SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, JK=8, JS=10, ∠KPL=28°, ∠SLK=24.5°, S is the center of the inscribed circle of triangle JPL, PM⊥SM, SK⊥JK, SQ⊥PQ. Find the length of line SQ.
问题：如图所示，JK=8，JS=10，∠KPL=28°，∠SLK=24.5°，△JPL的内切圆圆心为S，PM垂直于SM，SK垂直于JK，SQ垂直于PQ。求直线SQ的长度。
Construction (CDL):
Shape(JQ,QS,SJ)
Shape(JS,SK,KJ)
Shape(SQ,QP,PS)
Shape(SP,PM,MS)
Shape(SM,ML,LS)
Shape(KS,SL,LK)
Collinear(JQP)
Collinear(JKL)
Collinear(QSL)
Collinear(KSP)
Collinear(PML)
Collinear(JSM)
Conditions (CDL):
Equal(LengthOfLine(JK),8)
Equal(LengthOfLine(JS),10)
Equal(MeasureOfAngle(KPL),28)
Equal(MeasureOfAngle(SLK),24.5)
IsIncenterOfTriangle(S,JPL)
PerpendicularBetweenLine(PM,SM)
PerpendicularBetweenLine(SK,JK)
PerpendicularBetweenLine(SQ,PQ)
Goal: Value(LengthOfLine(SQ))
Answer: 6
Theorem sequence:
adjacent_complementary_angle(1,JQS,SQP)
mirror_congruent_triangle_judgment_aas(1,SJQ,SKJ)
mirror_congruent_triangle_property_line_equal(1,SJQ,SKJ)
right_triangle_judgment_angle(1,JQS)
right_triangle_property_pythagorean(1,JQS)
Problem: As shown in the diagram, ∠TSR=42°, the center of circle S is S. Find the measure of arc SKT.
问题：如图所示，∠TSR=42°，S是圆S的圆心。求⌒SKT的角度。
Construction (CDL):
Shape(TS,SR,SRT)
Shape(KS,ST,STK)
Shape(ES,SK,SKE)
Shape(RS,SE,SER)
Collinear(TSE)
Collinear(RSK)
Cocircular(S,TKER)
Conditions (CDL):
Equal(MeasureOfAngle(TSR),42)
IsCentreOfCircle(S,S)
Goal: Value(MeasureOfArc(SKT))
Answer: 222
Theorem sequence:
vertical_angle(1,TSR,ESK)
flat_angle(1,TSE)
angle_addition(1,TSE,ESK)
arc_property_center_angle(1,SKT,S)
Problem: As shown in the diagram, AD=m, BD=6, CD=m+5, AB is perpendicular to CB, CD is perpendicular to BD. Find the value of m.
问题：如图所示，AD=m，BD=6，CD=m+5，AB⊥CB，CD垂直于BD。求m的值。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(LengthOfLine(AD),m)
Equal(LengthOfLine(BD),6)
Equal(LengthOfLine(CD),m+5)
PerpendicularBetweenLine(AB,CB)
PerpendicularBetweenLine(CD,BD)
Goal: Value(m)
Answer: 4
Theorem sequence:
adjacent_complementary_angle(1,CDB,BDA)
line_addition(1,AD,DC)
right_triangle_judgment_angle(1,BDA)
right_triangle_judgment_angle(1,CDB)
right_triangle_judgment_angle(1,ABC)
right_triangle_property_pythagorean(1,BDA)
right_triangle_property_pythagorean(1,CDB)
right_triangle_property_pythagorean(1,ABC)
Problem: As shown in the diagram, ∠BCA=71°, ∠BDA=92°. Find the measure of ∠DBC.
问题：如图所示，∠BCA=71°，∠BDA=92°。求∠DBC的大小。
Construction (CDL):
Shape(BD,DA)
Shape(DB,BC,CD)
Collinear(ADC)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),71)
Equal(MeasureOfAngle(BDA),92)
Goal: Value(MeasureOfAngle(DBC))
Answer: 21
Theorem sequence:
flat_angle(1,CDA)
angle_addition(1,CDB,BDA)
triangle_property_angle_sum(1,DBC)
Problem: As shown in the diagram, ∠FJE=75°, DG is parallel to EC, DG∥MI. Find the measure of ∠TJC.
