SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AC=AN, BC=BM, ∠ACB=90°. Find the measure of ∠MCN.
问题：如图所示，AC=AN，BC=BM，∠ACB=90°。求∠MCN的大小。
Construction (CDL):
Shape(AC,CM,MA)
Shape(MC,CN,NM)
Shape(NC,CB,BN)
Collinear(AMNB)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(AN))
Equal(LengthOfLine(BC),LengthOfLine(BM))
Equal(MeasureOfAngle(ACB),90)
Goal: Value(MeasureOfAngle(MCN))
Answer: 45
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ACN)
isosceles_triangle_judgment_line_equal(1,BMC)
isosceles_triangle_property_angle_equal(1,ACN)
isosceles_triangle_property_angle_equal(1,BMC)
triangle_property_angle_sum(1,ACN)
triangle_property_angle_sum(1,BMC)
triangle_property_angle_sum(1,ACB)
triangle_property_angle_sum(1,MCN)
Problem: As shown in the diagram, AB=5, AB=AC, the perimeter of triangle BCE is 8, ED perpendicular bisects AB. Find the perimeter of △ABC.
问题：如图所示，AB=5，AB=AC，△BCE的周长为8，ED是AB的垂直平分线。求三角形ABC的周长。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BE,ED)
Shape(EB,BC,CE)
Collinear(ADB)
Collinear(AEC)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(PerimeterOfTriangle(BCE),8)
IsPerpendicularBisectorOfLine(ED,AB)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 13
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,ED,AB)
triangle_perimeter_formula(1,EBC)
triangle_perimeter_formula(1,ABC)
line_addition(1,AE,EC)
Problem: As shown in the diagram, AB=AC, ∠BEG=40°, ∠HDB=23°, GF is parallel to HI. Find the measure of ∠BCA.
问题：如图所示，AB=AC，∠BEG=40°，∠HDB=23°，GF∥HI。求∠BCA的大小。
Construction (CDL):
Shape(AE,EF,FA)
Shape(EG,GH,HB,BE)
Shape(BH,HD,DB)
Shape(EB,BD,DF,FE)
Shape(FD,DI,IF)
Shape(ID,DC,CI)
Collinear(AEBH)
Collinear(AFIC)
Collinear(GEF)
Collinear(BDC)
Collinear(HDI)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(BEG),40)
Equal(MeasureOfAngle(HDB),23)
ParallelBetweenLine(GF,HI)
Goal: Value(MeasureOfAngle(BCA))
Answer: 63
Theorem sequence:
parallel_property_collinear_extend(3,GF,HI,E)
parallel_property_alternate_interior_angle(2,GE,HI)
triangle_property_angle_sum(1,BHD)
adjacent_complementary_angle(1,ABD,DBH)
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
Problem: As shown in the diagram, BA=BE, ∠ABC=34°, ∠ECA=50°, F bisects segment EA. Find the measure of ∠BDA.
问题：如图所示，BA=BE，∠ABC=34°，∠ECA=50°，F平分线段EA。求∠BDA的大小。
Construction (CDL):
Shape(AB,BF,FA)
Shape(AF,FD,DA)
Shape(FB,BE,EF)
Shape(DF,FE,EC,CD)
Collinear(AFE)
Collinear(ADC)
Collinear(BEC)
Collinear(BFD)
Conditions (CDL):
Equal(LengthOfLine(BA),LengthOfLine(BE))
Equal(MeasureOfAngle(ABC),34)
Equal(MeasureOfAngle(ECA),50)
IsMidpointOfLine(F,EA)
Goal: Value(MeasureOfAngle(BDA))
Answer: 67
Theorem sequence:
median_of_triangle_judgment(1,BF,BEA)
isosceles_triangle_judgment_line_equal(1,BEA)
isosceles_triangle_property_line_coincidence(2,BEA,F)
angle_addition(1,ABF,FBE)
triangle_property_angle_sum(1,DBC)
adjacent_complementary_angle(1,CDF,FDA)
Problem: As shown in the diagram, AB=AC, ∠CAB=30°, ∠FDA=145°, DE∥GC. Find the measure of ∠GCB.
