SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AD=20, the center of circle O is O, AD is the tangent to ⊙O, the tangent to ⊙O is AE, BF is the tangent to circle O, the tangent to circle O is CF. Find the perimeter of △ABC.
问题：如图所示，AD=20，圆O的圆心为O，圆O的切线为AD，⊙O的切线为AE，⊙O的切线为BF，CF是圆O的切线。求△ABC的周长。
Construction (CDL):
Shape(OED,DO,OE)
Shape(ODF,OFE,EO,OD)
Shape(ODF,DC,CF)
Shape(OFE,FB,BE)
Shape(AB,BF,FC,CA)
Collinear(ACD)
Collinear(ABE)
Collinear(BFC)
Cocircular(O,DFE)
Conditions (CDL):
Equal(LengthOfLine(AD),20)
IsCentreOfCircle(O,O)
IsTangentOfCircle(AD,O)
IsTangentOfCircle(AE,O)
IsTangentOfCircle(BF,O)
IsTangentOfCircle(CF,O)
Goal: Value(PerimeterOfTriangle(ABC))
Answer: 40
Theorem sequence:
tangent_of_circle_property_perpendicular(1,AD,O,O)
tangent_of_circle_property_perpendicular(1,AE,O,O)
tangent_of_circle_judgment_perpendicular(1,CD,O,O)
tangent_of_circle_judgment_perpendicular(1,BE,O,O)
tangent_of_circle_property_length_equal(1,AD,AE,O)
tangent_of_circle_property_length_equal(1,CF,CD,O)
tangent_of_circle_property_length_equal(1,BF,BE,O)
line_addition(1,AC,CD)
line_addition(1,AB,BE)
line_addition(1,BF,FC)
triangle_perimeter_formula(1,ABC)
Problem: As shown in the diagram, AB=AC, BC=10, ∠DAB=90°, BD is the angle bisector of ∠ABE, DA⊥BA, DE is perpendicular to CE. Find the perimeter of △DEC.
问题：如图所示，AB=AC，BC=10，∠DAB=90°，BD平分∠ABE，DA⊥BA，DE⊥CE。求三角形DEC的周长。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BE,ED)
Shape(DE,EC,CD)
Collinear(ADC)
Collinear(BEC)
Conditions (CDL):
Equal(LengthOfLine(AB),LengthOfLine(AC))
Equal(LengthOfLine(BC),10)
Equal(MeasureOfAngle(DAB),90)
IsBisectorOfAngle(BD,ABE)
PerpendicularBetweenLine(DA,BA)
PerpendicularBetweenLine(DE,CE)
Goal: Value(PerimeterOfTriangle(DEC))
Answer: 10
Theorem sequence:
adjacent_complementary_angle(1,BED,DEC)
bisector_of_angle_property_distance_equal(1,BD,ABE)
mirror_congruent_triangle_judgment_hl(1,BED,BDA)
mirror_congruent_triangle_property_line_equal(1,DBE,DAB)
line_addition(1,BE,EC)
line_addition(1,AD,DC)
triangle_perimeter_formula(1,DEC)
Problem: As shown in the diagram, ∠FAD=115°, ED is the perpendicular bisector of AB, GF perpendicular bisects AC. Find the measure of ∠GAE.
问题：如图所示，∠FAD=115°，ED垂直平分AB，GF垂直平分AB。求∠GAE的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BE,ED)
Shape(AE,EG,GA)
Shape(AG,GF,FA)
Shape(FG,GC,CF)
Collinear(ADB)
Collinear(AFC)
Collinear(BEGC)
Conditions (CDL):
Equal(MeasureOfAngle(FAD),115)
IsPerpendicularBisectorOfLine(ED,AB)
IsPerpendicularBisectorOfLine(GF,AC)
Goal: Value(MeasureOfAngle(GAE))
Answer: 50
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,ED,AB)
perpendicular_bisector_property_distance_equal(1,GF,AC)
isosceles_triangle_judgment_line_equal(1,EAB)
isosceles_triangle_judgment_line_equal(1,GCA)
isosceles_triangle_property_angle_equal(1,EAB)
isosceles_triangle_property_angle_equal(1,GCA)
triangle_property_angle_sum(1,ABC)
angle_addition(1,GAE,EAD)
angle_addition(1,FAG,GAD)
Problem: As shown in the diagram, EF=15, the center of circle O is O, EF is the midsegment of quadrilateral ABCD, the tangent to circle O is AG, AM is the tangent to circle O, the tangent to circle O is BG, BN is the tangent to ⊙O, CH is the tangent to circle O, CN is the tangent to ⊙O, DH is the tangent to ⊙O, the tangent to circle O is DM, AD and BC are the parallel sides of isosceles trapezoid ABCD. Find the perimeter of ABCD.
