SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, NM=9, NP=10, QM=8, QP=7, WX=4, QPNM is similar to WZYX. Find the perimeter of quadrilateral WZYX.
问题：如图所示，NM=9，NP=10，QM=8，QP=7，WX=4，QPNM相似于WZYX。求WZYX的周长。
Construction (CDL):
Shape(WZ,ZY,YX,XW)
Shape(NM,MQ,QP,PN)
Conditions (CDL):
Equal(LengthOfLine(NM),9)
Equal(LengthOfLine(NP),10)
Equal(LengthOfLine(QM),8)
Equal(LengthOfLine(QP),7)
Equal(LengthOfLine(WX),4)
SimilarBetweenQuadrilateral(QPNM,WZYX)
Goal: Value(PerimeterOfQuadrilateral(WZYX))
Answer: 17
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,QPNM,WZYX)
similar_quadrilateral_property_line_ratio(1,PNMQ,ZYXW)
similar_quadrilateral_property_line_ratio(1,NMQP,YXWZ)
similar_quadrilateral_property_line_ratio(1,MQPN,XWZY)
quadrilateral_perimeter_formula(1,WZYX)
Problem: As shown in the diagram, QR=20, TS=12, ∠QRS=45°, ∠STQ=120°, the midsegment of TQRS is AB, QT and RS are the parallel sides of trapezoid QRST. Find the length of line AB.
问题：如图所示，QR=20，TS=12，∠QRS=45°，∠STQ=120°，AB是四边形TQRS的中位线，四边形QRST是梯形。求直线AB的长度。
Construction (CDL):
Shape(TA,AB,BS,ST)
Shape(AQ,QR,RB,BA)
Collinear(TAQ)
Collinear(SBR)
Conditions (CDL):
Equal(LengthOfLine(QR),20)
Equal(LengthOfLine(TS),12)
Equal(MeasureOfAngle(QRS),45)
Equal(MeasureOfAngle(STQ),120)
IsMidsegmentOfQuadrilateral(AB,TQRS)
Trapezoid(QRST)
Goal: Value(LengthOfLine(AB))
Answer: 16
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,AB,TQRS)
Problem: As shown in the diagram, BC=33, DE=10, ∠CDE=45°, quadrilateral CDAB is a parallelogram, CE is perpendicular to AE. Find the area of quadrilateral CDAB.
问题：如图所示，BC=33，DE=10，∠CDE=45°，四边形CDAB是▱，CE⊥AE。求四边形CDAB的面积。
Construction (CDL):
Shape(CD,DE,EC)
Shape(CE,EA,AB,BC)
Collinear(DEA)
Conditions (CDL):
Equal(LengthOfLine(BC),33)
Equal(LengthOfLine(DE),10)
Equal(MeasureOfAngle(CDE),45)
Parallelogram(CDAB)
PerpendicularBetweenLine(CE,AE)
Goal: Value(AreaOfQuadrilateral(CDAB))
Answer: 330
Theorem sequence:
adjacent_complementary_angle(1,DEC,CEA)
triangle_property_angle_sum(1,DEC)
isosceles_triangle_judgment_angle_equal(1,ECD)
altitude_of_quadrilateral_judgment_left_vertex(1,CE,CDAB)
parallelogram_property_opposite_line_equal(1,BCDA)
parallelogram_area_formula_common(1,CDAB)
Problem: As shown in the diagram, ∠CGH=42°, ∠DGC=3*x°, ∠GCD=z°, ∠GHC=y°, ∠HCG=48°, GD is parallel to HC, HG∥CD. Find the value of y.
问题：如图所示，∠CGH=42°，∠DGC=3*x°，∠GCD=z°，∠GHC=y°，∠HCG=48°，GD平行于HC，HG∥CD。求y的值。
Construction (CDL):
Shape(GH,HC,CG)
Shape(CD,DG,GC)
Conditions (CDL):
Equal(MeasureOfAngle(CGH),42)
Equal(MeasureOfAngle(DGC),3*x)
Equal(MeasureOfAngle(GCD),z)
Equal(MeasureOfAngle(GHC),y)
Equal(MeasureOfAngle(HCG),48)
ParallelBetweenLine(GD,HC)
ParallelBetweenLine(HG,CD)
Goal: Value(y)
Answer: 90
Theorem sequence:
triangle_property_angle_sum(1,GHC)
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 x.
问题：如图所示，AB=4*y-2，AB=AC，AC=2*y+2，∠BCA=6*x+8°，∠CAB=80°。求x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(x)
Answer: 7
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
Problem: As shown in the diagram, ∠ABE=57°, ∠BCE=42°, ∠EAB=28°, ∠ECD=36°. Find the measure of ∠CEB.