问题：如图所示，AB=AC，∠CAB=30°，∠FDA=145°，DE平行于GC。求∠GCB的大小。
Construction (CDL):
Shape(AF,FD,DA)
Shape(AD,DE,EA)
Shape(BG,GC,CB)
Shape(DB,BC,CE,ED)
Collinear(FDE)
Collinear(AEC)
Collinear(ADBG)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(MeasureOfAngle(CAB),30)
Equal(MeasureOfAngle(FDA),145)
ParallelBetweenLine(DE,GC)
Goal: Value(MeasureOfAngle(GCB))
Answer: 40
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ABC)
isosceles_triangle_property_angle_equal(1,ABC)
triangle_property_angle_sum(1,ABC)
adjacent_complementary_angle(1,FDA,ADE)
triangle_property_angle_sum(1,ADE)
parallel_property_corresponding_angle(2,CG,ED,A)
angle_addition(1,GCB,BCE)
Problem: As shown in the diagram, AB=3, BC=5, ED perpendicular bisects CA, ∠ABC and ∠BCA are the base angles of the isosceles triangle ABC. Find the perimeter of △ABE.
问题：如图所示，AB=3，BC=5，ED垂直平分AB，∠ABC和∠BCA是等腰三角形ABC的底角。求三角形ABE的周长。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,ED,DA)
Shape(DE,EC,CD)
Collinear(ADC)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(BC),5)
IsPerpendicularBisectorOfLine(ED,CA)
IsoscelesTriangle(ABC)
Goal: Value(PerimeterOfTriangle(ABE))
Answer: 8
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,ED,CA)
line_addition(1,BE,EC)
triangle_perimeter_formula(1,ABE)
Problem: As shown in the diagram, AB=AC, BC=10, the perimeter of △BCD is 24, DE perpendicular bisects CA. Find the length of line AC.
问题：如图所示，AB=AC，BC=10，△BCD的周长为24，DE垂直平分AB。求直线AC的长度。
Construction (CDL):
Shape(EA,AD,DE)
Shape(CE,ED,DC)
Shape(CD,DB,BC)
Collinear(CEA)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(BC),10)
Equal(PerimeterOfTriangle(BCD),24)
IsPerpendicularBisectorOfLine(DE,CA)
Goal: Value(LengthOfLine(AC))
Answer: 14
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,DE,CA)
line_addition(1,AD,DB)
triangle_perimeter_formula(1,BCD)
Problem: As shown in the diagram, AB=AC, CE=CD, ∠ECF=52°, CF is the median of triangle CDE. Find the measure of ∠BAC.
问题：如图所示，AB=AC，CE=CD，∠ECF=52°，CF是△CDE的中线。求∠BAC的大小。
Construction (CDL):
Shape(BA,AC,CE,EB)
Shape(FE,EC,CF)
Shape(FC,CD,DF)
Collinear(ACD)
Collinear(BEC)
Collinear(EFD)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(CE),LengthOfLine(CD))
Equal(MeasureOfAngle(ECF),52)
IsMedianOfTriangle(CF,CDE)
Goal: Value(MeasureOfAngle(BAC))
Answer: 28
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,ACB)
isosceles_triangle_judgment_line_equal(1,CDE)
isosceles_triangle_property_line_coincidence(2,CDE,F)
adjacent_complementary_angle(1,ACE,ECD)
isosceles_triangle_property_angle_equal(1,ACB)
triangle_property_angle_sum(1,BAC)
triangle_property_angle_sum(1,FEC)
triangle_property_angle_sum(1,FCD)
triangle_property_angle_sum(1,ECD)
Problem: As shown in the diagram, ∠AOB=90°, the radius of ⊙O is 2, Sub(AreaOfSector(OBA)=AreaOfTriangle(AOB)), O is the center of ⊙O. Find the value of x.
