SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, PS=3, RY=5, WX=10, WY=8, XY=6, RP∥XW, RQ⊥PQ, WY⊥XY, YS⊥PS. Find the length of line PQ.
问题：如图所示，PS=3，RY=5，WX=10，WY=8，XY=6，RP∥XW，RQ⊥PQ，WY垂直于XY，YS⊥PS。求直线PQ的长度。
Construction (CDL):
Shape(YR,RA,AY)
Shape(PY,YS,SP)
Shape(YA,AQ,QS,SY)
Shape(AX,XQ,QA)
Shape(SQ,QW,WS)
Collinear(YAX)
Collinear(YSW)
Collinear(PSQ)
Collinear(RAQ)
Collinear(RYP)
Collinear(XQW)
Conditions (CDL):
Equal(LengthOfLine(PS),3)
Equal(LengthOfLine(RY),5)
Equal(LengthOfLine(WX),10)
Equal(LengthOfLine(WY),8)
Equal(LengthOfLine(XY),6)
ParallelBetweenLine(RP,XW)
PerpendicularBetweenLine(RQ,PQ)
PerpendicularBetweenLine(WY,XY)
PerpendicularBetweenLine(YS,PS)
Goal: Value(LengthOfLine(PQ))
Answer: 6
Theorem sequence:
parallel_property_collinear_extend(3,RP,XW,Y)
parallel_property_collinear_extend(3,WX,PY,Q)
adjacent_complementary_angle(1,QSY,YSP)
adjacent_complementary_angle(1,RAY,YAQ)
parallel_judgment_ipsilateral_internal_angle(1,QA,SY)
parallel_judgment_ipsilateral_internal_angle(1,SQ,YA)
parallel_judgment_ipsilateral_internal_angle(1,QR,SY)
parallel_property_ipsilateral_internal_angle(1,YS,AQ)
parallel_property_alternate_interior_angle(1,YP,QW)
parallel_property_corresponding_angle(1,YS,RQ,P)
similar_triangle_judgment_aa(1,XWY,YRA)
similar_triangle_property_line_ratio(1,WYX,RAY)
similar_triangle_property_line_ratio(1,YXW,AYR)
parallelogram_judgment_parallel_and_parallel(1,YAQS)
parallelogram_property_opposite_line_equal(1,YAQS)
line_addition(1,PS,SQ)
Problem: As shown in the diagram, YW=18, ZY=18, the measure of arc OYW is 143, the measure of arc OZY is 2*x-1. Find the value of x.
问题：如图所示，YW=18，ZY=18，弧OYW的角度为143，⌒OZY的角度为2*x-1。求x的值。
Construction (CDL):
Shape(WY,OYW)
Shape(ZY,YW,OWZ)
Shape(YZ,OZY)
Cocircular(O,YWZ)
Conditions (CDL):
Equal(LengthOfLine(YW),18)
Equal(LengthOfLine(ZY),18)
Equal(MeasureOfArc(OYW),143)
Equal(MeasureOfArc(OZY),2*x-1)
Goal: Value(x)
Answer: 72
Theorem sequence:
congruent_arc_judgment_chord_equal(1,OYW,OZY)
congruent_arc_property_measure_equal(1,OYW,OZY)
Problem: As shown in the diagram, AC=12, BA=13, BC=15. Find the measure of ∠ACB.
问题：如图所示，AC=12，BA=13，BC=15。求∠ACB的大小。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(BA),13)
Equal(LengthOfLine(BC),15)
Goal: Value(MeasureOfAngle(ACB))
Answer: 180*acos(5/9)/pi
Theorem sequence:
cosine_theorem(1,CBA)
Problem: As shown in the diagram, ∠JQR=131°, QR is parallel to TS, TQ∥SR. Find the measure of ∠STC.
问题：如图所示，∠JQR=131°，QR∥TS，TQ平行于SR。求∠STC的大小。
Construction (CDL):
Shape(JQ,QR)
Shape(QR,RH)
Shape(QT,TS,SR,RQ)
Shape(ST,TC)
Shape(BS,ST)
Collinear(JQTC)
Collinear(HRSB)
Conditions (CDL):
Equal(MeasureOfAngle(JQR),131)
ParallelBetweenLine(QR,TS)
ParallelBetweenLine(TQ,SR)
Goal: Value(MeasureOfAngle(STC))
Answer: 49
Theorem sequence:
parallel_property_corresponding_angle(1,QR,TS,J)
adjacent_complementary_angle(1,QTS,STC)
Problem: As shown in the diagram, ∠ACB=125°, ∠BCD=x°, ∠DCA=121°. Find the value of x.
