SunnyLin/FormalGeo7k_v1 huggingface dataset:
Problem: As shown in the diagram, AC=12, ∠BCA=60°, AB⊥CB. Find the length of line AB.
问题：如图所示，AC=12，∠BCA=60°，AB垂直于CB。求直线AB的长度。
Construction (CDL):
Shape(BC,CA,AB)
Conditions (CDL):
Equal(LengthOfLine(AC),12)
Equal(MeasureOfAngle(BCA),60)
PerpendicularBetweenLine(AB,CB)
Goal: Value(LengthOfLine(AB))
Answer: 6*sqrt(3)
Theorem sequence:
sine_theorem(1,ABC)
Problem: As shown in the diagram, DB=4, DE=5, FH=3, ∠GDF=32°, AD⊥HD. Find the measure of ∠EDG.
问题：如图所示，DB=4，DE=5，FH=3，∠GDF=32°，AD垂直于HD。求∠EDG的大小。
Construction (CDL):
Shape(GD,DF)
Shape(FD,DA)
Shape(AD,DH)
Shape(ED,DG)
Shape(HD,DE)
Collinear(FDH)
Collinear(ADE)
Conditions (CDL):
Equal(LengthOfLine(DB),4)
Equal(LengthOfLine(DE),5)
Equal(LengthOfLine(FH),3)
Equal(MeasureOfAngle(GDF),32)
PerpendicularBetweenLine(AD,HD)
Goal: Value(MeasureOfAngle(EDG))
Answer: 58
Theorem sequence:
vertical_angle(1,EDF,ADH)
angle_addition(1,EDG,GDF)
Problem: As shown in the diagram, PS=4*x+8, QP=3*x+11, quadrilateral QTSR is a rectangle. Find the length of line QS.
问题：如图所示，PS=4*x+8，QP=3*x+11，四边形QTSR是矩形。求直线QS的长度。
Construction (CDL):
Shape(QP,PR,RQ)
Shape(TP,PQ,QT)
Shape(PT,TS,SP)
Shape(PS,SR,RP)
Collinear(QPS)
Collinear(TPR)
Conditions (CDL):
Equal(LengthOfLine(PS),4*x+8)
Equal(LengthOfLine(QP),3*x+11)
Rectangle(QTSR)
Goal: Value(LengthOfLine(QS))
Answer: 40
Theorem sequence:
parallelogram_property_diagonal_bisection(1,QTSR,P)
line_addition(1,QP,PS)
Problem: As shown in the diagram, ∠BDF=78°, ∠EGF=120°, ∠FBD=50°, ∠GEC=56°, ∠GFD=x°. Find the value of x.
问题：如图所示，∠BDF=78°，∠EGF=120°，∠FBD=50°，∠GEC=56°，∠GFD=x°。求x的值。
Construction (CDL):
Shape(BD,DF,FB)
Shape(FG,GA,AF)
Shape(GE,EC,CG)
Shape(GF,FD)
Shape(EG,GF)
Collinear(BFGC)
Collinear(DFA)
Collinear(AGE)
Conditions (CDL):
Equal(MeasureOfAngle(BDF),78)
Equal(MeasureOfAngle(EGF),120)
Equal(MeasureOfAngle(FBD),50)
Equal(MeasureOfAngle(GEC),56)
Equal(MeasureOfAngle(GFD),x)
Goal: Value(x)
Answer: 128
Theorem sequence:
triangle_property_angle_sum(1,BDF)
flat_angle(1,GFB)
angle_addition(1,GFD,DFB)
Problem: As shown in the diagram, ∠MJN=35°, KLJM is a parallelogram, PNML is a rectangle. Find the measure of ∠MPN.