问题：如图所示，EF=15，O是圆O的圆心，EF是ABCD的中位线，⊙O的切线为AG，⊙O的切线为AM，BG是⊙O的切线，圆O的切线为BN，圆O的切线为CH，CN是圆O的切线，⊙O的切线为DH，⊙O的切线为DM，AD和BC是等腰梯形ABCD的平行边。求ABCD的周长。
Construction (CDL):
Shape(OQH,OHM,OMG,OGP,PO,OQ)
Shape(OPN,ONQ,QO,OP)
Shape(OQH,QF,FH)
Shape(OHM,HD,DM)
Shape(OMG,MA,AG)
Shape(OGP,GE,EP)
Shape(OPN,PE,EB,BN)
Shape(ONQ,NC,CF,FQ)
Collinear(AMD)
Collinear(BNC)
Collinear(AGEB)
Collinear(DHFC)
Collinear(EPOQF)
Cocircular(O,MGPNQH)
Conditions (CDL):
Equal(LengthOfLine(EF),15)
IsCentreOfCircle(O,O)
IsMidsegmentOfQuadrilateral(EF,ABCD)
IsTangentOfCircle(AG,O)
IsTangentOfCircle(AM,O)
IsTangentOfCircle(BG,O)
IsTangentOfCircle(BN,O)
IsTangentOfCircle(CH,O)
IsTangentOfCircle(CN,O)
IsTangentOfCircle(DH,O)
IsTangentOfCircle(DM,O)
IsoscelesTrapezoid(ABCD)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 60
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,EF,ABCD)
tangent_of_circle_property_length_equal(1,AM,AG,O)
tangent_of_circle_property_length_equal(1,DM,DH,O)
tangent_of_circle_property_length_equal(1,BG,BN,O)
tangent_of_circle_property_length_equal(1,CH,CN,O)
line_addition(1,AM,MD)
line_addition(1,BN,NC)
line_addition(1,AG,GB)
line_addition(1,DH,HC)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, AB=7, ∠CAE=75°, ∠EBP=60°, AB is perpendicular to EB, CA⊥BA. Find the length of line BP.
问题：如图所示，AB=7，∠CAE=75°，∠EBP=60°，AB垂直于EB，CA垂直于BA。求直线BP的长度。
Construction (CDL):
Shape(AB,BE,EA)
Shape(PE,EB,BP)
Shape(CA,AE)
Shape(AE,ED)
Shape(DE,EP)
Collinear(AEP)
Collinear(DEB)
Conditions (CDL):
Equal(LengthOfLine(AB),7)
Equal(MeasureOfAngle(CAE),75)
Equal(MeasureOfAngle(EBP),60)
PerpendicularBetweenLine(AB,EB)
PerpendicularBetweenLine(CA,BA)
Goal: Value(LengthOfLine(BP))
Answer: 7
Theorem sequence:
angle_addition(1,ABE,EBP)
angle_addition(1,CAE,EAB)
triangle_property_angle_sum(1,PAB)
sine_theorem(1,BPA)
Problem: As shown in the diagram, CB=4, DB=7, D is the midpoint of segment AC. Find the length of line AC.
问题：如图所示，CB=4，DB=7，D是线段AC的中点。求直线AC的长度。
Construction (CDL):
Collinear(ADCB)
Conditions (CDL):
Equal(LengthOfLine(CB),4)
Equal(LengthOfLine(DB),7)
IsMidpointOfLine(D,AC)
Goal: Value(LengthOfLine(AC))
Answer: 6
Theorem sequence:
line_addition(1,AD,DC)
line_addition(1,DC,CB)
Problem: As shown in the diagram, ∠BCG=120°, ∠EAD=26°, triangle DFE is mirror congruent to triangle DEA, DE is parallel to BC. Find the measure of ∠GDB.
问题：如图所示，∠BCG=120°，∠EAD=26°，三角形DFE与三角形DEA是镜像全等三角形，DE平行于BC。求∠GDB的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(FE,EG,GF)
Shape(ED,DG,GE)
Shape(DB,BC,CG,GD)
Collinear(ADB)
Collinear(AEGC)
Collinear(DGF)
Conditions (CDL):
Equal(MeasureOfAngle(BCG),120)
Equal(MeasureOfAngle(EAD),26)
MirrorCongruentBetweenTriangle(DFE,DEA)
ParallelBetweenLine(DE,BC)
Goal: Value(MeasureOfAngle(GDB))
Answer: 112
Theorem sequence:
mirror_congruent_triangle_property_angle_equal(1,DFE,DEA)
parallel_property_corresponding_angle(2,CB,ED,A)
triangle_property_angle_sum(1,ADE)
angle_addition(1,ADE,EDG)
adjacent_complementary_angle(1,ADG,GDB)
Problem: As shown in the diagram, CE=6, DE=3/2, EA=2, BC⊥EC, DE⊥AE. Find the length of line BC.