问题：如图所示，∠FJE=75°，DG平行于EC，DG平行于MI。求∠TJC的大小。
Construction (CDL):
Shape(LB,BD)
Shape(DB,BF)
Shape(FB,BG)
Shape(GB,BL)
Shape(BF,FM)
Shape(MF,FJ)
Shape(JF,FI)
Shape(IF,FB)
Shape(FJ,JE)
Shape(EJ,JT)
Shape(TJ,JC)
Shape(CJ,JF)
Collinear(DBG)
Collinear(MFI)
Collinear(EJC)
Collinear(LBFJT)
Conditions (CDL):
Equal(MeasureOfAngle(FJE),75)
ParallelBetweenLine(DG,EC)
ParallelBetweenLine(DG,MI)
Goal: Value(MeasureOfAngle(TJC))
Answer: 75
Theorem sequence:
vertical_angle(1,LJE,TJC)
Problem: As shown in the diagram, AB=15, AC=x, BC=y, ∠ACB=60°, BA⊥CA. Find the value of y.
问题：如图所示，AB=15，AC=x，BC=y，∠ACB=60°，BA⊥CA。求y的值。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(AC),x)
Equal(LengthOfLine(BC),y)
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, ∠ACD=62°, ∠ADS=125°, ∠CBD=32°. Find the measure of ∠DAC.
问题：如图所示，∠ACD=62°，∠ADS=125°，∠CBD=32°。求∠DAC的大小。
Construction (CDL):
Shape(AC,CD,DA)
Shape(DC,CB,BD)
Shape(AD,DS)
Shape(SD,DB)
Collinear(CDS)
Collinear(ADB)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),62)
Equal(MeasureOfAngle(ADS),125)
Equal(MeasureOfAngle(CBD),32)
Goal: Value(MeasureOfAngle(DAC))
Answer: 63
Theorem sequence:
adjacent_complementary_angle(1,CDA,ADS)
triangle_property_angle_sum(1,ACD)
Problem: As shown in the diagram, AB=7, CD=7/2, AD⊥CD. Find the area of triangle CAB.
问题：如图所示，AB=7，CD=7/2，AD垂直于CD。求三角形CAB的面积。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AB),7)
Equal(LengthOfLine(CD),7/2)
PerpendicularBetweenLine(AD,CD)
Goal: Value(AreaOfTriangle(CAB))
Answer: 49/4
Theorem sequence:
altitude_of_triangle_judgment(1,CD,CAB)
triangle_area_formula_common(1,CAB)
Problem: As shown in the diagram, AB=x, AC=12, ∠ACB=39°, ∠BAC=118°. Find the value of x.
问题：如图所示，AB=x，AC=12，∠ACB=39°，∠BAC=118°。求x的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AB),x)
Equal(LengthOfLine(AC),12)
Equal(MeasureOfAngle(ACB),39)
Equal(MeasureOfAngle(BAC),118)
Goal: Value(x)
Answer: (-3*sqrt(15)*sqrt(sqrt(5)+5)-3*sqrt(sqrt(5)+5)+3*sqrt(2)+3*sqrt(6)+3*sqrt(10)+3*sqrt(3)*sqrt(sqrt(5)+5)+3*sqrt(30)+3*sqrt(5)*sqrt(sqrt(5)+5))/(4*sin(23*pi/180))
Theorem sequence:
triangle_property_angle_sum(1,BAC)
sine_theorem(1,ACB)
Problem: As shown in the diagram, JK=8, JS=10, ∠KPL=28°, ∠SLK=24.5°, the center of the inscribed circle of triangle JPL is S, PM⊥SM, SK is perpendicular to JK, SQ⊥PQ. Find the length of line QJ.
问题：如图所示，JK=8，JS=10，∠KPL=28°，∠SLK=24.5°，三角形JPL的内心为S，PM垂直于SM，SK⊥JK，SQ垂直于PQ。求直线QJ的长度。
Construction (CDL):
Shape(JQ,QS,SJ)
Shape(JS,SK,KJ)
Shape(SQ,QP,PS)
Shape(SP,PM,MS)
Shape(SM,ML,LS)
Shape(KS,SL,LK)
Collinear(JQP)
Collinear(JKL)
Collinear(QSL)
Collinear(KSP)
Collinear(PML)
Collinear(JSM)
Conditions (CDL):
Equal(LengthOfLine(JK),8)
Equal(LengthOfLine(JS),10)
Equal(MeasureOfAngle(KPL),28)
Equal(MeasureOfAngle(SLK),24.5)
IsIncenterOfTriangle(S,JPL)
PerpendicularBetweenLine(PM,SM)
PerpendicularBetweenLine(SK,JK)
PerpendicularBetweenLine(SQ,PQ)
Goal: Value(LengthOfLine(QJ))
Answer: 8
Theorem sequence:
adjacent_complementary_angle(1,JQS,SQP)
mirror_congruent_triangle_judgment_aas(1,SJQ,SKJ)
mirror_congruent_triangle_property_line_equal(1,SJQ,SKJ)
Problem: As shown in the diagram, ∠ABX=6*y-57°, ∠BXY=3*x+9°, ∠XYA=3*x+36°, ∠YAB=5*x-19°, AY and BX are opposite sides of the parallelogram ABXY. Find the value of x.
