SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AB=y, AD=z, BD=4, CB=x, CD=10, AB⊥CB, DC is perpendicular to AC. Find the value of z.
问题：如图所示，AB=y，AD=z，BD=4，CB=x，CD=10，AB⊥CB，DC垂直于AC。求z的值。
Construction (CDL):
Shape(CA,AB,BC)
Shape(CB,BD,DC)
Collinear(ABD)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AD),z)
Equal(LengthOfLine(BD),4)
Equal(LengthOfLine(CB),x)
Equal(LengthOfLine(CD),10)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(DC,AC)
Goal: Value(z)
Answer: 25
Theorem sequence:
adjacent_complementary_angle(1,ABC,CBD)
mirror_similar_triangle_judgment_aa(1,CBD,ADC)
mirror_similar_triangle_property_line_ratio(1,CBD,ADC)
mirror_similar_triangle_property_line_ratio(1,BDC,CAD)
Problem: As shown in the diagram, AC=20, BA=48, BC=x, CA⊥BA. Find the value of x.
问题：如图所示，AC=20，BA=48，BC=x，CA⊥BA。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(BA),48)
Equal(LengthOfLine(BC),x)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 52
Theorem sequence:
right_triangle_judgment_angle(1,CAB)
right_triangle_property_pythagorean(1,CAB)
Problem: As shown in the diagram, BA=18, BC=x, ∠BAC=45°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，BA=18，BC=x，∠BAC=45°，AC垂直于BC。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(BA),18)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(BAC),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 9*sqrt(2)
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, JM=MK, ∠JLM=6*x+8°, ∠MLK=9*x-4°, LK is perpendicular to MK, MJ⊥LJ. Find the measure of ∠MLK.
问题：如图所示，JM=MK，∠JLM=6*x+8°，∠MLK=9*x-4°，LK垂直于MK，MJ⊥LJ。求∠MLK的大小。
Construction (CDL):
Shape(JL,LM,MJ)
Shape(ML,LK,KM)
Conditions (CDL):
Equal(LengthOfLine(JM),LengthOfLine(MK))
Equal(MeasureOfAngle(JLM),6*x+8)
Equal(MeasureOfAngle(MLK),9*x-4)
PerpendicularBetweenLine(LK,MK)
PerpendicularBetweenLine(MJ,LJ)
Goal: Value(MeasureOfAngle(MLK))
Answer: 32
Theorem sequence:
mirror_congruent_triangle_judgment_hl(1,LKM,LMJ)
mirror_congruent_triangle_property_angle_equal(1,LKM,LMJ)
Problem: As shown in the diagram, PQ=6, PS=x, RP=15, TP=4. Find the value of x.
问题：如图所示，PQ=6，PS=x，RP=15，TP=4。求x的值。
Construction (CDL):
Shape(TP,PQ,AQT)
Shape(SP,PT,ATS)
Shape(RP,PS,ASR)
Shape(QP,PR,ARQ)
Collinear(QPS)
Collinear(TPR)
Cocircular(A,RQTS)
Conditions (CDL):
Equal(LengthOfLine(PQ),6)
Equal(LengthOfLine(PS),x)
Equal(LengthOfLine(RP),15)
Equal(LengthOfLine(TP),4)
Goal: Value(x)
Answer: 10
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,QPS,RPT,A)
Problem: As shown in the diagram, WY=6*x+3, XW=9*x, XW=LengthOfLine(WY),LengthOfLine(XY), XY=4*x+5. Find the length of line XW.
问题：如图所示，WY=6*x+3，XW=9*x，XW=LengthOfLine(WY),LengthOfLine(XY)，XY=4*x+5。求直线XW的长度。
Construction (CDL):
Shape(XW,WY,YX)
Conditions (CDL):
Equal(LengthOfLine(WY),6*x+3)
Equal(LengthOfLine(XW),9*x)
Equal(LengthOfLine(XW),LengthOfLine(WY),LengthOfLine(XY))
Equal(LengthOfLine(XY),4*x+5)
Goal: Value(LengthOfLine(XW))
Answer: 9
Theorem sequence:
Problem: As shown in the diagram, AD=7, AE=14, BD=x, the tangent to ⊙O is AE. Find the value of x.
