a+40°，∠WXE=2*a+25°，∠XZK=5*b-26°，WL∥XE，XN平行于ZK。求a的值。
Construction (CDL):
Shape(GW,WL)
Shape(LW,WX)
Shape(WX,XE)
Shape(EX,XN)
Shape(NX,XZ)
Shape(XZ,ZK)
Shape(KZ,ZH)
Shape(HZ,ZY)
Shape(ZY,YM)
Shape(MY,YI)
Shape(IY,YW)
Shape(YW,WG)
Shape(WY,YZ,ZX,XW)
Collinear(GWXN)
Collinear(IYZK)
Collinear(LWYM)
Collinear(EXZH)
Conditions (CDL):
Equal(MeasureOfAngle(LWX),3*a+40)
Equal(MeasureOfAngle(WXE),2*a+25)
Equal(MeasureOfAngle(XZK),5*b-26)
ParallelBetweenLine(WL,XE)
ParallelBetweenLine(XN,ZK)
Goal: Value(a)
Answer: 23
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,WL,XE)
Problem: As shown in the diagram, ∠AGC=40°, ∠DGF=53°, CB⊥GB, FG is perpendicular to CG, GF is perpendicular to DF. Find the measure of ∠FDG.
问题：如图所示，∠AGC=40°，∠DGF=53°，CB垂直于GB，FG垂直于CG，GF⊥DF。求∠FDG的大小。
Construction (CDL):
Shape(DG,GF,FD)
Shape(FG,GA,AF)
Shape(BG,GC,CB)
Shape(AB,BC,CA)
Collinear(DFA)
Collinear(ABG)
Conditions (CDL):
Equal(MeasureOfAngle(AGC),40)
Equal(MeasureOfAngle(DGF),53)
PerpendicularBetweenLine(CB,GB)
PerpendicularBetweenLine(FG,CG)
PerpendicularBetweenLine(GF,DF)
Goal: Value(MeasureOfAngle(FDG))
Answer: 37
Theorem sequence:
triangle_property_angle_sum(1,DGF)
Problem: As shown in the diagram, ∠FOE=45°, the center of circle O is O, CO⊥AO, EO⊥BO. Find the measure of arc OAE.
问题：如图所示，∠FOE=45°，O是⊙O的圆心，CO⊥AO，EO垂直于BO。求弧OAE的角度。
Construction (CDL):
Shape(OE,OEF,FO)
Shape(OF,OFA,AO)
Shape(OA,OAC,CO)
Shape(OC,OCB,BO)
Shape(OB,OBE,EO)
Collinear(EOC)
Collinear(AOB)
Cocircular(O,EFACB)
Conditions (CDL):
Equal(MeasureOfAngle(FOE),45)
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(CO,AO)
PerpendicularBetweenLine(EO,BO)
Goal: Value(MeasureOfArc(OAE))
Answer: 270
Theorem sequence:
adjacent_complementary_angle(1,EOB,BOC)
angle_addition(1,EOB,BOC)
angle_addition(1,EOC,COA)
arc_property_center_angle(1,OAE,O)
Problem: As shown in the diagram, BC=8, BH=12, the center of circle O is O, quadrilateral DCBH is a rectangle. Find the area of quadrilateral DCBH minus the area of the sector ODH.
问题：如图所示，BC=8，BH=12，圆O的圆心为O，四边形DCBH是矩形。求四边形DCBH的面积减去扇形ODH的面积。
Construction (CDL):
Shape(DC,CB,BH,ODH)
Shape(OD,ODH,HO)
Collinear(DOH)
Cocircular(O,DH)
Conditions (CDL):
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(BH),12)
IsCentreOfCircle(O,O)
Rectangle(DCBH)
Goal: Value(Sub(AreaOfQuadrilateral(DCBH),AreaOfSector(ODH)))
Answer: 96-8*pi
Theorem sequence:
parallelogram_area_formula_sine(1,CBHD)
diameter_of_circle_judgment_pass_centre(1,DOH,O)
parallelogram_property_opposite_line_equal(1,CBHD)
diameter_of_circle_property_length_equal(1,DH,O)
circle_property_length_of_radius_and_diameter(1,O)
flat_angle(1,HOD)
arc_property_center_angle(1,ODH,O)
sector_area_formula(1,ODH)
Problem: As shown in the diagram, AB=10, AC=6, BC=x, AC⊥BC. Find the value of x.