问题：如图所示，∠ABX=6*y-57°，∠BXY=3*x+9°，∠XYA=3*x+36°，∠YAB=5*x-19°，BA和XY是▱ABXY的一组对边。求x的值。
Construction (CDL):
Shape(AB,BX,XY,YA)
Conditions (CDL):
Equal(MeasureOfAngle(ABX),6*y-57)
Equal(MeasureOfAngle(BXY),3*x+9)
Equal(MeasureOfAngle(XYA),3*x+36)
Equal(MeasureOfAngle(YAB),5*x-19)
Parallelogram(ABXY)
Goal: Value(x)
Answer: 14
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,ABXY)
Problem: As shown in the diagram, ∠ACD=50°, ∠CDA=70°, ∠DBA=21°. Find the measure of ∠DAC.
问题：如图所示，∠ACD=50°，∠CDA=70°，∠DBA=21°。求∠DAC的大小。
Construction (CDL):
Shape(AC,CD,DA)
Shape(AD,DB,BA)
Collinear(CDB)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),50)
Equal(MeasureOfAngle(CDA),70)
Equal(MeasureOfAngle(DBA),21)
Goal: Value(MeasureOfAngle(DAC))
Answer: 60
Theorem sequence:
triangle_property_angle_sum(1,DAC)
Problem: As shown in the diagram, ∠GKM=62°, ∠HBF=38°, NO is parallel to MH. Find the measure of ∠BCK.
问题：如图所示，∠GKM=62°，∠HBF=38°，NO平行于MH。求∠BCK的大小。
Construction (CDL):
Shape(NC,CJ)
Shape(JC,CI)
Shape(IC,CO)
Shape(OC,CB)
Shape(BC,CK,KB)
Shape(KC,CN)
Shape(MK,KC)
Shape(BK,KG)
Shape(GK,KM)
Shape(CB,BH)
Shape(HB,BF)
Shape(FB,BK)
Collinear(NCO)
Collinear(MKBH)
Collinear(JCBF)
Collinear(ICKG)
Conditions (CDL):
Equal(MeasureOfAngle(GKM),62)
Equal(MeasureOfAngle(HBF),38)
ParallelBetweenLine(NO,MH)
Goal: Value(MeasureOfAngle(BCK))
Answer: 80
Theorem sequence:
vertical_angle(1,MBJ,HBF)
vertical_angle(1,IKB,GKM)
triangle_property_angle_sum(1,CKB)
Problem: As shown in the diagram, AC=b, BC=3, ∠BAC=37°, ∠CBA=68°. Find the value of b.
问题：如图所示，AC=b，BC=3，∠BAC=37°，∠CBA=68°。求b的值。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(AC),b)
Equal(LengthOfLine(BC),3)
Equal(MeasureOfAngle(BAC),37)
Equal(MeasureOfAngle(CBA),68)
Goal: Value(b)
Answer: 3*sin(17*pi/45)/sin(37*pi/180)
Theorem sequence:
sine_theorem(1,CBA)
Problem: As shown in the diagram, AB=19, DC=11, EC=11, EC is perpendicular to DC, BC and DA are the non-parallel sides (legs) of trapezoid BCDA. Find the area of quadrilateral BCDA.
问题：如图所示，AB=19，DC=11，EC=11，EC⊥DC，BCDA是梯形。求BCDA的面积。
Construction (CDL):
Shape(EB,BC,CE)
Shape(AE,EC,CD,DA)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AB),19)
Equal(LengthOfLine(DC),11)
Equal(LengthOfLine(EC),11)
PerpendicularBetweenLine(EC,DC)
Trapezoid(BCDA)
Goal: Value(AreaOfQuadrilateral(BCDA))
Answer: 165
Theorem sequence:
parallel_property_collinear_extend(3,BA,CD,E)
parallel_property_ipsilateral_internal_angle(1,EA,CD)
altitude_of_quadrilateral_judgment_right_vertex(2,CE,DABC)
trapezoid_area_formula(1,DABC)
Problem: As shown in the diagram, PQ=3*x-4, RS=14, TH=8, UH=8, the center of circle H is H, HU is perpendicular to RU, QT is perpendicular to HT. Find the value of x.
