SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, the perimeter of ABCD is 28, Sub(PerimeterOfTriangle(ABO)=PerimeterOfTriangle(OBC)), BA and CD are opposite sides of the parallelogram ABCD. Find the length of line AB.
问题：如图所示，四边形ABCD的周长为28，Sub(PerimeterOfTriangle(ABO)=PerimeterOfTriangle(OBC))，四边形ABCD是平行四边形。求直线AB的长度。
Construction (CDL):
Shape(AO,OD,DA)
Shape(DO,OC,CD)
Shape(OA,AB,BO)
Shape(CO,OB,BC)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),28)
Equal(Sub(PerimeterOfTriangle(ABO),PerimeterOfTriangle(OBC)),4)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(AB))
Answer: 9
Theorem sequence:
quadrilateral_perimeter_formula(1,DABC)
triangle_perimeter_formula(1,ABO)
triangle_perimeter_formula(1,OBC)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
Problem: As shown in the diagram, AB=3, AD=5, EO perpendicular bisects BD, AB and DC are opposite sides of the parallelogram ADCB. Find the perimeter of △CED.
问题：如图所示，AB=3，AD=5，EO垂直平分AB，AB和DC是▱ADCB的一组对边。求△CED的周长。
Construction (CDL):
Shape(BA,AF,FO,OB)
Shape(BO,OE,EB)
Shape(EO,OD,DE)
Shape(CE,ED,DC)
Shape(OF,FD,DO)
Collinear(BEC)
Collinear(AFD)
Collinear(BOD)
Collinear(EOF)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AD),5)
IsPerpendicularBisectorOfLine(EO,BD)
Parallelogram(ADCB)
Goal: Value(PerimeterOfTriangle(CED))
Answer: 8
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,EO,BD)
line_addition(1,BE,EC)
parallelogram_property_opposite_line_equal(1,ADCB)
parallelogram_property_opposite_line_equal(1,DCBA)
triangle_perimeter_formula(1,CED)
Problem: As shown in the diagram, AB=3, BC=4, quadrilateral AECD is a ▱, CD is perpendicular to OD, △CBA is a right △. Find the length of line DE.
问题：如图所示，AB=3，BC=4，EA和CD是平行四边形AECD的一组对边，CD垂直于OD，∠CBA是直角三角形CBA的直角。求直线DE的长度。
Construction (CDL):
Shape(EC,CO,OE)
Shape(EO,OA,AE)
Shape(CD,DO,OC)
Shape(OD,DA,AO)
Shape(AD,DB,BA)
Collinear(CDB)
Collinear(AOC)
Collinear(DOE)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(BC),4)
Parallelogram(AECD)
PerpendicularBetweenLine(CD,OD)
RightTriangle(CBA)
Goal: Value(LengthOfLine(DE))
Answer: 3
Theorem sequence:
similar_triangle_judgment_aa(1,OCD,ACB)
parallelogram_property_diagonal_bisection(1,AECD,O)
parallelogram_property_diagonal_bisection(1,ECDA,O)
line_addition(1,CO,OA)
similar_triangle_property_line_ratio(1,CDO,CBA)
similar_triangle_property_line_ratio(1,DOC,BAC)
line_addition(1,DO,OE)
Problem: As shown in the diagram, AD=8, CD=10, DE bisects ∠CDA, AD and BC are opposite sides of the parallelogram DABC. Find the length of line BE.
问题：如图所示，AD=8，CD=10，DE平分∠CDA，AD和BC是平行四边形DABC的一组对边。求直线BE的长度。
Construction (CDL):
Shape(DA,AE,ED)
Shape(DE,EB,BC,CD)
Collinear(AEB)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(CD),10)
IsBisectorOfAngle(DE,CDA)
Parallelogram(DABC)
Goal: Value(LengthOfLine(BE))
Answer: 2
Theorem sequence:
parallel_property_collinear_extend(3,BA,CD,E)
parallel_property_alternate_interior_angle(1,DC,AE)
isosceles_triangle_judgment_angle_equal(1,AED)
line_addition(1,AE,EB)
parallelogram_property_opposite_line_equal(1,ABCD)
Problem: As shown in the diagram, OE=1, E is the midpoint of segment BC, ABCD is a parallelogram. Find the length of line AB.