问题：如图所示，AB=10，AC=6，BC=x，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),6)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 8
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, AB=10.9, EP=14.9, PA=13, ∠DCP=28.5°, ∠PAE=33°, P is the center of the inscribed circle of triangle AEC, ED is perpendicular to PD, PB is perpendicular to AB, PF is perpendicular to EF. Find the measure of ∠CAD.
问题：如图所示，AB=10.9，EP=14.9，PA=13，∠DCP=28.5°，∠PAE=33°，P是三角形AEC内切圆的圆心，ED垂直于PD，PB垂直于AB，PF⊥EF。求∠CAD的大小。
Construction (CDL):
Shape(AF,FP,PA)
Shape(FE,EP,PF)
Shape(AP,PB,BA)
Shape(PE,ED,DP)
Shape(PD,DC,CP)
Shape(BP,PC,CB)
Collinear(AFE)
Collinear(EDC)
Collinear(CBA)
Collinear(EPB)
Collinear(FPC)
Collinear(DPA)
Conditions (CDL):
Equal(LengthOfLine(AB),10.9)
Equal(LengthOfLine(EP),14.9)
Equal(LengthOfLine(PA),13)
Equal(MeasureOfAngle(DCP),28.5)
Equal(MeasureOfAngle(PAE),33)
IsIncenterOfTriangle(P,AEC)
PerpendicularBetweenLine(ED,PD)
PerpendicularBetweenLine(PB,AB)
PerpendicularBetweenLine(PF,EF)
Goal: Value(MeasureOfAngle(CAD))
Answer: 33
Theorem sequence:
Problem: As shown in the diagram, AB=14, DP is perpendicular to AP, ADCB is a rhombus. Find the length of line BC.
问题：如图所示，AB=14，DP⊥AP，四边形ADCB是菱形。求直线BC的长度。
Construction (CDL):
Shape(AD,DP,PA)
Shape(PD,DC,CP)
Shape(PC,CB,BP)
Shape(AP,PB,BA)
Collinear(APC)
Collinear(DPB)
Conditions (CDL):
Equal(LengthOfLine(AB),14)
PerpendicularBetweenLine(DP,AP)
Rhombus(ADCB)
Goal: Value(LengthOfLine(BC))
Answer: 14
Theorem sequence:
Problem: As shown in the diagram, ∠BCE=∠EBC, ∠DAG=136°, ∠DEA=47°, ∠EFB=63°, ∠FED=69°. Find the measure of ∠EBC.
问题：如图所示，∠BCE=∠EBC，∠DAG=136°，∠DEA=47°，∠EFB=63°，∠FED=69°。求∠EBC的大小。
Construction (CDL):
Shape(FB,BE,EF)
Shape(DE,EA,AD)
Shape(BC,CE,EB)
Shape(FE,ED)
Shape(FE,EA)
Shape(DA,AG)
Shape(AE,EC)
Collinear(BEAG)
Collinear(FEC)
Conditions (CDL):
Equal(MeasureOfAngle(BCE),MeasureOfAngle(EBC))
Equal(MeasureOfAngle(DAG),136)
Equal(MeasureOfAngle(DEA),47)
Equal(MeasureOfAngle(EFB),63)
Equal(MeasureOfAngle(FED),69)
Goal: Value(MeasureOfAngle(EBC))
Answer: 32
Theorem sequence:
angle_addition(1,FED,DEA)
vertical_angle(1,FEA,CEB)
triangle_property_angle_sum(1,EBC)
Problem: As shown in the diagram, ∠LWX=53°, WL∥XE, XN is parallel to ZK. Find the measure of ∠XZK.