问题：如图所示，∠MJN=35°，KLJM是平行四边形，PNML是矩形。求∠MPN的大小。
Construction (CDL):
Shape(JP,PN,NJ)
Shape(PJ,JL,LP)
Shape(LJ,JM,ML)
Shape(JN,NM,MJ)
Shape(LM,MK,KL)
Collinear(PJM)
Collinear(NJL)
Conditions (CDL):
Equal(MeasureOfAngle(MJN),35)
Parallelogram(KLJM)
Rectangle(PNML)
Goal: Value(MeasureOfAngle(MPN))
Answer: 35/2
Theorem sequence:
rectangle_property_diagonal_equal(1,PNML)
parallelogram_property_diagonal_bisection(1,PNML,J)
parallelogram_property_diagonal_bisection(1,NMLP,J)
line_addition(1,PJ,JM)
line_addition(1,NJ,JL)
adjacent_complementary_angle(1,MJN,NJP)
isosceles_triangle_judgment_line_equal(1,JPN)
isosceles_triangle_property_angle_equal(1,JPN)
triangle_property_angle_sum(1,JPN)
Problem: As shown in the diagram, CD=22, ∠DCX=18°, ∠XDC=105°. Find the length of line XD.
问题：如图所示，CD=22，∠DCX=18°，∠XDC=105°。求直线XD的长度。
Construction (CDL):
Shape(CX,XD,DC)
Conditions (CDL):
Equal(LengthOfLine(CD),22)
Equal(MeasureOfAngle(DCX),18)
Equal(MeasureOfAngle(XDC),105)
Goal: Value(LengthOfLine(XD))
Answer: 88*(-1+sqrt(5))/(-sqrt(30)-sqrt(2)+sqrt(5-sqrt(5))+sqrt(6)+sqrt(3)*sqrt(5-sqrt(5))+sqrt(10)+sqrt(5)*sqrt(5-sqrt(5))+sqrt(15)*sqrt(5-sqrt(5)))
Theorem sequence:
triangle_property_angle_sum(1,CXD)
sine_theorem(1,CXD)
sine_theorem(1,XDC)
Problem: As shown in the diagram, the area of ADBC is 400, the area of quadrilateral EFGH is 64, CB=40, GH=x, ADBC is similar to EFGH. Find the value of x.
问题：如图所示，ADBC的面积为400，EFGH的面积为64，CB=40，GH=x，ADBC相似于EFGH。求x的值。
Construction (CDL):
Shape(AD,DB,BC,CA)
Shape(EF,FG,GH,HE)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ADBC),400)
Equal(AreaOfQuadrilateral(EFGH),64)
Equal(LengthOfLine(CB),40)
Equal(LengthOfLine(GH),x)
SimilarBetweenQuadrilateral(ADBC,EFGH)
Goal: Value(x)
Answer: 16
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,ADBC,EFGH)
similar_quadrilateral_property_line_ratio(1,BCAD,GHEF)
Problem: As shown in the diagram, the area of ADCB is 150, the area of quadrilateral FJHG is 54, CD=10, JH=x, AB and DC are opposite sides of the parallelogram ADCB, JF and HG are opposite sides of the parallelogram FJHG, quadrilateral FJHG is similar to quadrilateral ADCB. Find the value of x.
问题：如图所示，四边形ADCB的面积为150，四边形FJHG的面积为54，CD=10，JH=x，AB和DC是▱ADCB的一组对边，JF和HG是▱FJHG的一组对边，四边形FJHG与四边形ADCB相似。求x的值。
Construction (CDL):
Shape(FJ,JH,HG,GF)
Shape(AD,DC,CB,BA)
Conditions (CDL):
Equal(AreaOfQuadrilateral(ADCB),150)
Equal(AreaOfQuadrilateral(FJHG),54)
Equal(LengthOfLine(CD),10)
Equal(LengthOfLine(JH),x)
Parallelogram(ADCB)
Parallelogram(FJHG)
SimilarBetweenQuadrilateral(FJHG,ADCB)
Goal: Value(x)
Answer: 6
Theorem sequence:
similar_quadrilateral_property_area_square_ratio(1,FJHG,ADCB)
similar_quadrilateral_property_line_ratio(1,JHGF,DCBA)
Problem: As shown in the diagram, AB=16, AC=x-3, BE=20, CD=x+5, BC is parallel to ED. Find the length of line AC.