问题：如图所示，AD=7，AE=14，BD=x，AE是圆O的切线。求x的值。
Construction (CDL):
Shape(EA,AD,OED)
Shape(OED,DB,OBE)
Shape(BD,ODB)
Collinear(ADB)
Collinear(AEC)
Cocircular(O,EDB)
Conditions (CDL):
Equal(LengthOfLine(AD),7)
Equal(LengthOfLine(AE),14)
Equal(LengthOfLine(BD),x)
IsTangentOfCircle(AE,O)
Goal: Value(x)
Answer: 21
Theorem sequence:
line_addition(1,AD,DB)
circle_property_circular_power_tangent_and_segment_line(1,AE,ADB,O)
Problem: As shown in the diagram, AB=26, AD=12, DG=9/2, EF=8, GF=14, ∠AGF=108°, AGFE is similar to ADCB. Find the perimeter of quadrilateral AGFE.
问题：如图所示，AB=26，AD=12，DG=9/2，EF=8，GF=14，∠AGF=108°，四边形AGFE相似于四边形ADCB。求AGFE的周长。
Construction (CDL):
Shape(AG,GF,FE,EA)
Shape(GD,DC,CB,BE,EF,FG)
Collinear(AEB)
Collinear(AGD)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DG),9/2)
Equal(LengthOfLine(EF),8)
Equal(LengthOfLine(GF),14)
Equal(MeasureOfAngle(AGF),108)
SimilarBetweenQuadrilateral(AGFE,ADCB)
Goal: Value(PerimeterOfQuadrilateral(AGFE))
Answer: 183/4
Theorem sequence:
line_addition(1,AG,GD)
similar_quadrilateral_property_line_ratio(1,AGFE,ADCB)
similar_quadrilateral_property_line_ratio(1,EAGF,BADC)
quadrilateral_perimeter_formula(1,AGFE)
Problem: As shown in the diagram, ∠YZX=56°, quadrilateral WYZX is a rhombus. Find the measure of ∠TWY.
问题：如图所示，∠YZX=56°，四边形WYZX是菱形。求∠TWY的大小。
Construction (CDL):
Shape(WY,YT,TW)
Shape(TY,YZ,ZT)
Shape(XW,WT,TX)
Shape(TZ,ZX,XT)
Collinear(WTZ)
Collinear(XTY)
Conditions (CDL):
Equal(MeasureOfAngle(YZX),56)
Rhombus(WYZX)
Goal: Value(MeasureOfAngle(TWY))
Answer: 28
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,ZXWY,T)
isosceles_triangle_judgment_line_equal(1,ZXY)
altitude_of_triangle_judgment(1,ZT,ZXY)
isosceles_triangle_property_line_coincidence(1,ZXY,T)
angle_addition(1,YZT,TZX)
isosceles_triangle_judgment_line_equal(1,YZW)
isosceles_triangle_property_angle_equal(1,YZW)
Problem: As shown in the diagram, BA=5, CB=8, CD=12, NE=5, BA is perpendicular to DA, NE⊥BE. Find the sum of the area of △NCB and the area of triangle BCD.
问题：如图所示，BA=5，CB=8，CD=12，NE=5，BA垂直于DA，NE垂直于BE。求△NCB的面积与△BCD的面积之和。
Construction (CDL):
Shape(NE,EB,BN)
Shape(NC,CE,EN)
Shape(BC,CA,AB)
Shape(BA,AD,DB)
Collinear(CEB)
Collinear(CAD)
Conditions (CDL):
Equal(LengthOfLine(BA),5)
Equal(LengthOfLine(CB),8)
Equal(LengthOfLine(CD),12)
Equal(LengthOfLine(NE),5)
PerpendicularBetweenLine(BA,DA)
PerpendicularBetweenLine(NE,BE)
Goal: Value(Add(AreaOfTriangle(NCB),AreaOfTriangle(BCD)))
Answer: 50
Theorem sequence:
adjacent_complementary_angle(1,CEN,NEB)
adjacent_complementary_angle(1,CAB,BAD)
altitude_of_triangle_judgment(1,NE,NCB)
altitude_of_triangle_judgment(1,BA,BCD)
triangle_area_formula_common(1,NCB)
triangle_area_formula_common(1,BCD)
Problem: As shown in the diagram, BA=32, BC=x, CA=y, ∠BAC=60°, AC is perpendicular to BC. Find the value of y.
问题：如图所示，BA=32，BC=x，CA=y，∠BAC=60°，AC垂直于BC。求y的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(BA),32)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CA),y)
Equal(MeasureOfAngle(BAC),60)
PerpendicularBetweenLine(AC,BC)
Goal: Value(y)
Answer: 16
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,ACB)
Problem: As shown in the diagram, AE=x, BC=4, BE=3, CD=2. Find the value of x.