问题：如图所示，∠LWX=53°，WL∥XE，XN平行于ZK。求∠XZK的大小。
Construction (CDL):
Shape(GW,WL)
Shape(LW,WX)
Shape(WX,XE)
Shape(EX,XN)
Shape(NX,XZ)
Shape(XZ,ZK)
Shape(KZ,ZH)
Shape(HZ,ZY)
Shape(ZY,YM)
Shape(MY,YI)
Shape(IY,YW)
Shape(YW,WG)
Shape(WY,YZ,ZX,XW)
Collinear(GWXN)
Collinear(IYZK)
Collinear(LWYM)
Collinear(EXZH)
Conditions (CDL):
Equal(MeasureOfAngle(LWX),53)
ParallelBetweenLine(WL,XE)
ParallelBetweenLine(XN,ZK)
Goal: Value(MeasureOfAngle(XZK))
Answer: 53
Theorem sequence:
parallel_property_corresponding_angle(2,WL,XE,N)
parallel_property_corresponding_angle(1,XN,ZK,E)
Problem: As shown in the diagram, AC=x, AD=4, BC=y, BD=9, CD=z, AC⊥BC, BD⊥CD. Find the value of z.
问题：如图所示，AC=x，AD=4，BC=y，BD=9，CD=z，AC⊥BC，BD⊥CD。求z的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(DC,CB,BD)
Collinear(BDA)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),4)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(BD),9)
Equal(LengthOfLine(CD),z)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BD,CD)
Goal: Value(z)
Answer: 6
Theorem sequence:
line_addition(1,AD,DB)
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,ACB)
right_triangle_judgment_angle(1,BDC)
right_triangle_judgment_angle(1,CDA)
right_triangle_property_pythagorean(1,ACB)
right_triangle_property_pythagorean(1,BDC)
right_triangle_property_pythagorean(1,CDA)
Problem: As shown in the diagram, AB=8, CA=10, CB=a, ∠BAC=60°. Find the value of a.
问题：如图所示，AB=8，CA=10，CB=a，∠BAC=60°。求a的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(CA),10)
Equal(LengthOfLine(CB),a)
Equal(MeasureOfAngle(BAC),60)
Goal: Value(a)
Answer: 2*sqrt(21)
Theorem sequence:
cosine_theorem(1,ACB)
Problem: As shown in the diagram, AB=BC, DF=3*x-7, FE=x+9, the center of ⊙F is F, CE is perpendicular to FE, FD⊥AD. Find the value of x.
问题：如图所示，AB=BC，DF=3*x-7，FE=x+9，圆F的圆心为F，CE垂直于FE，FD垂直于AD。求x的值。
Construction (CDL):
Shape(DB,FBA,AD)
Shape(BD,DF,FB)
Shape(BF,FE,EB)
Shape(FD,DA,FAC,CE,EF)
Shape(BE,EC,FCB)
Collinear(BEC)
Collinear(BDA)
Cocircular(F,BAC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(BC))
Equal(LengthOfLine(DF),3*x-7)
Equal(LengthOfLine(FE),x+9)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(CE,FE)
PerpendicularBetweenLine(FD,AD)
Goal: Value(x)
Answer: 8
Theorem sequence:
adjacent_complementary_angle(1,BDF,FDA)
circle_property_chord_perpendicular_bisect_chord(1,F,FD,BA)
circle_property_chord_perpendicular_bisect_chord(1,F,FE,CB)
line_addition(1,BD,DA)
line_addition(1,BE,EC)
mirror_congruent_triangle_judgment_hl(2,BDF,BFE)
mirror_congruent_triangle_property_line_equal(1,BDF,BFE)
Problem: As shown in the diagram, ∠CFD=x+36°, ∠DEC=2*y°, ∠ECF=78°, ∠FDE=110°, CE is parallel to FD. Find the value of y.