问题：如图所示，∠ACB=125°，∠BCD=x°，∠DCA=121°。求x的值。
Construction (CDL):
Shape(BC,CD,CDB)
Shape(AC,CB,CBA)
Shape(DC,CA,CAD)
Cocircular(C,ADB)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),125)
Equal(MeasureOfAngle(BCD),x)
Equal(MeasureOfAngle(DCA),121)
Goal: Value(x)
Answer: 114
Theorem sequence:
angle_addition(1,DCA,ACB)
round_angle(1,DCB,BCD)
Problem: As shown in the diagram, CA=10, CD=2, CE=t-2, EB=t+1, AB is parallel to DE. Find the value of t.
问题：如图所示，CA=10，CD=2，CE=t-2，EB=t+1，AB∥DE。求t的值。
Construction (CDL):
Shape(AD,DE,EB,BA)
Shape(DC,CE,ED)
Collinear(ADC)
Collinear(CEB)
Conditions (CDL):
Equal(LengthOfLine(CA),10)
Equal(LengthOfLine(CD),2)
Equal(LengthOfLine(CE),t-2)
Equal(LengthOfLine(EB),t+1)
ParallelBetweenLine(AB,DE)
Goal: Value(t)
Answer: 3
Theorem sequence:
line_addition(1,CE,EB)
parallel_property_corresponding_angle(2,AB,DE,C)
similar_triangle_judgment_aa(1,EDC,BAC)
similar_triangle_property_line_ratio(1,EDC,BAC)
similar_triangle_property_line_ratio(1,DCE,ACB)
Problem: As shown in the diagram, AC=CB, AD=DC, ∠ADC=92°. Find the measure of ∠DCA.
问题：如图所示，AC=CB，AD=DC，∠ADC=92°。求∠DCA的大小。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AC,CB,BA)
Collinear(DCB)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(CB))
Equal(LengthOfLine(AD),LengthOfLine(DC))
Equal(MeasureOfAngle(ADC),92)
Goal: Value(MeasureOfAngle(DCA))
Answer: 44
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,DCA)
isosceles_triangle_property_angle_equal(1,DCA)
triangle_property_angle_sum(1,ADC)
Problem: As shown in the diagram, the measure of ⌒DQS is 238, D is the center of ⊙D, the tangent to ⊙D is RQ. Find the measure of ∠RQS.
问题：如图所示，弧DQS的角度为238，⊙D的圆心为D，⊙O的切线为RQ。求∠RQS的大小。
Construction (CDL):
Shape(QS,DSQ)
Shape(SQ,QD,DS)
Shape(SD,DQ,DQT,DTS)
Shape(RQ,QS)
Collinear(QR)
Cocircular(D,SQT)
Conditions (CDL):
Equal(MeasureOfArc(DQS),238)
IsCentreOfCircle(D,D)
IsTangentOfCircle(RQ,D)
Goal: Value(MeasureOfAngle(RQS))
Answer: 61
Theorem sequence:
arc_property_center_angle(1,DQS,D)
round_angle(1,QDS,SDQ)
radius_of_circle_property_length_equal(1,DS,D)
radius_of_circle_property_length_equal(1,DQ,D)
isosceles_triangle_judgment_line_equal(1,DSQ)
isosceles_triangle_property_angle_equal(1,DSQ)
triangle_property_angle_sum(1,DSQ)
tangent_of_circle_property_perpendicular(2,RQ,D,D)
angle_addition(1,RQS,SQD)
Problem: As shown in the diagram, AC=5, BA=12, CB=x, ∠BAC=60°. Find the value of x.
问题：如图所示，AC=5，BA=12，CB=x，∠BAC=60°。求x的值。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AC),5)
Equal(LengthOfLine(BA),12)
Equal(LengthOfLine(CB),x)
Equal(MeasureOfAngle(BAC),60)
Goal: Value(x)
Answer: sqrt(109)
Theorem sequence:
cosine_theorem(1,ACB)
Problem: As shown in the diagram, BA=y, BF=x, FA=z, LA=sqrt(3), LF=2*sqrt(3), FB⊥AB, LA⊥FA. Find the value of x.
