SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, triangle RST is congruent to triangle XYZ, TR=x+21, ZX=2*x-14, ∠TRS=4*y-10°, ∠ZXY=3*y+5°. Find the value of y.
问题：如图所示，三角形RST与三角形XYZ是全等三角形，TR=x+21，ZX=2*x-14，∠TRS=4*y-10°，∠ZXY=3*y+5°。求y的值。
Construction (CDL):
Shape(RS,ST,TR)
Shape(XY,YZ,ZX)
Conditions (CDL):
CongruentBetweenTriangle(RST,XYZ)
Equal(LengthOfLine(TR),x+21)
Equal(LengthOfLine(ZX),2*x-14)
Equal(MeasureOfAngle(TRS),4*y-10)
Equal(MeasureOfAngle(ZXY),3*y+5)
Goal: Value(y)
Answer: 15
Theorem sequence:
congruent_triangle_property_angle_equal(1,RST,XYZ)
Problem: As shown in the diagram, MN=3*x-4, NQ=15, PN=2*y+5, PQ=12, RM=18, RP=20, RQ=3*z-3, ∠MRQ=38°, ∠NQP=83°, ∠QNM=33°, MN and RP are opposite sides of the ▱ MRPN. Find the value of y.
问题：如图所示，MN=3*x-4，NQ=15，PN=2*y+5，PQ=12，RM=18，RP=20，RQ=3*z-3，∠MRQ=38°，∠NQP=83°，∠QNM=33°，RM和PN是▱MRPN的一组对边。求y的值。
Construction (CDL):
Shape(MR,RQ,QM)
Shape(QR,RP,PQ)
Shape(MQ,QN,NM)
Shape(QP,PN,NQ)
Collinear(MQP)
Collinear(RQN)
Conditions (CDL):
Equal(LengthOfLine(MN),3*x-4)
Equal(LengthOfLine(NQ),15)
Equal(LengthOfLine(PN),2*y+5)
Equal(LengthOfLine(PQ),12)
Equal(LengthOfLine(RM),18)
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(y)
Answer: 13/2
Theorem sequence:
parallelogram_property_opposite_line_equal(1,MRPN)
Problem: As shown in the diagram, BT=SB, QA=RA, QT=86, RS=54, the midsegment of quadrilateral AQTB is JK. Find the length of line JK.
问题：如图所示，BT=SB，QA=RA，QT=86，RS=54，JK是四边形AQTB的中位线。求直线JK的长度。
Construction (CDL):
Shape(RA,AB,BS,SR)
Shape(AJ,JK,KB,BA)
Shape(JQ,QT,TK,KJ)
Collinear(RAJQ)
Collinear(SBKT)
Conditions (CDL):
Equal(LengthOfLine(BT),LengthOfLine(SB))
Equal(LengthOfLine(QA),LengthOfLine(RA))
Equal(LengthOfLine(QT),86)
Equal(LengthOfLine(RS),54)
IsMidsegmentOfQuadrilateral(JK,AQTB)
Goal: Value(LengthOfLine(JK))
Answer: 78
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,AB,RQTS)
midsegment_of_quadrilateral_property_length(1,AB,RQTS)
midsegment_of_quadrilateral_property_length(1,JK,AQTB)
Problem: As shown in the diagram, WZ=23, XY=23, ∠ZWX=112°, XW is parallel to YZ, XYZW is a trapezoid. Find the measure of ∠YZW.
问题：如图所示，WZ=23，XY=23，∠ZWX=112°，XW平行于YZ，XY和ZW是梯形XYZW的腰。求∠YZW的大小。
Construction (CDL):
Shape(WX,XY,YZ,ZW)
Conditions (CDL):
Equal(LengthOfLine(WZ),23)
Equal(LengthOfLine(XY),23)
Equal(MeasureOfAngle(ZWX),112)
ParallelBetweenLine(XW,YZ)
Trapezoid(XYZW)
Goal: Value(MeasureOfAngle(YZW))
Answer: 68
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,ZY,WX)
Problem: As shown in the diagram, AB=y, AD=z, BC=x, BD=4, CD=10, AB⊥CB, DC⊥AC. Find the value of x.
问题：如图所示，AB=y，AD=z，BC=x，BD=4，CD=10，AB⊥CB，DC垂直于AC。求x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(BC),x)
Equal(LengthOfLine(BD),4)
Equal(LengthOfLine(CD),10)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(DC,AC)
Goal: Value(x)
Answer: 2*sqrt(21)
Theorem sequence:
adjacent_complementary_angle(1,ABC,CBD)
right_triangle_judgment_angle(1,CBD)
right_triangle_property_pythagorean(1,CBD)
Problem: As shown in the diagram, AC=20, CM=12, MB=30, CM is perpendicular to AM. Find the perimeter of △AMB.