问题：如图所示，∠AOB=90°，⊙O的半径为2，Sub(AreaOfSector(OBA)=AreaOfTriangle(AOB))，O是⊙O的圆心。求x的值。
Construction (CDL):
Shape(AO,OB,BA)
Shape(AB,OBA)
Shape(BO,OA,OAB)
Cocircular(O,AB)
Conditions (CDL):
Equal(MeasureOfAngle(AOB),90)
Equal(RadiusOfCircle(O),2)
Equal(Sub(AreaOfSector(OBA),AreaOfTriangle(AOB)),x)
IsCentreOfCircle(O,O)
Goal: Value(x)
Answer: -2+pi
Theorem sequence:
arc_property_center_angle(1,OBA,O)
sector_area_formula(1,OBA)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OA,O)
triangle_area_formula_sine(1,OBA)
Problem: As shown in the diagram, OA=6, ∠CBA=30°, the center of circle O is O, AB is the diameter of ⊙O. Find the area of the sector OAC.
问题：如图所示，OA=6，∠CBA=30°，⊙O的圆心为O，AB是圆O的直径。求扇形OAC的面积。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OA,AC,CO)
Shape(CA,OAC)
Shape(BO,OA,CB)
Shape(BC,OCB)
Collinear(AOB)
Cocircular(O,ACB)
Conditions (CDL):
Equal(LengthOfLine(OA),6)
Equal(MeasureOfAngle(CBA),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
Goal: Value(AreaOfSector(OAC))
Answer: 6*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
arc_property_circumference_angle_external(1,OAC,B)
sector_area_formula(1,OAC)
Problem: As shown in the diagram, PO=12, the radius of ⊙O is 6, the center of ⊙O is O, PA is the tangent to circle O, PB is the tangent to ⊙O. Find the area of the sector OBA.
问题：如图所示，PO=12，圆O的半径为6，圆O的圆心为O，⊙O的切线为PA，PB是圆O的切线。求扇形OBA的面积。
Construction (CDL):
Shape(CO,OB,OBC)
Shape(AO,OC,OCA)
Shape(BP,PC,OBC)
Shape(CP,PA,OCA)
Shape(BO,OA,OAB)
Collinear(OCP)
Cocircular(O,BCA)
Conditions (CDL):
Equal(LengthOfLine(PO),12)
Equal(RadiusOfCircle(O),6)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(AreaOfSector(OBA))
Answer: 12*pi
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
tangent_of_circle_property_perpendicular(1,PB,O,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
sine_theorem(1,OPA)
triangle_property_angle_sum(1,OBP)
triangle_property_angle_sum(1,AOP)
sine_theorem(1,OBP)
angle_addition(1,AOC,COB)
arc_property_center_angle(1,OBA,O)
sector_area_formula(1,OBA)
Problem: As shown in the diagram, AB=4, ∠BCA=60°, C is the center of circle C, AD⊥CD. Find the area of the sector CAB.
问题：如图所示，AB=4，∠BCA=60°，⊙C的圆心为C，AD垂直于CD。求扇形CAB的面积。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Shape(BA,CAB)
Shape(AC,CB,CBA)
Collinear(ADB)
Cocircular(C,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(MeasureOfAngle(BCA),60)
IsCentreOfCircle(C,C)
PerpendicularBetweenLine(AD,CD)
Goal: Value(AreaOfSector(CAB))
Answer: 8*pi/3
Theorem sequence:
radius_of_circle_property_length_equal(1,CA,C)
radius_of_circle_property_length_equal(1,CB,C)
isosceles_triangle_judgment_line_equal(1,CAB)
altitude_of_triangle_judgment(1,CD,CAB)
isosceles_triangle_property_line_coincidence(1,CAB,D)
line_addition(1,AD,DB)
angle_addition(1,BCD,DCA)
sine_theorem(1,ADC)
arc_property_center_angle(1,CAB,C)
sector_area_formula(1,CAB)
Problem: As shown in the diagram, the length of ⌒OAB is 20*pi, the radius of circle O is 24, O is the center of circle O. Find the area of the sector OAB.
问题：如图所示，⌒OAB的长度为20*pi，圆O的半径为24，圆O的圆心为O。求扇形OAB的面积。
Construction (CDL):
Shape(OA,OAB,BO)
Shape(OB,OBA,AO)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfArc(OAB),20*pi)
Equal(RadiusOfCircle(O),24)
IsCentreOfCircle(O,O)
Goal: Value(AreaOfSector(OAB))
Answer: 240*pi
Theorem sequence:
arc_length_formula(1,OAB)
sector_area_formula(1,OAB)
Problem: As shown in the diagram, PA=3, ∠APB=60°, ∠PBO=90°, triangle APB is an equilateral triangle, the center of circle O is O, AC is the diameter of circle O, the tangent to ⊙O is PA, PB is the tangent to circle O. Find the sum of the area of △ABC and the area of the sector OBC.
