SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, ∠DFB=110°, the center of circle F is F, CB is the tangent to ⊙F, the tangent to ⊙F is CD. Find the measure of ∠BCD.
问题：如图所示，∠DFB=110°，圆F的圆心为F，圆O的切线为CB，CD是⊙O的切线。求∠BCD的大小。
Construction (CDL):
Shape(FBD,BC,CD)
Shape(FBD,DF,FB)
Shape(FDB,BF,FD)
Cocircular(F,BDA)
Conditions (CDL):
Equal(MeasureOfAngle(DFB),110)
IsCentreOfCircle(F,F)
IsTangentOfCircle(CB,F)
IsTangentOfCircle(CD,F)
Goal: Value(MeasureOfAngle(BCD))
Answer: 70
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CB,F,F)
tangent_of_circle_property_perpendicular(2,CD,F,F)
quadrilateral_property_angle_sum(1,BCDF)
Problem: As shown in the diagram, AB=13, AC=5, BC=x, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=13，AC=5，BC=x，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),13)
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 12
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
Problem: As shown in the diagram, SU=20, YW=20, the measure of arc RYX is 45, R is the center of circle R, RV is perpendicular to SV, YZ is perpendicular to RZ. Find the measure of ⌒RTS.
问题：如图所示，SU=20，YW=20，弧RYX的角度为45，R是圆R的圆心，RV垂直于SV，YZ垂直于RZ。求弧RTS的角度。
Construction (CDL):
Shape(RTS,SV,VT)
Shape(VS,SR,RV)
Shape(RSY,YR,RS)
Shape(RY,YZ,ZR)
Shape(RYX,XZ,ZY)
Shape(RXW,WZ,ZX)
Shape(RZ,ZW,WU,UV,VR)
Shape(RUT,TV,VU)
Collinear(SVU)
Collinear(YZW)
Collinear(RZX)
Collinear(RVT)
Cocircular(R,TSYXWU)
Conditions (CDL):
Equal(LengthOfLine(SU),20)
Equal(LengthOfLine(YW),20)
Equal(MeasureOfArc(RYX),45)
IsCentreOfCircle(R,R)
PerpendicularBetweenLine(RV,SV)
PerpendicularBetweenLine(YZ,RZ)
Goal: Value(MeasureOfArc(RTS))
Answer: 45
Theorem sequence:
radius_of_circle_property_length_equal(1,RS,R)
radius_of_circle_property_length_equal(1,RY,R)
arc_property_center_angle(1,RYX,R)
adjacent_complementary_angle(1,UVR,RVS)
circle_property_chord_perpendicular_bisect_chord(1,R,RV,US)
circle_property_chord_perpendicular_bisect_chord(1,R,RZ,YW)
line_addition(1,SV,VU)
line_addition(1,YZ,ZW)
mirror_congruent_triangle_judgment_hl(2,YZR,SRV)
mirror_congruent_triangle_property_angle_equal(1,RYZ,RVS)
arc_property_center_angle(1,RTS,R)
Problem: As shown in the diagram, 720=Add(MeasureOfAngle(CBA), ∠AFE=x-4°, ∠BAF=x+2°, ∠CBA=x-8°, ∠DCB=x+7°, ∠EDC=x-3°, ∠FED=x+6°. Find the measure of ∠DCB.
问题：如图所示，720=Add(MeasureOfAngle(CBA)，∠AFE=x-4°，∠BAF=x+2°，∠CBA=x-8°，∠DCB=x+7°，∠EDC=x-3°，∠FED=x+6°。求∠DCB的大小。
Construction (CDL):
Shape(AF,FC,CB,BA)
Shape(FE,ED,DC,CF)
Conditions (CDL):
Equal(720,Add(MeasureOfAngle(CBA),MeasureOfAngle(BAF),MeasureOfAngle(DCB),MeasureOfAngle(FED),MeasureOfAngle(CBA),MeasureOfAngle(EDC)))
Equal(MeasureOfAngle(AFE),x-4)
Equal(MeasureOfAngle(BAF),x+2)
Equal(MeasureOfAngle(CBA),x-8)
Equal(MeasureOfAngle(DCB),x+7)
Equal(MeasureOfAngle(EDC),x-3)
Equal(MeasureOfAngle(FED),x+6)
Goal: Value(MeasureOfAngle(DCB))
Answer: 383/3
Theorem sequence:
Problem: As shown in the diagram, CA=16, CX=x, XB=x+16, the tangent to ⊙D is CA. Find the value of x.