问题：如图所示，∠ABE=57°，∠BCE=42°，∠EAB=28°，∠ECD=36°。求∠CEB的大小。
Construction (CDL):
Shape(DE,EC,CD)
Shape(EA,AB,BE)
Shape(EB,BC,CE)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(MeasureOfAngle(ABE),57)
Equal(MeasureOfAngle(BCE),42)
Equal(MeasureOfAngle(EAB),28)
Equal(MeasureOfAngle(ECD),36)
Goal: Value(MeasureOfAngle(CEB))
Answer: 85
Theorem sequence:
triangle_property_angle_sum(1,EBC)
triangle_property_angle_sum(1,ABC)
angle_addition(1,ABE,EBC)
Problem: As shown in the diagram, AB=y, AC=x, BC=20, ∠ACB=60°, BA⊥CA. Find the value of y.
问题：如图所示，AB=y，AC=x，BC=20，∠ACB=60°，BA⊥CA。求y的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),20)
Equal(MeasureOfAngle(ACB),60)
PerpendicularBetweenLine(BA,CA)
Goal: Value(y)
Answer: 10*sqrt(3)
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, KL=3*y, KN=36, LJ=7, LM=9*x, NJ=y+15, KL and NM are opposite sides of the parallelogram KNML. Find the value of y.
问题：如图所示，KL=3*y，KN=36，LJ=7，LM=9*x，NJ=y+15，KNML是▱。求y的值。
Construction (CDL):
Shape(KN,NJ,JK)
Shape(JN,NM,MJ)
Shape(JL,LK,KJ)
Shape(JM,ML,LJ)
Collinear(KJM)
Collinear(NJL)
Conditions (CDL):
Equal(LengthOfLine(KL),3*y)
Equal(LengthOfLine(KN),36)
Equal(LengthOfLine(LJ),7)
Equal(LengthOfLine(LM),9*x)
Equal(LengthOfLine(NJ),y+15)
Parallelogram(KNML)
Goal: Value(y)
Answer: -8
Theorem sequence:
parallelogram_property_diagonal_bisection(1,NMLK,J)
Problem: As shown in the diagram, ∠SVT=75°, ∠TVP=72°, V is the center of circle V, RV⊥SV. Find the measure of ⌒VTQ.
问题：如图所示，∠SVT=75°，∠TVP=72°，圆V的圆心为V，RV⊥SV。求⌒VTQ的角度。
Construction (CDL):
Shape(VR,VRQ,QV)
Shape(VS,VSR,RV)
Shape(VT,VTS,SV)
Shape(VP,VPT,TV)
Shape(VQ,VQP,PV)
Collinear(QVS)
Cocircular(V,SRQPT)
Conditions (CDL):
Equal(MeasureOfAngle(SVT),75)
Equal(MeasureOfAngle(TVP),72)
IsCentreOfCircle(V,V)
PerpendicularBetweenLine(RV,SV)
Goal: Value(MeasureOfArc(VTQ))
Answer: 255
Theorem sequence:
flat_angle(1,QVS)
angle_addition(1,QVS,SVT)
arc_property_center_angle(1,VTQ,V)
Problem: As shown in the diagram, OW=x-9, OY=x, XO=x+4, ZO=x-12. Find the value of x.
问题：如图所示，OW=x-9，OY=x，XO=x+4，ZO=x-12。求x的值。
Construction (CDL):
Shape(OZ,AZW,WO)
Shape(OW,AWX,XA,AO)
Shape(OA,AX,AXY,YO)
Shape(OY,OYZ,ZO)
Collinear(XAOZ)
Collinear(YOW)
Cocircular(A,WXYZ)
Conditions (CDL):
Equal(LengthOfLine(OW),x-9)
Equal(LengthOfLine(OY),x)
Equal(LengthOfLine(XO),x+4)
Equal(LengthOfLine(ZO),x-12)
Goal: Value(x)
Answer: 48
Theorem sequence:
circle_property_circular_power_chord_and_chord(1,XOZ,YOW,A)
Problem: As shown in the diagram, AB=6, AE=9, BC=4, EB is parallel to DC. Find the length of line ED.