问题：如图所示，CE=6，DE=3/2，EA=2，BC⊥EC，DE⊥AE。求直线BC的长度。
Construction (CDL):
Shape(AD,DE,EA)
Shape(DB,BC,CE,ED)
Collinear(AEC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(CE),6)
Equal(LengthOfLine(DE),3/2)
Equal(LengthOfLine(EA),2)
PerpendicularBetweenLine(BC,EC)
PerpendicularBetweenLine(DE,AE)
Goal: Value(LengthOfLine(BC))
Answer: 6
Theorem sequence:
line_addition(1,AE,EC)
similar_triangle_judgment_aa(1,DEA,BCA)
similar_triangle_property_line_ratio(1,DEA,BCA)
similar_triangle_property_line_ratio(1,ADE,ABC)
Problem: As shown in the diagram, ∠EDC=52°, O is the center of ⊙O, DE is the tangent to circle O, BH is perpendicular to AH. Find the measure of ∠ABE.
问题：如图所示，∠EDC=52°，O是⊙O的圆心，DE是⊙O的切线，BH⊥AH。求∠ABE的大小。
Construction (CDL):
Shape(OCA,AH,HC)
Shape(OAB,BA)
Shape(AB,BH,HA)
Shape(HB,BE,EO,OH)
Shape(OEC,CH,HO,OE)
Shape(OEC,ED,DC)
Shape(OBE,EB)
Collinear(BHCD)
Collinear(AHO)
Cocircular(O,ABEC)
Conditions (CDL):
Equal(MeasureOfAngle(EDC),52)
IsCentreOfCircle(O,O)
IsTangentOfCircle(DE,O)
PerpendicularBetweenLine(BH,AH)
Goal: Value(MeasureOfAngle(ABE))
Answer: 64
Theorem sequence:
tangent_of_circle_property_perpendicular(1,DE,O,O)
vertical_angle(1,BHA,CHO)
quadrilateral_property_angle_sum(1,DHOE)
arc_property_center_angle(1,OEA,O)
arc_property_circumference_angle_external(1,OEA,B)
Problem: As shown in the diagram, ∠GBD=120°, ED perpendicular bisects BA, FG is the perpendicular bisector of CB. Find the measure of ∠FBE.
问题：如图所示，∠GBD=120°，ED是BA的垂直平分线，FG垂直平分AB。求∠FBE的大小。
Construction (CDL):
Shape(BD,DE,EB)
Shape(DA,AE,ED)
Shape(BE,EF,FB)
Shape(BF,FG,GB)
Shape(CG,GF,FC)
Collinear(ADB)
Collinear(BGC)
Collinear(AEFC)
Conditions (CDL):
Equal(MeasureOfAngle(GBD),120)
IsPerpendicularBisectorOfLine(ED,BA)
IsPerpendicularBisectorOfLine(FG,CB)
Goal: Value(MeasureOfAngle(FBE))
Answer: 60
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,ED,BA)
perpendicular_bisector_property_distance_equal(1,FG,CB)
isosceles_triangle_judgment_line_equal(1,EBA)
isosceles_triangle_judgment_line_equal(1,FCB)
isosceles_triangle_property_angle_equal(1,EBA)
isosceles_triangle_property_angle_equal(1,FCB)
triangle_property_angle_sum(1,BAC)
angle_addition(1,FBE,EBD)
angle_addition(1,GBF,FBD)
Problem: As shown in the diagram, BF=BE, DF=DE, ∠FBE=50°, BD is the angle bisector of ∠FBE, DE∥AB. Find the measure of ∠DFB.
问题：如图所示，BF=BE，DF=DE，∠FBE=50°，BD是∠FBE的角平分线，DE平行于AB。求∠DFB的大小。
Construction (CDL):
Shape(DF,FB,BD)
Shape(DB,BE,ED)
Shape(AF,FD)
Shape(DE,EC)
Collinear(AFB)
Collinear(BEC)
Collinear(BPG)
Conditions (CDL):
Equal(LengthOfLine(BF),LengthOfLine(BE))
Equal(LengthOfLine(DF),LengthOfLine(DE))
Equal(MeasureOfAngle(FBE),50)
IsBisectorOfAngle(BD,FBE)
ParallelBetweenLine(DE,AB)
Goal: Value(MeasureOfAngle(DFB))
Answer: 130
Theorem sequence:
parallel_property_ipsilateral_internal_angle(1,BA,ED)
mirror_congruent_triangle_judgment_sas(1,BDF,BED)
mirror_congruent_triangle_property_angle_equal(1,FBD,EDB)
Problem: As shown in the diagram, AB=16, D bisects segment AC, E is the midpoint of segment BC. Find the length of line DE.