问题：如图所示，AE=x，BC=4，BE=3，CD=2。求x的值。
Construction (CDL):
Shape(DC,XCD)
Shape(EB,BC,XEC)
Shape(AE,XEC,CD,XDA)
Shape(EA,XAE)
Collinear(DCB)
Collinear(AEB)
Cocircular(X,AECD)
Conditions (CDL):
Equal(LengthOfLine(AE),x)
Equal(LengthOfLine(BC),4)
Equal(LengthOfLine(BE),3)
Equal(LengthOfLine(CD),2)
Goal: Value(x)
Answer: 5
Theorem sequence:
line_addition(1,DC,CB)
line_addition(1,AE,EB)
circle_property_circular_power_segment_and_segment_line(1,BEA,BCD,X)
Problem: As shown in the diagram, the diameter of ⊙R is 20, the diameter of ⊙S is 30, DS=9, R is the center of circle R, the center of ⊙S is S. Find the length of line CD.
问题：如图所示，⊙R的直径为20，⊙S的直径为30，DS=9，R是圆R的圆心，⊙S的圆心为S。求直线CD的长度。
Construction (CDL):
Shape(SEC,CD,RDE)
Shape(SCF,RFD,DC)
Shape(REF,SEC,SCF)
Shape(SFB,BS,SD,RFD)
Shape(DS,SB,SBE,RDE)
Collinear(RCDSB)
Cocircular(R,FDE)
Cocircular(S,ECFB)
Conditions (CDL):
Equal(DiameterOfCircle(R),20)
Equal(DiameterOfCircle(S),30)
Equal(LengthOfLine(DS),9)
IsCentreOfCircle(R,R)
IsCentreOfCircle(S,S)
Goal: Value(LengthOfLine(CD))
Answer: 6
Theorem sequence:
circle_property_length_of_radius_and_diameter(1,S)
radius_of_circle_property_length_equal(1,SC,S)
line_addition(1,CD,DS)
Problem: As shown in the diagram, FA=10, FW=x, WA=4, quadrilateral BFWC is similar to quadrilateral NFAD. Find The ratio of similar quadrilaterals BFWC and NFAD.
问题：如图所示，FA=10，FW=x，WA=4，BFWC相似于NFAD。求四边形BFWC和四边形NFAD的相似比。
Construction (CDL):
Shape(NB,BC,CW,WA,AD,DN)
Shape(BF,FW,WC,CB)
Collinear(NBF)
Collinear(FWA)
Conditions (CDL):
Equal(LengthOfLine(FA),10)
Equal(LengthOfLine(FW),x)
Equal(LengthOfLine(WA),4)
SimilarBetweenQuadrilateral(BFWC,NFAD)
Goal: Value(RatioOfSimilarQuadrilateral(BFWC,NFAD))
Answer: 3/5
Theorem sequence:
line_addition(1,FW,WA)
similar_quadrilateral_property_line_ratio(1,FWCB,FADN)
Problem: As shown in the diagram, BA=c, BC=a, BE=x, CA=b, EA=y, ∠CAE=30°, ∠EBC=60°, x=7*sqrt(3), AE is perpendicular to CE, BC is perpendicular to AC. Find the value of y.
问题：如图所示，BA=c，BC=a，BE=x，CA=b，EA=y，∠CAE=30°，∠EBC=60°，x=7*sqrt(3)，AE垂直于CE，BC垂直于AC。求y的值。
Construction (CDL):
Shape(BC,CE,EB)
Shape(EC,CA,AE)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(BA),c)
Equal(LengthOfLine(BC),a)
Equal(LengthOfLine(BE),x)
Equal(LengthOfLine(CA),b)
Equal(LengthOfLine(EA),y)
Equal(MeasureOfAngle(CAE),30)
Equal(MeasureOfAngle(EBC),60)
Equal(x,7*sqrt(3))
PerpendicularBetweenLine(AE,CE)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 21*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,AEC,CEB)
triangle_property_angle_sum(1,AEC)
sine_theorem(1,AEC)
sine_theorem(1,CAE)
triangle_property_angle_sum(1,CEB)
sine_theorem(1,CEB)
sine_theorem(1,BCE)
Problem: As shown in the diagram, AB=26, AD=10, CB=16, DC=8, AD is the altitude of triangle ACB. Find the area of △ACB.