问题：如图所示，CA=16，CX=x，XB=x+16，CA是⊙O的切线。求x的值。
Construction (CDL):
Shape(DXB,BX)
Shape(DBX,XB)
Shape(DAX,AC,CX)
Collinear(BXC)
Cocircular(D,BAX)
Conditions (CDL):
Equal(LengthOfLine(CA),16)
Equal(LengthOfLine(CX),x)
Equal(LengthOfLine(XB),x+16)
IsTangentOfCircle(CA,D)
Goal: Value(x)
Answer: 8
Theorem sequence:
line_addition(1,BX,XC)
circle_property_circular_power_tangent_and_segment_line(1,CA,CXB,D)
Problem: As shown in the diagram, AF=CA, EA=13, EC=CA, EC=FE, EC=x, FE=AF, ∠EAF=y°, EC⊥AC. Find the value of x.
问题：如图所示，AF=CA，EA=13，EC=CA，EC=FE，EC=x，FE=AF，∠EAF=y°，EC⊥AC。求x的值。
Construction (CDL):
Shape(CA,AE,EC)
Shape(EA,AF,FE)
Conditions (CDL):
Equal(LengthOfLine(AF),LengthOfLine(CA))
Equal(LengthOfLine(EA),13)
Equal(LengthOfLine(EC),LengthOfLine(CA))
Equal(LengthOfLine(EC),LengthOfLine(FE))
Equal(LengthOfLine(EC),x)
Equal(LengthOfLine(FE),LengthOfLine(AF))
Equal(MeasureOfAngle(EAF),y)
PerpendicularBetweenLine(EC,AC)
Goal: Value(x)
Answer: 13*sqrt(2)/2
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,CAE)
isosceles_triangle_property_angle_equal(1,CAE)
triangle_property_angle_sum(1,CAE)
sine_theorem(1,ECA)
Problem: As shown in the diagram, AD=DB, AF=FC, CA=15, CE=EB, EF=9.2, ∠BDE=82°. Find the length of line DE.
问题：如图所示，AD=DB，AF=FC，CA=15，CE=EB，EF=9.2，∠BDE=82°。求直线DE的长度。
Construction (CDL):
Shape(AF,FE,ED,DA)
Shape(FC,CE,EF)
Shape(DE,EB,BD)
Collinear(CFA)
Collinear(CEB)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(DB))
Equal(LengthOfLine(AF),LengthOfLine(FC))
Equal(LengthOfLine(CA),15)
Equal(LengthOfLine(CE),LengthOfLine(EB))
Equal(LengthOfLine(EF),9.2)
Equal(MeasureOfAngle(BDE),82)
Goal: Value(LengthOfLine(DE))
Answer: 15/2
Theorem sequence:
midsegment_of_triangle_judgment_midpoint(1,DE,BAC)
midsegment_of_triangle_property_length(1,DE,BAC)
Problem: As shown in the diagram, ∠YBX=106°, the center of ⊙B is B, the tangent to circle B is WX, WY is the tangent to circle B. Find the measure of ∠XWY.
问题：如图所示，∠YBX=106°，圆B的圆心为B，⊙O的切线为WX，圆O的切线为WY。求∠XWY的大小。
Construction (CDL):
Shape(XW,WY,BXY)
Shape(BXY,YB,BX)
Shape(BYX,XB,BY)
Cocircular(B,XYZ)
Conditions (CDL):
Equal(MeasureOfAngle(YBX),106)
IsCentreOfCircle(B,B)
IsTangentOfCircle(WX,B)
IsTangentOfCircle(WY,B)
Goal: Value(MeasureOfAngle(XWY))
Answer: 74
Theorem sequence:
tangent_of_circle_property_perpendicular(1,WX,B,B)
tangent_of_circle_property_perpendicular(2,WY,B,B)
quadrilateral_property_angle_sum(1,WYBX)
Problem: As shown in the diagram, the center of ⊙O is O. Find the measure of arc OAB.
问题：如图所示，O是圆O的圆心。求弧OAB的角度。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OBA,AO,OB)
Collinear(BOA)
Cocircular(O,BAC)
Conditions (CDL):
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfArc(OAB))
Answer: 180
Theorem sequence:
flat_angle(1,BOA)
arc_property_center_angle(1,OAB,O)
Problem: As shown in the diagram, AD=8, AE=6, AF=a, AG=b, BF=4, BG=10, BG is the tangent to ⊙H. Find the value of b.