问题：如图所示，AC=20，CM=12，MB=30，CM⊥AM。求三角形AMB的周长。
Construction (CDL):
Shape(AC,CM,MA)
Shape(AM,MB,BA)
Collinear(CMB)
Conditions (CDL):
Equal(LengthOfLine(AC),20)
Equal(LengthOfLine(CM),12)
Equal(LengthOfLine(MB),30)
PerpendicularBetweenLine(CM,AM)
Goal: Value(PerimeterOfTriangle(AMB))
Answer: 80
Theorem sequence:
adjacent_complementary_angle(1,CMA,AMB)
right_triangle_judgment_angle(1,CMA)
right_triangle_judgment_angle(1,AMB)
right_triangle_property_pythagorean(1,CMA)
right_triangle_property_pythagorean(1,AMB)
triangle_perimeter_formula(1,AMB)
Problem: As shown in the diagram, TR=7*sqrt(2), TS=3*sqrt(2), RS is perpendicular to TS. Find the measure of ∠STR.
问题：如图所示，TR=7*sqrt(2)，TS=3*sqrt(2)，RS垂直于TS。求∠STR的大小。
Construction (CDL):
Shape(TR,RS,ST)
Conditions (CDL):
Equal(LengthOfLine(TR),7*sqrt(2))
Equal(LengthOfLine(TS),3*sqrt(2))
PerpendicularBetweenLine(RS,TS)
Goal: Value(MeasureOfAngle(STR))
Answer: 180*asin(2*sqrt(10)/7)/pi
Theorem sequence:
right_triangle_judgment_angle(1,RST)
right_triangle_property_pythagorean(1,RST)
sine_theorem(1,RST)
Problem: As shown in the diagram, AB=4/7, AC=x, BC=5/7, BA⊥CA. Find the value of x.
问题：如图所示，AB=4/7，AC=x，BC=5/7，BA垂直于CA。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),4/7)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),5/7)
PerpendicularBetweenLine(BA,CA)
Goal: Value(x)
Answer: 3/7
Theorem sequence:
right_triangle_judgment_angle(1,BAC)
right_triangle_property_pythagorean(1,BAC)
Problem: As shown in the diagram, JK=10, MN=16, the measure of ⌒KMN is 98, the center of circle K is K, MP⊥KP. Find the length of line LN.
问题：如图所示，JK=10，MN=16，⌒KMN的角度为98，⊙K的圆心为K，MP垂直于KP。求直线LN的长度。
Construction (CDL):
Shape(KJM,MK,KJ)
Shape(MP,PK,KM)
Shape(KML,LP,PM)
Shape(PL,LN,NP)
Shape(KLN,NL)
Shape(PN,KNJ,JK,KP)
Collinear(JKPL)
Collinear(MPN)
Cocircular(K,JMLN)
Conditions (CDL):
Equal(LengthOfLine(JK),10)
Equal(LengthOfLine(MN),16)
Equal(MeasureOfArc(KMN),98)
IsCentreOfCircle(K,K)
PerpendicularBetweenLine(MP,KP)
Goal: Value(LengthOfLine(LN))
Answer: 4*sqrt(5)
Theorem sequence:
circle_property_chord_perpendicular_bisect_chord(1,K,KP,MN)
line_addition(1,MP,PN)
radius_of_circle_property_length_equal(1,KJ,K)
radius_of_circle_property_length_equal(1,KM,K)
radius_of_circle_property_length_equal(1,KL,K)
right_triangle_judgment_angle(1,MPK)
right_triangle_property_pythagorean(1,MPK)
line_addition(1,LP,PK)
vertical_angle(1,MPK,NPL)
right_triangle_judgment_angle(1,NPL)
right_triangle_property_pythagorean(1,NPL)
Problem: As shown in the diagram, ∠FCD=x°, the measure of arc EDB is 10*x, the measure of arc EFD is 40, CD is the tangent to circle E. Find the value of x.
问题：如图所示，∠FCD=x°，弧EDB的角度为10*x，弧EFD的角度为40，圆O的切线为CD。求x的值。
Construction (CDL):
Shape(EDB,BF,EFD)
Shape(CD,EFD,FC)
Shape(FB,EBF)
Collinear(BFC)
Cocircular(E,FDB)
Conditions (CDL):
Equal(MeasureOfAngle(FCD),x)
Equal(MeasureOfArc(EDB),10*x)
Equal(MeasureOfArc(EFD),40)
IsTangentOfCircle(CD,E)
Goal: Value(x)
Answer: 5
Theorem sequence:
circle_property_circular_power_tangent_and_segment_angle(2,CD,CFB,E)
Problem: As shown in the diagram, ∠CFK=28°, ∠GKF=35°, ∠KAC=25°, ∠KHC=51°, EG is perpendicular to FG, HC is perpendicular to KC, KF is perpendicular to EF. Find the measure of ∠FEK.