问题：如图所示，AB=6，AE=9，BC=4，EB∥DC。求直线ED的长度。
Construction (CDL):
Shape(BA,AE,EB)
Shape(BE,ED,DC,CB)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(AE),9)
Equal(LengthOfLine(BC),4)
ParallelBetweenLine(EB,DC)
Goal: Value(LengthOfLine(ED))
Answer: 6
Theorem sequence:
parallel_property_corresponding_angle(1,EB,DC,A)
parallel_property_corresponding_angle(2,CD,BE,A)
similar_triangle_judgment_aa(1,ADC,AEB)
line_addition(1,AB,BC)
line_addition(1,AE,ED)
similar_triangle_property_line_ratio(1,DCA,EBA)
similar_triangle_property_line_ratio(1,CAD,BAE)
Problem: As shown in the diagram, MN=3*x-4, NQ=15.4, PN=2*y+5, PQ=11.1, RM=17.9, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, MRPN is a ▱. Find the measure of ∠NRP.
问题：如图所示，MN=3*x-4，NQ=15.4，PN=2*y+5，PQ=11.1，RM=17.9，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，MRPN是平行四边形。求∠NRP的大小。
Construction (CDL):
Shape(QN,NM,MQ)
Shape(QM,MR,RQ)
Shape(QR,RP,PQ)
Shape(QP,PN,NQ)
Collinear(MQP)
Collinear(RQN)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(NQ),15.4)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),11.1)
Equal(LengthOfLine(RM),17.9)
Equal(LengthOfLine(RP),20)
Equal(LengthOfLine(RQ),3*z-3)
Equal(MeasureOfAngle(MRQ),38)
Equal(MeasureOfAngle(NQP),83)
Equal(MeasureOfAngle(QNM),33)
Parallelogram(MRPN)
Goal: Value(MeasureOfAngle(NRP))
Answer: 33
Theorem sequence:
parallel_property_alternate_interior_angle(2,MN,RP)
Problem: As shown in the diagram, WZ=25, XY=YZ, YZ=22, WY⊥XY. Find the length of line XY.
问题：如图所示，WZ=25，XY=YZ，YZ=22，WY垂直于XY。求直线XY的长度。
Construction (CDL):
Shape(XW,WY,YX)
Shape(WZ,ZY,YW)
Collinear(DWY)
Collinear(XYZ)
Collinear(WYT)
Conditions (CDL):
Equal(LengthOfLine(WZ),25)
Equal(LengthOfLine(XY),LengthOfLine(YZ))
Equal(LengthOfLine(YZ),22)
PerpendicularBetweenLine(WY,XY)
Goal: Value(LengthOfLine(XY))
Answer: 22
Theorem sequence:
Problem: As shown in the diagram, AB=DC, AC=188, AD=BC, BD=214, ∠BPC=70°, P bisects segment AC. Find the perimeter of ADCB.
问题：如图所示，AB=DC，AC=188，AD=BC，BD=214，∠BPC=70°，P是线段AC的中点。求ADCB的周长。
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),LengthOfLine(DC))
Equal(LengthOfLine(AC),188)
Equal(LengthOfLine(AD),LengthOfLine(BC))
Equal(LengthOfLine(BD),214)
Equal(MeasureOfAngle(BPC),70)
IsMidpointOfLine(P,AC)
Goal: Value(PerimeterOfQuadrilateral(ADCB))
Answer: 2*sqrt(20285-20116*sin(pi/9))+2*sqrt(20116*sin(pi/9)+20285)
Theorem sequence:
parallelogram_judgment_equal_and_equal(1,ADCB)
parallelogram_property_diagonal_bisection(1,DCBA,P)
adjacent_complementary_angle(1,APB,BPC)
line_addition(1,AP,PC)
line_addition(1,DP,PB)
cosine_theorem(1,PCB)
cosine_theorem(1,PBA)
quadrilateral_perimeter_formula(1,ADCB)
Problem: As shown in the diagram, KJ=3*f-6, LM=2*f+8, ∠JML=3*d-2°, ∠KJM=56°, JMLK is a ▱. Find the value of f.
问题：如图所示，KJ=3*f-6，LM=2*f+8，∠JML=3*d-2°，∠KJM=56°，JMLK是平行四边形。求f的值。
Construction (CDL):
Shape(JM,ML,LK,KJ)
Conditions (CDL):
Equal(LengthOfLine(KJ),3*f-6)
Equal(LengthOfLine(LM),2*f+8)
Equal(MeasureOfAngle(JML),3*d-2)
Equal(MeasureOfAngle(KJM),56)
Parallelogram(JMLK)
Goal: Value(f)
Answer: 14
Theorem sequence:
parallelogram_property_opposite_line_equal(1,KJML)
Problem: As shown in the diagram, AB=12, AC=h, ∠ABC=30°, BC⊥AC. Find the value of h.