问题：如图所示，AD=8，AE=6，AF=a，AG=b，BF=4，BG=10，圆O的切线为BG。求b的值。
Construction (CDL):
Shape(GB,BF,HGF)
Shape(HGF,FA,AG)
Shape(HFD,DA,AF)
Shape(HDE,EA,AD)
Shape(HEG,GA,AE)
Collinear(GAD)
Collinear(BFAE)
Cocircular(H,EGFD)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(AE),6)
Equal(LengthOfLine(AF),a)
Equal(LengthOfLine(AG),b)
Equal(LengthOfLine(BF),4)
Equal(LengthOfLine(BG),10)
IsTangentOfCircle(BG,H)
Goal: Value(b)
Answer: 45/4
Theorem sequence:
circle_property_circular_power_tangent_and_segment_line(1,BG,BFE,H)
line_addition(1,BF,FE)
line_addition(1,FA,AE)
circle_property_circular_power_chord_and_chord(1,EAF,GAD,H)
Problem: As shown in the diagram, AD=17, CA=22, AD⊥BD, BC is perpendicular to AC, CA is perpendicular to DA, DB⊥CB. Find the area of quadrilateral CADB.
问题：如图所示，AD=17，CA=22，AD垂直于BD，BC⊥AC，CA⊥DA，DB垂直于CB。求CADB的面积。
Construction (CDL):
Shape(AD,DB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AD),17)
Equal(LengthOfLine(CA),22)
PerpendicularBetweenLine(AD,BD)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(CA,DA)
PerpendicularBetweenLine(DB,CB)
Goal: Value(AreaOfQuadrilateral(CADB))
Answer: 374
Theorem sequence:
parallel_judgment_ipsilateral_internal_angle(1,DA,BC)
parallel_judgment_ipsilateral_internal_angle(1,BD,CA)
parallelogram_judgment_parallel_and_parallel(1,CADB)
parallelogram_area_formula_sine(1,CADB)
Problem: As shown in the diagram, VW=72, XV=75, XW=21, VW⊥XW. Find the value of tan(WX).
问题：如图所示，VW=72，XV=75，XW=21，VW垂直于XW。求tan(WX)的值。
Construction (CDL):
Shape(VW,WX,XV)
Conditions (CDL):
Equal(LengthOfLine(VW),72)
Equal(LengthOfLine(XV),75)
Equal(LengthOfLine(XW),21)
PerpendicularBetweenLine(VW,XW)
Goal: Value(Tan(MeasureOfAngle(WXV)))
Answer: 24/7
Theorem sequence:
cosine_theorem(1,XVW)
Problem: As shown in the diagram, AC=10, AD=x, BD=y, CD=4, BD⊥AD, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AC=10，AD=x，BD=y，CD=4，BD垂直于AD，CA⊥BA。求x的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AC),10)
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(BD),y)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 2*sqrt(21)
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,ADC)
right_triangle_property_pythagorean(1,ADC)
Problem: As shown in the diagram, AB=2*sqrt(3), AY=x, CY=y, ∠ACB=60°, ∠BAC=30°, AY⊥BY, CB is perpendicular to AB. Find the value of y.
问题：如图所示，AB=2*sqrt(3)，AY=x，CY=y，∠ACB=60°，∠BAC=30°，AY⊥BY，CB⊥AB。求y的值。
Construction (CDL):
Shape(AY,YB,BA)
Shape(BY,YC,CB)
Collinear(AYC)
Conditions (CDL):
Equal(LengthOfLine(AB),2*sqrt(3))
Equal(LengthOfLine(AY),x)
Equal(LengthOfLine(CY),y)
Equal(MeasureOfAngle(ACB),60)
Equal(MeasureOfAngle(BAC),30)
PerpendicularBetweenLine(AY,BY)
PerpendicularBetweenLine(CB,AB)
Goal: Value(y)
Answer: 1
Theorem sequence:
adjacent_complementary_angle(1,AYB,BYC)
sine_theorem(1,BAY)
triangle_property_angle_sum(1,BYC)
sine_theorem(1,BYC)
sine_theorem(1,CBY)
Problem: As shown in the diagram, AD=x, CD=6, DB=3, DF=2. Find the value of x.