问题：如图所示，PA=3，∠APB=60°，∠PBO=90°，三角形APB为等边三角形，O是圆O的圆心，⊙O的直径为AC，PA是⊙O的切线，⊙O的切线为PB。求三角形ABC的面积与扇形OBC的面积之和。
Construction (CDL):
Shape(AP,PB,OAB)
Shape(OAB,BA)
Shape(OA,AB,BO)
Shape(OB,BC,CO)
Shape(OBC,CB)
Shape(OC,OCA,AO)
Collinear(AOC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(PA),3)
Equal(MeasureOfAngle(APB),60)
Equal(MeasureOfAngle(PBO),90)
EquilateralTriangle(APB)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AC,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(Add(AreaOfTriangle(ABC),AreaOfSector(OBC)))
Answer: pi/2+3*sqrt(3)/2
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PA,O,O)
quadrilateral_property_angle_sum(1,APBO)
flat_angle(1,COA)
angle_addition(1,COB,BOA)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OAB)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OAB)
isosceles_triangle_property_angle_equal(1,OBC)
triangle_property_angle_sum(1,OAB)
triangle_property_angle_sum(1,OBC)
triangle_property_angle_sum(1,ABC)
sine_theorem(1,BOA)
sine_theorem(1,ABC)
sine_theorem(1,BCA)
triangle_area_formula_sine(1,CAB)
arc_property_center_angle(1,OBC,O)
sector_area_formula(1,OBC)
Problem: As shown in the diagram, BD=6, ∠BDO=90°, ∠OBD=30°, △ABC is an equilateral △, O is the center of circle O, O is the center of the inscribed circle of triangle ABC. Find the area of the sector OED.
问题：如图所示，BD=6，∠BDO=90°，∠OBD=30°，△ABC为等边△，O是⊙O的圆心，三角形ABC的内心为O。求扇形OED的面积。
Construction (CDL):
Shape(OE,OED,DO)
Shape(OD,ODE,EO)
Shape(DC,CF,ODF)
Shape(FA,AG,OFG)
Shape(BE,OGE,GB)
Shape(EB,BD,OED)
Collinear(AGB)
Collinear(AFC)
Collinear(BDC)
Collinear(BEO)
Cocircular(O,FGED)
Conditions (CDL):
Equal(LengthOfLine(BD),6)
Equal(MeasureOfAngle(BDO),90)
Equal(MeasureOfAngle(OBD),30)
EquilateralTriangle(ABC)
IsCentreOfCircle(O,O)
IsIncenterOfTriangle(O,ABC)
Goal: Value(AreaOfSector(OED))
Answer: 2*pi
Theorem sequence:
triangle_property_angle_sum(1,DOB)
sine_theorem(1,DOB)
arc_property_center_angle(1,OED,O)
radius_of_circle_property_length_equal(1,OD,O)
sector_area_formula(1,OED)
Problem: As shown in the diagram, OA=3, PO=6, the center of ⊙O is O, PA is the tangent to ⊙O. Find the measure of ∠OPA.
问题：如图所示，OA=3，PO=6，O是圆O的圆心，⊙O的切线为PA。求∠OPA的大小。
Construction (CDL):
Shape(PA,OBA,BP)
Shape(AO,OB,OBA)
Shape(OA,OAB,BO)
Collinear(OBP)
Cocircular(O,BA)
Conditions (CDL):
Equal(LengthOfLine(OA),3)
Equal(LengthOfLine(PO),6)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(OPA))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
cosine_theorem(1,PAO)
cosine_theorem(1,AOP)
Problem: As shown in the diagram, AB=6, ∠BCA=60°, C is the center of ⊙C. Find the area of the sector CAB.