问题：如图所示，AB=16，D是线段AC的中点，E是线段BC的中点。求直线DE的长度。
Construction (CDL):
Collinear(ADCEB)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
IsMidpointOfLine(D,AC)
IsMidpointOfLine(E,BC)
Goal: Value(LengthOfLine(DE))
Answer: 8
Theorem sequence:
line_addition(1,AD,DC)
line_addition(1,AD,DE)
line_addition(1,AC,CE)
line_addition(1,AE,EB)
Problem: As shown in the diagram, AC=4, ∠BAC=37°, ∠ACB is the right angle of the right triangle ACB. Find the length of line BC.
问题：如图所示，AC=4，∠BAC=37°，AC和CB是直角△ACB的两个直角边。求直线BC的长度。
Construction (CDL):
Shape(AC,CB,BA)
Conditions (CDL):
Equal(LengthOfLine(AC),4)
Equal(MeasureOfAngle(BAC),37)
RightTriangle(ACB)
Goal: Value(LengthOfLine(BC))
Answer: 4*tan(37*pi/180)
Theorem sequence:
triangle_property_angle_sum(1,ACB)
sine_theorem(1,CBA)
Problem: As shown in the diagram, ∠EAD=50°, DE is the perpendicular bisector of CA. Find the measure of ∠CDB.
问题：如图所示，∠EAD=50°，DE垂直平分AB。求∠CDB的大小。
Construction (CDL):
Shape(AD,DE,EA)
Shape(ED,DC,CE)
Shape(DB,BC,CD)
Collinear(AEC)
Collinear(ADB)
Conditions (CDL):
Equal(MeasureOfAngle(EAD),50)
IsPerpendicularBisectorOfLine(DE,CA)
Goal: Value(MeasureOfAngle(CDB))
Answer: 100
Theorem sequence:
perpendicular_bisector_property_bisector(1,DE,CA)
triangle_property_angle_sum(1,ADE)
angle_addition(1,ADE,EDC)
adjacent_complementary_angle(1,ADC,CDB)
Problem: As shown in the diagram, ∠GHD=80°, ∠MGA=∠HGM, AB is parallel to CD. Find the measure of ∠GMH.
问题：如图所示，∠GHD=80°，∠MGA=∠HGM，AB平行于CD。求∠GMH的大小。
Construction (CDL):
Shape(GM,MH,HG)
Shape(AG,GE)
Shape(EG,GB)
Shape(MG,GA)
Shape(BG,GH)
Shape(CM,MG)
Shape(NM,MC)
Shape(HM,MN)
Shape(FH,HM)
Shape(DH,HF)
Shape(GH,HD)
Collinear(AGB)
Collinear(CMHD)
Collinear(EHF)
Collinear(GMN)
Conditions (CDL):
Equal(MeasureOfAngle(GHD),80)
Equal(MeasureOfAngle(MGA),MeasureOfAngle(HGM))
ParallelBetweenLine(AB,CD)
Goal: Value(MeasureOfAngle(GMH))
Answer: 40
Theorem sequence:
parallel_property_collinear_extend(3,AB,CD,G)
parallel_property_collinear_extend(3,DC,GA,M)
parallel_property_collinear_extend(3,DC,GA,H)
parallel_property_alternate_interior_angle(2,AG,HD)
angle_addition(1,HGM,MGA)
parallel_property_alternate_interior_angle(2,AG,MD)
Problem: As shown in the diagram, ∠DEC=80°, BD is the angle bisector of ∠ABE, DE is parallel to AB. Find the measure of ∠EDB.
问题：如图所示，∠DEC=80°，BD平分∠ABE，DE∥AB。求∠EDB的大小。
Construction (CDL):
Shape(AB,BD,DA)
Shape(DB,BE,ED)
Shape(CD,DE,EC)
Collinear(ADC)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(DEC),80)
IsBisectorOfAngle(BD,ABE)
ParallelBetweenLine(DE,AB)
Goal: Value(MeasureOfAngle(EDB))
Answer: 40
Theorem sequence:
parallel_property_corresponding_angle(2,BA,ED,C)
angle_addition(1,ABD,DBE)
parallel_property_alternate_interior_angle(1,DE,AB)
Problem: As shown in the diagram, ∠DMA=40°, DE is parallel to FG, △ABC is a right △. Find the measure of ∠NBG.