问题：如图所示，OE=1，E平分线段BC，ABCD是▱。求直线AB的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BE,EO)
Shape(CO,OE,EC)
Shape(AO,OD,DA)
Shape(DO,OC,CD)
Collinear(AOC)
Collinear(BOD)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(OE),1)
IsMidpointOfLine(E,BC)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(AB))
Answer: 2
Theorem sequence:
parallelogram_property_diagonal_bisection(1,ABCD,O)
midsegment_of_triangle_judgment_midpoint(1,OE,CAB)
midsegment_of_triangle_property_length(1,OE,CAB)
Problem: As shown in the diagram, OE=5/2, OE is parallel to BC, quadrilateral ABCD is a parallelogram. Find the length of line AD.
问题：如图所示，OE=5/2，OE∥BC，BA和CD是▱ABCD的一组对边。求直线AD的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(OB,BC,CO)
Shape(DO,OE,ED)
Shape(EO,OC,CE)
Collinear(AOC)
Collinear(BOD)
Collinear(CED)
Conditions (CDL):
Equal(LengthOfLine(OE),5/2)
ParallelBetweenLine(OE,BC)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(AD))
Answer: 5
Theorem sequence:
parallelogram_property_diagonal_bisection(1,BCDA,O)
midsegment_of_triangle_judgment_parallel(1,OE,DBC)
midsegment_of_triangle_property_length(1,OE,DBC)
parallelogram_property_opposite_line_equal(1,BCDA)
Problem: As shown in the diagram, ∠CDA=55°, AD and BC are opposite sides of the parallelogram ABCD, CE is perpendicular to BE. Find the measure of ∠BCE.
问题：如图所示，∠CDA=55°，BA和CD是▱ABCD的一组对边，CE垂直于BE。求∠BCE的大小。
Construction (CDL):
Shape(AE,EC,CD,DA)
Shape(EB,BC,CE)
Collinear(AEB)
Conditions (CDL):
Equal(MeasureOfAngle(CDA),55)
Parallelogram(ABCD)
PerpendicularBetweenLine(CE,BE)
Goal: Value(MeasureOfAngle(BCE))
Answer: 35
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BCDA)
triangle_property_angle_sum(1,EBC)
Problem: As shown in the diagram, AB=6, the perimeter of △BOA is 15, ABCD is a ▱. Find the sum of the length of line AC and the length of line BD.
问题：如图所示，AB=6，△BOA的周长为15，四边形ABCD是▱。求直线AC的长度与直线BD的长度之和。
Construction (CDL):
Shape(AB,BO,OA)
Shape(AO,OD,DA)
Shape(OB,BC,CO)
Shape(DO,OC,CD)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(PerimeterOfTriangle(BOA),15)
Parallelogram(ABCD)
Goal: Value(Add(LengthOfLine(AC),LengthOfLine(BD)))
Answer: 18
Theorem sequence:
triangle_perimeter_formula(1,ABO)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
line_addition(1,AO,OC)
line_addition(1,BO,OD)
Problem: As shown in the diagram, EA=EB, EB=EC, ∠CDA=50°, quadrilateral ABCD is a ▱. Find the measure of ∠AEC.
问题：如图所示，EA=EB，EB=EC，∠CDA=50°，AD和BC是平行四边形ABCD的一组对边。求∠AEC的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BC,CE)
Shape(AE,EC,CD,DA)
Conditions (CDL):
Equal(LengthOfLine(EA),LengthOfLine(EB))
Equal(LengthOfLine(EB),LengthOfLine(EC))
Equal(MeasureOfAngle(CDA),50)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(AEC))
Answer: 100
Theorem sequence:
parallelogram_property_opposite_angle_equal(1,BCDA)
isosceles_triangle_judgment_line_equal(1,EAB)
isosceles_triangle_judgment_line_equal(1,EBC)
isosceles_triangle_property_angle_equal(1,EAB)
isosceles_triangle_property_angle_equal(1,EBC)
triangle_property_angle_sum(1,ABE)
triangle_property_angle_sum(1,EBC)
angle_addition(1,ABE,EBC)
angle_addition(1,CEB,BEA)
round_angle(1,AEC,CEA)
Problem: As shown in the diagram, AB=3, AC=4, BC=5, BA and CD are opposite sides of the ▱ ABCD. Find the area of quadrilateral ABCD.