问题：如图所示，AB=6，∠BCA=60°，C是⊙C的圆心。求扇形CAB的面积。
Construction (CDL):
Shape(CA,AB,BC)
Shape(CAB,BA)
Shape(CB,CBA,AC)
Cocircular(C,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(MeasureOfAngle(BCA),60)
IsCentreOfCircle(C,C)
Goal: Value(AreaOfSector(CAB))
Answer: 6*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,CA,C)
radius_of_circle_property_length_equal(1,CB,C)
isosceles_triangle_judgment_line_equal(1,CAB)
isosceles_triangle_property_angle_equal(1,CAB)
triangle_property_angle_sum(1,CAB)
isosceles_triangle_judgment_angle_equal(1,BCA)
arc_property_center_angle(1,CAB,C)
sector_area_formula(1,CAB)
Problem: As shown in the diagram, the area of the sector ACD is 6*pi, ∠ACB=90°, ∠BAC=60°, A is the center of ⊙A. Find the length of line AB.
问题：如图所示，扇形ACD的面积为6*pi，∠ACB=90°，∠BAC=60°，A是圆A的圆心。求直线AB的长度。
Construction (CDL):
Shape(DA,AC,ACD)
Shape(AD,ADC,CA)
Shape(CB,BD,ACD)
Collinear(ADB)
Cocircular(A,CD)
Conditions (CDL):
Equal(AreaOfSector(ACD),6*pi)
Equal(MeasureOfAngle(ACB),90)
Equal(MeasureOfAngle(BAC),60)
IsCentreOfCircle(A,A)
Goal: Value(LengthOfLine(AB))
Answer: 12
Theorem sequence:
arc_property_center_angle(1,ACD,A)
sector_area_formula(1,ACD)
radius_of_circle_property_length_equal(1,AC,A)
triangle_property_angle_sum(1,ACB)
sine_theorem(1,ACB)
Problem: As shown in the diagram, AB=sqrt(3), BC=1, the center of circle O is O, ABCD is a rectangle. Find the area of the sector ODA.
问题：如图所示，AB=sqrt(3)，BC=1，O是⊙O的圆心，ABCD是长方形。求扇形ODA的面积。
Construction (CDL):
Shape(OD,DA,AO)
Shape(ODA,AD)
Shape(OAB,BA)
Shape(OA,AB,BO)
Shape(OB,BC,CO)
Shape(OBC,CB)
Shape(DO,OC,CD)
Shape(OCD,DC)
Collinear(AOC)
Collinear(BOD)
Cocircular(O,DABC)
Conditions (CDL):
Equal(LengthOfLine(AB),sqrt(3))
Equal(LengthOfLine(BC),1)
IsCentreOfCircle(O,O)
Rectangle(ABCD)
Goal: Value(AreaOfSector(ODA))
Answer: pi/6
Theorem sequence:
cosine_theorem(1,BCA)
cosine_theorem(1,ABC)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
angle_addition(1,DAC,CAB)
isosceles_triangle_judgment_line_equal(1,ODA)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,ODA)
isosceles_triangle_property_angle_equal(1,OBC)
triangle_property_angle_sum(1,ODA)
triangle_property_angle_sum(1,OBC)
triangle_property_angle_sum(1,ABC)
isosceles_triangle_judgment_angle_equal(1,COB)
arc_property_center_angle(1,ODA,O)
sector_area_formula(1,ODA)
Problem: As shown in the diagram, the radius of circle A is 2, the radius of ⊙B is 2, the radius of ⊙C is 2, the center of ⊙A is A, the center of circle B is B, C is the center of circle C. Find the sum of the area of the sector CDE, the area of the sector AGF and the area of the sector BIH.
问题：如图所示，⊙A的半径为2，⊙B的半径为2，圆C的半径为2，A是圆A的圆心，⊙B的圆心为B，圆C的圆心为C。求扇形CDE的面积、扇形AGF的面积与扇形BIH的面积之和。
Construction (CDL):
Shape(CD,CDE,EC)
Shape(DF,AGF,GH,BIH,IE,CDE)
Shape(FA,AG,AGF)
Shape(BI,BIH,HB)
Collinear(AFDC)
Collinear(AGHB)
Collinear(CEIB)
Cocircular(C,DE)
Cocircular(A,GF)
Cocircular(B,IH)
Conditions (CDL):
Equal(RadiusOfCircle(A),2)
Equal(RadiusOfCircle(B),2)
Equal(RadiusOfCircle(C),2)
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
Goal: Value(Add(AreaOfSector(CDE),AreaOfSector(AGF),AreaOfSector(BIH)))
Answer: 2*pi
Theorem sequence:
triangle_property_angle_sum(1,CAB)
arc_property_center_angle(1,CDE,C)
arc_property_center_angle(1,AGF,A)
arc_property_center_angle(1,BIH,B)
sector_area_formula(1,CDE)
sector_area_formula(1,AGF)
sector_area_formula(1,BIH)
Problem: As shown in the diagram, AB=3, ∠CAB=120°, A is the center of ⊙A. Find the area of the sector ABC.