问题：如图所示，AD=x，CD=6，DB=3，DF=2。求x的值。
Construction (CDL):
Shape(ECF,FD,DC)
Shape(EFB,BD,DF)
Shape(EBA,AD,DB)
Shape(EAC,CD,DA)
Collinear(FDA)
Collinear(CDB)
Cocircular(E,ACFB)
Conditions (CDL):
Equal(LengthOfLine(AD),x)
Equal(LengthOfLine(CD),6)
Equal(LengthOfLine(DB),3)
Equal(LengthOfLine(DF),2)
Goal: Value(x)
Answer: 9
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,ADF,CDB,E)
Problem: As shown in the diagram, AE=12, AJ=15, ∠JBA=34°, ∠JCB=32°, J is the center of the inscribed circle of triangle ACB, CD is perpendicular to JD, JE⊥AE, JF⊥CF. Find the length of line JF.
问题：如图所示，AE=12，AJ=15，∠JBA=34°，∠JCB=32°，J是三角形ACB的内心，CD垂直于JD，JE垂直于AE，JF⊥CF。求直线JF的长度。
Construction (CDL):
Shape(JD,DB,BJ)
Shape(JB,BE,EJ)
Shape(JE,EA,AJ)
Shape(JA,AF,FJ)
Shape(JF,FC,CJ)
Shape(JC,CD,DJ)
Collinear(BDC)
Collinear(BEA)
Collinear(CFA)
Conditions (CDL):
Equal(LengthOfLine(AE),12)
Equal(LengthOfLine(AJ),15)
Equal(MeasureOfAngle(JBA),34)
Equal(MeasureOfAngle(JCB),32)
IsIncenterOfTriangle(J,ACB)
PerpendicularBetweenLine(CD,JD)
PerpendicularBetweenLine(JE,AE)
PerpendicularBetweenLine(JF,CF)
Goal: Value(LengthOfLine(JF))
Answer: 9
Theorem sequence:
adjacent_complementary_angle(1,AFJ,JFC)
right_triangle_judgment_angle(1,JEA)
right_triangle_property_pythagorean(1,JEA)
bisector_of_angle_judgment_angle_equal(1,AJ,EAF)
bisector_of_angle_property_distance_equal(1,AJ,EAF)
Problem: As shown in the diagram, ∠AID=112°, ∠BED=47°, ∠EDH=65°. Find the measure of ∠EBH.
问题：如图所示，∠AID=112°，∠BED=47°，∠EDH=65°。求∠EBH的大小。
Construction (CDL):
Shape(BE,ED,DB)
Shape(DI,IF,FD)
Shape(EB,BH)
Collinear(HBDI)
Collinear(EDF)
Collinear(FIA)
Conditions (CDL):
Equal(MeasureOfAngle(AID),112)
Equal(MeasureOfAngle(BED),47)
Equal(MeasureOfAngle(EDH),65)
Goal: Value(MeasureOfAngle(EBH))
Answer: 112
Theorem sequence:
triangle_property_angle_sum(1,DBE)
adjacent_complementary_angle(1,DBE,EBH)
Problem: As shown in the diagram, ∠PSR=x+10°, ∠QPS=x°, ∠RQP=2*x-16°, ∠SRQ=2*x°. Find the measure of ∠QPS.
问题：如图所示，∠PSR=x+10°，∠QPS=x°，∠RQP=2*x-16°，∠SRQ=2*x°。求∠QPS的大小。
Construction (CDL):
Shape(QP,PS,SR,RQ)
Conditions (CDL):
Equal(MeasureOfAngle(PSR),x+10)
Equal(MeasureOfAngle(QPS),x)
Equal(MeasureOfAngle(RQP),2*x-16)
Equal(MeasureOfAngle(SRQ),2*x)
Goal: Value(MeasureOfAngle(QPS))
Answer: 61
Theorem sequence:
quadrilateral_property_angle_sum(1,QPSR)
Problem: As shown in the diagram, PS=21, RS=7, QS⊥RS, RQ⊥PQ. Find the height of triangle QPR.