问题：如图所示，AB=3，AC=4，BC=5，ABCD是▱。求ABCD的面积。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(LengthOfLine(AB),3)
Equal(LengthOfLine(AC),4)
Equal(LengthOfLine(BC),5)
Parallelogram(ABCD)
Goal: Value(AreaOfQuadrilateral(ABCD))
Answer: 12
Theorem sequence:
cosine_theorem(1,BCA)
parallelogram_area_formula_sine(1,ABCD)
Problem: As shown in the diagram, ∠ABC=80°, AE is the angle bisector of ∠DAB, EA∥CF, AD and BC are opposite sides of the parallelogram ABCD. Find the measure of ∠ECF.
问题：如图所示，∠ABC=80°，AE平分∠DAB，EA∥CF，AD和BC是平行四边形ABCD的一组对边。求∠ECF的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(AE,EC,CF,FA)
Shape(FC,CD,DF)
Collinear(AFD)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(ABC),80)
IsBisectorOfAngle(AE,DAB)
ParallelBetweenLine(EA,CF)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(ECF))
Answer: 50
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,AD,BC)
angle_addition(1,FAE,EAB)
triangle_property_angle_sum(1,ABE)
parallel_property_corresponding_angle(1,EA,CF,B)
Problem: As shown in the diagram, AB=7, AC=10, the perimeter of △ABO is 16, AD and BC are opposite sides of the ▱ ABCD. Find the length of line BD.
问题：如图所示，AB=7，AC=10，△ABO的周长为16，BA和CD是▱ABCD的一组对边。求直线BD的长度。
Construction (CDL):
Shape(AB,BO,OA)
Shape(OB,BC,CO)
Shape(AO,OD,DA)
Shape(DO,OC,CD)
Collinear(AOC)
Collinear(BOD)
Conditions (CDL):
Equal(LengthOfLine(AB),7)
Equal(LengthOfLine(AC),10)
Equal(PerimeterOfTriangle(ABO),16)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(BD))
Answer: 8
Theorem sequence:
triangle_perimeter_formula(1,ABO)
parallelogram_property_diagonal_bisection(1,ABCD,O)
parallelogram_property_diagonal_bisection(1,BCDA,O)
line_addition(1,AO,OC)
line_addition(1,BO,OD)
Problem: As shown in the diagram, AB=4, DG=1, AE bisects ∠DAB, F is the midpoint of segment DC, DC and AB are opposite sides of the ▱ DABC, DG⊥FG. Find the length of line AE.
问题：如图所示，AB=4，DG=1，AE平分∠DAB，F平分线段DC，DC和AB是平行四边形DABC的一组对边，DG⊥FG。求直线AE的长度。
Construction (CDL):
Shape(DA,AG,GD)
Shape(DG,GF,FD)
Shape(FG,GA,AB,BC,CF)
Shape(EF,FC,CE)
Collinear(DFC)
Collinear(AGFE)
Collinear(BCE)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(DG),1)
IsBisectorOfAngle(AE,DAB)
IsMidpointOfLine(F,DC)
Parallelogram(DABC)
PerpendicularBetweenLine(DG,FG)
Goal: Value(LengthOfLine(AE))
Answer: 4*sqrt(3)
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
line_addition(1,DF,FC)
right_triangle_judgment_angle(1,DGF)
right_triangle_property_pythagorean(1,DGF)
parallel_property_collinear_extend(3,DC,AB,F)
parallel_property_alternate_interior_angle(2,DF,AB)
isosceles_triangle_judgment_angle_equal(1,DAF)
adjacent_complementary_angle(1,AGD,DGF)
mirror_congruent_triangle_judgment_hl(1,AGD,FDG)
mirror_congruent_triangle_property_line_equal(1,DAG,DGF)
line_addition(1,AG,GF)
parallel_property_corresponding_angle(1,FC,AB,E)
similar_triangle_judgment_aa(1,CEF,BEA)
similar_triangle_property_line_ratio(1,CEF,BEA)
similar_triangle_property_line_ratio(1,EFC,EAB)
line_addition(1,EF,FA)
Problem: As shown in the diagram, AB=5, AD=8, AF bisects ∠DAB, DE is the angle bisector of ∠CDA, BA and CD are opposite sides of the parallelogram ABCD. Find the length of line EF.