问题：如图所示，PQ=3*x-4，RS=14，TH=8，UH=8，圆H的圆心为H，HU垂直于RU，QT垂直于HT。求x的值。
Construction (CDL):
Shape(PT,TQ,HQP)
Shape(QT,TH,HQ)
Shape(RH,HU,UR)
Shape(RU,US,HSR)
Shape(QH,HR,HRQ)
Shape(SU,UH,HT,TP,HPS)
Collinear(QTP)
Collinear(RUS)
Cocircular(H,QPRS)
Conditions (CDL):
Equal(LengthOfLine(PQ),3*x-4)
Equal(LengthOfLine(RS),14)
Equal(LengthOfLine(TH),8)
Equal(LengthOfLine(UH),8)
IsCentreOfCircle(H,H)
PerpendicularBetweenLine(HU,RU)
PerpendicularBetweenLine(QT,HT)
Goal: Value(x)
Answer: 6
Theorem sequence:
adjacent_complementary_angle(1,SUH,HUR)
radius_of_circle_property_length_equal(1,HQ,H)
radius_of_circle_property_length_equal(1,HR,H)
right_triangle_judgment_angle(1,QTH)
right_triangle_judgment_angle(1,HUR)
circle_property_chord_perpendicular_bisect_chord(1,H,HT,QP)
circle_property_chord_perpendicular_bisect_chord(1,H,HU,SR)
line_addition(1,QT,TP)
line_addition(1,RU,US)
right_triangle_property_pythagorean(1,QTH)
right_triangle_property_pythagorean(1,HUR)
Problem: As shown in the diagram, ∠GCH=2*x°, ∠HCD=6*x+28°, C is the center of circle C, FC is perpendicular to GC. Find the measure of ⌒CHF.
问题：如图所示，∠GCH=2*x°，∠HCD=6*x+28°，⊙C的圆心为C，FC垂直于GC。求弧CHF的角度。
Construction (CDL):
Shape(HGC,CH,CHG)
Shape(FC,CG,CGF)
Shape(EC,CF,CFE)
Shape(DC,CE,CED)
Shape(HC,CD,CDH)
Collinear(GCD)
Collinear(HCE)
Cocircular(C,HGFED)
Conditions (CDL):
Equal(MeasureOfAngle(GCH),2*x)
Equal(MeasureOfAngle(HCD),6*x+28)
IsCentreOfCircle(C,C)
PerpendicularBetweenLine(FC,GC)
Goal: Value(MeasureOfArc(CHF))
Answer: 128
Theorem sequence:
adjacent_complementary_angle(1,GCH,HCD)
angle_addition(1,FCG,GCH)
arc_property_center_angle(1,CHF,C)
Problem: As shown in the diagram, ∠ABD=3*x-60°, ∠ECF=2*x+15°, BD is parallel to CE. Find the measure of ∠DBC.
问题：如图所示，∠ABD=3*x-60°，∠ECF=2*x+15°，BD∥CE。求∠DBC的大小。
Construction (CDL):
Shape(RB,BA)
Shape(AB,BD)
Shape(DB,BC)
Shape(CB,BR)
Shape(VC,CB)
Shape(BC,CE)
Shape(EC,CF)
Shape(FC,CV)
Collinear(ABCF)
Collinear(RBD)
Collinear(VCE)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),3*x-60)
Equal(MeasureOfAngle(ECF),2*x+15)
ParallelBetweenLine(BD,CE)
Goal: Value(MeasureOfAngle(DBC))
Answer: 105
Theorem sequence:
adjacent_complementary_angle(1,ABD,DBC)
adjacent_complementary_angle(1,FCV,VCA)
adjacent_complementary_angle(1,VCB,BCE)
adjacent_complementary_angle(1,ECF,FCV)
parallel_property_corresponding_angle(1,BD,CE,A)
Problem: As shown in the diagram, ∠ADC=23*x+12°, ∠CBA=21*x-8°. Find the measure of ∠CBA.
问题：如图所示，∠ADC=23*x+12°，∠CBA=21*x-8°。求∠CBA的大小。
Construction (CDL):
Shape(AB,GBA)
Shape(DA,GAD)
Shape(CD,GDC)
Shape(BC,GCB)
Shape(BA,AD,DC,CB)
Cocircular(G,ADCB)
Conditions (CDL):
Equal(MeasureOfAngle(ADC),23*x+12)
Equal(MeasureOfAngle(CBA),21*x-8)
Goal: Value(MeasureOfAngle(CBA))
Answer: 76
Theorem sequence:
arc_property_circumference_angle_external(1,GAC,B)
arc_property_circumference_angle_internal(1,GAC,D)
Problem: As shown in the diagram, RQ=4*x, RS=2*x+1, SQ=6*x-1, △RQS is an equilateral △. Find the length of line QS.