问题：如图所示，PS=21，RS=7，QS⊥RS，RQ垂直于PQ。求三角形QPR的高。
Construction (CDL):
Shape(QP,PS,SQ)
Shape(QS,SR,RQ)
Collinear(PSR)
Conditions (CDL):
Equal(LengthOfLine(PS),21)
Equal(LengthOfLine(RS),7)
PerpendicularBetweenLine(QS,RS)
PerpendicularBetweenLine(RQ,PQ)
Goal: Value(HeightOfTriangle(QPR))
Answer: 7*sqrt(3)
Theorem sequence:
line_addition(1,PS,SR)
adjacent_complementary_angle(1,PSQ,QSR)
altitude_of_triangle_judgment(1,QS,QPR)
right_triangle_judgment_angle(1,PSQ)
right_triangle_judgment_angle(1,QSR)
right_triangle_judgment_angle(1,RQP)
right_triangle_property_pythagorean(1,PSQ)
right_triangle_property_pythagorean(1,QSR)
right_triangle_property_pythagorean(1,RQP)
Problem: As shown in the diagram, WX=WY, ∠WYX=4*x+20°, ∠XWY=52°, ∠YXW=6*y-2°. Find the value of x.
问题：如图所示，WX=WY，∠WYX=4*x+20°，∠XWY=52°，∠YXW=6*y-2°。求x的值。
Construction (CDL):
Shape(WY,YX,XW)
Conditions (CDL):
Equal(LengthOfLine(WX),LengthOfLine(WY))
Equal(MeasureOfAngle(WYX),4*x+20)
Equal(MeasureOfAngle(XWY),52)
Equal(MeasureOfAngle(YXW),6*y-2)
Goal: Value(x)
Answer: 11
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,WYX)
isosceles_triangle_property_angle_equal(1,WYX)
triangle_property_angle_sum(1,WYX)
Problem: As shown in the diagram, HP=2*y, IH=4/3*y+1, IH=HP, JB=6-x, LJ=2*x+3, LJ=JR, JH∥BP, LI∥BP, LI∥JH. Find the value of y.
问题：如图所示，HP=2*y，IH=4/3*y+1，IH=HP，JB=6-x，LJ=2*x+3，LJ=JR，JH平行于BP，LI平行于BP，LI平行于JH。求y的值。
Construction (CDL):
Shape(IL,LJ,JH,HI)
Shape(JB,BP,PH,HJ)
Collinear(GLIK)
Collinear(MJHC)
Collinear(DBPN)
Collinear(ALJRO)
Collinear(EIHPF)
Conditions (CDL):
Equal(LengthOfLine(HP),2*y)
Equal(LengthOfLine(IH),4/3*y+1)
Equal(LengthOfLine(IH),LengthOfLine(HP))
Equal(LengthOfLine(JB),6-x)
Equal(LengthOfLine(LJ),2*x+3)
Equal(LengthOfLine(LJ),LengthOfLine(JR))
ParallelBetweenLine(JH,BP)
ParallelBetweenLine(LI,BP)
ParallelBetweenLine(LI,JH)
Goal: Value(y)
Answer: 3/2
Theorem sequence:
Problem: As shown in the diagram, ∠ADC=x+34°, ∠BAD=82°, ∠DCB=83°, ∠PRS=3*y-13°, ∠QPR=97°, ∠RSQ=98°, quadrilateral ADCB is similar to quadrilateral QPRS. Find the value of x.
问题：如图所示，∠ADC=x+34°，∠BAD=82°，∠DCB=83°，∠PRS=3*y-13°，∠QPR=97°，∠RSQ=98°，ADCB相似于QPRS。求x的值。
Construction (CDL):
Shape(AD,DC,CB,BA)
Shape(PR,RS,SQ,QP)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),x+34)
Equal(MeasureOfAngle(BAD),82)
Equal(MeasureOfAngle(DCB),83)
Equal(MeasureOfAngle(PRS),3*y-13)
Equal(MeasureOfAngle(QPR),97)
Equal(MeasureOfAngle(RSQ),98)
SimilarBetweenQuadrilateral(ADCB,QPRS)
Goal: Value(x)
Answer: 63
Theorem sequence:
similar_quadrilateral_property_angle_equal(1,DCBA,PRSQ)
Problem: As shown in the diagram, AP=6*r+4, PH=22+3*r, ∠AHW=8*q+17°, ∠HWP=10+q°, WP perpendicular bisects AH. Find the measure of ∠HWP.