问题：如图所示，∠DMA=40°，DE∥FG，∠ABC是直角△ABC的直角。求∠NBG的大小。
Construction (CDL):
Shape(AM,MN,NC,CA)
Shape(MB,BN,NM)
Shape(DM,MA)
Shape(CN,NE)
Shape(FB,BM)
Shape(NB,BG)
Collinear(AMB)
Collinear(CNB)
Collinear(DMNE)
Collinear(FBG)
Conditions (CDL):
Equal(MeasureOfAngle(DMA),40)
ParallelBetweenLine(DE,FG)
RightTriangle(ABC)
Goal: Value(MeasureOfAngle(NBG))
Answer: 50
Theorem sequence:
vertical_angle(1,DMA,NMB)
triangle_property_angle_sum(1,MBN)
parallel_property_collinear_extend(3,DE,FG,N)
parallel_property_collinear_extend(3,GF,ND,B)
parallel_property_alternate_interior_angle(2,DN,BG)
Problem: As shown in the diagram, the perimeter of quadrilateral ABCD is 22, AD and BC are opposite sides of the ▱ ABCD, AO is perpendicular to EO. Find the perimeter of △CDE.
问题：如图所示，ABCD的周长为22，四边形ABCD是▱，AO⊥EO。求△CDE的周长。
Construction (CDL):
Shape(EA,AO,OE)
Shape(DE,EF,FD)
Shape(FE,EO,OF)
Shape(DF,FC,CD)
Shape(FO,OC,CF)
Shape(OA,AB,BO)
Shape(OB,BC,CO)
Collinear(AED)
Collinear(AOC)
Collinear(BOFD)
Collinear(EFC)
Conditions (CDL):
Equal(PerimeterOfQuadrilateral(ABCD),22)
Parallelogram(ABCD)
PerpendicularBetweenLine(AO,EO)
Goal: Value(PerimeterOfTriangle(CDE))
Answer: 11
Theorem sequence:
quadrilateral_perimeter_formula(1,ABCD)
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
parallelogram_property_diagonal_bisection(1,ABCD,O)
adjacent_complementary_angle(1,AOE,EOC)
mirror_congruent_triangle_judgment_sas(1,OEA,OCE)
mirror_congruent_triangle_property_line_equal(1,OEA,OCE)
line_addition(1,AE,ED)
triangle_perimeter_formula(1,CDE)
Problem: As shown in the diagram, triangle CBA is congruent to triangle DEA, ∠BAE=90°, ∠EDC=33°. Find the measure of ∠CBA.
问题：如图所示，△CBA全等于△DEA，∠BAE=90°，∠EDC=33°。求∠CBA的大小。
Construction (CDL):
Shape(CB,BA,AE,EC)
Shape(DE,EA,AD)
Shape(DC,CE,ED)
Collinear(CEA)
Collinear(BAD)
Conditions (CDL):
CongruentBetweenTriangle(CBA,DEA)
Equal(MeasureOfAngle(BAE),90)
Equal(MeasureOfAngle(EDC),33)
Goal: Value(MeasureOfAngle(CBA))
Answer: 78
Theorem sequence:
congruent_triangle_property_line_equal(1,BAC,EAD)
isosceles_triangle_judgment_line_equal(1,ADC)
isosceles_triangle_property_angle_equal(1,ADC)
adjacent_complementary_angle(1,CED,DEA)
adjacent_complementary_angle(1,BAE,EAD)
triangle_property_angle_sum(1,CAD)
triangle_property_angle_sum(1,CED)
congruent_triangle_property_angle_equal(1,BAC,EAD)
Problem: As shown in the diagram, ∠BMD=55°, DE∥FG, FG is parallel to HI, AB and BC are the legs of the right triangle ABC. Find the measure of ∠HPB.