问题：如图所示，AB=12，AC=h，∠ABC=30°，BC垂直于AC。求h的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(AC),h)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(BC,AC)
Goal: Value(h)
Answer: 6
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, ∠ACD=50°, ∠BFE=56°, ∠CDE=78°, ∠FGA=120°. Find the measure of ∠EAG.
问题：如图所示，∠ACD=50°，∠BFE=56°，∠CDE=78°，∠FGA=120°。求∠EAG的大小。
Construction (CDL):
Shape(CD,DA,AC)
Shape(EA,AG,GA)
Shape(GF,FB,BG)
Collinear(CAGBH)
Collinear(DAE)
Collinear(EGF)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),50)
Equal(MeasureOfAngle(BFE),56)
Equal(MeasureOfAngle(CDE),78)
Equal(MeasureOfAngle(FGA),120)
Goal: Value(MeasureOfAngle(EAG))
Answer: 52
Theorem sequence:
triangle_property_angle_sum(1,CDA)
vertical_angle(1,DAC,EAG)
Problem: As shown in the diagram, ∠EBK=120°, ∠HKB=54°, ∠JKH=36°, BK is parallel to HJ. Find the measure of ∠BHK.
问题：如图所示，∠EBK=120°，∠HKB=54°，∠JKH=36°，BK平行于HJ。求∠BHK的大小。
Construction (CDL):
Shape(KB,BH,HK)
Shape(KH,HJ,JK)
Shape(EB,BK)
Collinear(EBH)
Collinear(BHD)
Conditions (CDL):
Equal(MeasureOfAngle(EBK),120)
Equal(MeasureOfAngle(HKB),54)
Equal(MeasureOfAngle(JKH),36)
ParallelBetweenLine(BK,HJ)
Goal: Value(MeasureOfAngle(BHK))
Answer: 66
Theorem sequence:
flat_angle(1,EBH)
angle_addition(1,EBK,KBH)
triangle_property_angle_sum(1,KBH)
Problem: As shown in the diagram, AB=17, BC=8, CA=15, BC⊥AC. Find the value of tan(CA).
问题：如图所示，AB=17，BC=8，CA=15，BC⊥AC。求tan(CA)的值。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),17)
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(CA),15)
PerpendicularBetweenLine(BC,AC)
Goal: Value(Tan(MeasureOfAngle(CAB)))
Answer: 8/15
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, AD=BD, AF=CF, CA=15, CE=EB, EF=9, ∠BDE=82°. Find the length of line DB.
问题：如图所示，AD=BD，AF=CF，CA=15，CE=EB，EF=9，∠BDE=82°。求直线DB的长度。
Construction (CDL):
Shape(FC,CE,EF)
Shape(AF,FE,ED,DA)
Shape(DE,EB,BD)
Collinear(AFC)
Collinear(CEB)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),LengthOfLine(BD))
Equal(LengthOfLine(AF),LengthOfLine(CF))
Equal(LengthOfLine(CA),15)
Equal(LengthOfLine(CE),LengthOfLine(EB))
Equal(LengthOfLine(EF),9)
Equal(MeasureOfAngle(BDE),82)
Goal: Value(LengthOfLine(DB))
Answer: 9
Theorem sequence:
line_addition(1,AF,FC)
line_addition(1,CE,EB)
line_addition(1,AD,DB)
similar_triangle_judgment_sas(1,CEF,CBA)
similar_triangle_property_line_ratio(1,EFC,BAC)
similar_triangle_property_line_ratio(1,CEF,CBA)
Problem: As shown in the diagram, AC=6, ∠CAB=45°, BC⊥AC. Find the length of line AB.
问题：如图所示，AC=6，∠CAB=45°，BC垂直于AC。求直线AB的长度。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AC),6)
Equal(MeasureOfAngle(CAB),45)
PerpendicularBetweenLine(BC,AC)
Goal: Value(LengthOfLine(AB))
Answer: 6*sqrt(2)
Theorem sequence:
triangle_property_angle_sum(1,BCA)
sine_theorem(1,ABC)
Problem: As shown in the diagram, MQ=18, QP=18, ∠PNQ=4*x-8°, ∠QNM=3*x+5°, NM⊥QM, QP⊥NP. Find the measure of ∠PNM.