问题：如图所示，RQ=4*x，RS=2*x+1，SQ=6*x-1，△RQS为等边△。求直线QS的长度。
Construction (CDL):
Shape(RQ,QS,SR)
Conditions (CDL):
Equal(LengthOfLine(RQ),4*x)
Equal(LengthOfLine(RS),2*x+1)
Equal(LengthOfLine(SQ),6*x-1)
EquilateralTriangle(RQS)
Goal: Value(LengthOfLine(QS))
Answer: 2
Theorem sequence:
Problem: As shown in the diagram, ∠LWX=53°, ZI is parallel to WG. Find the measure of ∠IYW.
问题：如图所示，∠LWX=53°，ZI平行于WG。求∠IYW的大小。
Construction (CDL):
Shape(GW,WL)
Shape(LW,WX)
Shape(XW,WY)
Shape(YW,WG)
Shape(IY,YW)
Shape(WY,YZ)
Shape(ZY,YM)
Shape(MY,YI)
Collinear(LWYM)
Collinear(GWX)
Collinear(IYZ)
Conditions (CDL):
Equal(MeasureOfAngle(LWX),53)
ParallelBetweenLine(ZI,WG)
Goal: Value(MeasureOfAngle(IYW))
Answer: 127
Theorem sequence:
parallel_property_collinear_extend(3,ZI,WG,Y)
vertical_angle(1,LWX,YWG)
parallel_property_ipsilateral_internal_angle(1,YI,WG)
Problem: As shown in the diagram, AB=12, BC=y, CA=x, ∠BCA=60°, CA⊥BA. Find the value of y.
问题：如图所示，AB=12，BC=y，CA=x，∠BCA=60°，CA垂直于BA。求y的值。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),12)
Equal(LengthOfLine(BC),y)
Equal(LengthOfLine(CA),x)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(CA,BA)
Goal: Value(y)
Answer: 8*sqrt(3)
Theorem sequence:
sine_theorem(1,BCA)
Problem: As shown in the diagram, VZ=a-9, WV=6, WX=7, WY=a, VX is parallel to ZY. Find the length of line WY.
问题：如图所示，VZ=a-9，WV=6，WX=7，WY=a，VX平行于ZY。求直线WY的长度。
Construction (CDL):
Shape(WV,VX,XW)
Shape(VZ,ZY,YX,XV)
Collinear(WVZ)
Collinear(WXY)
Conditions (CDL):
Equal(LengthOfLine(VZ),a-9)
Equal(LengthOfLine(WV),6)
Equal(LengthOfLine(WX),7)
Equal(LengthOfLine(WY),a)
ParallelBetweenLine(VX,ZY)
Goal: Value(LengthOfLine(WY))
Answer: 21
Theorem sequence:
line_addition(1,WV,VZ)
parallel_property_corresponding_angle(1,VX,ZY,W)
parallel_property_corresponding_angle(2,YZ,XV,W)
similar_triangle_judgment_aa(1,WVX,WZY)
similar_triangle_property_line_ratio(1,VXW,ZYW)
similar_triangle_property_line_ratio(1,XWV,YWZ)
Problem: As shown in the diagram, BC=15, BE=6, DA=10, DE=3*x-2, EC=12, ∠ADE=∠CBE. Find the length of line DE.
问题：如图所示，BC=15，BE=6，DA=10，DE=3*x-2，EC=12，∠ADE=∠CBE。求直线DE的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(BE,EC,CB)
Collinear(AEC)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(BC),15)
Equal(LengthOfLine(BE),6)
Equal(LengthOfLine(DA),10)
Equal(LengthOfLine(DE),3*x-2)
Equal(LengthOfLine(EC),12)
Equal(MeasureOfAngle(ADE),MeasureOfAngle(CBE))
Goal: Value(LengthOfLine(DE))
Answer: 4
Theorem sequence:
vertical_angle(1,DEA,BEC)
similar_triangle_judgment_aa(1,ADE,CBE)
similar_triangle_property_line_ratio(1,EAD,ECB)
similar_triangle_property_line_ratio(1,ADE,CBE)
Problem: As shown in the diagram, AB=36, AC=24, BD=CD, ∠CBA=x°, ∠DAC=y°, AC is perpendicular to DC. Find the value of y.
问题：如图所示，AB=36，AC=24，BD=CD，∠CBA=x°，∠DAC=y°，AC⊥DC。求y的值。
Construction (CDL):
Shape(BA,AD,DB)
Shape(DA,AC,CD)
Collinear(BDC)
Conditions (CDL):
Equal(LengthOfLine(AB),36)
Equal(LengthOfLine(AC),24)
Equal(LengthOfLine(BD),LengthOfLine(CD))
Equal(MeasureOfAngle(CBA),x)
Equal(MeasureOfAngle(DAC),y)
PerpendicularBetweenLine(AC,DC)
Goal: Value(y)
Answer: 180*asin(sqrt(105)/21)/pi
Theorem sequence:
right_triangle_judgment_angle(1,ACB)
right_triangle_property_pythagorean(1,ACB)
line_addition(1,BD,DC)
right_triangle_judgment_angle(1,ACD)
right_triangle_property_pythagorean(1,ACD)
sine_theorem(1,DAC)
Problem: As shown in the diagram, ∠ADK=96°, ∠HGJ=42°, KH∥AJ. Find the measure of ∠AGH.