问题：如图所示，BA=y，BF=x，FA=z，LA=sqrt(3)，LF=2*sqrt(3)，FB⊥AB，LA⊥FA。求x的值。
Construction (CDL):
Shape(FB,BA,AF)
Shape(BL,LA,AB)
Collinear(LBF)
Conditions (CDL):
Equal(LengthOfLine(BA),y)
Equal(LengthOfLine(BF),x)
Equal(LengthOfLine(FA),z)
Equal(LengthOfLine(LA),sqrt(3))
Equal(LengthOfLine(LF),2*sqrt(3))
PerpendicularBetweenLine(FB,AB)
PerpendicularBetweenLine(LA,FA)
Goal: Value(x)
Answer: 3*sqrt(3)/2
Theorem sequence:
adjacent_complementary_angle(1,FBA,ABL)
mirror_similar_triangle_judgment_aa(1,ABL,FLA)
mirror_similar_triangle_property_line_ratio(1,ABL,FLA)
mirror_similar_triangle_property_line_ratio(1,BLA,AFL)
line_addition(1,LB,BF)
Problem: As shown in the diagram, AB=x, AC=16, CB=30, BC⊥AC. Find the value of x.
问题：如图所示，AB=x，AC=16，CB=30，BC垂直于AC。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),16)
Equal(LengthOfLine(CB),30)
PerpendicularBetweenLine(BC,AC)
Goal: Value(x)
Answer: 34
Theorem sequence:
right_triangle_judgment_angle(1,BCA)
right_triangle_property_pythagorean(1,BCA)
Problem: As shown in the diagram, ∠POM=43°, FB∥QM, KC∥GJ. Find the measure of ∠BPO.
问题：如图所示，∠POM=43°，FB平行于QM，KC∥GJ。求∠BPO的大小。
Construction (CDL):
Shape(KP,PB)
Shape(BP,PO)
Shape(PO,OM)
Shape(MO,OC)
Shape(KP,PH)
Shape(GH,HP)
Shape(PH,HN,NO,OP)
Shape(CO,ON)
Shape(ON,NJ)
Shape(FH,HG)
Shape(NH,HF)
Shape(QN,NH)
Shape(JN,NQ)
Collinear(KPOC)
Collinear(GHNJ)
Collinear(BPHF)
Collinear(MONQ)
Conditions (CDL):
Equal(MeasureOfAngle(POM),43)
ParallelBetweenLine(FB,QM)
ParallelBetweenLine(KC,GJ)
Goal: Value(MeasureOfAngle(BPO))
Answer: 137
Theorem sequence:
parallel_property_collinear_extend(3,FB,QM,P)
parallel_property_collinear_extend(3,MQ,BP,O)
parallel_property_ipsilateral_internal_angle(1,PB,OM)
Problem: As shown in the diagram, YX=24, ZY=28, ∠XWZ=105°, WX and ZY are opposite sides of the parallelogram WZYX. Find the length of line WZ.
问题：如图所示，YX=24，ZY=28，∠XWZ=105°，WX和ZY是平行四边形WZYX的一组对边。求直线WZ的长度。
Construction (CDL):
Shape(WZ,ZY,YX,XW)
Conditions (CDL):
Equal(LengthOfLine(YX),24)
Equal(LengthOfLine(ZY),28)
Equal(MeasureOfAngle(XWZ),105)
Parallelogram(WZYX)
Goal: Value(LengthOfLine(WZ))
Answer: 24
Theorem sequence:
parallelogram_property_opposite_line_equal(1,WZYX)
Problem: As shown in the diagram, BC=4*x-17, DF=2*x-1, ∠CBF=3*y+5°, ∠DFB=5*y-13°, DB and FC are opposite sides of the parallelogram BDFC. Find the value of y.
问题：如图所示，BC=4*x-17，DF=2*x-1，∠CBF=3*y+5°，∠DFB=5*y-13°，DB和FC是平行四边形BDFC的一组对边。求y的值。
Construction (CDL):
Shape(BD,DF,FB)
Shape(FC,CB,BF)
Conditions (CDL):
Equal(LengthOfLine(BC),4*x-17)
Equal(LengthOfLine(DF),2*x-1)
Equal(MeasureOfAngle(CBF),3*y+5)
Equal(MeasureOfAngle(DFB),5*y-13)
Parallelogram(BDFC)
Goal: Value(y)
Answer: 9
Theorem sequence:
parallel_property_alternate_interior_angle(1,BC,DF)
Problem: As shown in the diagram, AC=21, AD=17, DE=8, quadrilateral BCAD is a parallelogram, CA is perpendicular to EA, DE⊥AE. Find the area of quadrilateral BCAD.