问题：如图所示，∠CFK=28°，∠GKF=35°，∠KAC=25°，∠KHC=51°，EG⊥FG，HC垂直于KC，KF垂直于EF。求∠FEK的大小。
Construction (CDL):
Shape(KA,AH,HK)
Shape(KH,HC,CK)
Shape(KC,CF,FK)
Shape(KF,FG,GK)
Shape(GF,FE,EG)
Collinear(AHCF)
Collinear(EGK)
Conditions (CDL):
Equal(MeasureOfAngle(CFK),28)
Equal(MeasureOfAngle(GKF),35)
Equal(MeasureOfAngle(KAC),25)
Equal(MeasureOfAngle(KHC),51)
PerpendicularBetweenLine(EG,FG)
PerpendicularBetweenLine(HC,KC)
PerpendicularBetweenLine(KF,EF)
Goal: Value(MeasureOfAngle(FEK))
Answer: 55
Theorem sequence:
triangle_property_angle_sum(1,FEK)
Problem: As shown in the diagram, AB=30, CD=30, the measure of arc XCZ is 40, the center of ⊙X is X, AM is perpendicular to YM, DN⊥ZN. Find the measure of arc XBA.
问题：如图所示，AB=30，CD=30，⌒XCZ的角度为40，⊙X的圆心为X，AM垂直于YM，DN⊥ZN。求弧XBA的角度。
Construction (CDL):
Shape(XBY,YM,MB)
Shape(XYA,AM,MY)
Shape(XM,MA,XAC,CN,NX)
Shape(NC,XCZ,ZN)
Shape(NZ,XZD,DN)
Shape(MX,XN,ND,XDB,BM)
Collinear(AMB)
Collinear(YMX)
Collinear(XNZ)
Collinear(CND)
Cocircular(X,ACZDBY)
Conditions (CDL):
Equal(LengthOfLine(AB),30)
Equal(LengthOfLine(CD),30)
Equal(MeasureOfArc(XCZ),40)
IsCentreOfCircle(X,X)
PerpendicularBetweenLine(AM,YM)
PerpendicularBetweenLine(DN,ZN)
Goal: Value(MeasureOfArc(XBA))
Answer: 80
Theorem sequence:
vertical_angle(1,CNX,DNZ)
circle_property_chord_perpendicular_bisect_arc(1,XCD,XNZ)
arc_addition_length(1,XCZ,XZD)
similar_arc_judgment_cocircular(1,XCZ,XCD)
similar_arc_property_length_ratio(1,XCZ,XCD)
similar_arc_property_measure_ratio(1,XCZ,XCD)
congruent_arc_judgment_chord_equal(1,XCD,XBA)
congruent_arc_property_measure_equal(1,XCD,XBA)
Problem: As shown in the diagram, AB=5, BD=x, CA=15/2, EC=9/2. Find the value of x.
问题：如图所示，AB=5，BD=x，CA=15/2，EC=9/2。求x的值。
Construction (CDL):
Shape(XDB,BD)
Shape(BA,AC,XBC)
Shape(DB,XBC,CE,XED)
Shape(XCE,EC)
Collinear(ABXD)
Collinear(ACE)
Cocircular(X,BCED)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(BD),x)
Equal(LengthOfLine(CA),15/2)
Equal(LengthOfLine(EC),9/2)
Goal: Value(x)
Answer: 13
Theorem sequence:
circle_property_circular_power_segment_and_segment_line(1,ABD,ACE,X)
line_addition(1,AB,BD)
line_addition(1,AC,CE)
Problem: As shown in the diagram, AJ=2*x+3, CJ=8*y-36, JB=5*x, JD=4*y, quadrilateral ACBD is a parallelogram. Find the value of y.
问题：如图所示，AJ=2*x+3，CJ=8*y-36，JB=5*x，JD=4*y，AD和CB是▱ACBD的一组对边。求y的值。
Construction (CDL):
Shape(AC,CJ,JA)
Shape(JC,CB,BJ)
Shape(AJ,JD,DA)
Shape(JB,BD,DJ)
Collinear(AJB)
Collinear(CJD)
Conditions (CDL):
Equal(LengthOfLine(AJ),2*x+3)
Equal(LengthOfLine(CJ),8*y-36)
Equal(LengthOfLine(JB),5*x)
Equal(LengthOfLine(JD),4*y)
Parallelogram(ACBD)
Goal: Value(y)
Answer: 9
Theorem sequence:
parallelogram_property_diagonal_bisection(1,CBDA,J)
Problem: As shown in the diagram, IB=1/2*x-7, JI=1/4*x+5, LD=66-2/3*y, NL=1/3*y-6, NL=LD, CJ⊥NJ, IB is perpendicular to DB, JI is perpendicular to LI. Find the value of x.