问题：如图所示，AB=26，AD=10，CB=16，DC=8，AD是△ACB的高。求三角形ACB的面积。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AC,CB,BA)
Collinear(DCB)
Conditions (CDL):
Equal(LengthOfLine(AB),26)
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(CB),16)
Equal(LengthOfLine(DC),8)
IsAltitudeOfTriangle(AD,ACB)
Goal: Value(AreaOfTriangle(ACB))
Answer: 80
Theorem sequence:
triangle_area_formula_common(1,ACB)
Problem: As shown in the diagram, PT=8, SR=3, TQ=3, ST∥RQ. Find the length of line PS.
问题：如图所示，PT=8，SR=3，TQ=3，ST∥RQ。求直线PS的长度。
Construction (CDL):
Shape(PS,ST,TP)
Shape(SR,RQ,QT,TS)
Collinear(PSR)
Collinear(PTQ)
Conditions (CDL):
Equal(LengthOfLine(PT),8)
Equal(LengthOfLine(SR),3)
Equal(LengthOfLine(TQ),3)
ParallelBetweenLine(ST,RQ)
Goal: Value(LengthOfLine(PS))
Answer: 8
Theorem sequence:
line_addition(1,PS,SR)
line_addition(1,PT,TQ)
parallel_property_corresponding_angle(1,ST,RQ,P)
similar_triangle_judgment_aa(1,TPS,QPR)
similar_triangle_property_line_ratio(1,TPS,QPR)
similar_triangle_property_line_ratio(1,STP,RQP)
Problem: As shown in the diagram, ST=TU, SY=YZ, UJ=9, VJ=3, ZT=18, ZV=VU. Find the length of line JT.
问题：如图所示，ST=TU，SY=YZ，UJ=9，VJ=3，ZT=18，ZV=VU。求直线JT的长度。
Construction (CDL):
Shape(SY,YJ,JS)
Shape(SJ,JT,TS)
Shape(YZ,ZJ,JY)
Shape(TJ,JU,UT)
Shape(JZ,ZV,VJ)
Shape(JV,VU,UJ)
Collinear(SYZ)
Collinear(STU)
Collinear(ZVU)
Collinear(SJV)
Collinear(YJU)
Collinear(ZJT)
Conditions (CDL):
Equal(LengthOfLine(ST),LengthOfLine(TU))
Equal(LengthOfLine(SY),LengthOfLine(YZ))
Equal(LengthOfLine(UJ),9)
Equal(LengthOfLine(VJ),3)
Equal(LengthOfLine(ZT),18)
Equal(LengthOfLine(ZV),LengthOfLine(VU))
Goal: Value(LengthOfLine(JT))
Answer: 6
Theorem sequence:
median_of_triangle_judgment(1,ZT,ZUS)
median_of_triangle_judgment(1,UY,USZ)
line_addition(1,ZJ,JT)
centroid_of_triangle_judgment_intersection(1,J,SZU,Y,T)
centroid_of_triangle_property_line_ratio(1,J,ZUS,T)
Problem: As shown in the diagram, ∠EAH=38°, ∠GEB=52°, AH is perpendicular to EH. Find the measure of ∠CHA.
问题：如图所示，∠EAH=38°，∠GEB=52°，AH⊥EH。求∠CHA的大小。
Construction (CDL):
Shape(IA,AF)
Shape(FA,AE)
Shape(AE,EG)
Shape(GE,EB)
Shape(CH,HA)
Shape(HA,AI)
Shape(DH,HC)
Shape(BE,EH)
Shape(EH,HD)
Shape(AH,HE,EA)
Collinear(IAEB)
Collinear(FAHD)
Collinear(CHEG)
Conditions (CDL):
Equal(MeasureOfAngle(EAH),38)
Equal(MeasureOfAngle(GEB),52)
PerpendicularBetweenLine(AH,EH)
Goal: Value(MeasureOfAngle(CHA))
Answer: 90
Theorem sequence:
adjacent_complementary_angle(1,CHA,AHE)
Problem: As shown in the diagram, AD=10, BC=15, BE=6, EC=12, ED=3*x-2, ∠ADE=∠CBE. Find the value of x.
问题：如图所示，AD=10，BC=15，BE=6，EC=12，ED=3*x-2，∠ADE=∠CBE。求x的值。
Construction (CDL):
Shape(AD,DE,EA)
Shape(BE,EC,CB)
Collinear(AEC)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(BE),6)
Equal(LengthOfLine(EC),12)
Equal(LengthOfLine(ED),3*x-2)
Equal(MeasureOfAngle(ADE),MeasureOfAngle(CBE))
Goal: Value(x)
Answer: 2
Theorem sequence:
vertical_angle(1,BEC,DEA)
similar_triangle_judgment_aa(1,ADE,CBE)
similar_triangle_property_line_ratio(1,ADE,CBE)
similar_triangle_property_line_ratio(1,EAD,ECB)
Problem: As shown in the diagram, the diameter of circle A is 8, the diameter of circle B is 18, the diameter of ⊙C is 11, the center of ⊙A is A, the center of ⊙B is B, C is the center of circle C. Find the length of line FB.