问题：如图所示，AC=21，AD=17，DE=8，四边形BCAD是平行四边形，CA垂直于EA，DE⊥AE。求BCAD的面积。
Construction (CDL):
Shape(BC,CA,AE,EB)
Shape(EA,AD,DE)
Collinear(BED)
Conditions (CDL):
Equal(LengthOfLine(AC),21)
Equal(LengthOfLine(AD),17)
Equal(LengthOfLine(DE),8)
Parallelogram(BCAD)
PerpendicularBetweenLine(CA,EA)
PerpendicularBetweenLine(DE,AE)
Goal: Value(AreaOfQuadrilateral(BCAD))
Answer: 315
Theorem sequence:
right_triangle_judgment_angle(1,DEA)
right_triangle_property_pythagorean(1,DEA)
altitude_of_quadrilateral_judgment_left_vertex(1,AE,ADBC)
parallelogram_property_opposite_line_equal(1,DBCA)
parallelogram_area_formula_common(1,ADBC)
Problem: As shown in the diagram, JN=14-x, KL=3*x+2*y, MK=6*x, quadrilateral LJMK is a parallelogram, quadrilateral PNML is a rectangle. Find the value of y.
问题：如图所示，JN=14-x，KL=3*x+2*y，MK=6*x，LJMK是平行四边形，四边形PNML是长方形。求y的值。
Construction (CDL):
Shape(PN,NJ,JP)
Shape(JN,NM,MJ)
Shape(JM,ML,LJ)
Shape(JL,LP,PJ)
Shape(MK,KL,LM)
Collinear(PJM)
Collinear(NJL)
Conditions (CDL):
Equal(LengthOfLine(JN),14-x)
Equal(LengthOfLine(KL),3*x+2*y)
Equal(LengthOfLine(MK),6*x)
Parallelogram(LJMK)
Rectangle(PNML)
Goal: Value(y)
Answer: 3
Theorem sequence:
line_addition(1,PJ,JM)
line_addition(1,NJ,JL)
rectangle_property_diagonal_equal(1,PNML)
parallelogram_property_diagonal_bisection(1,PNML,J)
parallelogram_property_diagonal_bisection(1,NMLP,J)
parallelogram_property_opposite_line_equal(1,LJMK)
parallelogram_property_opposite_line_equal(1,JMKL)
Problem: As shown in the diagram, ∠EDF=39°, ∠FBA=48°, FC∥ED, AF is perpendicular to BF, DC⊥FC, FE is perpendicular to DE. Find the measure of ∠FDC.
问题：如图所示，∠EDF=39°，∠FBA=48°，FC∥ED，AF⊥BF，DC⊥FC，FE⊥DE。求∠FDC的大小。
Construction (CDL):
Shape(AF,FB,BA)
Shape(FE,ED,DF)
Shape(FD,DC,CB,BF)
Collinear(FBC)
Collinear(AFE)
Conditions (CDL):
Equal(MeasureOfAngle(EDF),39)
Equal(MeasureOfAngle(FBA),48)
ParallelBetweenLine(FC,ED)
PerpendicularBetweenLine(AF,BF)
PerpendicularBetweenLine(DC,FC)
PerpendicularBetweenLine(FE,DE)
Goal: Value(MeasureOfAngle(FDC))
Answer: 51
Theorem sequence:
parallel_property_alternate_interior_angle(1,FC,ED)
triangle_property_angle_sum(1,FDC)
Problem: As shown in the diagram, AN=21, BL=6, BN=18, SC=4, BL is perpendicular to NL, SC⊥NC. Find the sum of the area of triangle SBN and the area of triangle NBA.
问题：如图所示，AN=21，BL=6，BN=18，SC=4，BL垂直于NL，SC⊥NC。求△SBN的面积与△NBA的面积之和。
Construction (CDL):
Shape(BA,AL,LB)
Shape(BL,LN,NB)
Shape(SB,BC,CS)
Shape(SC,CN,NS)
Collinear(ALN)
Collinear(BCN)
Conditions (CDL):
Equal(LengthOfLine(AN),21)
Equal(LengthOfLine(BL),6)
Equal(LengthOfLine(BN),18)
Equal(LengthOfLine(SC),4)
PerpendicularBetweenLine(BL,NL)
PerpendicularBetweenLine(SC,NC)
Goal: Value(Add(AreaOfTriangle(SBN),AreaOfTriangle(NBA)))
Answer: 99
Theorem sequence:
adjacent_complementary_angle(1,ALB,BLN)
adjacent_complementary_angle(1,BCS,SCN)
altitude_of_triangle_judgment(1,BL,BAN)
altitude_of_triangle_judgment(1,SC,SBN)
triangle_area_formula_common(1,BAN)
triangle_area_formula_common(1,SBN)
Problem: As shown in the diagram, RD=3, SA=3, TA=x, UC=x, the perimeter of quadrilateral TSRU is 18, RC is the tangent to ⊙J, RD is the tangent to circle J, SA is the tangent to ⊙J, the tangent to ⊙J is SD, TA is the tangent to circle J, the tangent to circle J is TB, the tangent to ⊙J is UB, UC is the tangent to ⊙J. Find the value of x.