问题：如图所示，∠CFD=x+36°，∠DEC=2*y°，∠ECF=78°，∠FDE=110°，CE平行于FD。求y的值。
Construction (CDL):
Shape(EC,CF,FD,DE)
Conditions (CDL):
Equal(MeasureOfAngle(CFD),x+36)
Equal(MeasureOfAngle(DEC),2*y)
Equal(MeasureOfAngle(ECF),78)
Equal(MeasureOfAngle(FDE),110)
ParallelBetweenLine(CE,FD)
Goal: Value(y)
Answer: 35
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,DF,EC)
Problem: As shown in the diagram, ∠AJH=x°, ∠HGB=2*x°, GB⊥HB, JH is perpendicular to GH. Find the measure of ∠BHG.
问题：如图所示，∠AJH=x°，∠HGB=2*x°，GB垂直于HB，JH⊥GH。求∠BHG的大小。
Construction (CDL):
Shape(AGF,FB,BG)
Shape(AB,BF,AFJ,JA)
Shape(BA,AJ,JH,HB)
Shape(AJH,HJ)
Shape(BH,HG,GB)
Shape(AHG,GH)
Collinear(FBH)
Collinear(GBAJ)
Cocircular(A,GFJH)
Conditions (CDL):
Equal(MeasureOfAngle(AJH),x)
Equal(MeasureOfAngle(HGB),2*x)
PerpendicularBetweenLine(GB,HB)
PerpendicularBetweenLine(JH,GH)
Goal: Value(MeasureOfAngle(BHG))
Answer: 30
Theorem sequence:
triangle_property_angle_sum(1,GJH)
triangle_property_angle_sum(1,GBH)
Problem: As shown in the diagram, ∠CBD=55°, ∠FBG=35°, B is the center of circle B. Find the measure of arc BCD.
问题：如图所示，∠CBD=55°，∠FBG=35°，圆B的圆心为B。求⌒BCD的角度。
Construction (CDL):
Shape(BC,BCA,AB)
Shape(BA,BAG,GB)
Shape(BG,BGF,FB)
Shape(BF,BFD,DB)
Shape(BD,BDC,CB)
Collinear(CBG)
Collinear(ABD)
Cocircular(B,AGFDC)
Conditions (CDL):
Equal(MeasureOfAngle(CBD),55)
Equal(MeasureOfAngle(FBG),35)
IsCentreOfCircle(B,B)
Goal: Value(MeasureOfArc(BCD))
Answer: 305
Theorem sequence:
round_angle(1,CBD,DBC)
arc_property_center_angle(1,BCD,B)
Problem: As shown in the diagram, BA=3*x-13, BC=2*x+5, ∠BCD=∠CDB, ∠BDA=∠DAB, ∠DBC=60°. Find the value of x.
问题：如图所示，BA=3*x-13，BC=2*x+5，∠BCD=∠CDB，∠BDA=∠DAB，∠DBC=60°。求x的值。
Construction (CDL):
Shape(BC,CD,DB)
Shape(BD,DA,AB)
Conditions (CDL):
Equal(LengthOfLine(BA),3*x-13)
Equal(LengthOfLine(BC),2*x+5)
Equal(MeasureOfAngle(BCD),MeasureOfAngle(CDB))
Equal(MeasureOfAngle(BDA),MeasureOfAngle(DAB))
Equal(MeasureOfAngle(DBC),60)
Goal: Value(x)
Answer: 18
Theorem sequence:
isosceles_triangle_judgment_angle_equal(1,BCD)
isosceles_triangle_judgment_angle_equal(1,BDA)
Problem: As shown in the diagram, QT=x, RQ=6, SQ=18, SR=14, ∠TRQ=∠SRT. Find the value of x.
问题：如图所示，QT=x，RQ=6，SQ=18，SR=14，∠TRQ=∠SRT。求x的值。
Construction (CDL):
Shape(QT,TR,RQ)
Shape(TS,SR,RT)
Collinear(QTS)
Conditions (CDL):
Equal(LengthOfLine(QT),x)
Equal(LengthOfLine(RQ),6)
Equal(LengthOfLine(SQ),18)
Equal(LengthOfLine(SR),14)
Equal(MeasureOfAngle(TRQ),MeasureOfAngle(SRT))
Goal: Value(x)
Answer: 27/5
Theorem sequence:
line_addition(1,QT,TS)
bisector_of_angle_judgment_angle_equal(1,RT,SRQ)
bisector_of_angle_property_line_ratio(1,RT,SRQ)
Problem: As shown in the diagram, AD=5, BC=32, CD=12, AC⊥BC, BD⊥CD, triangle ACB is similar to triangle CDB.. Find the perimeter of triangle ACB.