问题：如图所示，圆A的直径为8，⊙B的直径为18，圆C的直径为11，A是⊙A的圆心，⊙B的圆心为B，圆C的圆心为C。求直线FB的长度。
Construction (CDL):
Shape(AF,AFI,BIA)
Shape(FA,BAJ,AJF)
Shape(BIJ,AIJ)
Shape(FB,BM,CKM,BKI,AFI)
Shape(MB,BF,AJF,BJL,CML)
Shape(MG,BGK,CKM)
Shape(GM,CML,BLG)
Shape(BLG,BGK,CLK)
Collinear(AFBMGC)
Cocircular(A,JFI)
Cocircular(B,IAJLGK)
Cocircular(C,KML)
Conditions (CDL):
Equal(DiameterOfCircle(A),8)
Equal(DiameterOfCircle(B),18)
Equal(DiameterOfCircle(C),11)
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
Goal: Value(LengthOfLine(FB))
Answer: 5
Theorem sequence:
circle_property_length_of_radius_and_diameter(1,A)
circle_property_length_of_radius_and_diameter(1,B)
radius_of_circle_property_length_equal(1,AF,A)
radius_of_circle_property_length_equal(1,BA,B)
line_addition(1,AF,FB)
Problem: As shown in the diagram, AP=3*y+11, HP=7*y-5, ∠HWQ=4*x-16°, ∠HWX=x+12°, ∠QAP=3*x-2°, WX is the angle bisector of ∠HWQ, WP is the median of △ WAH. Find the value of y.
问题：如图所示，AP=3*y+11，HP=7*y-5，∠HWQ=4*x-16°，∠HWX=x+12°，∠QAP=3*x-2°，WX平分∠HWQ，WP是△WAH的中线。求y的值。
Construction (CDL):
Shape(HW,WX,XH)
Shape(HX,XP,PH)
Shape(XW,WQ,QX)
Shape(PX,XQ,QA,AP)
Collinear(HXQ)
Collinear(WXP)
Collinear(HPA)
Collinear(WQA)
Conditions (CDL):
Equal(LengthOfLine(AP),3*y+11)
Equal(LengthOfLine(HP),7*y-5)
Equal(MeasureOfAngle(HWQ),4*x-16)
Equal(MeasureOfAngle(HWX),x+12)
Equal(MeasureOfAngle(QAP),3*x-2)
IsBisectorOfAngle(WX,HWQ)
IsMedianOfTriangle(WP,WAH)
Goal: Value(y)
Answer: 4
Theorem sequence:
Problem: As shown in the diagram, ∠ADK=96°, ∠HGJ=42°, DH∥AG. Find the measure of ∠EAD.
问题：如图所示，∠ADK=96°，∠HGJ=42°，DH∥AG。求∠EAD的大小。
Construction (CDL):
Shape(KD,DL)
Shape(LD,DH)
Shape(DH,HI)
Shape(IH,HC)
Shape(AD,DK)
Shape(HD,DA)
Shape(GH,HD)
Shape(CH,HG)
Shape(EA,AD)
Shape(DA,AG)
Shape(AG,GH)
Shape(HG,GJ)
Shape(MA,AE)
Shape(GA,AM)
Shape(FG,GA)
Shape(JG,GF)
Collinear(KDHC)
Collinear(EAGJ)
Collinear(LDAM)
Collinear(IHGF)
Conditions (CDL):
Equal(MeasureOfAngle(ADK),96)
Equal(MeasureOfAngle(HGJ),42)
ParallelBetweenLine(DH,AG)
Goal: Value(MeasureOfAngle(EAD))
Answer: 84
Theorem sequence:
parallel_property_collinear_extend(1,DH,AG,K)
parallel_property_collinear_extend(2,GA,DK,E)
parallel_property_ipsilateral_internal_angle(1,AE,DK)
Problem: As shown in the diagram, AC=13, CD=6, DB=29, CD⊥AD. Find the area of triangle ADB.