问题：如图所示，RD=3，SA=3，TA=x，UC=x，四边形TSRU的周长为18，RC是⊙O的切线，圆O的切线为RD，SA是⊙O的切线，⊙O的切线为SD，圆O的切线为TA，TB是⊙O的切线，UB是圆O的切线，⊙O的切线为UC。求x的值。
Construction (CDL):
Shape(BT,TA,JBA)
Shape(AS,SD,JAD)
Shape(DR,RC,JDC)
Shape(CU,UB,JCB)
Shape(JAD,JDC,JCB,JBA)
Collinear(TBU)
Collinear(TAS)
Collinear(SDR)
Collinear(UCR)
Cocircular(J,BADC)
Conditions (CDL):
Equal(LengthOfLine(RD),3)
Equal(LengthOfLine(SA),3)
Equal(LengthOfLine(TA),x)
Equal(LengthOfLine(UC),x)
Equal(PerimeterOfQuadrilateral(TSRU),18)
IsTangentOfCircle(RC,J)
IsTangentOfCircle(RD,J)
IsTangentOfCircle(SA,J)
IsTangentOfCircle(SD,J)
IsTangentOfCircle(TA,J)
IsTangentOfCircle(TB,J)
IsTangentOfCircle(UB,J)
IsTangentOfCircle(UC,J)
Goal: Value(x)
Answer: 3/2
Theorem sequence:
tangent_of_circle_property_length_equal(1,SA,SD,J)
tangent_of_circle_property_length_equal(1,TB,TA,J)
tangent_of_circle_property_length_equal(1,UC,UB,J)
tangent_of_circle_property_length_equal(1,RD,RC,J)
line_addition(1,TB,BU)
line_addition(1,TA,AS)
line_addition(1,SD,DR)
line_addition(1,UC,CR)
quadrilateral_perimeter_formula(1,TSRU)
Problem: As shown in the diagram, AB=24, BC=8, ∠ADC=45°, ∠CBA=60°, AB is parallel to DC, BF⊥CF, DE is perpendicular to AE. Find the perimeter of quadrilateral ADCB.
问题：如图所示，AB=24，BC=8，∠ADC=45°，∠CBA=60°，AB∥DC，BF⊥CF，DE⊥AE。求四边形ADCB的周长。
Construction (CDL):
Shape(AD,DE,EA)
Shape(AE,EC,CF,FA)
Shape(CB,BF,FC)
Collinear(AFB)
Collinear(DEC)
Conditions (CDL):
Equal(LengthOfLine(AB),24)
Equal(LengthOfLine(BC),8)
Equal(MeasureOfAngle(ADC),45)
Equal(MeasureOfAngle(CBA),60)
ParallelBetweenLine(AB,DC)
PerpendicularBetweenLine(BF,CF)
PerpendicularBetweenLine(DE,AE)
Goal: Value(PerimeterOfQuadrilateral(ADCB))
Answer: 4*sqrt(3)+4*sqrt(6)+52
Theorem sequence:
adjacent_complementary_angle(1,BFC,CFA)
adjacent_complementary_angle(1,DEA,AEC)
parallel_property_collinear_extend(3,AB,DC,F)
parallel_property_collinear_extend(3,CD,FA,E)
parallel_property_ipsilateral_internal_angle(1,CE,FA)
parallel_judgment_ipsilateral_internal_angle(1,EA,CF)
parallelogram_judgment_parallel_and_parallel(1,AECF)
parallelogram_property_opposite_line_equal(1,AECF)
parallelogram_property_opposite_line_equal(1,ECFA)
triangle_property_angle_sum(1,ADE)
triangle_property_angle_sum(1,BFC)
sine_theorem(1,CBF)
sine_theorem(1,BFC)
sine_theorem(1,ADE)
sine_theorem(1,EAD)
line_addition(1,AF,FB)
line_addition(1,DE,EC)
quadrilateral_perimeter_formula(1,ADCB)
Problem: As shown in the diagram, AB=3, B is the center of circle B. Find the circumference of the ⊙B.
问题：如图所示，AB=3，⊙B的圆心为B。求圆B的周长。
Construction (CDL):
Shape(BA)
Cocircular(B,A)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
IsCentreOfCircle(B,B)
Goal: Value(PerimeterOfCircle(B))
Answer: 6*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,BA,B)
circle_perimeter_formula(1,B)
Problem: As shown in the diagram, AC=x, BA=18, ∠ABC=25°, CA⊥BA. Find the value of x.