问题：如图所示，AB=3，∠CAB=120°，圆A的圆心为A。求扇形ABC的面积。
Construction (CDL):
Shape(AB,ABC,CA)
Shape(AC,ACB,BA)
Cocircular(A,BC)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(MeasureOfAngle(CAB),120)
IsCentreOfCircle(A,A)
Goal: Value(AreaOfSector(ABC))
Answer: 3*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,AB,A)
arc_property_center_angle(1,ABC,A)
sector_area_formula(1,ABC)
Problem: As shown in the diagram, OA=3, QC=1, ∠AOB=90°, ∠CQD=90°, the center of circle O is O, Q is the center of ⊙Q. Find the area of the sector OBA minus the area of the sector QDC.
问题：如图所示，OA=3，QC=1，∠AOB=90°，∠CQD=90°，O是⊙O的圆心，⊙Q的圆心为Q。求扇形OBA的面积减去扇形QDC的面积。
Construction (CDL):
Shape(OB,OBA,AO)
Shape(OA,OAB,BO)
Shape(QD,QDC,CQ)
Shape(QC,QCD,DQ)
Cocircular(O,BA)
Cocircular(Q,DC)
Conditions (CDL):
Equal(LengthOfLine(OA),3)
Equal(LengthOfLine(QC),1)
Equal(MeasureOfAngle(AOB),90)
Equal(MeasureOfAngle(CQD),90)
IsCentreOfCircle(O,O)
IsCentreOfCircle(Q,Q)
Goal: Value(Sub(AreaOfSector(OBA),AreaOfSector(QDC)))
Answer: 2*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,QC,Q)
arc_property_center_angle(1,OBA,O)
arc_property_center_angle(1,QDC,Q)
sector_area_formula(1,OBA)
sector_area_formula(1,QDC)
Problem: As shown in the diagram, the radius of circle A is 2, the radius of circle B is 2, the radius of ⊙C is 2, A is the center of ⊙A, B is the center of ⊙B, the center of circle C is C. Find the sum of the area of the sector CDE, the area of the sector AGF and the area of the sector BIH.
问题：如图所示，⊙A的半径为2，⊙B的半径为2，圆C的半径为2，圆A的圆心为A，圆B的圆心为B，⊙C的圆心为C。求扇形CDE的面积、扇形AGF的面积与扇形BIH的面积之和。
Construction (CDL):
Shape(CD,CDE,EC)
Shape(DF,AGF,GH,BIH,IE,CDE)
Shape(FA,AG,AGF)
Shape(BI,BIH,HB)
Collinear(AFDC)
Collinear(AGHB)
Collinear(CEIB)
Cocircular(C,DE)
Cocircular(A,GF)
Cocircular(B,IH)
Conditions (CDL):
Equal(RadiusOfCircle(A),2)
Equal(RadiusOfCircle(B),2)
Equal(RadiusOfCircle(C),2)
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
Goal: Value(Add(AreaOfSector(CDE),AreaOfSector(AGF),AreaOfSector(BIH)))
Answer: 2*pi
Theorem sequence:
triangle_property_angle_sum(1,CAB)
arc_property_center_angle(1,CDE,C)
arc_property_center_angle(1,AGF,A)
arc_property_center_angle(1,BIH,B)
sector_area_formula(1,CDE)
sector_area_formula(1,AGF)
sector_area_formula(1,BIH)
Problem: As shown in the diagram, BC=4, ∠CAB=90°, ∠EAD=90°, the center of circle A is A, AD is the median of △ ABC, CA and AB are the legs of the right △ CAB. Find the area of the sector ADE.