问题：如图所示，AC=13，CD=6，DB=29，CD垂直于AD。求三角形ADB的面积。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(LengthOfLine(AC),13)
Equal(LengthOfLine(CD),6)
Equal(LengthOfLine(DB),29)
PerpendicularBetweenLine(CD,AD)
Goal: Value(AreaOfTriangle(ADB))
Answer: 29*sqrt(133)/2
Theorem sequence:
adjacent_complementary_angle(1,CDA,ADB)
right_triangle_judgment_angle(1,CDA)
right_triangle_property_pythagorean(1,CDA)
triangle_area_formula_sine(1,DBA)
Problem: As shown in the diagram, AB=15, DA=9, DB=12, DE=x, quadrilateral ADCB is a ▱, BE is perpendicular to DE, CB⊥DB. Find the value of x.
问题：如图所示，AB=15，DA=9，DB=12，DE=x，ADCB是平行四边形，BE⊥DE，CB垂直于DB。求x的值。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DB,BE)
Shape(BD,DC,CB)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(DA),9)
Equal(LengthOfLine(DB),12)
Equal(LengthOfLine(DE),x)
Parallelogram(ADCB)
PerpendicularBetweenLine(BE,DE)
PerpendicularBetweenLine(CB,DB)
Goal: Value(x)
Answer: 36/5
Theorem sequence:
line_addition(1,AE,EB)
adjacent_complementary_angle(1,BED,DEA)
right_triangle_judgment_angle(1,BED)
right_triangle_judgment_angle(1,DEA)
right_triangle_property_pythagorean(1,BED)
right_triangle_property_pythagorean(1,DEA)
Problem: As shown in the diagram, AB=15, PB=12, ∠PBA=24°, quadrilateral ADCB is a rhombus. Find the length of line AP.
问题：如图所示，AB=15，PB=12，∠PBA=24°，ADCB是菱形。求直线AP的长度。
Construction (CDL):
Shape(AD,DP,PA)
Shape(PD,DC,CP)
Shape(PC,CB,BP)
Shape(PB,BA,AP)
Collinear(APC)
Collinear(DPB)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(PB),12)
Equal(MeasureOfAngle(PBA),24)
Rhombus(ADCB)
Goal: Value(LengthOfLine(AP))
Answer: 9
Theorem sequence:
kite_property_diagonal_perpendicular_bisection(1,BADC,P)
right_triangle_judgment_angle(1,APB)
right_triangle_property_pythagorean(1,APB)
Problem: As shown in the diagram, PQ=UQ, PR=RT, ∠UQP=40°. Find the measure of ∠SRQ.
问题：如图所示，PQ=UQ，PR=RT，∠UQP=40°。求∠SRQ的大小。
Construction (CDL):
Shape(RQ,QS,SR)
Shape(QP,PU,UQ)
Shape(QU,US,SQ)
Shape(SU,UT,TS)
Collinear(RQP)
Collinear(RST)
Collinear(PUT)
Conditions (CDL):
Equal(LengthOfLine(PQ),LengthOfLine(UQ))
Equal(LengthOfLine(PR),LengthOfLine(RT))
Equal(MeasureOfAngle(UQP),40)
Goal: Value(MeasureOfAngle(SRQ))
Answer: 40
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,QPU)
isosceles_triangle_judgment_line_equal(1,RPT)
isosceles_triangle_property_angle_equal(1,QPU)
isosceles_triangle_property_angle_equal(1,RPT)
triangle_property_angle_sum(1,QPU)
triangle_property_angle_sum(1,TRP)
Problem: As shown in the diagram, GK=14, MF=8, the measure of arc FKG is 142, F is the center of circle F, HJ is perpendicular to KJ. Find the measure of arc FMK.
问题：如图所示，GK=14，MF=8，⌒FKG的角度为142，F是圆F的圆心，HJ⊥KJ。求弧FMK的角度。
Construction (CDL):
Shape(GJ,JH,FHG)
Shape(HJ,JK,FKH)
Shape(FJ,JG,GF)
Shape(MF,FG,FGM)
Shape(JF,FK,KJ)
Shape(KF,FM,FMK)
Collinear(GJK)
Collinear(MFJH)
Cocircular(F,GMKH)
Conditions (CDL):
Equal(LengthOfLine(GK),14)
Equal(LengthOfLine(MF),8)
Equal(MeasureOfArc(FKG),142)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(HJ,KJ)
Goal: Value(MeasureOfArc(FMK))
Answer: 109
Theorem sequence:
radius_of_circle_property_length_equal(1,FK,F)
radius_of_circle_property_length_equal(1,FG,F)
adjacent_complementary_angle(1,JFK,KFM)
adjacent_complementary_angle(1,HJK,KJM)
arc_property_center_angle(1,FKG,F)
isosceles_triangle_judgment_line_equal(1,FKG)
altitude_of_triangle_judgment(1,FJ,FKG)
isosceles_triangle_property_line_coincidence(1,FKG,J)
angle_addition(1,GFJ,JFK)
arc_property_center_angle(1,FMK,F)
Problem: As shown in the diagram, AB=2*sqrt(3), AY=x, YC=y, ∠BAY=30°, ∠YCB=60°, AY⊥BY, CB is perpendicular to AB. Find the value of x.