问题：如图所示，AC=x，BA=18，∠ABC=25°，CA垂直于BA。求x的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BA),18)
Equal(MeasureOfAngle(ABC),25)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 18*tan(5*pi/36)
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,ABC)
Problem: As shown in the diagram, EH=9, HG=15, ∠EHD=45°, EHGF is a parallelogram, ED⊥GD. Find the area of EHGF.
问题：如图所示，EH=9，HG=15，∠EHD=45°，EF和HG是▱EHGF的一组对边，ED⊥GD。求四边形EHGF的面积。
Construction (CDL):
Shape(EH,HD,DE)
Shape(ED,DG,GF,FE)
Collinear(HDG)
Conditions (CDL):
Equal(LengthOfLine(EH),9)
Equal(LengthOfLine(HG),15)
Equal(MeasureOfAngle(EHD),45)
Parallelogram(EHGF)
PerpendicularBetweenLine(ED,GD)
Goal: Value(AreaOfQuadrilateral(EHGF))
Answer: 135*sqrt(2)/2
Theorem sequence:
parallelogram_area_formula_sine(1,EHGF)
Problem: As shown in the diagram, ∠ADG=36°, ∠AGF=104°, ∠EFC=40°, AB⊥CB. Find the measure of ∠BCF.
问题：如图所示，∠ADG=36°，∠AGF=104°，∠EFC=40°，AB垂直于CB。求∠BCF的大小。
Construction (CDL):
Shape(AD,DG,GA)
Shape(FG,GB,BF)
Shape(FB,BC,CF)
Shape(AG,GF)
Shape(EF,FC)
Shape(BG,GD)
Collinear(DGFC)
Collinear(AGB)
Collinear(BFE)
Conditions (CDL):
Equal(MeasureOfAngle(ADG),36)
Equal(MeasureOfAngle(AGF),104)
Equal(MeasureOfAngle(EFC),40)
PerpendicularBetweenLine(AB,CB)
Goal: Value(MeasureOfAngle(BCF))
Answer: 14
Theorem sequence:
vertical_angle(1,EFC,BFG)
adjacent_complementary_angle(1,AGF,FGB)
adjacent_complementary_angle(1,EFC,CFB)
triangle_property_angle_sum(1,FGB)
triangle_property_angle_sum(1,CFB)
angle_addition(1,GBF,FBC)
Problem: As shown in the diagram, PT=6, QR=12, SP=4, PT∥QR. Find the length of line SQ.
问题：如图所示，PT=6，QR=12，SP=4，PT平行于QR。求直线SQ的长度。
Construction (CDL):
Shape(SP,PT,TS)
Shape(PQ,QR,RT,TP)
Collinear(SPQ)
Collinear(STR)
Conditions (CDL):
Equal(LengthOfLine(PT),6)
Equal(LengthOfLine(QR),12)
Equal(LengthOfLine(SP),4)
ParallelBetweenLine(PT,QR)
Goal: Value(LengthOfLine(SQ))
Answer: 8
Theorem sequence:
parallel_property_corresponding_angle(1,PT,QR,S)
similar_triangle_judgment_aa(1,TSP,RSQ)
similar_triangle_property_line_ratio(1,TSP,RSQ)
similar_triangle_property_line_ratio(1,SPT,SQR)
Problem: As shown in the diagram, ∠BEC=2*x°, ∠BFA=3*x-15°, ∠ECB=y**2°, ∠FAE=68°, AF is parallel to EB. Find the value of x.
问题：如图所示，∠BEC=2*x°，∠BFA=3*x-15°，∠ECB=y**2°，∠FAE=68°，AF平行于EB。求x的值。
Construction (CDL):
Shape(FA,AE,EB,BF)
Shape(BE,EC,CB)
Collinear(FBC)
Collinear(AEC)
Conditions (CDL):
Equal(MeasureOfAngle(BEC),2*x)
Equal(MeasureOfAngle(BFA),3*x-15)
Equal(MeasureOfAngle(ECB),y**2)
Equal(MeasureOfAngle(FAE),68)
ParallelBetweenLine(AF,EB)
Goal: Value(x)
Answer: 34
Theorem sequence:
parallel_property_corresponding_angle(2,AF,EB,C)
Problem: As shown in the diagram, AB=6, AF=8, BC=x, CD=y, DE=2*y-3, FE=x+10/3, BF is parallel to CD, CB∥DF. Find the length of line CD.