问题：如图所示，AB=5，AD=8，AF是∠DAB的角平分线，DE是∠CDA的角平分线，AD和BC是▱ABCD的一组对边。求直线EF的长度。
Construction (CDL):
Shape(AB,BE,EM,MA)
Shape(AM,MD,DA)
Shape(ME,EF,FM)
Shape(DM,MF,FC,CD)
Collinear(BEFC)
Collinear(AMF)
Collinear(EMD)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(AD),8)
IsBisectorOfAngle(AF,DAB)
IsBisectorOfAngle(DE,CDA)
Parallelogram(ABCD)
Goal: Value(LengthOfLine(EF))
Answer: 2
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
parallel_property_collinear_extend(3,CB,DA,E)
parallel_property_collinear_extend(3,CB,DA,F)
parallel_property_alternate_interior_angle(2,AD,EC)
parallel_property_alternate_interior_angle(1,AD,BF)
isosceles_triangle_judgment_angle_equal(1,BFA)
isosceles_triangle_judgment_angle_equal(1,CDE)
line_addition(1,EF,FC)
line_addition(1,BF,FC)
Problem: As shown in the diagram, DE=2, DF=3, E is the midpoint of segment DC, quadrilateral DABC is a ▱. Find the perimeter of quadrilateral DABC.
问题：如图所示，DE=2，DF=3，E平分线段DC，AD和BC是平行四边形DABC的一组对边。求四边形DABC的周长。
Construction (CDL):
Shape(FD,DE,EF)
Shape(DA,AB,BE,ED)
Shape(EB,BC,CE)
Collinear(FDA)
Collinear(DEC)
Collinear(FEB)
Conditions (CDL):
Equal(LengthOfLine(DE),2)
Equal(LengthOfLine(DF),3)
IsMidpointOfLine(E,DC)
Parallelogram(DABC)
Goal: Value(PerimeterOfQuadrilateral(DABC))
Answer: 14
Theorem sequence:
line_addition(1,DE,EC)
parallelogram_property_opposite_line_equal(1,ABCD)
parallel_property_corresponding_angle(1,DC,AB,F)
similar_triangle_judgment_aa(1,EFD,BFA)
similar_triangle_property_line_ratio(1,EFD,BFA)
similar_triangle_property_line_ratio(1,FDE,FAB)
line_addition(1,FD,DA)
parallelogram_property_opposite_line_equal(1,DABC)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, AB=10, BC=6, DH=4, the height of ABCD is DF, the height of quadrilateral DABC is DH, DABC is a parallelogram. Find the length of line DF.
问题：如图所示，AB=10，BC=6，DH=4，四边形ABCD的高为DF，四边形DABC的高为DH，AD和BC是▱DABC的一组对边。求直线DF的长度。
Construction (CDL):
Shape(DA,AH,HD)
Shape(DH,HB,BF,FD)
Shape(DF,FC,CD)
Collinear(AHB)
Collinear(BFC)
Conditions (CDL):
Equal(LengthOfLine(AB),10)
Equal(LengthOfLine(BC),6)
Equal(LengthOfLine(DH),4)
IsAltitudeOfQuadrilateral(DF,ABCD)
IsAltitudeOfQuadrilateral(DH,DABC)
Parallelogram(DABC)
Goal: Value(LengthOfLine(DF))
Answer: 20/3
Theorem sequence:
parallelogram_area_formula_common(1,DABC)
parallelogram_area_formula_common(1,ABCD)
Problem: As shown in the diagram, ∠BCD=108°, BE is the angle bisector of ∠ABC, quadrilateral ABCD is a ▱. Find the measure of ∠BEA.