问题：如图所示，AB=16，AC=x-3，BE=20，CD=x+5，BC∥ED。求直线AC的长度。
Construction (CDL):
Shape(AB,BC,CA)
Shape(CB,BE,ED,DC)
Collinear(ABE)
Collinear(ACD)
Conditions (CDL):
Equal(LengthOfLine(AB),16)
Equal(LengthOfLine(AC),x-3)
Equal(LengthOfLine(BE),20)
Equal(LengthOfLine(CD),x+5)
ParallelBetweenLine(BC,ED)
Goal: Value(LengthOfLine(AC))
Answer: 32
Theorem sequence:
parallel_property_corresponding_angle(1,BC,ED,A)
similar_triangle_judgment_aa(1,CAB,DAE)
line_addition(1,AB,BE)
line_addition(1,AC,CD)
similar_triangle_property_line_ratio(1,CAB,DAE)
similar_triangle_property_line_ratio(1,BCA,EDA)
Problem: As shown in the diagram, AB=15, RC=26, the midsegment of RVTC is AB, RVTC is a trapezoid. Find the length of line VT.
问题：如图所示，AB=15，RC=26，AB是RVTC的中位线，RV和TC是梯形RVTC的腰。求直线VT的长度。
Construction (CDL):
Shape(RA,AB,BC,CR)
Shape(AV,VT,TB,BA)
Collinear(RAV)
Collinear(TBC)
Conditions (CDL):
Equal(LengthOfLine(AB),15)
Equal(LengthOfLine(RC),26)
IsMidsegmentOfQuadrilateral(AB,RVTC)
Trapezoid(RVTC)
Goal: Value(LengthOfLine(VT))
Answer: 4
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,AB,RVTC)
Problem: As shown in the diagram, AB=y, AR=3*x+6, RB=5*x-14, CP is the altitude of triangle CAB, CQ bisects ∠CCB, R is the midpoint of segment AB. Find the value of y.
问题：如图所示，AB=y，AR=3*x+6，RB=5*x-14，CP是三角形CAB的高，CQ平分∠CCB，R是线段AB的中点。求y的值。
Construction (CDL):
Shape(CA,AP,PC)
Shape(CP,PQ,QC)
Shape(CQ,QR,RC)
Shape(CR,RB,BC)
Collinear(APQRB)
Conditions (CDL):
Equal(LengthOfLine(AB),y)
Equal(LengthOfLine(AR),3*x+6)
Equal(LengthOfLine(RB),5*x-14)
IsAltitudeOfTriangle(CP,CAB)
IsBisectorOfAngle(CQ,CAB)
IsMidpointOfLine(R,AB)
Goal: Value(y)
Answer: 72
Theorem sequence:
line_addition(1,AR,RB)
Problem: As shown in the diagram, QZ=12, XQ=15, YR=20, XY∥QR. Find the length of line RZ.
问题：如图所示，QZ=12，XQ=15，YR=20，XY平行于QR。求直线RZ的长度。
Construction (CDL):
Shape(QZ,ZR,RQ)
Shape(XQ,QR,RY,YX)
Collinear(XQZ)
Collinear(ZRY)
Conditions (CDL):
Equal(LengthOfLine(QZ),12)
Equal(LengthOfLine(XQ),15)
Equal(LengthOfLine(YR),20)
ParallelBetweenLine(XY,QR)
Goal: Value(LengthOfLine(RZ))
Answer: 16
Theorem sequence:
parallel_property_corresponding_angle(1,RQ,YX,Z)
similar_triangle_judgment_aa(1,QZR,XZY)
line_addition(1,ZR,RY)
line_addition(1,ZQ,QX)
similar_triangle_property_line_ratio(1,QZR,XZY)
similar_triangle_property_line_ratio(1,RQZ,YXZ)
Problem: As shown in the diagram, ∠AGB=30°, the center of circle G is G, CG is perpendicular to DG. Find the measure of ⌒GFC.
问题：如图所示，∠AGB=30°，⊙G的圆心为G，CG垂直于DG。求弧GFC的角度。
Construction (CDL):
Shape(AG,GB,GBA)
Shape(BG,GC,GCB)
Shape(CG,GD,GDC)
Shape(DG,GF,GFD)
Shape(FG,GA,GAF)
Collinear(AGD)
Collinear(BGF)
Cocircular(G,AFDCB)
Conditions (CDL):
Equal(MeasureOfAngle(AGB),30)
IsCentreOfCircle(G,G)
PerpendicularBetweenLine(CG,DG)
Goal: Value(MeasureOfArc(GFC))
Answer: 120
Theorem sequence:
vertical_angle(1,AGB,DGF)
arc_property_center_angle(1,GFD,G)
arc_property_center_angle(1,GDC,G)
arc_addition_measure(1,GFD,GDC)
Problem: As shown in the diagram, ∠AEC=74°, ∠EAB=15°, BC⊥AC. Find the measure of ∠CAE.