问题：如图所示，MQ=18，QP=18，∠PNQ=4*x-8°，∠QNM=3*x+5°，NM⊥QM，QP⊥NP。求∠PNM的大小。
Construction (CDL):
Shape(NM,MQ,QN)
Shape(NQ,QP,PN)
Collinear(BMN)
Collinear(CPN)
Conditions (CDL):
Equal(LengthOfLine(MQ),18)
Equal(LengthOfLine(QP),18)
Equal(MeasureOfAngle(PNQ),4*x-8)
Equal(MeasureOfAngle(QNM),3*x+5)
PerpendicularBetweenLine(NM,QM)
PerpendicularBetweenLine(QP,NP)
Goal: Value(MeasureOfAngle(PNM))
Answer: 88
Theorem sequence:
mirror_congruent_triangle_judgment_hl(1,NMQ,NQP)
mirror_congruent_triangle_property_angle_equal(1,NMQ,NQP)
angle_addition(1,PNQ,QNM)
Problem: As shown in the diagram, ∠JQR=131°, QR is parallel to TS, TQ is parallel to SR. Find the measure of ∠TSR.
问题：如图所示，∠JQR=131°，QR∥TS，TQ平行于SR。求∠TSR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(TSR))
Answer: 49
Theorem sequence:
parallelogram_judgment_parallel_and_parallel(1,QTSR)
adjacent_complementary_angle(1,JQR,RQT)
parallelogram_property_opposite_angle_equal(1,QTSR)
Problem: As shown in the diagram, AF=2*x, AG=y, BD=x, BF=24, CG=12, DC=9, the center of circle E is E, the tangent to ⊙E is DC. Find the value of y.
问题：如图所示，AF=2*x，AG=y，BD=x，BF=24，CG=12，DC=9，E是⊙E的圆心，⊙O的切线为DC。求y的值。
Construction (CDL):
Shape(EFB,BF)
Shape(BD,DC,EBC)
Shape(GA,AF,EGF)
Shape(EBC,CG,EGF,FB)
Shape(ECG,GC)
Collinear(DBFA)
Collinear(AGC)
Cocircular(E,BCGF)
Conditions (CDL):
Equal(LengthOfLine(AF),2*x)
Equal(LengthOfLine(AG),y)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(BF),24)
Equal(LengthOfLine(CG),12)
Equal(LengthOfLine(DC),9)
IsCentreOfCircle(E,E)
IsTangentOfCircle(DC,E)
Goal: Value(y)
Answer: -6+6*sqrt(6)
Theorem sequence:
line_addition(1,DB,BF)
circle_property_circular_power_tangent_and_segment_line(1,DC,DBF,E)
line_addition(1,AF,FB)
line_addition(1,AG,GC)
circle_property_circular_power_segment_and_segment_line(1,AFB,AGC,E)
Problem: As shown in the diagram, BE=7*sqrt(3), ∠ABC=60°, ∠CAB=30°, AE⊥CE, BC⊥AC. Find the length of line CE.
问题：如图所示，BE=7*sqrt(3)，∠ABC=60°，∠CAB=30°，AE⊥CE，BC垂直于AC。求直线CE的长度。
Construction (CDL):
Shape(BC,CE,EB)
Shape(EC,CA,AE)
Collinear(BEA)
Conditions (CDL):
Equal(LengthOfLine(BE),7*sqrt(3))
Equal(MeasureOfAngle(ABC),60)
Equal(MeasureOfAngle(CAB),30)
PerpendicularBetweenLine(AE,CE)
PerpendicularBetweenLine(BC,AC)
Goal: Value(LengthOfLine(CE))
Answer: 21
Theorem sequence:
adjacent_complementary_angle(1,AEC,CEB)
triangle_property_angle_sum(1,CEB)
sine_theorem(1,CEB)
sine_theorem(1,BCE)
Problem: As shown in the diagram, KN=9, LN=16, PM=Mul(LengthOfLine(KP),2), PR∥KL, KN is perpendicular to MN, LM⊥KM. Find the length of line KP.
问题：如图所示，KN=9，LN=16，PM=Mul(LengthOfLine(KP),2)，PR平行于KL，KN垂直于MN，LM垂直于KM。求直线KP的长度。
Construction (CDL):
Shape(LR,RQ,QN,NL)
Shape(RM,MQ,QR)
Shape(QM,MP,PQ)
Shape(NQ,QP,PK,KN)
Collinear(LNK)
Collinear(RQP)
Collinear(LRM)
Collinear(MPK)
Collinear(NQM)
Conditions (CDL):
Equal(LengthOfLine(KN),9)
Equal(LengthOfLine(LN),16)
Equal(LengthOfLine(PM),Mul(LengthOfLine(KP),2))
ParallelBetweenLine(PR,KL)
PerpendicularBetweenLine(KN,MN)
PerpendicularBetweenLine(LM,KM)
Goal: Value(LengthOfLine(KP))
Answer: 5
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,MKN,LMK)
line_addition(1,LN,NK)
line_addition(1,MP,PK)
mirror_similar_triangle_property_line_ratio(1,MKN,LMK)
mirror_similar_triangle_property_line_ratio(1,NMK,MKL)
Problem: As shown in the diagram, AB=20, AC=x, AD=10, BC=z, BD=y, CA is perpendicular to BA, DB is perpendicular to CB. Find the length of line CB.