问题：如图所示，∠BCD=108°，BE是∠ABC的角平分线，BA和CD是平行四边形ABCD的一组对边。求∠BEA的大小。
Construction (CDL):
Shape(AB,BE,EA)
Shape(EB,BC,CD,DE)
Collinear(AED)
Conditions (CDL):
Equal(MeasureOfAngle(BCD),108)
IsBisectorOfAngle(BE,ABC)
Parallelogram(ABCD)
Goal: Value(MeasureOfAngle(BEA))
Answer: 36
Theorem sequence:
parallel_property_collinear_extend(3,AD,BC,E)
parallel_property_alternate_interior_angle(2,AE,BC)
parallelogram_property_opposite_angle_equal(1,ABCD)
triangle_property_angle_sum(1,ABE)
Problem: As shown in the diagram, ∠CAP=40°, ∠PCA=100°, the center of circle O is O, the diameter of ⊙O is AD, PA is the tangent to ⊙O. Find the measure of ∠BAC.
问题：如图所示，∠CAP=40°，∠PCA=100°，圆O的圆心为O，⊙O的直径为AD，PA是⊙O的切线。求∠BAC的大小。
Construction (CDL):
Shape(ODB,BE,ED)
Shape(AO,OE,EB,BA)
Shape(OBA,AB)
Shape(DE,EC,CD)
Shape(EO,OA,AC,CE)
Shape(OAC,CA)
Shape(OCD,DC)
Shape(AP,PC,OAC)
Collinear(AOED)
Collinear(BECP)
Cocircular(O,ACDB)
Conditions (CDL):
Equal(MeasureOfAngle(CAP),40)
Equal(MeasureOfAngle(PCA),100)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AD,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(BAC))
Answer: 60
Theorem sequence:
diameter_of_circle_property_right_angle(1,ACD,O)
adjacent_complementary_angle(1,PCA,ACE)
angle_addition(1,ACE,ECD)
arc_property_circumference_angle_external(1,ODB,C)
arc_property_circumference_angle_external(1,ODB,A)
tangent_of_circle_property_perpendicular(1,PA,O,O)
angle_addition(1,BAD,DAP)
triangle_property_angle_sum(1,CAP)
triangle_property_angle_sum(1,BAP)
triangle_property_angle_sum(1,BAC)
Problem: As shown in the diagram, BE=2, ∠ABC=30°, the center of circle O is O, BD is the tangent to ⊙O, CA is the tangent to circle O, CD is the tangent to ⊙O, CA and AB are the legs of the right triangle CAB. Find the length of line DC.
问题：如图所示，BE=2，∠ABC=30°，O是圆O的圆心，BD是⊙O的切线，CA是圆O的切线，圆O的切线为CD，△CAB为直角△。求直线DC的长度。
Construction (CDL):
Shape(CA,ODA,DC)
Shape(ODA,AO,OD)
Shape(DO,OE,OED)
Shape(BD,OED,EB)
Shape(EO,OA,OAE)
Collinear(AOEB)
Collinear(CDB)
Cocircular(O,AED)
Conditions (CDL):
Equal(LengthOfLine(BE),2)
Equal(MeasureOfAngle(ABC),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BD,O)
IsTangentOfCircle(CA,O)
IsTangentOfCircle(CD,O)
RightTriangle(CAB)
Goal: Value(LengthOfLine(DC))
Answer: 2*sqrt(3)
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BD,O,O)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OE,O)
sine_theorem(1,OBD)
line_addition(1,OE,EB)
radius_of_circle_property_length_equal(1,OA,O)
line_addition(1,AO,OB)
triangle_property_angle_sum(1,CAB)
sine_theorem(1,ABC)
tangent_of_circle_property_length_equal(1,CD,CA,O)
Problem: As shown in the diagram, AB=4, CD=9, O is the center of circle O, AD is the diameter of circle O, BA is the tangent to ⊙O, BE is the tangent to circle O, CD is the tangent to ⊙O, CE is the tangent to circle O, DH⊥BH. Find the radius of ⊙O.