问题：如图所示，∠AEC=74°，∠EAB=15°，BC垂直于AC。求∠CAE的大小。
Construction (CDL):
Shape(BE,EA,AB)
Shape(EC,CA,AE)
Collinear(BEC)
Conditions (CDL):
Equal(MeasureOfAngle(AEC),74)
Equal(MeasureOfAngle(EAB),15)
PerpendicularBetweenLine(BC,AC)
Goal: Value(MeasureOfAngle(CAE))
Answer: 16
Theorem sequence:
triangle_property_angle_sum(1,ECA)
Problem: As shown in the diagram, GH=HK, HJ=JK, ∠GJK=100°. Find the measure of ∠KGJ.
问题：如图所示，GH=HK，HJ=JK，∠GJK=100°。求∠KGJ的大小。
Construction (CDL):
Shape(KG,GH,HK)
Shape(KH,HJ,JK)
Collinear(GHJ)
Conditions (CDL):
Equal(LengthOfLine(GH),LengthOfLine(HK))
Equal(LengthOfLine(HJ),LengthOfLine(JK))
Equal(MeasureOfAngle(GJK),100)
Goal: Value(MeasureOfAngle(KGJ))
Answer: 20
Theorem sequence:
isosceles_triangle_judgment_line_equal(1,HKG)
isosceles_triangle_property_angle_equal(1,HKG)
isosceles_triangle_judgment_line_equal(1,JKH)
isosceles_triangle_property_angle_equal(1,JKH)
triangle_property_angle_sum(1,KGH)
triangle_property_angle_sum(1,KHJ)
adjacent_complementary_angle(1,GHK,KHJ)
Problem: As shown in the diagram, ML=17, MN=2*x+7, NL=3*x-4, NL=MN. Find the length of line MN.
问题：如图所示，ML=17，MN=2*x+7，NL=3*x-4，NL=MN。求直线MN的长度。
Construction (CDL):
Shape(ML,LN,NM)
Conditions (CDL):
Equal(LengthOfLine(ML),17)
Equal(LengthOfLine(MN),2*x+7)
Equal(LengthOfLine(NL),3*x-4)
Equal(LengthOfLine(NL),LengthOfLine(MN))
Goal: Value(LengthOfLine(MN))
Answer: 29
Theorem sequence:
Problem: As shown in the diagram, ∠WCY=128°, the measure of arc AXZ is x, the measure of arc AYW is 154. Find the value of x.
问题：如图所示，∠WCY=128°，⌒AXZ的角度为x，⌒AYW的角度为154。求x的值。
Construction (CDL):
Shape(CY,YW,WC)
Shape(CW,WX,XC)
Shape(CX,XZ,ZC)
Shape(WY,AYW)
Shape(XW,AWX)
Shape(ZX,AXZ)
Shape(YC,CZ,AZY)
Collinear(WCZ)
Collinear(YCX)
Cocircular(A,WXZY)
Conditions (CDL):
Equal(MeasureOfAngle(WCY),128)
Equal(MeasureOfArc(AXZ),x)
Equal(MeasureOfArc(AYW),154)
Goal: Value(x)
Answer: 102
Theorem sequence:
arc_property_circumference_angle_external(1,AYW,X)
arc_property_circumference_angle_external(1,AXZ,W)
adjacent_complementary_angle(1,XCW,WCY)
triangle_property_angle_sum(1,XCW)
Problem: As shown in the diagram, FH=12, GH=12, FE is perpendicular to GE, GH⊥EH. Find the length of line EH.