问题：如图所示，AB=20，AC=x，AD=10，BC=z，BD=y，CA垂直于BA，DB垂直于CB。求直线CB的长度。
Construction (CDL):
Shape(CA,AB,BC)
Shape(AD,DB,BA)
Collinear(CAD)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(AD),10)
Equal(LengthOfLine(BC),z)
Equal(LengthOfLine(BD),y)
PerpendicularBetweenLine(CA,BA)
PerpendicularBetweenLine(DB,CB)
Goal: Value(LengthOfLine(CB))
Answer: 20*sqrt(5)
Theorem sequence:
adjacent_complementary_angle(1,CAB,BAD)
right_triangle_judgment_angle(1,BAD)
right_triangle_property_pythagorean(1,BAD)
mirror_similar_triangle_judgment_aa(1,BAD,CDB)
mirror_similar_triangle_property_line_ratio(1,BAD,CDB)
mirror_similar_triangle_property_line_ratio(1,DBA,DBC)
Problem: As shown in the diagram, ∠WZY=4*x°, ∠XWZ=10*x°, ∠YXW=120°, ZW and ZY are one pair of adjacent sides of the kite XWZY. Find the measure of ∠ZYX.
问题：如图所示，∠WZY=4*x°，∠XWZ=10*x°，∠YXW=120°，XWZY是风筝形。求∠ZYX的大小。
Construction (CDL):
Shape(WZ,ZU,UW)
Shape(ZY,YU,UZ)
Shape(UY,YX,XU)
Shape(WU,UX,XW)
Collinear(ZUX)
Collinear(WUY)
Conditions (CDL):
Equal(MeasureOfAngle(WZY),4*x)
Equal(MeasureOfAngle(XWZ),10*x)
Equal(MeasureOfAngle(YXW),120)
Kite(XWZY)
Goal: Value(MeasureOfAngle(ZYX))
Answer: 100
Theorem sequence:
kite_property_opposite_angle_equal(1,XWZY)
quadrilateral_property_angle_sum(1,WZYX)
Problem: As shown in the diagram, ∠GKM=62°, ∠HBF=38°, NO∥MH. Find the measure of ∠JCI.
问题：如图所示，∠GKM=62°，∠HBF=38°，NO平行于MH。求∠JCI的大小。
Construction (CDL):
Shape(NC,CJ)
Shape(JC,CI)
Shape(IC,CO)
Shape(KC,CN)
Shape(OC,CB)
Shape(MK,KC)
Shape(CB,BH)
Shape(GK,KM)
Shape(BK,KG)
Shape(FB,BK)
Shape(HB,BF)
Shape(CK,KB,BC)
Collinear(JCBF)
Collinear(ICKG)
Collinear(NCO)
Collinear(MKBH)
Conditions (CDL):
Equal(MeasureOfAngle(GKM),62)
Equal(MeasureOfAngle(HBF),38)
ParallelBetweenLine(NO,MH)
Goal: Value(MeasureOfAngle(JCI))
Answer: 80
Theorem sequence:
vertical_angle(1,GKM,CKB)
vertical_angle(1,KBC,HBF)
triangle_property_angle_sum(1,CKB)
vertical_angle(1,BCK,JCI)
Problem: As shown in the diagram, KJ=2, LK=4, LP=2, MP=3, PQ=6, RS=6, QK is parallel to PL, RJ∥QK, SH is parallel to RJ. Find the length of line QK.
问题：如图所示，KJ=2，LK=4，LP=2，MP=3，PQ=6，RS=6，QK平行于PL，RJ平行于QK，SH平行于RJ。求直线QK的长度。
Construction (CDL):
Shape(HS,SR,RJ,JH)
Shape(JR,RQ,QK,KJ)
Shape(KQ,QP,PL,LK)
Shape(LP,PM,ML)
Collinear(HJKLM)
Collinear(SRQPM)
Conditions (CDL):
Equal(LengthOfLine(KJ),2)
Equal(LengthOfLine(LK),4)
Equal(LengthOfLine(LP),2)
Equal(LengthOfLine(MP),3)
Equal(LengthOfLine(PQ),6)
Equal(LengthOfLine(RS),6)
ParallelBetweenLine(QK,PL)
ParallelBetweenLine(RJ,QK)
ParallelBetweenLine(SH,RJ)
Goal: Value(LengthOfLine(QK))
Answer: 6
Theorem sequence:
parallel_property_corresponding_angle(1,LP,KQ,M)
similar_triangle_judgment_aa(1,QMK,PML)
line_addition(1,QP,PM)
similar_triangle_property_line_ratio(1,KQM,LPM)
similar_triangle_property_line_ratio(1,MKQ,MLP)
Problem: As shown in the diagram, ∠EDF=39°, ∠FBA=48°, FC∥ED, AF is perpendicular to BF, DC is perpendicular to FC, FE⊥DE. Find the measure of ∠FDC.