问题：如图所示，AB=2*sqrt(3)，AY=x，YC=y，∠BAY=30°，∠YCB=60°，AY⊥BY，CB⊥AB。求x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(YC),y)
Equal(MeasureOfAngle(BAY),30)
Equal(MeasureOfAngle(YCB),60)
PerpendicularBetweenLine(AY,BY)
PerpendicularBetweenLine(CB,AB)
Goal: Value(x)
Answer: 3
Theorem sequence:
triangle_property_angle_sum(1,AYB)
sine_theorem(1,AYB)
Problem: As shown in the diagram, AB=x+1, CD=x-1, FE=8, HG=5, ∠ADC=∠EHG, ∠BAD=∠FEH, ∠CBA=∠GFE, ∠DCB=∠HGF, BADC is similar to FEHG. Find the length of line CD.
问题：如图所示，AB=x+1，CD=x-1，FE=8，HG=5，∠ADC=∠EHG，∠BAD=∠FEH，∠CBA=∠GFE，∠DCB=∠HGF，四边形BADC与四边形FEHG相似。求直线CD的长度。
Construction (CDL):
Shape(BA,AD,DC,CB)
Shape(EH,HG,GF,FE)
Conditions (CDL):
Equal(LengthOfLine(AB),x+1)
Equal(LengthOfLine(CD),x-1)
Equal(LengthOfLine(FE),8)
Equal(LengthOfLine(HG),5)
Equal(MeasureOfAngle(ADC),MeasureOfAngle(EHG))
Equal(MeasureOfAngle(BAD),MeasureOfAngle(FEH))
Equal(MeasureOfAngle(CBA),MeasureOfAngle(GFE))
Equal(MeasureOfAngle(DCB),MeasureOfAngle(HGF))
SimilarBetweenQuadrilateral(BADC,FEHG)
Goal: Value(LengthOfLine(CD))
Answer: 10/3
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,BADC,FEHG)
similar_quadrilateral_property_line_ratio(1,DCBA,HGFE)
Problem: As shown in the diagram, the area of CBTA is 104, AB=16, CE=ET, ET=x, CE is perpendicular to AE. Find the value of x.
问题：如图所示，CBTA的面积为104，AB=16，CE=ET，ET=x，CE⊥AE。求x的值。
Construction (CDL):
Shape(CB,BE,EC)
Shape(CE,EA,AC)
Shape(BT,TE,EB)
Shape(ET,TA,AE)
Collinear(CET)
Collinear(BEA)
Conditions (CDL):
Equal(AreaOfQuadrilateral(CBTA),104)
Equal(LengthOfLine(AB),16)
Equal(LengthOfLine(CE),LengthOfLine(ET))
Equal(LengthOfLine(ET),x)
PerpendicularBetweenLine(CE,AE)
Goal: Value(x)
Answer: 13/2
Theorem sequence:
adjacent_complementary_angle(1,CEA,AET)
adjacent_complementary_angle(1,AET,TEB)
perpendicular_bisector_judgment_per_and_mid(1,BE,TC)
perpendicular_bisector_judgment_per_and_mid(1,AE,CT)
perpendicular_bisector_property_distance_equal(1,BE,TC)
perpendicular_bisector_property_distance_equal(1,AE,CT)
kite_judgment_equal_and_equal(1,BTAC)
kite_area_formula_diagonal(1,BTAC)
line_addition(1,CE,ET)
Problem: As shown in the diagram, KJ=11, KL=11, ML=5.5, ∠KJM=60°, KM is perpendicular to LM. Find the measure of ∠LKJ.