问题：如图所示，AD=5，BC=32，CD=12，AC垂直于BC，BD垂直于CD，三角形ACB相似于三角形CDB。求△ACB的周长。
Construction (CDL):
Shape(AC,CD,DA)
Shape(DC,CB,BD)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),5)
Equal(LengthOfLine(BC),32)
Equal(LengthOfLine(CD),12)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(BD,CD)
SimilarBetweenTriangle(ACB,CDB)
Goal: Value(PerimeterOfTriangle(ACB))
Answer: sqrt(1193)+45
Theorem sequence:
adjacent_complementary_angle(1,BDC,CDA)
right_triangle_judgment_angle(1,CDA)
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,CDA)
right_triangle_property_pythagorean(1,ACB)
triangle_perimeter_formula(1,ACB)
Problem: As shown in the diagram, ∠DEF=25°, ∠GFD=65°, EG⊥DG. Find the measure of ∠FDG.
问题：如图所示，∠DEF=25°，∠GFD=65°，EG垂直于DG。求∠FDG的大小。
Construction (CDL):
Shape(DE,EG,GD)
Shape(DG,GF,FD)
Collinear(EGF)
Conditions (CDL):
Equal(MeasureOfAngle(DEF),25)
Equal(MeasureOfAngle(GFD),65)
PerpendicularBetweenLine(EG,DG)
Goal: Value(MeasureOfAngle(FDG))
Answer: 25
Theorem sequence:
adjacent_complementary_angle(1,EGD,DGF)
triangle_property_angle_sum(1,DGF)
Problem: As shown in the diagram, JK=12, LK=2, ML=x, MN=6. Find the value of x.
问题：如图所示，JK=12，LK=2，ML=x，MN=6。求x的值。
Construction (CDL):
Shape(AKJ,JK)
Shape(KJ,AJN,NM,AMK)
Shape(AMK,ML,LK)
Shape(ANM,MN)
Collinear(LKJ)
Collinear(LMN)
Cocircular(A,KJNM)
Conditions (CDL):
Equal(LengthOfLine(JK),12)
Equal(LengthOfLine(LK),2)
Equal(LengthOfLine(ML),x)
Equal(LengthOfLine(MN),6)
Goal: Value(x)
Answer: -3+sqrt(37)
Theorem sequence:
circle_property_circular_power_segment_and_segment_line(1,LKJ,LMN,A)
line_addition(1,LK,KJ)
line_addition(1,LM,MN)
Problem: As shown in the diagram, ∠FHE=15*x°, △EFG is an equilateral △, EH is the angle bisector of ∠GEF. Find the value of x.
问题：如图所示，∠FHE=15*x°，三角形EFG为等边三角形，EH平分∠GEF。求x的值。
Construction (CDL):
Shape(EF,FH,HE)
Shape(EH,HG,GE)
Collinear(FHG)
Conditions (CDL):
Equal(MeasureOfAngle(FHE),15*x)
EquilateralTriangle(EFG)
IsBisectorOfAngle(EH,GEF)
Goal: Value(x)
Answer: 6
Theorem sequence:
isosceles_triangle_property_line_coincidence(3,EFG,H)
Problem: As shown in the diagram, ∠TYZ=52°, ∠ZYX=38°, TY is perpendicular to XY, XW is perpendicular to TW, YX⊥WX, YZ is perpendicular to TZ. Find the measure of ∠WTZ.