问题：如图所示，AP=6*r+4，PH=22+3*r，∠AHW=8*q+17°，∠HWP=10+q°，WP垂直平分AB。求∠HWP的大小。
Construction (CDL):
Shape(HW,WX,XH)
Shape(XW,WQ,QX)
Shape(HX,XP,PH)
Shape(XQ,QA,AP,PX)
Collinear(WXP)
Collinear(WQA)
Collinear(HPA)
Collinear(HXQ)
Conditions (CDL):
Equal(LengthOfLine(AP),6*r+4)
Equal(LengthOfLine(PH),22+3*r)
Equal(MeasureOfAngle(AHW),8*q+17)
Equal(MeasureOfAngle(HWP),10+q)
IsPerpendicularBisectorOfLine(WP,AH)
Goal: Value(MeasureOfAngle(HWP))
Answer: 17
Theorem sequence:
triangle_property_angle_sum(1,HWP)
Problem: As shown in the diagram, KJ=7*x, KJ=LJ, KL=11*x-8, KL=KJ, LJ=x+12. Find the length of line KL.
问题：如图所示，KJ=7*x，KJ=LJ，KL=11*x-8，KL=KJ，LJ=x+12。求直线KL的长度。
Construction (CDL):
Shape(KJ,JL,LK)
Conditions (CDL):
Equal(LengthOfLine(KJ),7*x)
Equal(LengthOfLine(KJ),LengthOfLine(LJ))
Equal(LengthOfLine(KL),11*x-8)
Equal(LengthOfLine(KL),LengthOfLine(KJ))
Equal(LengthOfLine(LJ),x+12)
Goal: Value(LengthOfLine(KL))
Answer: 14
Theorem sequence:
Problem: As shown in the diagram, AB=25, AC=22, BC=x, BC is perpendicular to AC. Find the value of x.
问题：如图所示，AB=25，AC=22，BC=x，BC垂直于AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),25)
Equal(LengthOfLine(AC),22)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: sqrt(141)
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, DE=6*x-4, EF=2*x, the tangent to ⊙H is ED, the tangent to ⊙H is EF. Find the value of x.
问题：如图所示，DE=6*x-4，EF=2*x，⊙O的切线为ED，⊙O的切线为EF。求x的值。
Construction (CDL):
Shape(HFD,DE,EF)
Cocircular(H,FD)
Conditions (CDL):
Equal(LengthOfLine(DE),6*x-4)
Equal(LengthOfLine(EF),2*x)
IsTangentOfCircle(ED,H)
IsTangentOfCircle(EF,H)
Goal: Value(x)
Answer: 1
Theorem sequence:
tangent_of_circle_property_length_equal(1,ED,EF,H)
Problem: As shown in the diagram, ∠BEC=57°, AL is parallel to BG, HC∥FJ. Find the measure of ∠IDH.
问题：如图所示，∠BEC=57°，AL∥BG，HC平行于FJ。求∠IDH的大小。
Construction (CDL):
Shape(BE,EC)
Shape(KE,ED)
Shape(ID,DH)
Collinear(HDEC)
Collinear(FIKJ)
Collinear(ADIL)
Collinear(BEKG)
Conditions (CDL):
Equal(MeasureOfAngle(BEC),57)
ParallelBetweenLine(AL,BG)
ParallelBetweenLine(HC,FJ)
Goal: Value(MeasureOfAngle(IDH))
Answer: 57
Theorem sequence:
vertical_angle(1,BEC,KED)
parallel_property_collinear_extend(3,AL,BG,D)
parallel_property_collinear_extend(3,GB,DA,E)
parallel_property_corresponding_angle(2,EB,DA,H)
angle_addition(1,ADI,IDH)
angle_addition(1,BEK,KED)
flat_angle(1,ADI)
flat_angle(1,BEK)
Problem: As shown in the diagram, AB=x, AC=12, AD=y, BD=z, CD=4, AD is perpendicular to CD, CA⊥BA. Find the value of x.
问题：如图所示，AB=x，AC=12，AD=y，BD=z，CD=4，AD⊥CD，CA垂直于BA。求x的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(AD),y)
Equal(LengthOfLine(BD),z)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 24*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,ADC)
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,ADC)
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,BDA)
right_triangle_property_pythagorean(1,BDA)
line_addition(1,BD,DC)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, AB=12, AC=x, BC=y, ∠ABC=30°, ∠BCA=60°, CA⊥BA. Find the value of x.
问题：如图所示，AB=12，AC=x，BC=y，∠ABC=30°，∠BCA=60°，CA⊥BA。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
Equal(MeasureOfAngle(ABC),30)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 4*sqrt(3)
Theorem sequence:
sine_theorem(1,CAB)
sine_theorem(1,BCA)
Problem: As shown in the diagram, AB=y, AC=x, AD=z, BD=9, CD=4, BD is perpendicular to AD, CA⊥BA. Find the value of x.