问题：如图所示，AB=4，CD=9，⊙O的圆心为O，圆O的直径为AD，BA是圆O的切线，⊙O的切线为BE，CD是圆O的切线，圆O的切线为CE，DH垂直于BH。求圆O的半径。
Construction (CDL):
Shape(CE,OGE,GH,HC)
Shape(OGE,OEF,FG)
Shape(EB,BF,OEF)
Shape(FB,BA,OFA)
Shape(GF,OFA,AO,OD,ODG)
Shape(DH,HG,ODG)
Collinear(BEC)
Collinear(BFGH)
Collinear(CHD)
Collinear(AOD)
Cocircular(O,DGEFA)
Conditions (CDL):
Equal(LengthOfLine(AB),4)
Equal(LengthOfLine(CD),9)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AD,O)
IsTangentOfCircle(BA,O)
IsTangentOfCircle(BE,O)
IsTangentOfCircle(CD,O)
IsTangentOfCircle(CE,O)
PerpendicularBetweenLine(DH,BH)
Goal: Value(RadiusOfCircle(O))
Answer: 6
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BA,O,O)
tangent_of_circle_property_perpendicular(1,CD,O,O)
parallel_judgment_ipsilateral_internal_angle(1,AB,DH)
parallel_judgment_ipsilateral_internal_angle(1,DA,HB)
parallelogram_judgment_parallel_and_parallel(1,BADH)
parallelogram_property_opposite_line_equal(1,BADH)
line_addition(1,CH,HD)
tangent_of_circle_property_length_equal(1,BA,BE,O)
tangent_of_circle_property_length_equal(1,CE,CD,O)
line_addition(1,BE,EC)
adjacent_complementary_angle(1,DHB,BHC)
right_triangle_judgment_angle(1,BHC)
right_triangle_property_pythagorean(1,BHC)
parallelogram_property_opposite_line_equal(1,ADHB)
diameter_of_circle_property_length_equal(1,AD,O)
circle_property_length_of_radius_and_diameter(1,O)
Problem: As shown in the diagram, ∠CAB=55°, O is the center of ⊙O, CA is the tangent to ⊙O, CB is the tangent to circle O. Find the measure of ∠AOB.
问题：如图所示，∠CAB=55°，圆O的圆心为O，圆O的切线为CA，CB是圆O的切线。求∠AOB的大小。
Construction (CDL):
Shape(OA,OAB,BO)
Shape(AO,OB,BA)
Shape(OBA,AB)
Shape(CA,OBA,BC)
Cocircular(O,AB)
Conditions (CDL):
Equal(MeasureOfAngle(CAB),55)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(AOB))
Answer: 110
Theorem sequence:
tangent_of_circle_property_perpendicular(2,CA,O,O)
angle_addition(1,CAB,BAO)
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OBA)
isosceles_triangle_property_angle_equal(1,OBA)
triangle_property_angle_sum(1,OBA)
Problem: As shown in the diagram, ∠ABD=30°, the center of circle O is O, the tangent to circle O is BD. Find the measure of ∠BDA.
问题：如图所示，∠ABD=30°，圆O的圆心为O，BD是圆O的切线。求∠BDA的大小。
Construction (CDL):
Shape(ODA,AD)
Shape(OD,DA,AO)
Shape(OCD,DO,OC)
Shape(CO,OA,OAC)
Shape(CB,BD,OCD)
Collinear(AOCB)
Cocircular(O,CDA)
Conditions (CDL):
Equal(MeasureOfAngle(ABD),30)
IsCentreOfCircle(O,O)
IsTangentOfCircle(BD,O)
Goal: Value(MeasureOfAngle(BDA))
Answer: 120
Theorem sequence:
tangent_of_circle_property_perpendicular(2,BD,O,O)
triangle_property_angle_sum(1,DOB)
adjacent_complementary_angle(1,AOD,DOB)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OA,O)
isosceles_triangle_judgment_line_equal(1,ODA)
isosceles_triangle_property_angle_equal(1,ODA)
triangle_property_angle_sum(1,ODA)
triangle_property_angle_sum(1,DAB)
Problem: As shown in the diagram, AN=6, NB=4, ∠MAC=∠ABC, AC bisects ∠MAB, MA is the tangent to ⊙O, NA is the tangent to ⊙O. Find the length of line AC.