问题：如图所示，∠ADK=96°，∠HGJ=42°，KH∥AJ。求∠AGH的大小。
Construction (CDL):
Shape(KD,DL)
Shape(LD,DH)
Shape(HD,DA)
Shape(AD,DK)
Shape(AG,GH)
Shape(HG,GJ)
Shape(JG,GF)
Shape(FG,GA)
Collinear(KDH)
Collinear(AGJ)
Conditions (CDL):
Equal(MeasureOfAngle(ADK),96)
Equal(MeasureOfAngle(HGJ),42)
ParallelBetweenLine(KH,AJ)
Goal: Value(MeasureOfAngle(AGH))
Answer: 138
Theorem sequence:
adjacent_complementary_angle(1,AGH,HGJ)
Problem: As shown in the diagram, ∠FHE=15*x°, triangle EFG is an equilateral triangle, EH bisects ∠GEF. Find the measure of ∠HEF.
问题：如图所示，∠FHE=15*x°，△EFG为等边△，EH是∠GEF的角平分线。求∠HEF的大小。
Construction (CDL):
Shape(EF,FH,HE)
Shape(EH,HG,GE)
Collinear(FHG)
Conditions (CDL):
Equal(MeasureOfAngle(FHE),15*x)
EquilateralTriangle(EFG)
IsBisectorOfAngle(EH,GEF)
Goal: Value(MeasureOfAngle(HEF))
Answer: 30
Theorem sequence:
equilateral_triangle_property_angle(1,EFG)
angle_addition(1,GEH,HEF)
Problem: As shown in the diagram, AB=17, BC=8, CA=15, BC is perpendicular to AC. Find the value of cos(CA).
问题：如图所示，AB=17，BC=8，CA=15，BC⊥AC。求cos(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(Cos(MeasureOfAngle(CAB)))
Answer: 15/17
Theorem sequence:
cosine_theorem(1,ABC)
Problem: As shown in the diagram, AD=8, AE=6, AF=a, AG=b, BF=4, BG=10, H is the center of ⊙H, BG is the tangent to circle H. Find the value of a.
问题：如图所示，AD=8，AE=6，AF=a，AG=b，BF=4，BG=10，H是⊙H的圆心，BG是⊙O的切线。求a的值。
Construction (CDL):
Shape(GA,AE,AEG)
Shape(EA,AD,ADE)
Shape(FA,AG,AGF)
Shape(DA,AF,AFD)
Shape(GB,BF,AGF)
Collinear(GAD)
Collinear(BFAE)
Cocircular(H,GFDE)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(AE),6)
Equal(LengthOfLine(AF),a)
Equal(LengthOfLine(AG),b)
Equal(LengthOfLine(BF),4)
Equal(LengthOfLine(BG),10)
IsCentreOfCircle(H,H)
IsTangentOfCircle(BG,H)
Goal: Value(a)
Answer: 15
Theorem sequence:
line_addition(1,BF,FE)
line_addition(1,FA,AE)
circle_property_circular_power_tangent_and_segment_line(1,BG,BFE,H)
Problem: As shown in the diagram, AB=11, EP=15, PA=13, ∠DCP=30°, ∠PAF=33°, P is the incenter of △AEC, ED is perpendicular to PD, PB⊥AB, PF is perpendicular to EF. Find the length of line PB.
问题：如图所示，AB=11，EP=15，PA=13，∠DCP=30°，∠PAF=33°，P是三角形AEC的内心，ED垂直于PD，PB垂直于AB，PF⊥EF。求直线PB的长度。
Construction (CDL):
Shape(AF,FP,PA)
Shape(FE,EP,PF)
Shape(ED,DP,PE)
Shape(PD,DC,CP)
Shape(PC,CB,BP)
Shape(AP,PB,BA)
Collinear(AFE)
Collinear(EDC)
Collinear(ABC)
Conditions (CDL):
Equal(LengthOfLine(AB),11)
Equal(LengthOfLine(EP),15)
Equal(LengthOfLine(PA),13)
Equal(MeasureOfAngle(DCP),30)
Equal(MeasureOfAngle(PAF),33)
IsIncenterOfTriangle(P,AEC)
PerpendicularBetweenLine(ED,PD)
PerpendicularBetweenLine(PB,AB)
PerpendicularBetweenLine(PF,EF)
Goal: Value(LengthOfLine(PB))
Answer: 4*sqrt(3)
Theorem sequence:
right_triangle_judgment_angle(1,PBA)
right_triangle_property_pythagorean(1,PBA)
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 DE.