问题：如图所示，AB=6，AF=8，BC=x，CD=y，DE=2*y-3，FE=x+10/3，BF∥CD，CB∥DF。求直线CD的长度。
Construction (CDL):
Shape(AB,BF,FA)
Shape(BC,CD,DF,FB)
Shape(FD,DE,EF)
Collinear(ABC)
Collinear(CDE)
Collinear(AFE)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AF),8)
Equal(LengthOfLine(BC),x)
Equal(LengthOfLine(CD),y)
Equal(LengthOfLine(DE),2*y-3)
Equal(LengthOfLine(FE),x+10/3)
ParallelBetweenLine(BF,CD)
ParallelBetweenLine(CB,DF)
Goal: Value(LengthOfLine(CD))
Answer: 9
Theorem sequence:
line_addition(1,AB,BC)
line_addition(1,CD,DE)
line_addition(1,AF,FE)
parallel_property_ipsilateral_internal_angle(1,BF,CD)
parallel_property_ipsilateral_internal_angle(1,CB,DF)
flat_angle(1,ABC)
flat_angle(1,CDE)
angle_addition(1,ABF,FBC)
angle_addition(1,CDF,FDE)
similar_triangle_judgment_aa(1,FAB,EAC)
similar_triangle_judgment_aa(1,FDE,ACE)
similar_triangle_property_line_ratio(1,BFA,CEA)
similar_triangle_property_line_ratio(1,FAB,EAC)
similar_triangle_property_line_ratio(1,DEF,CEA)
similar_triangle_property_line_ratio(1,FDE,ACE)
Problem: As shown in the diagram, PS=RS, ∠QSR=48°, ∠SQP=∠RQS, QP is perpendicular to SP, SR⊥QR. Find the measure of ∠SQP.
问题：如图所示，PS=RS，∠QSR=48°，∠SQP=∠RQS，QP垂直于SP，SR⊥QR。求∠SQP的大小。
Construction (CDL):
Shape(QP,PS,SQ)
Shape(QS,SR,RQ)
Conditions (CDL):
Equal(LengthOfLine(PS),LengthOfLine(RS))
Equal(MeasureOfAngle(QSR),48)
Equal(MeasureOfAngle(SQP),MeasureOfAngle(RQS))
PerpendicularBetweenLine(QP,SP)
PerpendicularBetweenLine(SR,QR)
Goal: Value(MeasureOfAngle(SQP))
Answer: 42
Theorem sequence:
triangle_property_angle_sum(1,QSR)
Problem: As shown in the diagram, Add(PerimeterOfCircle(K)=PerimeterOfCircle(J), CJ=2*x, HA=x, HC=x, KA=4*x, H is the center of circle H, the center of ⊙J is J, the center of ⊙K is K. Find the length of line KJ.
问题：如图所示，Add(PerimeterOfCircle(K)=PerimeterOfCircle(J)，CJ=2*x，HA=x，HC=x，KA=4*x，⊙H的圆心为H，J是圆J的圆心，圆K的圆心为K。求直线KJ的长度。
Construction (CDL):
Shape(KA,KAB,BK)
Shape(AK,KB,KBA)
Shape(KBA,JBC,HCA)
Shape(CH,HA,HAC)
Shape(HC,HCA,AH)
Shape(BJ,JC,JCB)
Shape(CJ,JB,JBC)
Collinear(KAH)
Collinear(HCJ)
Collinear(KBJ)
Cocircular(K,BA)
Cocircular(J,CB)
Cocircular(H,AC)
Conditions (CDL):
Equal(Add(PerimeterOfCircle(K),PerimeterOfCircle(J),PerimeterOfCircle(H)),56*pi)
Equal(LengthOfLine(CJ),2*x)
Equal(LengthOfLine(HA),x)
Equal(LengthOfLine(HC),x)
Equal(LengthOfLine(KA),4*x)
IsCentreOfCircle(H,H)
IsCentreOfCircle(J,J)
IsCentreOfCircle(K,K)
Goal: Value(LengthOfLine(KJ))
Answer: 24
Theorem sequence:
radius_of_circle_property_length_equal(1,KA,K)
radius_of_circle_property_length_equal(1,KB,K)
radius_of_circle_property_length_equal(1,HC,H)
radius_of_circle_property_length_equal(1,JC,J)
radius_of_circle_property_length_equal(1,JB,J)
circle_perimeter_formula(1,K)
circle_perimeter_formula(1,J)
circle_perimeter_formula(1,H)
line_addition(1,KB,BJ)
Problem: As shown in the diagram, ∠CBA=17°, ∠DCA=29°, AC⊥CC. Find the measure of ∠BAC.