问题：如图所示，∠BMD=55°，DE平行于FG，FG平行于HI，∠ABC是直角三角形ABC的直角。求∠HPB的大小。
Construction (CDL):
Shape(AM,MN,NA)
Shape(MB,BK,KN,NM)
Shape(BP,PQ,QK,KB)
Shape(PC,CQ,QP)
Shape(DM,MA)
Shape(AN,NE)
Shape(BM,MD)
Shape(EN,NK)
Shape(FB,BM)
Shape(NK,KG)
Shape(HP,PB)
Shape(KQ,QI)
Shape(GK,KQ)
Shape(PB,BF)
Shape(CP,PH)
Shape(IQ,QC)
Collinear(DMNE)
Collinear(FBKG)
Collinear(HPQI)
Collinear(AMB)
Collinear(ANKQC)
Collinear(BPC)
Conditions (CDL):
Equal(MeasureOfAngle(BMD),55)
ParallelBetweenLine(DE,FG)
ParallelBetweenLine(FG,HI)
RightTriangle(ABC)
Goal: Value(MeasureOfAngle(HPB))
Answer: 35
Theorem sequence:
parallel_property_collinear_extend(3,DE,FG,M)
parallel_property_collinear_extend(3,FG,HI,B)
parallel_property_collinear_extend(3,GF,MD,B)
parallel_property_collinear_extend(3,IH,GB,P)
parallel_property_alternate_interior_angle(1,PH,GB)
parallel_property_alternate_interior_angle(2,GB,MD)
angle_addition(1,MBK,KBP)
Problem: As shown in the diagram, AB=6, BC=8, ABCD is a rectangle. Find the perimeter of quadrilateral ABCD.
问题：如图所示，AB=6，BC=8，ABCD是长方形。求四边形ABCD的周长。
Construction (CDL):
Shape(AB,BC,CA)
Shape(AC,CD,DA)
Conditions (CDL):
Equal(LengthOfLine(AB),6)
Equal(LengthOfLine(BC),8)
Rectangle(ABCD)
Goal: Value(PerimeterOfQuadrilateral(ABCD))
Answer: 28
Theorem sequence:
parallelogram_property_opposite_line_equal(1,ABCD)
parallelogram_property_opposite_line_equal(1,BCDA)
quadrilateral_perimeter_formula(1,ABCD)
Problem: As shown in the diagram, ∠AOC=35°. Find the measure of ∠COB.
问题：如图所示，∠AOC=35°。求∠COB的大小。
Construction (CDL):
Shape(AO,OC)
Shape(CO,OB)
Collinear(AOB)
Conditions (CDL):
Equal(MeasureOfAngle(AOC),35)
Goal: Value(MeasureOfAngle(COB))
Answer: 145
Theorem sequence:
adjacent_complementary_angle(1,AOC,COB)
Problem: As shown in the diagram, ∠DCE=62°, O is the center of circle O. Find the measure of ∠DOB.
问题：如图所示，∠DCE=62°，⊙O的圆心为O。求∠DOB的大小。
Construction (CDL):
Shape(ODA,AD)
Shape(OAB,BA)
Shape(DA,AB,BO,OD)
Shape(OBC,CB)
Shape(OCD,DC)
Shape(OB,BC,CD,DO)
Shape(DC,CE)
Collinear(BCE)
Cocircular(O,ABCD)
Conditions (CDL):
Equal(MeasureOfAngle(DCE),62)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(DOB))
Answer: 124
Theorem sequence:
adjacent_complementary_angle(1,BCD,DCE)
arc_property_circumference_angle_internal(1,OBD,C)
arc_property_center_angle(1,OBD,O)
Problem: As shown in the diagram, ∠BOD=80°, the center of circle O is O. Find the measure of ∠DCA.
问题：如图所示，∠BOD=80°，圆O的圆心为O。求∠DCA的大小。
Construction (CDL):
Shape(OBC,CD,DO,OB)
Shape(OCA,AC)
Shape(OAB,BD,DA)
Shape(DC,CA,AD)
Shape(OD,DB,BO)
Collinear(ADO)
Collinear(CDB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(BOD),80)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(DCA))
Answer: 40
Theorem sequence:
arc_property_center_angle(1,OAB,O)
arc_property_circumference_angle_external(1,OAB,C)
Problem: As shown in the diagram, ∠CBA=100°. Find the measure of ∠EDA.
问题：如图所示，∠CBA=100°。求∠EDA的大小。
Construction (CDL):
Shape(OAD,DA)
Shape(ODC,CD)
Shape(OCB,BC)
Shape(OBA,AB)
Shape(AD,DC,CB,BA)
Shape(ED,DA)
Collinear(EDC)
Cocircular(O,ADCB)
Conditions (CDL):
Equal(MeasureOfAngle(CBA),100)
Goal: Value(MeasureOfAngle(EDA))
Answer: 100
Theorem sequence:
arc_property_circumference_angle_external(1,OAC,B)
arc_property_circumference_angle_internal(1,OAC,D)
adjacent_complementary_angle(1,EDA,ADC)
Problem: As shown in the diagram, Add(MeasureOfAngle(CBQ)=MeasureOfAngle(DCG). Find the measure of ∠BAE.