问题：如图所示，IB=1/2*x-7，JI=1/4*x+5，LD=66-2/3*y，NL=1/3*y-6，NL=LD，CJ⊥NJ，IB⊥DB，JI垂直于LI。求x的值。
Construction (CDL):
Shape(JI,IL,LN,NJ)
Shape(IB,BD,DL,LI)
Shape(CJ,JN)
Collinear(CJIB)
Collinear(NLD)
Conditions (CDL):
Equal(LengthOfLine(IB),1/2*x-7)
Equal(LengthOfLine(JI),1/4*x+5)
Equal(LengthOfLine(LD),66-2/3*y)
Equal(LengthOfLine(NL),1/3*y-6)
Equal(LengthOfLine(NL),LengthOfLine(LD))
PerpendicularBetweenLine(CJ,NJ)
PerpendicularBetweenLine(IB,DB)
PerpendicularBetweenLine(JI,LI)
Goal: Value(x)
Answer: 48
Theorem sequence:
parallel_judgment_corresponding_angle(1,JN,BD,C)
trapezoid_judgment_parallel(1,JBDN)
parallel_judgment_corresponding_angle(1,IL,BD,J)
midsegment_of_quadrilateral_judgment_parallel(3,IL,JBDN)
Problem: As shown in the diagram, AB=10, AC=y, BC=x, ∠ABC=60°, BC is perpendicular to AC. Find the value of y.
问题：如图所示，AB=10，AC=y，BC=x，∠ABC=60°，BC垂直于AC。求y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(AC),y)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(ABC),60)
PerpendicularBetweenLine(BC,AC)
Goal: Value(y)
Answer: 5*sqrt(3)
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, AB=8, AD=27, AE=12, CD is parallel to BE. Find the length of line BC.
问题：如图所示，AB=8，AD=27，AE=12，CD平行于BE。求直线BC的长度。
Construction (CDL):
Shape(CB,BE,ED,DC)
Shape(BA,AE,EB)
Collinear(ABC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(AD),27)
Equal(LengthOfLine(AE),12)
ParallelBetweenLine(CD,BE)
Goal: Value(LengthOfLine(BC))
Answer: 10
Theorem sequence:
parallel_property_corresponding_angle(1,EB,DC,A)
similar_triangle_judgment_aa(1,BAE,CAD)
similar_triangle_property_line_ratio(1,BAE,CAD)
similar_triangle_property_line_ratio(1,EBA,DCA)
line_addition(1,AB,BC)
Problem: As shown in the diagram, ∠NMQ=10*x°, ∠PNM=20*x°, ∠PNM=∠MQP, ∠QPN=∠NMQ, MQPN is a parallelogram. Find the measure of ∠MQP.
问题：如图所示，∠NMQ=10*x°，∠PNM=20*x°，∠PNM=∠MQP，∠QPN=∠NMQ，四边形MQPN是平行四边形。求∠MQP的大小。
Construction (CDL):
Shape(MQ,QP,PN,NM)
Conditions (CDL):
Equal(MeasureOfAngle(NMQ),10*x)
Equal(MeasureOfAngle(PNM),20*x)
Equal(MeasureOfAngle(PNM),MeasureOfAngle(MQP))
Equal(MeasureOfAngle(QPN),MeasureOfAngle(NMQ))
Parallelogram(MQPN)
Goal: Value(MeasureOfAngle(MQP))
Answer: 120
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,NP,MQ)
Problem: As shown in the diagram, AB=4*x-17, CD=2*x-1, ∠BCD=4*y-19°, ∠CBA=3*y+3°, quadrilateral ACDB is a ▱. Find the value of y.
问题：如图所示，AB=4*x-17，CD=2*x-1，∠BCD=4*y-19°，∠CBA=3*y+3°，AB和CD是▱ACDB的一组对边。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Shape(BC,CD,DB)
Conditions (CDL):
Equal(LengthOfLine(AB),4*x-17)
Equal(LengthOfLine(CD),2*x-1)
Equal(MeasureOfAngle(BCD),4*y-19)
Equal(MeasureOfAngle(CBA),3*y+3)
Parallelogram(ACDB)
Goal: Value(y)
Answer: 22
Theorem sequence:
parallel_property_alternate_interior_angle(2,AB,CD)
Problem: As shown in the diagram, AE=BE, BA=8, BC=8, BE=CE, DA=10, DC=10, ∠ADE=x°, DE⊥AE. Find sin(x).