问题：如图所示，AN=6，NB=4，∠MAC=∠ABC，AC是∠MAB的角平分线，⊙O的切线为MA，NA是⊙O的切线。求直线AC的长度。
Construction (CDL):
Shape(AN,NB,OAB)
Shape(OAB,BA)
Shape(CA,AB,BC)
Shape(OBC,CB)
Shape(OCA,AC)
Shape(MA,AC)
Collinear(NAM)
Collinear(NBC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(LengthOfLine(AN),6)
Equal(LengthOfLine(NB),4)
Equal(MeasureOfAngle(MAC),MeasureOfAngle(ABC))
IsBisectorOfAngle(AC,MAB)
IsTangentOfCircle(MA,O)
IsTangentOfCircle(NA,O)
Goal: Value(LengthOfLine(AC))
Answer: 5
Theorem sequence:
circle_property_circular_power_tangent_and_segment_line(1,NA,NBC,O)
line_addition(1,NB,BC)
isosceles_triangle_judgment_angle_equal(1,CAB)
Problem: As shown in the diagram, ∠BCA=65°, the center of circle O is O, the tangent to circle O is PA, PB is the tangent to ⊙O. Find the measure of ∠APB.
问题：如图所示，∠BCA=65°，圆O的圆心为O，⊙O的切线为PA，圆O的切线为PB。求∠APB的大小。
Construction (CDL):
Shape(AP,PB,OAB)
Shape(OA,OAB,BO)
Shape(AO,OB,BC,CA)
Shape(OCA,AC)
Shape(OBC,CB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BCA),65)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(APB))
Answer: 50
Theorem sequence:
arc_property_circumference_angle_external(1,OAB,C)
arc_property_center_angle(1,OAB,O)
tangent_of_circle_property_perpendicular(1,PA,O,O)
tangent_of_circle_property_perpendicular(2,PB,O,O)
quadrilateral_property_angle_sum(1,APBO)
Problem: As shown in the diagram, ∠BPA=68°, the center of ⊙O is O, PA is the tangent to circle O, the tangent to circle O is PB. Find the measure of ∠ACB.
问题：如图所示，∠BPA=68°，O是圆O的圆心，PA是⊙O的切线，PB是⊙O的切线。求∠ACB的大小。
Construction (CDL):
Shape(OAC,CA)
Shape(AC,CB,BO,OA)
Shape(OCB,BC)
Shape(AO,OB,OBA)
Shape(OBA,BP,PA)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BPA),68)
IsCentreOfCircle(O,O)
IsTangentOfCircle(PA,O)
IsTangentOfCircle(PB,O)
Goal: Value(MeasureOfAngle(ACB))
Answer: 56
Theorem sequence:
tangent_of_circle_property_perpendicular(1,PB,O,O)
tangent_of_circle_property_perpendicular(2,PA,O,O)
quadrilateral_property_angle_sum(1,AOBP)
arc_property_center_angle(1,OBA,O)
arc_property_circumference_angle_external(1,OBA,C)
Problem: As shown in the diagram, ∠OPA=30°, the center of ⊙O is O, AB is the diameter of circle O, the tangent to circle O is PA. Find the measure of ∠ABC.
问题：如图所示，∠OPA=30°，圆O的圆心为O，AB是⊙O的直径，PA是圆O的切线。求∠ABC的大小。
Construction (CDL):
Shape(BO,OA,OAB)
Shape(AO,OC,OCA)
Shape(OB,BC,CO)
Shape(OBC,CB)
Shape(CP,PA,OCA)
Collinear(AOB)
Collinear(OCP)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(OPA),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(PA,O)
Goal: Value(MeasureOfAngle(ABC))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(2,PA,O,O)
triangle_property_angle_sum(1,AOP)
adjacent_complementary_angle(1,AOP,POB)
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OB,O)
isosceles_triangle_judgment_line_equal(1,OBC)
isosceles_triangle_property_angle_equal(1,OBC)
triangle_property_angle_sum(1,OBC)
Problem: As shown in the diagram, ∠BAC=20°, the center of circle O is O, AB is the diameter of ⊙O, the tangent to ⊙O is DC. Find the measure of ∠CDB.