问题：如图所示，∠TYZ=52°，∠ZYX=38°，TY垂直于XY，XW垂直于TW，YX⊥WX，YZ垂直于TZ。求∠WTZ的大小。
Construction (CDL):
Shape(TY,YZ,ZT)
Shape(TZ,ZW,WT)
Shape(ZY,YX,XW,WZ)
Collinear(YZW)
Conditions (CDL):
Equal(MeasureOfAngle(TYZ),52)
Equal(MeasureOfAngle(ZYX),38)
PerpendicularBetweenLine(TY,XY)
PerpendicularBetweenLine(XW,TW)
PerpendicularBetweenLine(YX,WX)
PerpendicularBetweenLine(YZ,TZ)
Goal: Value(MeasureOfAngle(WTZ))
Answer: 52
Theorem sequence:
triangle_property_angle_sum(1,WYX)
angle_addition(1,XWY,YWT)
adjacent_complementary_angle(1,YZT,TZW)
triangle_property_angle_sum(1,WTZ)
Problem: As shown in the diagram, PQ=y, QR=x, SP=2, SR=4, PQ⊥RQ, RS is perpendicular to QS. Find the value of y.
问题：如图所示，PQ=y，QR=x，SP=2，SR=4，PQ⊥RQ，RS⊥QS。求y的值。
Construction (CDL):
Shape(RS,SQ,QR)
Shape(SP,PQ,QS)
Collinear(RSP)
Conditions (CDL):
Equal(LengthOfLine(PQ),y)
Equal(LengthOfLine(QR),x)
Equal(LengthOfLine(SP),2)
Equal(LengthOfLine(SR),4)
PerpendicularBetweenLine(PQ,RQ)
PerpendicularBetweenLine(RS,QS)
Goal: Value(y)
Answer: 2*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,RSQ,QSP)
right_triangle_judgment_angle(1,RSQ)
right_triangle_judgment_angle(1,QSP)
right_triangle_judgment_angle(1,PQR)
right_triangle_property_pythagorean(1,RSQ)
right_triangle_property_pythagorean(1,QSP)
right_triangle_property_pythagorean(1,PQR)
line_addition(1,RS,SP)
Problem: As shown in the diagram, AB=y, AC=12, AD=BD, ∠CAB=x°, DB is perpendicular to CB, ADBC is a square. Find the value of x.
问题：如图所示，AB=y，AC=12，AD=BD，∠CAB=x°，DB垂直于CB，四边形ADBC是正方形。求x的值。
Construction (CDL):
Shape(AD,DB,BA)
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(AD),LengthOfLine(BD))
Equal(MeasureOfAngle(CAB),x)
PerpendicularBetweenLine(DB,CB)
Square(ADBC)
Goal: Value(x)
Answer: 45
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,CAB)
isosceles_triangle_property_angle_equal(1,CAB)
triangle_property_angle_sum(1,CAB)
Problem: As shown in the diagram, AN=10, BC=30, CN=5, AC is perpendicular to NC. Find the area of △ABC.
问题：如图所示，AN=10，BC=30，CN=5，AC⊥NC。求△ABC的面积。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CN,NA)
Collinear(BCN)
Conditions (CDL):
Equal(LengthOfLine(AN),10)
Equal(LengthOfLine(BC),30)
Equal(LengthOfLine(CN),5)
PerpendicularBetweenLine(AC,NC)
Goal: Value(AreaOfTriangle(ABC))
Answer: 75*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,BCA,ACN)
right_triangle_judgment_angle(1,ACN)
right_triangle_property_pythagorean(1,ACN)
triangle_area_formula_sine(1,CAB)
Problem: As shown in the diagram, QT=5*y, RT=x, ST=2*y+12, TP=5*x-28, quadrilateral RQPS is a ▱. Find the value of x.
问题：如图所示，QT=5*y，RT=x，ST=2*y+12，TP=5*x-28，QR和PS是▱RQPS的一组对边。求x的值。
Construction (CDL):
Shape(RQ,QT,TR)
Shape(TQ,QP,PT)
Shape(RT,TS,SR)
Shape(TP,PS,ST)
Collinear(RTP)
Collinear(QTS)
Conditions (CDL):
Equal(LengthOfLine(QT),5*y)
Equal(LengthOfLine(RT),x)
Equal(LengthOfLine(ST),2*y+12)
Equal(LengthOfLine(TP),5*x-28)
Parallelogram(RQPS)
Goal: Value(x)
Answer: 7
Theorem sequence:
parallelogram_property_diagonal_bisection(1,RQPS,T)
Problem: As shown in the diagram, ∠BGE=x°, ∠CGD=135°, ∠DGB=145°, ∠EGC=x°. Find the value of x.