问题：如图所示，AE=BE，BA=8，BC=8，BE=CE，DA=10，DC=10，∠ADE=x°，DE垂直于AE。求sin(x)。
Construction (CDL):
Shape(AD,DE,EA)
Shape(AE,EB,BA)
Shape(ED,DC,CE)
Shape(BE,EC,CB)
Collinear(DEB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AE),LengthOfLine(BE))
Equal(LengthOfLine(BA),8)
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(BE),LengthOfLine(CE))
Equal(LengthOfLine(DA),10)
Equal(LengthOfLine(DC),10)
Equal(MeasureOfAngle(ADE),x)
PerpendicularBetweenLine(DE,AE)
Goal: Value(Sin(x))
Answer: 2*sqrt(2)/5
Theorem sequence:
adjacent_complementary_angle(1,DEA,AEB)
right_triangle_judgment_angle(1,AEB)
right_triangle_property_pythagorean(1,AEB)
sine_theorem(1,ADE)
Problem: As shown in the diagram, ∠ABX=5*y-6°, ∠BXC=2*x+24°, ∠CAB=3*x-17°, ∠XCA=y+58°, BA and XC are opposite sides of the ▱ ABXC. Find the value of y.
问题：如图所示，∠ABX=5*y-6°，∠BXC=2*x+24°，∠CAB=3*x-17°，∠XCA=y+58°，BA和XC是▱ABXC的一组对边。求y的值。
Construction (CDL):
Shape(AB,BX,XC,CA)
Conditions (CDL):
Equal(MeasureOfAngle(ABX),5*y-6)
Equal(MeasureOfAngle(BXC),2*x+24)
Equal(MeasureOfAngle(CAB),3*x-17)
Equal(MeasureOfAngle(XCA),y+58)
Parallelogram(ABXC)
Goal: Value(y)
Answer: 16
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BXCA)
Problem: As shown in the diagram, ∠NJK=101°, JA is parallel to NF. Find the measure of ∠HNJ.
问题：如图所示，∠NJK=101°，JA∥NF。求∠HNJ的大小。
Construction (CDL):
Shape(KJ,JE)
Shape(EJ,JA)
Shape(AJ,JN)
Shape(NJ,JK)
Shape(HN,NJ)
Shape(JN,NF)
Shape(FN,NI)
Shape(IN,NH)
Collinear(KJA)
Collinear(HNF)
Collinear(EJNI)
Conditions (CDL):
Equal(MeasureOfAngle(NJK),101)
ParallelBetweenLine(JA,NF)
Goal: Value(MeasureOfAngle(HNJ))
Answer: 79
Theorem sequence:
parallel_property_collinear_extend(1,JA,NF,K)
parallel_property_collinear_extend(2,FN,JK,H)
parallel_property_ipsilateral_internal_angle(1,NH,JK)
Problem: As shown in the diagram, SR=3*x-5, TR=2*x+7, TS=22, RS and RT are the legs of the isosceles △ RST. Find the length of line RS.
问题：如图所示，SR=3*x-5，TR=2*x+7，TS=22，∠RST和∠STR是等腰△RST的底角。求直线RS的长度。
Construction (CDL):
Shape(RS,ST,TR)
Conditions (CDL):
Equal(LengthOfLine(SR),3*x-5)
Equal(LengthOfLine(TR),2*x+7)
Equal(LengthOfLine(TS),22)
IsoscelesTriangle(RST)
Goal: Value(LengthOfLine(RS))
Answer: 31
Theorem sequence:
Problem: As shown in the diagram, YX=24, YZ=28, ∠XWZ=105°, quadrilateral WZYX is a ▱. Find the measure of ∠WZY.
问题：如图所示，YX=24，YZ=28，∠XWZ=105°，WX和ZY是平行四边形WZYX的一组对边。求∠WZY的大小。
Construction (CDL):
Shape(WZ,ZY,YX,XW)
Conditions (CDL):
Equal(LengthOfLine(YX),24)
Equal(LengthOfLine(YZ),28)
Equal(MeasureOfAngle(XWZ),105)
Parallelogram(WZYX)
Goal: Value(MeasureOfAngle(WZY))
Answer: 75
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,WX,ZY)
Problem: As shown in the diagram, AB=x, AD=3*sqrt(3), BD=9, CD=y, ∠ABC=30°, ∠BCA=60°, AD is perpendicular to CD, CA is perpendicular to BA. Find the value of x.