问题：如图所示，∠BAC=20°，O是圆O的圆心，AB是圆O的直径，圆O的切线为DC。求∠CDB的大小。
Construction (CDL):
Shape(AO,OB,OBA)
Shape(OA,AC,CO)
Shape(OAC,CA)
Shape(BO,OC,OCB)
Shape(CD,DB,OCB)
Collinear(AOBD)
Cocircular(O,ACB)
Conditions (CDL):
Equal(MeasureOfAngle(BAC),20)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(DC,O)
Goal: Value(MeasureOfAngle(CDB))
Answer: 50
Theorem sequence:
radius_of_circle_property_length_equal(1,OA,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OAC)
isosceles_triangle_property_angle_equal(1,OAC)
triangle_property_angle_sum(1,OAC)
adjacent_complementary_angle(1,BOC,COA)
tangent_of_circle_property_perpendicular(1,DC,O,O)
triangle_property_angle_sum(1,OCD)
Problem: As shown in the diagram, ∠BAT=40°, the center of circle O is O, the diameter of circle O is CB, AT is the tangent to ⊙O. Find the measure of ∠TCB.
问题：如图所示，∠BAT=40°，O是⊙O的圆心，圆O的直径为CB，圆O的切线为AT。求∠TCB的大小。
Construction (CDL):
Shape(OTC,CT)
Shape(TC,CO,OT)
Shape(TO,OB,OBT)
Shape(BO,OC,OCB)
Shape(BA,AT,OBT)
Collinear(COBA)
Cocircular(O,CBT)
Conditions (CDL):
Equal(MeasureOfAngle(BAT),40)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(CB,O)
IsTangentOfCircle(AT,O)
Goal: Value(MeasureOfAngle(TCB))
Answer: 25
Theorem sequence:
tangent_of_circle_property_perpendicular(2,AT,O,O)
triangle_property_angle_sum(1,TOA)
adjacent_complementary_angle(1,COT,TOA)
radius_of_circle_property_length_equal(1,OT,O)
radius_of_circle_property_length_equal(1,OC,O)
isosceles_triangle_judgment_line_equal(1,OTC)
isosceles_triangle_property_angle_equal(1,OTC)
triangle_property_angle_sum(1,OTC)
Problem: As shown in the diagram, ∠BCO=30°, O is the center of circle O, AB is the diameter of ⊙O, the tangent to ⊙O is CB. Find the measure of ∠DAB.
问题：如图所示，∠BCO=30°，圆O的圆心为O，AB是圆O的直径，CB是圆O的切线。求∠DAB的大小。
Construction (CDL):
Shape(ODA,AD)
Shape(DA,AO,OD)
Shape(BO,OA,OAB)
Shape(DO,OB,OBD)
Shape(BC,CD,OBD)
Collinear(ODC)
Collinear(AOB)
Cocircular(O,ABD)
Conditions (CDL):
Equal(MeasureOfAngle(BCO),30)
IsCentreOfCircle(O,O)
IsDiameterOfCircle(AB,O)
IsTangentOfCircle(CB,O)
Goal: Value(MeasureOfAngle(DAB))
Answer: 30
Theorem sequence:
tangent_of_circle_property_perpendicular(1,CB,O,O)
triangle_property_angle_sum(1,COB)
adjacent_complementary_angle(1,AOD,DOB)
radius_of_circle_property_length_equal(1,OD,O)
radius_of_circle_property_length_equal(1,OA,O)
isosceles_triangle_judgment_line_equal(1,ODA)
isosceles_triangle_property_angle_equal(1,ODA)
triangle_property_angle_sum(1,ODA)