问题：如图所示，BC=4，∠CAB=90°，∠EAD=90°，A是⊙A的圆心，AD是△ABC的中线，∠CAB是直角△CAB的直角。求扇形ADE的面积。
Construction (CDL):
Shape(AB,BD,DA)
Shape(AD,DC,CA)
Shape(AD,ADE,EA)
Shape(AE,AED,DA)
Collinear(BDC)
Cocircular(A,DE)
Conditions (CDL):
Equal(LengthOfLine(BC),4)
Equal(MeasureOfAngle(CAB),90)
Equal(MeasureOfAngle(EAD),90)
IsCentreOfCircle(A,A)
IsMedianOfTriangle(AD,ABC)
RightTriangle(CAB)
Goal: Value(AreaOfSector(ADE))
Answer: pi
Theorem sequence:
right_triangle_property_length_of_median(1,CAB,D)
radius_of_circle_property_length_equal(1,AD,A)
arc_property_center_angle(1,ADE,A)
sector_area_formula(1,ADE)
Problem: As shown in the diagram, the radius of ⊙O is 1, the center of circle O is O. Find the area of the circle O.
问题：如图所示，圆O的半径为1，O是圆O的圆心。求圆O的面积。
Construction (CDL):
Shape(OA,OAB,BO)
Shape(OB,OBA,AO)
Cocircular(O,AB)
Conditions (CDL):
Equal(RadiusOfCircle(O),1)
IsCentreOfCircle(O,O)
Goal: Value(AreaOfCircle(O))
Answer: pi
Theorem sequence:
circle_area_formula(1,O)
Problem: As shown in the diagram, AB=6, HF=6, the diameter of circle O is HF, quadrilateral ABCD is a square. Find the area of quadrilateral ABCD minus the area of the circle O.
问题：如图所示，AB=6，HF=6，圆O的直径为HF，ABCD是正方形。求四边形ABCD的面积减去⊙O的面积。
Construction (CDL):
Shape(HA,AE,OHE)
Shape(EB,BF,OEF)
Shape(FC,CG,OFG)
Shape(GD,DH,OGH)
Shape(OHE,OEF,FH)
Shape(HF,OFG,OGH)
Collinear(AEB)
Collinear(BFC)
Collinear(CGD)
Collinear(DHA)
Cocircular(O,HEFG)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(HF),6)
IsDiameterOfCircle(HF,O)
Square(ABCD)
Goal: Value(Sub(AreaOfQuadrilateral(ABCD),AreaOfCircle(O)))
Answer: 36-9*pi
Theorem sequence:
diameter_of_circle_property_length_equal(1,HF,O)
circle_property_length_of_radius_and_diameter(1,O)
circle_area_formula(1,O)
parallelogram_area_formula_sine(1,ABCD)
Problem: As shown in the diagram, BA=6, BC=6, ∠BAC=30°, ∠BOC=60°, O is the center of circle O. Find the sum of the area of triangle BAC and the area of the sector OCB.
问题：如图所示，BA=6，BC=6，∠BAC=30°，∠BOC=60°，圆O的圆心为O。求三角形BAC的面积与扇形OCB的面积之和。
Construction (CDL):
Shape(OC,CB,BO)
Shape(OCB,BC)
Shape(CB,BA,AC)
Shape(OB,OBC,CO)
Cocircular(O,BC)
Conditions (CDL):
Equal(LengthOfLine(BA),6)
Equal(LengthOfLine(BC),6)
Equal(MeasureOfAngle(BAC),30)
Equal(MeasureOfAngle(BOC),60)
IsCentreOfCircle(O,O)
Goal: Value(Add(AreaOfTriangle(BAC),AreaOfSector(OCB)))
Answer: 9*sqrt(3)+6*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,OB,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OCB)
isosceles_triangle_judgment_line_equal(1,BAC)
isosceles_triangle_property_angle_equal(1,OCB)
isosceles_triangle_property_angle_equal(1,BAC)
triangle_property_angle_sum(1,OCB)
triangle_property_angle_sum(1,CBA)
isosceles_triangle_judgment_angle_equal(1,CBO)
arc_property_center_angle(1,OCB,O)
sector_area_formula(1,OCB)
triangle_area_formula_sine(1,BAC)
Problem: As shown in the diagram, ∠AOB=120°, the radius of ⊙O is 1, the center of ⊙O is O. Find the area of the sector OBA.