问题：如图所示，KJ=11，KL=11，ML=5.5，∠KJM=60°，KM垂直于LM。求∠LKJ的大小。
Construction (CDL):
Shape(KJ,JM,MK)
Shape(KM,ML,LK)
Collinear(JML)
Conditions (CDL):
Equal(LengthOfLine(KJ),11)
Equal(LengthOfLine(KL),11)
Equal(LengthOfLine(ML),5.5)
Equal(MeasureOfAngle(KJM),60)
PerpendicularBetweenLine(KM,LM)
Goal: Value(MeasureOfAngle(LKJ))
Answer: 60
Theorem sequence:
cosine_theorem(1,JLK)
cosine_theorem(1,KJL)
Problem: As shown in the diagram, ML=7, ∠MLK=92°, L is the center of circle L. Find the area of the sector LKM.
问题：如图所示，ML=7，∠MLK=92°，圆L的圆心为L。求扇形LKM的面积。
Construction (CDL):
Shape(ML,LK,LKM)
Cocircular(L,KM)
Conditions (CDL):
Equal(LengthOfLine(ML),7)
Equal(MeasureOfAngle(MLK),92)
IsCentreOfCircle(L,L)
Goal: Value(AreaOfSector(LKM))
Answer: 1127*pi/90
Theorem sequence:
radius_of_circle_property_length_equal(1,LM,L)
arc_property_center_angle(1,LKM,L)
sector_area_formula(1,LKM)
Problem: As shown in the diagram, ∠ILR=2*y+8°, ∠JME=z°, ∠KIL=4*x+6°, ∠RLP=142°, HK is parallel to BR, JN∥HK. Find the value of y.
问题：如图所示，∠ILR=2*y+8°，∠JME=z°，∠KIL=4*x+6°，∠RLP=142°，HK∥BR，JN∥HK。求y的值。
Construction (CDL):
Shape(JM,ME)
Shape(EM,MN)
Shape(IM,MJ)
Shape(NM,MI)
Shape(HI,IM)
Shape(MI,IK)
Shape(LI,IH)
Shape(KI,IL)
Shape(BL,LI)
Shape(IL,LR)
Shape(PL,LB)
Shape(RL,LP)
Collinear(JMN)
Collinear(HIK)
Collinear(BLR)
Collinear(PLIME)
Conditions (CDL):
Equal(MeasureOfAngle(ILR),2*y+8)
Equal(MeasureOfAngle(JME),z)
Equal(MeasureOfAngle(KIL),4*x+6)
Equal(MeasureOfAngle(RLP),142)
ParallelBetweenLine(HK,BR)
ParallelBetweenLine(JN,HK)
Goal: Value(y)
Answer: 15
Theorem sequence:
adjacent_complementary_angle(1,ILR,RLP)
Problem: As shown in the diagram, ∠ACE=25°, ∠AEG=51°, ∠DAB=35°, ∠GBA=28°, AB is perpendicular to FB, AG is perpendicular to BG, BD is perpendicular to FD. Find the measure of ∠BAG.
问题：如图所示，∠ACE=25°，∠AEG=51°，∠DAB=35°，∠GBA=28°，AB垂直于FB，AG垂直于BG，BD垂直于FD。求∠BAG的大小。
Construction (CDL):
Shape(AC,CE,EA)
Shape(AE,EG,GA)
Shape(AG,GB,BA)
Shape(AB,BD,DA)
Shape(DB,BF,FD)
Collinear(CEGB)
Collinear(ADF)
Conditions (CDL):
Equal(MeasureOfAngle(ACE),25)
Equal(MeasureOfAngle(AEG),51)
Equal(MeasureOfAngle(DAB),35)
Equal(MeasureOfAngle(GBA),28)
PerpendicularBetweenLine(AB,FB)
PerpendicularBetweenLine(AG,BG)
PerpendicularBetweenLine(BD,FD)
Goal: Value(MeasureOfAngle(BAG))
Answer: 62
Theorem sequence:
triangle_property_angle_sum(1,AGB)
Problem: As shown in the diagram, PQ=UQ, ∠QPU=32°. Find the measure of ∠PUQ.
问题：如图所示，PQ=UQ，∠QPU=32°。求∠PUQ的大小。
Construction (CDL):
Shape(RQ,QS,SR)
Shape(QP,PU,UQ)
Shape(QU,US,SQ)
Shape(SU,UT,TS)
Collinear(RQP)
Collinear(RST)
Collinear(PUT)
Conditions (CDL):
Equal(LengthOfLine(PQ),LengthOfLine(UQ))
Equal(MeasureOfAngle(QPU),32)
Goal: Value(MeasureOfAngle(PUQ))
Answer: 32
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,QPU)
isosceles_triangle_property_angle_equal(1,QPU)