问题：如图所示，∠CBA=17°，∠DCA=29°，AC⊥CC。求∠BAC的大小。
Construction (CDL):
Shape(AD,DC,CA)
Shape(AC,CB,BA)
Collinear(DCB)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),17)
Equal(MeasureOfAngle(DCA),29)
PerpendicularBetweenLine(AC,CD)
Goal: Value(MeasureOfAngle(BAC))
Answer: 12
Theorem sequence:
flat_angle(1,DCB)
triangle_property_angle_sum(1,ACB)
angle_addition(1,DCA,ACB)
Problem: As shown in the diagram, AC=12, AS=x, BC=9, SB=7, SR=6, RC is perpendicular to BC, RS⊥AS. Find the length of line AS.
问题：如图所示，AC=12，AS=x，BC=9，SB=7，SR=6，RC⊥BC，RS⊥AS。求直线AS的长度。
Construction (CDL):
Shape(AR,RS,SA)
Shape(SR,RC,CB,BS)
Collinear(ARC)
Collinear(ASB)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(LengthOfLine(AS),x)
Equal(LengthOfLine(BC),9)
Equal(LengthOfLine(SB),7)
Equal(LengthOfLine(SR),6)
PerpendicularBetweenLine(RC,BC)
PerpendicularBetweenLine(RS,AS)
Goal: Value(LengthOfLine(AS))
Answer: 8
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,RSA,BAC)
mirror_similar_triangle_property_line_ratio(1,RSA,BAC)
mirror_similar_triangle_property_line_ratio(1,ARS,ACB)
Problem: As shown in the diagram, AB=15, AC=9, BC=12, BC is perpendicular to AC. Find the perimeter of △ABC.
问题：如图所示，AB=15，AC=9，BC=12，BC⊥AC。求三角形ABC的周长。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AC),9)
Equal(LengthOfLine(BC),12)
PerpendicularBetweenLine(BC,AC)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 36
Theorem sequence:
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, ∠AFC=70°, ∠EBF=35°, AD is perpendicular to ED, EB⊥CB. Find the measure of ∠FEB.
问题：如图所示，∠AFC=70°，∠EBF=35°，AD⊥ED，EB垂直于CB。求∠FEB的大小。
Construction (CDL):
Shape(DE,EF,FA,AD)
Shape(EB,BF,FE)
Shape(FB,BC,CF)
Shape(AF,FC)
Collinear(DEB)
Collinear(AFB)
Collinear(EFC)
Conditions (CDL):
Equal(MeasureOfAngle(AFC),70)
Equal(MeasureOfAngle(EBF),35)
PerpendicularBetweenLine(AD,ED)
PerpendicularBetweenLine(EB,CB)
Goal: Value(MeasureOfAngle(FEB))
Answer: 75
Theorem sequence:
vertical_angle(1,AFC,BFE)
triangle_property_angle_sum(1,EBF)
Problem: As shown in the diagram, AD=x+2, BD=3*y-9, BD=DY, CA=AY, CB=3/2*x+11, DY=2*y+6. Find the value of x.
问题：如图所示，AD=x+2，BD=3*y-9，BD=DY，CA=AY，CB=3/2*x+11，DY=2*y+6。求x的值。
Construction (CDL):
Shape(AY,YD,DA)
Shape(BC,CA,AD,DB)
Collinear(CAY)
Collinear(BDY)
Conditions (CDL):
Equal(LengthOfLine(AD),x+2)
Equal(LengthOfLine(BD),3*y-9)
Equal(LengthOfLine(BD),LengthOfLine(DY))
Equal(LengthOfLine(CA),LengthOfLine(AY))
Equal(LengthOfLine(CB),3/2*x+11)
Equal(LengthOfLine(DY),2*y+6)
Goal: Value(x)
Answer: 14
Theorem sequence:
line_addition(1,CA,AY)
line_addition(1,BD,DY)
similar_triangle_judgment_sas(1,YDA,YBC)
similar_triangle_property_line_ratio(1,AYD,CYB)
similar_triangle_property_line_ratio(1,YDA,YBC)
Problem: As shown in the diagram, AB=17, AD=6, DC=15, CD⊥AD, DA⊥BA, DC and AB are the parallel sides of trapezoid DABC. Find the area of quadrilateral DABC.
问题：如图所示，AB=17，AD=6，DC=15，CD垂直于AD，DA垂直于BA，四边形DABC是梯形。求DABC的面积。
Construction (CDL):
Shape(DA,AB,BC,CD)
Conditions (CDL):
Equal(LengthOfLine(AB),17)
Equal(LengthOfLine(AD),6)
Equal(LengthOfLine(DC),15)
PerpendicularBetweenLine(CD,AD)
PerpendicularBetweenLine(DA,BA)
Trapezoid(DABC)
Goal: Value(AreaOfQuadrilateral(DABC))
Answer: 96
Theorem sequence:
right_trapezoid_judgment_right_angle(1,DABC)
right_trapezoid_area_formular(1,DABC)