问题：如图所示，AB=y，AC=x，AD=z，BD=9，CD=4，BD垂直于AD，CA⊥BA。求x的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(DA,AB,BD)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),z)
Equal(LengthOfLine(BD),9)
Equal(LengthOfLine(CD),4)
PerpendicularBetweenLine(BD,AD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 2*sqrt(13)
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADC)
right_triangle_judgment_angle(1,CAB)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,ADC)
line_addition(1,CD,DB)
right_triangle_property_pythagorean(1,CAB)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,ADC)
Problem: As shown in the diagram, ∠PQR=101°, ∠QRS=x°, ∠RSP=y°, QRSP is a ▱. Find the value of x.
问题：如图所示，∠PQR=101°，∠QRS=x°，∠RSP=y°，RQ和SP是平行四边形QRSP的一组对边。求x的值。
Construction (CDL):
Shape(QR,RS,SP,PQ)
Conditions (CDL):
Equal(MeasureOfAngle(PQR),101)
Equal(MeasureOfAngle(QRS),x)
Equal(MeasureOfAngle(RSP),y)
Parallelogram(QRSP)
Goal: Value(x)
Answer: 79
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,QRSP)
parallelogram_property_opposite_angle_equal(1,RSPQ)
quadrilateral_property_angle_sum(1,QRSP)
Problem: As shown in the diagram, ∠BCH=8*y-12°, ∠BHD=1/4*x°, ∠HBC=y-8°, ∠HDB=4*x-8°, BD and CH are opposite sides of the ▱ DBCH. Find the value of x.
问题：如图所示，∠BCH=8*y-12°，∠BHD=1/4*x°，∠HBC=y-8°，∠HDB=4*x-8°，DH和BC是▱DBCH的一组对边。求x的值。
Construction (CDL):
Shape(DB,BH,HD)
Shape(BC,CH,HB)
Conditions (CDL):
Equal(MeasureOfAngle(BCH),8*y-12)
Equal(MeasureOfAngle(BHD),1/4*x)
Equal(MeasureOfAngle(HBC),y-8)
Equal(MeasureOfAngle(HDB),4*x-8)
Parallelogram(DBCH)
Goal: Value(x)
Answer: 30
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,DBCH)
parallel_property_alternate_interior_angle(2,DH,BC)
Problem: As shown in the diagram, AD=28, CA=30, ∠CAE=30°, quadrilateral CADB is a ▱, BE⊥AE, EA is perpendicular to DA. Find the area of CADB.
问题：如图所示，AD=28，CA=30，∠CAE=30°，CADB是平行四边形，BE⊥AE，EA⊥DA。求四边形CADB的面积。
Construction (CDL):
Shape(EC,CA,AE)
Shape(BE,EA,AD,DB)
Collinear(CEB)
Conditions (CDL):
Equal(LengthOfLine(AD),28)
Equal(LengthOfLine(CA),30)
Equal(MeasureOfAngle(CAE),30)
Parallelogram(CADB)
PerpendicularBetweenLine(BE,AE)
PerpendicularBetweenLine(EA,DA)
Goal: Value(AreaOfQuadrilateral(CADB))
Answer: 420*sqrt(3)
Theorem sequence:
angle_addition(1,CAE,EAD)
parallelogram_area_formula_sine(1,CADB)
Problem: As shown in the diagram, AB=4*y-2, AB=AC, AC=2*y+2, ∠BCA=6*x+8°, ∠CAB=80°. Find the value of y.
问题：如图所示，AB=4*y-2，AB=AC，AC=2*y+2，∠BCA=6*x+8°，∠CAB=80°。求y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),4*y-2)
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(AC),2*y+2)
Equal(MeasureOfAngle(BCA),6*x+8)
Equal(MeasureOfAngle(CAB),80)
Goal: Value(y)
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, AE=EB, BE=12, CE=ED, ED=10, quadrilateral ACBD is a rhombus. Find the area of ACBD.