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平行于DF。求直线DE的长度。
Construction (CDL):
Shape(AB,BF,FA)
Shape(FD,DE,EF)
Shape(BC,CD,DF,FB)
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(DE))
Answer: 15
Theorem sequence:
line_addition(1,AB,BC)
line_addition(1,AF,FE)
line_addition(1,CD,DE)
parallelogram_judgment_parallel_and_parallel(1,BCDF)
parallelogram_property_opposite_line_equal(1,FBCD)
parallel_property_corresponding_angle(1,BF,CD,A)
similar_triangle_judgment_aa(1,FAB,EAC)
similar_triangle_property_line_ratio(1,FAB,EAC)
similar_triangle_property_line_ratio(1,BFA,CEA)
similar_triangle_property_line_ratio(1,ABF,ACE)
Problem: As shown in the diagram, ∠BFC=72°, ∠CFD=4*a-1°, ∠DFA=2*a-5°, the center of circle F is F. Find the measure of ∠DFA.
问题：如图所示，∠BFC=72°，∠CFD=4*a-1°，∠DFA=2*a-5°，⊙F的圆心为F。求∠DFA的大小。
Construction (CDL):
Shape(BF,FC,FCB)
Shape(AF,FB,FBA)
Shape(DF,FA,FAD)
Shape(CF,FD,FDC)
Collinear(AFC)
Cocircular(F,ADCB)
Conditions (CDL):
Equal(MeasureOfAngle(BFC),72)
Equal(MeasureOfAngle(CFD),4*a-1)
Equal(MeasureOfAngle(DFA),2*a-5)
IsCentreOfCircle(F,F)
Goal: Value(MeasureOfAngle(DFA))
Answer: 57
Theorem sequence:
adjacent_complementary_angle(1,CFD,DFA)
Problem: As shown in the diagram, BN=EN, CB=BN, CN=2*sqrt(2), EN=CE, NE is perpendicular to CE, CBNE is a square. Find the perimeter of CBNE.
问题：如图所示，BN=EN，CB=BN，CN=2*sqrt(2)，EN=CE，NE⊥CE，CBNE是正方形。求四边形CBNE的周长。
Construction (CDL):
Shape(EC,CN,NE)
Shape(CB,BN,NC)
Conditions (CDL):
Equal(LengthOfLine(BN),LengthOfLine(EN))
Equal(LengthOfLine(CB),LengthOfLine(BN))
Equal(LengthOfLine(CN),2*sqrt(2))
Equal(LengthOfLine(EN),LengthOfLine(CE))
PerpendicularBetweenLine(NE,CE)
Square(CBNE)
Goal: Value(PerimeterOfQuadrilateral(CBNE))
Answer: 8
Theorem sequence:
right_triangle_judgment_angle(1,NEC)
right_triangle_property_pythagorean(1,NEC)
quadrilateral_perimeter_formula(1,NECB)
Problem: As shown in the diagram, PR=3, the radius of circle P is 5, the center of circle P is P, PR⊥QR. Find the length of line QR.
问题：如图所示，PR=3，圆P的半径为5，⊙P的圆心为P，PR⊥QR。求直线QR的长度。
Construction (CDL):
Shape(SP,PQ,PQS)
Shape(QP,PR,RQ)
Shape(RP,PS,SR)
Shape(QR,RS,PSQ)
Collinear(QRS)
Cocircular(P,QS)
Conditions (CDL):
Equal(LengthOfLine(PR),3)
Equal(RadiusOfCircle(P),5)
IsCentreOfCircle(P,P)
PerpendicularBetweenLine(PR,QR)
Goal: Value(LengthOfLine(QR))
Answer: 4
Theorem sequence:
radius_of_circle_property_length_equal(1,PQ,P)
right_triangle_judgment_angle(1,PRQ)
right_triangle_property_pythagorean(1,PRQ)
Problem: As shown in the diagram, ∠ABD=75°, ∠ACD=28°, ∠BDC=35°. Find the measure of ∠BAC.
问题：如图所示，∠ABD=75°，∠ACD=28°，∠BDC=35°。求∠BAC的大小。
Construction (CDL):
Shape(BD,DE,EB)
Shape(EA,AC,CE)
Collinear(BEA)
Collinear(DEC)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),75)
Equal(MeasureOfAngle(ACD),28)
Equal(MeasureOfAngle(BDC),35)
Goal: Value(MeasureOfAngle(BAC))
Answer: 82
Theorem sequence:
triangle_property_angle_sum(1,BDE)
vertical_angle(1,DEB,CEA)
triangle_property_angle_sum(1,EAC)
Problem: As shown in the diagram, BC=4, △ABC is an equilateral △. Find the area of triangle ABC.