问题：如图所示，∠EDF=39°，∠FBA=48°，FC平行于ED，AF⊥BF，DC垂直于FC，FE垂直于DE。求∠FDC的大小。
Construction (CDL):
Shape(AF,FB,BA)
Shape(FE,ED,DF)
Shape(FD,DC,CB,BF)
Collinear(FBC)
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, NM=9, NP=10, QM=8, QP=7, WX=4, quadrilateral XWZY is similar to quadrilateral MQPN. Find the perimeter of XWZY.
问题：如图所示，NM=9，NP=10，QM=8，QP=7，WX=4，XWZY与MQPN相似。求四边形XWZY的周长。
Construction (CDL):
Shape(XW,WZ,ZY,YX)
Shape(MQ,QP,PN,NM)
Conditions (CDL):
Equal(LengthOfLine(NM),9)
Equal(LengthOfLine(NP),10)
Equal(LengthOfLine(QM),8)
Equal(LengthOfLine(QP),7)
Equal(LengthOfLine(WX),4)
SimilarBetweenQuadrilateral(XWZY,MQPN)
Goal: Value(PerimeterOfQuadrilateral(XWZY))
Answer: 17
Theorem sequence:
similar_quadrilateral_property_line_ratio(1,XWZY,MQPN)
quadrilateral_perimeter_formula(1,MQPN)
similar_quadrilateral_property_perimeter_ratio(1,XWZY,MQPN)
Problem: As shown in the diagram, ∠ACH=x-10°, ∠GMO=104°, JO is parallel to HD. Find the value of x.
问题：如图所示，∠ACH=x-10°，∠GMO=104°，JO平行于HD。求x的值。
Construction (CDL):
Shape(GM,MO)
Shape(OM,MC)
Shape(MC,CD)
Shape(DC,CA)
Shape(JM,MG)
Shape(CM,MJ)
Shape(HC,CM)
Shape(AC,CH)
Collinear(GMCA)
Collinear(OMJ)
Collinear(DCH)
Conditions (CDL):
Equal(MeasureOfAngle(ACH),x-10)
Equal(MeasureOfAngle(GMO),104)
ParallelBetweenLine(JO,HD)
Goal: Value(x)
Answer: 114
Theorem sequence:
parallel_property_collinear_extend(3,JO,HD,M)
parallel_property_collinear_extend(3,DH,OM,C)
parallel_property_corresponding_angle(1,MO,CD,G)
vertical_angle(1,MCD,ACH)
Problem: As shown in the diagram, PA=8, ∠PBA=∠CBP, BA is perpendicular to PA, PC is perpendicular to BC. Find the length of line CP.
问题：如图所示，PA=8，∠PBA=∠CBP，BA⊥PA，PC垂直于BC。求直线CP的长度。
Construction (CDL):
Shape(BA,AP,PB)
Shape(BP,PC,CB)
Conditions (CDL):
Equal(LengthOfLine(PA),8)
Equal(MeasureOfAngle(PBA),MeasureOfAngle(CBP))
PerpendicularBetweenLine(BA,PA)
PerpendicularBetweenLine(PC,BC)
Goal: Value(LengthOfLine(CP))
Answer: 8
Theorem sequence:
mirror_congruent_triangle_judgment_aas(3,PCB,PBA)
mirror_congruent_triangle_property_line_equal(1,BPC,BAP)
Problem: As shown in the diagram, ∠FJE=75°, GD is parallel to IM, IM is parallel to CE. Find the measure of ∠IFB.