问题：如图所示，AB=x，AD=3*sqrt(3)，BD=9，CD=y，∠ABC=30°，∠BCA=60°，AD⊥CD，CA垂直于BA。求x的值。
Construction (CDL):
Shape(CA,AD,DC)
Shape(DA,AB,BD)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AD),3*sqrt(3))
Equal(LengthOfLine(BD),9)
Equal(LengthOfLine(CD),y)
Equal(MeasureOfAngle(ABC),30)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(AD,CD)
PerpendicularBetweenLine(CA,BA)
Goal: Value(x)
Answer: 6*sqrt(3)
Theorem sequence:
adjacent_complementary_angle(1,BDA,ADC)
triangle_property_angle_sum(1,BDA)
sine_theorem(1,BDA)
Problem: As shown in the diagram, AB=y, AC=x, BC=18, ∠ABC=30°, CA is perpendicular to BA. Find the value of y.
问题：如图所示，AB=y，AC=x，BC=18，∠ABC=30°，CA⊥BA。求y的值。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),18)
Equal(MeasureOfAngle(ABC),30)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 9*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,ABC)
sine_theorem(1,BCA)
Problem: As shown in the diagram, the diameter of ⊙H is 18, LM=12, the measure of ⌒HML is 84, H is the center of circle H, MP⊥HP. Find the measure of ⌒HKL.
问题：如图所示，圆H的直径为18，LM=12，⌒HML的角度为84，⊙H的圆心为H，MP垂直于HP。求⌒HKL的角度。
Construction (CDL):
Shape(HLJ,JH,HP,PL)
Shape(HJM,MP,PH,HJ)
Shape(HMK,KP,PM)
Shape(HKL,LP,PK)
Collinear(JHPK)
Collinear(LPM)
Cocircular(H,JMKL)
Conditions (CDL):
Equal(DiameterOfCircle(H),18)
Equal(LengthOfLine(LM),12)
Equal(MeasureOfArc(HML),84)
IsCentreOfCircle(H,H)
PerpendicularBetweenLine(MP,HP)
Goal: Value(MeasureOfArc(HKL))
Answer: 42
Theorem sequence:
circle_property_chord_perpendicular_bisect_arc(1,HML,HPK)
congruent_arc_judgment_length_equal(1,HMK,HKL)
congruent_arc_property_measure_equal(1,HMK,HKL)
arc_addition_measure(1,HMK,HKL)
Problem: As shown in the diagram, ∠CHE=9*x-11°, ∠GDF=8*x+4°, BD is parallel to EH. Find the value of x.
问题：如图所示，∠CHE=9*x-11°，∠GDF=8*x+4°，BD∥EH。求x的值。
Construction (CDL):
Shape(GD,DF)
Shape(FD,DH)
Shape(DH,HA)
Shape(AH,HC)
Shape(CH,HE)
Shape(EH,HD)
Shape(HD,DB)
Shape(BD,DG)
Collinear(GDHC)
Collinear(FDB)
Collinear(AHE)
Conditions (CDL):
Equal(MeasureOfAngle(CHE),9*x-11)
Equal(MeasureOfAngle(GDF),8*x+4)
ParallelBetweenLine(BD,EH)
Goal: Value(x)
Answer: 15
Theorem sequence:
vertical_angle(1,GDF,HDB)
parallel_property_corresponding_angle(1,HE,DB,C)
Problem: As shown in the diagram, RS=5, TR=6, TS=3, ∠SRT=x°. Find the value of x.
问题：如图所示，RS=5，TR=6，TS=3，∠SRT=x°。求x的值。
Construction (CDL):
Shape(SR,RT,TS)
Conditions (CDL):
Equal(LengthOfLine(RS),5)
Equal(LengthOfLine(TR),6)
Equal(LengthOfLine(TS),3)
Equal(MeasureOfAngle(SRT),x)
Goal: Value(x)
Answer: 180*acos(13/15)/pi
Theorem sequence:
cosine_theorem(1,RTS)
Problem: As shown in the diagram, the area of BLAN is 72, the area of quadrilateral CDEF is 50, BN=6, CF=x, CDEF is similar to BLAN. Find the value of x.
问题：如图所示，四边形BLAN的面积为72，CDEF的面积为50，BN=6，CF=x，CDEF相似于BLAN。求x的值。
Construction (CDL):
Shape(CD,DE,EF,FC)
Shape(BL,LA,AN,NB)
Conditions (CDL):
Equal(AreaOfQuadrilateral(BLAN),72)
Equal(AreaOfQuadrilateral(CDEF),50)
Equal(LengthOfLine(BN),6)
Equal(LengthOfLine(CF),x)
SimilarBetweenQuadrilateral(CDEF,BLAN)
Goal: Value(x)
Answer: 5
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,CDEF,BLAN)
similar_quadrilateral_property_line_ratio(1,FCDE,NBLA)
Problem: As shown in the diagram, ∠ACD=x°, ∠ECA=2*x°, ∠GCE=x°. Find the value of x.