问题：如图所示，Add(MeasureOfAngle(CBQ)=MeasureOfAngle(DCG)。求∠BAE的大小。
Construction (CDL):
Shape(AB,BC,CD,DA)
Shape(BA,AE)
Shape(EA,AP)
Shape(PA,AD)
Shape(FD,DC)
Shape(CB,BQ)
Shape(DC,CG)
Collinear(EADF)
Collinear(PABQ)
Collinear(BCG)
Conditions (CDL):
Equal(Add(MeasureOfAngle(CBQ),MeasureOfAngle(DCG),MeasureOfAngle(FDC)),320)
Goal: Value(MeasureOfAngle(BAE))
Answer: 40
Theorem sequence:
adjacent_complementary_angle(1,DAB,BAE)
adjacent_complementary_angle(1,FDC,CDE)
adjacent_complementary_angle(1,PBC,CBQ)
adjacent_complementary_angle(1,BCD,DCG)
quadrilateral_property_angle_sum(1,ABCD)
Problem: As shown in the diagram, ∠AOC=130°, AO and OB are the legs of the right triangle AOB, DO and OC are the legs of the right △ DOC. Find the measure of ∠EOG.
问题：如图所示，∠AOC=130°，三角形AOB为直角三角形，∠DOC是直角三角形DOC的直角。求∠EOG的大小。
Construction (CDL):
Shape(AO,OE,EA)
Shape(DE,EF,FD)
Shape(FE,EO,OG,GF)
Shape(FG,GB,BF)
Shape(OC,CG,GO)
Collinear(AEFB)
Collinear(OGB)
Collinear(OED)
Collinear(DFGC)
Conditions (CDL):
Equal(MeasureOfAngle(AOC),130)
RightTriangle(AOB)
RightTriangle(DOC)
Goal: Value(MeasureOfAngle(EOG))
Answer: 50
Theorem sequence:
angle_addition(1,AOE,EOG)
angle_addition(1,AOE,EOC)
Problem: As shown in the diagram, AC=DC, ∠DBE=25°, DE is the perpendicular bisector of BC. Find the measure of ∠DCA.
问题：如图所示，AC=DC，∠DBE=25°，DE垂直平分AB。求∠DCA的大小。
Construction (CDL):
Shape(CA,AD,DC)
Shape(EC,CD,DE)
Shape(BE,ED,DB)
Collinear(CEB)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AC),LengthOfLine(DC))
Equal(MeasureOfAngle(DBE),25)
IsPerpendicularBisectorOfLine(DE,BC)
Goal: Value(MeasureOfAngle(DCA))
Answer: 80
Theorem sequence:
perpendicular_bisector_property_distance_equal(1,DE,BC)
isosceles_triangle_judgment_line_equal(1,CAD)
isosceles_triangle_judgment_line_equal(1,DBC)
isosceles_triangle_property_angle_equal(1,CAD)
isosceles_triangle_property_angle_equal(1,DBC)
triangle_property_angle_sum(1,CDB)
adjacent_complementary_angle(1,ADC,CDB)
triangle_property_angle_sum(1,CAD)
Problem: As shown in the diagram, ∠GFC=25°, ∠HFG=30°, ABCD is a rectangle, ∠GEF is the right angle of the right triangle GEF. Find the measure of ∠HGE.
问题：如图所示，∠GFC=25°，∠HFG=30°，四边形ABCD是长方形，∠GEF是直角三角形GEF的直角。求∠HGE的大小。
Construction (CDL):
Shape(AB,BF,FH,HA)
Shape(GH,HF,FG)
Shape(DG,GF,FC,CD)
Shape(EF,FG,GE)
Collinear(AHGD)
Collinear(BFC)
Collinear(FHE)
Conditions (CDL):
Equal(MeasureOfAngle(GFC),25)
Equal(MeasureOfAngle(HFG),30)
Rectangle(ABCD)
RightTriangle(GEF)
Goal: Value(MeasureOfAngle(HGE))
Answer: 35
Theorem sequence:
triangle_property_angle_sum(1,GEF)
parallel_property_collinear_extend(3,AD,BC,G)
parallel_property_collinear_extend(3,CB,GA,F)
parallel_property_alternate_interior_angle(2,CF,GA)
angle_addition(1,FGH,HGE)
Problem: As shown in the diagram, ∠ABO=25°, O is the center of circle O, CA is the tangent to circle O. Find the measure of ∠DCA.