问题：如图所示，∠BGE=x°，∠CGD=135°，∠DGB=145°，∠EGC=x°。求x的值。
Construction (CDL):
Shape(GDC,CG,GD)
Shape(GCE,EG,GC)
Shape(GEB,BG,GE)
Shape(GBD,DG,GB)
Cocircular(G,DCEB)
Conditions (CDL):
Equal(MeasureOfAngle(BGE),x)
Equal(MeasureOfAngle(CGD),135)
Equal(MeasureOfAngle(DGB),145)
Equal(MeasureOfAngle(EGC),x)
Goal: Value(x)
Answer: 40
Theorem sequence:
angle_addition(1,CGD,DGB)
angle_addition(1,BGE,EGC)
round_angle(1,BGC,CGB)
Problem: As shown in the diagram, DE=6*x-12, FE=4*y, GD=6*y-42, GF=2*x+36, EDGF is a ▱. Find the value of y.
问题：如图所示，DE=6*x-12，FE=4*y，GD=6*y-42，GF=2*x+36，四边形EDGF是▱。求y的值。
Construction (CDL):
Shape(ED,DG,GF,FE)
Conditions (CDL):
Equal(LengthOfLine(DE),6*x-12)
Equal(LengthOfLine(FE),4*y)
Equal(LengthOfLine(GD),6*y-42)
Equal(LengthOfLine(GF),2*x+36)
Parallelogram(EDGF)
Goal: Value(y)
Answer: 21
Theorem sequence:
parallelogram_property_opposite_line_equal(1,DGFE)
Problem: As shown in the diagram, BA=5, BC=12, DB=10, EB=x. Find the value of x.
问题：如图所示，BA=5，BC=12，DB=10，EB=x。求x的值。
Construction (CDL):
Shape(ODA,AB,BD)
Shape(OAE,EB,BA)
Shape(OEC,CB,BE)
Shape(OCD,DB,BC)
Collinear(ABC)
Collinear(EBD)
Cocircular(O,AECD)
Conditions (CDL):
Equal(LengthOfLine(BA),5)
Equal(LengthOfLine(BC),12)
Equal(LengthOfLine(DB),10)
Equal(LengthOfLine(EB),x)
Goal: Value(x)
Answer: 6
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,ABC,EBD,O)
Problem: As shown in the diagram, BC=x, BQ=6, QC=8. Find the value of x.
问题：如图所示，BC=x，BQ=6，QC=8。求x的值。
Construction (CDL):
Shape(AD,DB,BC,CA)
Shape(DQ,QB,BD)
Collinear(ADQ)
Collinear(QBC)
Conditions (CDL):
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(BQ),6)
Equal(LengthOfLine(QC),8)
Goal: Value(x)
Answer: 2
Theorem sequence:
line_addition(1,QB,BC)
Problem: As shown in the diagram, ∠DEF=5*x°, ∠FDE=5*x°, ∠GFE=9*x+7°. Find the measure of ∠GFE.
问题：如图所示，∠DEF=5*x°，∠FDE=5*x°，∠GFE=9*x+7°。求∠GFE的大小。
Construction (CDL):
Shape(GF,FE)
Shape(FD,DE,EF)
Collinear(GFD)
Conditions (CDL):
Equal(MeasureOfAngle(DEF),5*x)
Equal(MeasureOfAngle(FDE),5*x)
Equal(MeasureOfAngle(GFE),9*x+7)
Goal: Value(MeasureOfAngle(GFE))
Answer: 70
Theorem sequence:
adjacent_complementary_angle(1,GFE,EFD)
triangle_property_angle_sum(1,FDE)