问题：如图所示，∠ACD=x°，∠ECA=2*x°，∠GCE=x°。求x的值。
Construction (CDL):
Shape(GC,CE)
Shape(EC,CA)
Shape(GC,CA)
Shape(EC,CD)
Shape(AC,CD)
Collinear(GCD)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),x)
Equal(MeasureOfAngle(ECA),2*x)
Equal(MeasureOfAngle(GCE),x)
Goal: Value(x)
Answer: 45
Theorem sequence:
angle_addition(1,GCE,ECA)
adjacent_complementary_angle(1,GCA,ACD)
Problem: As shown in the diagram, ∠DFH=4*x°, ∠HFA=2*x-6°. Find the measure of ∠DFH.
问题：如图所示，∠DFH=4*x°，∠HFA=2*x-6°。求∠DFH的大小。
Construction (CDL):
Shape(DF,FH)
Shape(HF,FA)
Shape(AF,FB)
Shape(BF,FD)
Collinear(DFA)
Collinear(BFH)
Conditions (CDL):
Equal(MeasureOfAngle(DFH),4*x)
Equal(MeasureOfAngle(HFA),2*x-6)
Goal: Value(MeasureOfAngle(DFH))
Answer: 124
Theorem sequence:
adjacent_complementary_angle(1,DFH,HFA)
Problem: As shown in the diagram, AB=2*x+3, BC=5*x, ADCB is a rhombus. Find the value of x.
问题：如图所示，AB=2*x+3，BC=5*x，ADCB是菱形。求x的值。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DC,CE)
Shape(EC,CB,BE)
Shape(BA,AE,EB)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(AB),2*x+3)
Equal(LengthOfLine(BC),5*x)
Rhombus(ADCB)
Goal: Value(x)
Answer: 1
Theorem sequence:
Problem: As shown in the diagram, OA=3, ∠AOB=45°, the center of ⊙O is O. Find the length of ⌒OBA.
问题：如图所示，OA=3，∠AOB=45°，O是圆O的圆心。求弧OBA的长度。
Construction (CDL):
Shape(OA,OAB,BO)
Shape(AO,OB,OBA)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfLine(OA),3)
Equal(MeasureOfAngle(AOB),45)
IsCentreOfCircle(O,O)
Goal: Value(LengthOfArc(OBA))
Answer: 3*pi/4
Theorem sequence:
arc_property_center_angle(1,OBA,O)
radius_of_circle_property_length_equal(1,OA,O)
arc_length_formula(1,OBA)
Problem: As shown in the diagram, QR=2, VS=7, S is the midpoint of segment RT, V bisects segment QU, QU and TR are the non-parallel sides (legs) of trapezoid QUTR. Find the length of line UT.
问题：如图所示，QR=2，VS=7，S是线段RT的中点，V是线段QU的中点，QUTR是梯形。求直线UT的长度。
Construction (CDL):
Shape(QV,VS,SR,RQ)
Shape(VU,UT,TS,SV)
Collinear(QVU)
Collinear(RST)
Conditions (CDL):
Equal(LengthOfLine(QR),2)
Equal(LengthOfLine(VS),7)
IsMidpointOfLine(S,RT)
IsMidpointOfLine(V,QU)
Trapezoid(QUTR)
Goal: Value(LengthOfLine(UT))
Answer: 12
Theorem sequence:
midsegment_of_quadrilateral_judgment_midpoint(1,VS,QUTR)
midsegment_of_quadrilateral_property_length(1,VS,QUTR)
Problem: As shown in the diagram, ∠ABD=130°, ∠CBA=x°, ∠DBC=95°. Find the value of x.
问题：如图所示，∠ABD=130°，∠CBA=x°，∠DBC=95°。求x的值。
Construction (CDL):
Shape(BAC,CB,BA)
Shape(BDA,AB,BD)
Shape(BCD,DB,BC)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),130)
Equal(MeasureOfAngle(CBA),x)
Equal(MeasureOfAngle(DBC),95)
Goal: Value(x)
Answer: 135
Theorem sequence:
angle_addition(1,ABD,DBC)
round_angle(1,CBA,ABC)
Problem: As shown in the diagram, ∠DMN=56°, ∠GLI=3*y-11°, ∠HNK=4*x°, LI is parallel to MD, MD is parallel to NS. Find the value of x.