问题：如图所示，∠ABO=25°，O是圆O的圆心，CA是圆O的切线。求∠DCA的大小。
Construction (CDL):
Shape(ODA,AO,OD)
Shape(OAB,BA)
Shape(OBD,DO,OB)
Shape(ODA,DC,CA)
Shape(OA,AB,BO)
Collinear(BODC)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(ABO),25)
IsCentreOfCircle(O,O)
IsTangentOfCircle(CA,O)
Goal: Value(MeasureOfAngle(DCA))
Answer: 40
Theorem sequence:
arc_property_circumference_angle_external(1,OCA,B)
arc_property_center_angle(1,OCA,O)
tangent_of_circle_property_perpendicular(2,CA,O,O)
triangle_property_angle_sum(1,AOC)
Problem: As shown in the diagram, AB=20, O is the center of circle O, the center of ⊙P is O, the tangent to circle O is AC, BC is the tangent to circle O. Find the area of the ⊙P minus the area of the ⊙O.
问题：如图所示，AB=20，⊙O的圆心为O，O是⊙P的圆心，⊙O的切线为AC，BC是⊙O的切线。求⊙P的面积减去⊙O的面积。
Construction (CDL):
Shape(OCD,DO,OC)
Shape(ODC,CO,OD)
Shape(PBA,AC,ODC,DB)
Shape(OCD,CB,BD)
Shape(PAB,BC,CA)
Shape(O)
Shape(P)
Collinear(ACB)
Collinear(ODB)
Cocircular(O,CD)
Cocircular(P,AB)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
IsCentreOfCircle(O,O)
IsCentreOfCircle(O,P)
IsTangentOfCircle(AC,O)
IsTangentOfCircle(BC,O)
Goal: Value(Sub(AreaOfCircle(P),AreaOfCircle(O)))
Answer: 100*pi
Theorem sequence:
radius_of_circle_property_length_equal(1,OC,O)
radius_of_circle_property_length_equal(1,OB,P)
tangent_of_circle_property_perpendicular(2,AC,O,O)
circle_property_chord_perpendicular_bisect_chord(1,P,OC,AB)
line_addition(1,AC,CB)
right_triangle_judgment_angle(1,OCB)
right_triangle_property_pythagorean(1,OCB)
circle_area_formula(1,O)
circle_area_formula(1,P)
Problem: As shown in the diagram, ∠AOB=65°, ∠ECA=20°, triangle ODA is mirror congruent to triangle OBC. Find the measure of ∠EAO.
问题：如图所示，∠AOB=65°，∠ECA=20°，三角形ODA镜像全等于三角形OBC。求∠EAO的大小。
Construction (CDL):
Shape(OB,BE,EA,AO)
Shape(BD,DE,EB)
Shape(AE,EC,CA)
Collinear(OBD)
Collinear(OAC)
Collinear(BEC)
Collinear(DEA)
Conditions (CDL):
Equal(MeasureOfAngle(AOB),65)
Equal(MeasureOfAngle(ECA),20)
MirrorCongruentBetweenTriangle(ODA,OBC)
Goal: Value(MeasureOfAngle(EAO))
Answer: 95
Theorem sequence:
triangle_property_angle_sum(1,OBC)
mirror_congruent_triangle_property_angle_equal(1,AOD,BCO)
Problem: As shown in the diagram, ∠OCA=25°, the center of circle O is O. Find the measure of ∠COB.
问题：如图所示，∠OCA=25°，O是圆O的圆心。求∠COB的大小。
Construction (CDL):
Shape(OAB,BO,OA)
Shape(OCA,AC)
Shape(OBC,CO,OB)
Shape(AO,OC,CA)
Collinear(AOB)
Cocircular(O,ABC)
Conditions (CDL):
Equal(MeasureOfAngle(OCA),25)
IsCentreOfCircle(O,O)
Goal: Value(MeasureOfAngle(COB))
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,OCA)
isosceles_triangle_property_angle_equal(1,OCA)
arc_property_center_angle(1,OBC,O)
arc_property_circumference_angle_external(1,OBC,A)
Problem: As shown in the diagram, ∠ACD=34°, ∠ACE=88°, ∠FCE=34°. Find the measure of ∠DCF.
问题：如图所示，∠ACD=34°，∠ACE=88°，∠FCE=34°。求∠DCF的大小。
Construction (CDL):
Shape(BA,AC,CD,DF,FB)
Shape(DC,CF,FD)
Shape(FC,CE,EF)
Collinear(CFB)
Collinear(DFE)
Conditions (CDL):
Equal(MeasureOfAngle(ACD),34)
Equal(MeasureOfAngle(ACE),88)
Equal(MeasureOfAngle(FCE),34)
Goal: Value(MeasureOfAngle(DCF))
Answer: 20
Theorem sequence:
angle_addition(1,ACD,DCE)
angle_addition(1,DCF,FCE)