问题：如图所示，∠AOB=120°，⊙O的半径为1，圆O的圆心为O。求扇形OBA的面积。
Construction (CDL):
Shape(AO,OB,OBA)
Cocircular(O,BA)
Conditions (CDL):
Equal(MeasureOfAngle(AOB),120)
Equal(RadiusOfCircle(O),1)
IsCentreOfCircle(O,O)
Goal: Value(AreaOfSector(OBA))
Answer: pi/3
Theorem sequence:
arc_property_center_angle(1,OBA,O)
sector_area_formula(1,OBA)
Problem: As shown in the diagram, the length of ⌒OAB is 3, AC=4, the radius of ⊙O is 6, the center of ⊙O is O. Find the area of triangle OAC minus the area of the sector OAB.
问题：如图所示，⌒OAB的长度为3，AC=4，圆O的半径为6，O是圆O的圆心。求△OAC的面积减去扇形OAB的面积。
Construction (CDL):
Shape(OA,OAB,BO)
Shape(AC,CB,OAB)
Collinear(CBO)
Cocircular(O,AB)
Conditions (CDL):
Equal(LengthOfArc(OAB),3)
Equal(LengthOfLine(AC),4)
Equal(RadiusOfCircle(O),6)
IsCentreOfCircle(O,O)
Goal: Value(Sub(AreaOfTriangle(OAC),AreaOfSector(OAB)))
Answer: -9-3*sqrt(-2+18*cos(1))*sin(1/2)+9*sin(1)
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
arc_length_formula(1,OAB)
arc_property_center_angle(1,OAB,O)
sector_area_formula(1,OAB)
sine_theorem(1,ACO)
cosine_theorem(1,OAC)
triangle_area_formula_sine(1,COA)
Problem: As shown in the diagram, the radius of circle A is 1, the radius of circle B is 1, the radius of circle C is 1, the center of ⊙A is A, the center of circle B is B, the center of ⊙C is C. Find the sum of the area of the sector CDE, the area of the sector AGF and the area of the sector BIH.
问题：如图所示，⊙A的半径为1，圆B的半径为1，圆C的半径为1，⊙A的圆心为A，B是圆B的圆心，C是圆C的圆心。求扇形CDE的面积、扇形AGF的面积与扇形BIH的面积之和。
Construction (CDL):
Shape(CD,CDE,EC)
Shape(DF,AGF,GH,BIH,IE,CDE)
Shape(FA,AG,AGF)
Shape(BI,BIH,HB)
Collinear(AFDC)
Collinear(AGHB)
Collinear(CEIB)
Cocircular(C,DE)
Cocircular(A,GF)
Cocircular(B,IH)
Conditions (CDL):
Equal(RadiusOfCircle(A),1)
Equal(RadiusOfCircle(B),1)
Equal(RadiusOfCircle(C),1)
IsCentreOfCircle(A,A)
IsCentreOfCircle(B,B)
IsCentreOfCircle(C,C)
Goal: Value(Add(AreaOfSector(CDE),AreaOfSector(AGF),AreaOfSector(BIH)))
Answer: pi/2
Theorem sequence:
triangle_property_angle_sum(1,CAB)
arc_property_center_angle(1,CDE,C)
arc_property_center_angle(1,AGF,A)
arc_property_center_angle(1,BIH,B)
sector_area_formula(1,CDE)
sector_area_formula(1,AGF)
sector_area_formula(1,BIH)
Problem: As shown in the diagram, AB=1000, BC=800, △BCA is a right △. Find the length of line CA.
问题：如图所示，AB=1000，BC=800，∠BCA是直角三角形BCA的直角。求直线CA的长度。
Construction (CDL):
Shape(AB,BC,CA)
Conditions (CDL):
Equal(LengthOfLine(AB),1000)
Equal(LengthOfLine(BC),800)
RightTriangle(BCA)
Goal: Value(LengthOfLine(CA))
Answer: 600
Theorem sequence:
right_triangle_property_pythagorean(1,BCA)