问题：如图所示，∠FJE=75°，GD平行于IM，IM平行于CE。求∠IFB的大小。
Construction (CDL):
Shape(LB,BD)
Shape(GB,BL)
Shape(DB,BF)
Shape(FB,BG)
Shape(BF,FM)
Shape(IF,FB)
Shape(MF,FJ)
Shape(JF,FI)
Shape(FJ,JE)
Shape(CJ,JF)
Shape(EJ,JT)
Shape(TJ,JC)
Collinear(LBFJT)
Collinear(DBG)
Collinear(MFI)
Collinear(EJC)
Conditions (CDL):
Equal(MeasureOfAngle(FJE),75)
ParallelBetweenLine(GD,IM)
ParallelBetweenLine(IM,CE)
Goal: Value(MeasureOfAngle(IFB))
Answer: 105
Theorem sequence:
parallel_property_collinear_extend(3,IM,CE,F)
parallel_property_collinear_extend(3,EC,MF,J)
parallel_property_corresponding_angle(1,FM,JE,B)
adjacent_complementary_angle(1,IFB,BFM)
Problem: As shown in the diagram, ∠WXY=4*x+15°, ∠ZXV=2*x+65°. Find the measure of ∠YXZ.
问题：如图所示，∠WXY=4*x+15°，∠ZXV=2*x+65°。求∠YXZ的大小。
Construction (CDL):
Shape(XZ,XZY,YX)
Shape(XY,XYW,WX)
Shape(XW,XWV,VX)
Shape(XV,XVZ,ZX)
Collinear(ZXW)
Collinear(YXV)
Conditions (CDL):
Equal(MeasureOfAngle(WXY),4*x+15)
Equal(MeasureOfAngle(ZXV),2*x+65)
Goal: Value(MeasureOfAngle(YXZ))
Answer: 65
Theorem sequence:
vertical_angle(1,ZXV,WXY)
adjacent_complementary_angle(1,YXZ,ZXV)
Problem: As shown in the diagram, DE=60, FC=16, the center of circle C is C, the diameter of ⊙C is AB, CF is perpendicular to EF. Find the length of line AB.
问题：如图所示，DE=60，FC=16，C是⊙C的圆心，AB是⊙C的直径，CF垂直于EF。求直线AB的长度。
Construction (CDL):
Shape(FC,CD,DF)
Shape(FD,CDA,AF)
Shape(FA,CAE,EF)
Shape(CF,FE,CEB,BC)
Shape(CB,CBD,DC)
Collinear(AFCB)
Collinear(EFD)
Cocircular(C,AEBD)
Conditions (CDL):
Equal(LengthOfLine(DE),60)
Equal(LengthOfLine(FC),16)
IsCentreOfCircle(C,C)
IsDiameterOfCircle(AB,C)
PerpendicularBetweenLine(CF,EF)
Goal: Value(LengthOfLine(AB))
Answer: 68
Theorem sequence:
adjacent_complementary_angle(1,DFC,CFE)
right_triangle_judgment_angle(1,DFC)
circle_property_chord_perpendicular_bisect_chord(1,C,CF,DE)
line_addition(1,EF,FD)
right_triangle_property_pythagorean(1,DFC)
radius_of_circle_property_length_equal(1,CD,C)
circle_property_length_of_radius_and_diameter(1,C)
diameter_of_circle_property_length_equal(1,AB,C)
Problem: As shown in the diagram, EH=9, FL=6, HG=21, EF is parallel to HL. Find the length of line LG.
问题：如图所示，EH=9，FL=6，HG=21，EF∥HL。求直线LG的长度。
Construction (CDL):
Shape(EH,HL,LF,FE)
Shape(HG,GL,LH)
Collinear(EHG)
Collinear(FLG)
Conditions (CDL):
Equal(LengthOfLine(EH),9)
Equal(LengthOfLine(FL),6)
Equal(LengthOfLine(HG),21)
ParallelBetweenLine(EF,HL)
Goal: Value(LengthOfLine(LG))
Answer: 14
Theorem sequence:
parallel_property_corresponding_angle(1,LH,FE,G)
similar_triangle_judgment_aa(1,HGL,EGF)
line_addition(1,GL,LF)
line_addition(1,GH,HE)
similar_triangle_property_line_ratio(1,HGL,EGF)
similar_triangle_property_line_ratio(1,LHG,FEG)
Problem: As shown in the diagram, ∠DCF=53°. Find the measure of ∠ECB.
问题：如图所示，∠DCF=53°。求∠ECB的大小。
Construction (CDL):
Shape(FC,CE)
Shape(DC,CF)
Shape(EC,CB)
Shape(BC,CD)
Collinear(FCB)
Collinear(DCE)
Conditions (CDL):
Equal(MeasureOfAngle(DCF),53)
Goal: Value(MeasureOfAngle(ECB))
Answer: 53
Theorem sequence:
vertical_angle(1,DCF,ECB)