问题：如图所示，AE=EB，BE=12，CE=ED，ED=10，ACBD是菱形。求ACBD的面积。
Construction (CDL):
Shape(AC,CE,EA)
Shape(AE,ED,DA)
Shape(EC,CB,BE)
Shape(EB,BD,DE)
Collinear(AEB)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(AE),LengthOfLine(EB))
Equal(LengthOfLine(BE),12)
Equal(LengthOfLine(CE),LengthOfLine(ED))
Equal(LengthOfLine(ED),10)
Rhombus(ACBD)
Goal: Value(AreaOfQuadrilateral(ACBD))
Answer: 240
Theorem sequence:
line_addition(1,AE,EB)
line_addition(1,CE,ED)
kite_area_formula_diagonal(1,ACBD)
Problem: As shown in the diagram, ∠CDF=65°, ∠CEF=x°, ∠DFE=85°. Find the value of x.
问题：如图所示，∠CDF=65°，∠CEF=x°，∠DFE=85°。求x的值。
Construction (CDL):
Shape(ED,DF,FE)
Shape(CE,EF)
Collinear(CED)
Conditions (CDL):
Equal(MeasureOfAngle(CDF),65)
Equal(MeasureOfAngle(CEF),x)
Equal(MeasureOfAngle(DFE),85)
Goal: Value(x)
Answer: 150
Theorem sequence:
triangle_property_angle_sum(1,EDF)
adjacent_complementary_angle(1,CEF,FED)
Problem: As shown in the diagram, AB=19, EC=Add(LengthOfLine(DE),LengthOfLine(FA)), FD=6, the incenter of triangle DCA is O, AB is the tangent to ⊙O, the tangent to ⊙O is AF, the tangent to circle O is CB, CE is the tangent to ⊙O, DE is the tangent to circle O, DF is the tangent to circle O. Find the perimeter of triangle DCA.
问题：如图所示，AB=19，EC=Add(LengthOfLine(DE),LengthOfLine(FA))，FD=6，△DCA的内心为O，AB是圆O的切线，AF是⊙O的切线，CB是⊙O的切线，圆O的切线为CE，DE是圆O的切线，DF是圆O的切线。求△DCA的周长。
Construction (CDL):
Shape(EC,CB,OEB)
Shape(FD,DE,OFE)
Shape(BA,AF,OBF)
Shape(OFE,OEB,OBF)
Collinear(DEC)
Collinear(AFD)
Collinear(CBA)
Cocircular(O,EBF)
Conditions (CDL):
Equal(LengthOfLine(AB),19)
Equal(LengthOfLine(EC),Add(LengthOfLine(DE),LengthOfLine(FA)))
Equal(LengthOfLine(FD),6)
IsIncenterOfTriangle(O,DCA)
IsTangentOfCircle(AB,O)
IsTangentOfCircle(AF,O)
IsTangentOfCircle(CB,O)
IsTangentOfCircle(CE,O)
IsTangentOfCircle(DE,O)
IsTangentOfCircle(DF,O)
Goal: Value(PerimeterOfTriangle(DCA))
Answer: 100
Theorem sequence:
tangent_of_circle_property_length_equal(1,AB,AF,O)
tangent_of_circle_property_length_equal(1,CE,CB,O)
tangent_of_circle_property_length_equal(1,DF,DE,O)
line_addition(1,DE,EC)
line_addition(1,AF,FD)
line_addition(1,CB,BA)
triangle_perimeter_formula(1,CAD)
Problem: As shown in the diagram, ∠CBA=36°, ∠CDE=57°, ∠DBE=42°, ∠ECD=28°. Find the measure of ∠DEC.
问题：如图所示，∠CBA=36°，∠CDE=57°，∠DBE=42°，∠ECD=28°。求∠DEC的大小。
Construction (CDL):
Shape(AE,EB,BA)
Shape(BE,ED,DB)
Shape(EC,CD,DE)
Collinear(AED)
Collinear(CEB)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),36)
Equal(MeasureOfAngle(CDE),57)
Equal(MeasureOfAngle(DBE),42)
Equal(MeasureOfAngle(ECD),28)
Goal: Value(MeasureOfAngle(DEC))
Answer: 95
Theorem sequence:
triangle_property_angle_sum(1,CDE)
Problem: As shown in the diagram, AC=18, BC=15, ∠CAB=x°, BC⊥AC. Find the value of x.
问题：如图所示，AC=18，BC=15，∠CAB=x°，BC⊥AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),18)
Equal(LengthOfLine(BC),15)
Equal(MeasureOfAngle(CAB),x)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 180*acos(6*sqrt(61)/61)/pi
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
cosine_theorem(1,ABC)