问题：如图所示，BC=4，三角形ABC为等边三角形。求三角形ABC的面积。
Construction (CDL):
Shape(CA,AB,BC)
Conditions (CDL):
Equal(LengthOfLine(BC),4)
EquilateralTriangle(ABC)
Goal: Value(AreaOfTriangle(ABC))
Answer: 4*sqrt(3)
Theorem sequence:
equilateral_triangle_property_angle(1,ABC)
triangle_area_formula_sine(1,ABC)
Problem: As shown in the diagram, ∠FHM=94°, DM is parallel to CA. Find the measure of ∠ACI.
问题：如图所示，∠FHM=94°，DM平行于CA。求∠ACI的大小。
Construction (CDL):
Shape(FH,HM)
Shape(MH,HC)
Shape(CH,HD)
Shape(DH,HF)
Shape(HC,CA)
Shape(AC,CI)
Shape(IC,CE)
Shape(EC,CH)
Collinear(FHCI)
Collinear(MHD)
Collinear(ACE)
Conditions (CDL):
Equal(MeasureOfAngle(FHM),94)
ParallelBetweenLine(DM,CA)
Goal: Value(MeasureOfAngle(ACI))
Answer: 86
Theorem sequence:
parallel_property_collinear_extend(3,DM,CA,H)
adjacent_complementary_angle(1,FHM,MHC)
parallel_property_corresponding_angle(2,HM,CA,I)
Problem: As shown in the diagram, OK=8, RB=OB, the center of circle O is O, AB⊥OB. Find the length of line AP.
问题：如图所示，OK=8，RB=OB，O是圆O的圆心，AB⊥OB。求直线AP的长度。
Construction (CDL):
Shape(PB,BR,OPR)
Shape(RB,BA,OAR)
Shape(AB,BO,OA)
Shape(AO,OK,OKA)
Shape(KO,OB,BP,OPK)
Collinear(PBA)
Collinear(OBR)
Cocircular(O,ARPK)
Conditions (CDL):
Equal(LengthOfLine(OK),8)
Equal(LengthOfLine(RB),LengthOfLine(OB))
IsCentreOfCircle(O,O)
PerpendicularBetweenLine(AB,OB)
Goal: Value(LengthOfLine(AP))
Answer: 8*sqrt(3)
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OR,O)
radius_of_circle_property_length_equal(1,OK,O)
line_addition(1,OB,BR)
line_addition(1,PB,BA)
circle_property_chord_perpendicular_bisect_chord(1,O,OB,AP)
right_triangle_judgment_angle(1,ABO)
right_triangle_property_pythagorean(1,ABO)
Problem: As shown in the diagram, AD=7*y, CA=5*y+16, CE=8*x-18, CE=EB, EB=12-2*x, ∠BDA=∠EAC. Find the value of x.
问题：如图所示，AD=7*y，CA=5*y+16，CE=8*x-18，CE=EB，EB=12-2*x，∠BDA=∠EAC。求x的值。
Construction (CDL):
Shape(AC,CE,EA)
Shape(DA,AE,EB,BD)
Collinear(DAC)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AD),7*y)
Equal(LengthOfLine(CA),5*y+16)
Equal(LengthOfLine(CE),8*x-18)
Equal(LengthOfLine(CE),LengthOfLine(EB))
Equal(LengthOfLine(EB),12-2*x)
Equal(MeasureOfAngle(BDA),MeasureOfAngle(EAC))
Goal: Value(x)
Answer: 3
Theorem sequence:
Problem: As shown in the diagram, ∠BAE=58°, ∠CBE=26°, ∠EBA=47°, AB is parallel to EC. Find the measure of ∠AEB.
问题：如图所示，∠BAE=58°，∠CBE=26°，∠EBA=47°，AB∥EC。求∠AEB的大小。
Construction (CDL):
Shape(BA,AE,EB)
Shape(BE,EC,CB)
Shape(CE,ED,DC)
Collinear(AED)
Collinear(BCD)
Conditions (CDL):
Equal(MeasureOfAngle(BAE),58)
Equal(MeasureOfAngle(CBE),26)
Equal(MeasureOfAngle(EBA),47)
ParallelBetweenLine(AB,EC)
Goal: Value(MeasureOfAngle(AEB))
Answer: 75
Theorem sequence:
triangle_property_angle_sum(1,BAE)