问题：如图所示，∠DMN=56°，∠GLI=3*y-11°，∠HNK=4*x°，LI平行于MD，MD∥NS。求x的值。
Construction (CDL):
Shape(HN,NK)
Shape(KN,NM)
Shape(NM,ME)
Shape(EM,ML)
Shape(ML,LC)
Shape(CL,LG)
Shape(GL,LI)
Shape(IL,LM)
Shape(LM,MD)
Shape(DM,MN)
Shape(MN,NS)
Shape(SN,NH)
Collinear(HNMLG)
Collinear(SNK)
Collinear(DME)
Collinear(ILC)
Conditions (CDL):
Equal(MeasureOfAngle(DMN),56)
Equal(MeasureOfAngle(GLI),3*y-11)
Equal(MeasureOfAngle(HNK),4*x)
ParallelBetweenLine(LI,MD)
ParallelBetweenLine(MD,NS)
Goal: Value(x)
Answer: 31
Theorem sequence:
vertical_angle(1,HNK,MNS)
parallel_property_ipsilateral_internal_angle(1,MD,NS)
Problem: As shown in the diagram, AB=y, AC=8, BC=x, ∠BAC=60°, AC is perpendicular to BC. Find the value of x.
问题：如图所示，AB=y，AC=8，BC=x，∠BAC=60°，AC垂直于BC。求x的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AC),8)
Equal(LengthOfLine(BC),x)
Equal(MeasureOfAngle(BAC),60)
PerpendicularBetweenLine(AC,BC)
Goal: Value(x)
Answer: 8*sqrt(3)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,ACB)
sine_theorem(1,BAC)
Problem: As shown in the diagram, EJ=6, LK=7, ML=4, quadrilateral JMLK is a parallelogram, JE is perpendicular to LE. Find the perimeter of quadrilateral JMLK.
问题：如图所示，EJ=6，LK=7，ML=4，JMLK是平行四边形，JE⊥LE。求JMLK的周长。
Construction (CDL):
Shape(JM,ME,EJ)
Shape(JE,EL,LK,KJ)
Collinear(MEL)
Conditions (CDL):
Equal(LengthOfLine(EJ),6)
Equal(LengthOfLine(LK),7)
Equal(LengthOfLine(ML),4)
Parallelogram(JMLK)
PerpendicularBetweenLine(JE,LE)
Goal: Value(PerimeterOfQuadrilateral(JMLK))
Answer: 22
Theorem sequence:
parallelogram_property_opposite_line_equal(1,JMLK)
parallelogram_property_opposite_line_equal(1,MLKJ)
quadrilateral_perimeter_formula(1,JMLK)
Problem: As shown in the diagram, AB=8, AC=14, BC=8, BX=x, BX⊥CX. Find the value of x.
问题：如图所示，AB=8，AC=14，BC=8，BX=x，BX⊥CX。求x的值。
Construction (CDL):
Shape(BA,AX,XB)
Shape(BX,XC,CB)
Collinear(AXC)
Conditions (CDL):
Equal(LengthOfLine(AB),8)
Equal(LengthOfLine(AC),14)
Equal(LengthOfLine(BC),8)
Equal(LengthOfLine(BX),x)
PerpendicularBetweenLine(BX,CX)
Goal: Value(x)
Answer: sqrt(15)
Theorem sequence:
adjacent_complementary_angle(1,AXB,BXC)
line_addition(1,AX,XC)
right_triangle_judgment_angle(1,AXB)
right_triangle_judgment_angle(1,BXC)
right_triangle_property_pythagorean(1,AXB)
right_triangle_property_pythagorean(1,BXC)
Problem: As shown in the diagram, NM=4, ∠NLM=∠PLN, ∠PLN=25°, LM⊥NM, NP⊥LP. Find the measure of ∠MNP.
问题：如图所示，NM=4，∠NLM=∠PLN，∠PLN=25°，LM⊥NM，NP⊥LP。求∠MNP的大小。
Construction (CDL):
Shape(LM,MN,NL)
Shape(LN,NP,PL)
Conditions (CDL):
Equal(LengthOfLine(NM),4)
Equal(MeasureOfAngle(NLM),MeasureOfAngle(PLN))
Equal(MeasureOfAngle(PLN),25)
PerpendicularBetweenLine(LM,NM)
PerpendicularBetweenLine(NP,LP)
Goal: Value(MeasureOfAngle(MNP))
Answer: 130
Theorem sequence:
triangle_property_angle_sum(1,LMN)
triangle_property_angle_sum(1,LNP)
angle_addition(1,MNL,LNP)