问题：如图所示，FH=12，GH=12，FE垂直于GE，GH垂直于EH。求直线EH的长度。
Construction (CDL):
Shape(FE,EH,HF)
Shape(HE,EG,GH)
Collinear(FHG)
Conditions (CDL):
Equal(LengthOfLine(FH),12)
Equal(LengthOfLine(GH),12)
PerpendicularBetweenLine(FE,GE)
PerpendicularBetweenLine(GH,EH)
Goal: Value(LengthOfLine(EH))
Answer: 12
Theorem sequence:
mirror_similar_triangle_judgment_aa(1,EGH,FEG)
line_addition(1,FH,HG)
right_triangle_judgment_angle(1,GHE)
right_triangle_property_pythagorean(1,GHE)
mirror_similar_triangle_property_line_ratio(1,EGH,FEG)
mirror_similar_triangle_property_line_ratio(1,HEG,EGF)
Problem: As shown in the diagram, ∠TZY=30*x°, ∠TZY=∠ZYW, ∠WTZ=20*x°, ∠WTZ=∠YWT, quadrilateral TZYW is a isosceles trapezoid. Find the measure of ∠YWT.
问题：如图所示，∠TZY=30*x°，∠TZY=∠ZYW，∠WTZ=20*x°，∠WTZ=∠YWT，TW和ZY是等腰梯形TZYW的平行边。求∠YWT的大小。
Construction (CDL):
Shape(TZ,ZY,YW,WT)
Conditions (CDL):
Equal(MeasureOfAngle(TZY),30*x)
Equal(MeasureOfAngle(TZY),MeasureOfAngle(ZYW))
Equal(MeasureOfAngle(WTZ),20*x)
Equal(MeasureOfAngle(WTZ),MeasureOfAngle(YWT))
IsoscelesTrapezoid(TZYW)
Goal: Value(MeasureOfAngle(YWT))
Answer: 72
Theorem sequence:
quadrilateral_property_angle_sum(1,TZYW)
Problem: As shown in the diagram, ∠QRS=x°, ∠RST=2*x+7°, ∠STQ=x°, ∠TQR=2*x+5°. Find the measure of ∠TQR.
问题：如图所示，∠QRS=x°，∠RST=2*x+7°，∠STQ=x°，∠TQR=2*x+5°。求∠TQR的大小。
Construction (CDL):
Shape(QR,RS,ST,TQ)
Conditions (CDL):
Equal(MeasureOfAngle(QRS),x)
Equal(MeasureOfAngle(RST),2*x+7)
Equal(MeasureOfAngle(STQ),x)
Equal(MeasureOfAngle(TQR),2*x+5)
Goal: Value(MeasureOfAngle(TQR))
Answer: 121
Theorem sequence:
quadrilateral_property_angle_sum(1,QRST)
Problem: As shown in the diagram, GK=14, MF=8, the measure of ⌒FKG is 122, F is the center of circle F, HJ is perpendicular to KJ. Find the measure of ⌒FHG.
问题：如图所示，GK=14，MF=8，弧FKG的角度为122，圆F的圆心为F，HJ⊥KJ。求弧FHG的角度。
Construction (CDL):
Shape(GF,FJ,JG)
Shape(JH,FHG,GJ)
Shape(FG,FGM,MF)
Shape(JF,FM,FMK,KJ)
Shape(HJ,JK,FKH)
Collinear(GJK)
Collinear(MFJH)
Cocircular(F,GMKH)
Conditions (CDL):
Equal(LengthOfLine(GK),14)
Equal(LengthOfLine(MF),8)
Equal(MeasureOfArc(FKG),122)
IsCentreOfCircle(F,F)
PerpendicularBetweenLine(HJ,KJ)
Goal: Value(MeasureOfArc(FHG))
Answer: 180*asin(7/8)/pi
Theorem sequence:
adjacent_complementary_angle(1,HJK,KJM)
circle_property_chord_perpendicular_bisect_chord(1,F,FJ,KG)
line_addition(1,GJ,JK)
radius_of_circle_property_length_equal(1,FM,F)
radius_of_circle_property_length_equal(1,FG,F)
sine_theorem(1,GFJ)
arc_property_center_angle(1,FHG,F)
Problem: As shown in the diagram, DE=7, EX=24, the center of circle D is D, DA⊥XA, XE is perpendicular to DE. Find the length of line QX.
问题：如图所示，DE=7，EX=24，圆D的圆心为D，DA⊥XA，XE垂直于DE。求直线QX的长度。
Construction (CDL):
Shape(DA,DAQ,QD)
Shape(DQ,DQE,ED)
Shape(DE,DET,TD)
Shape(DT,DTA,AD)
Shape(AX,XQ,DAQ)
Shape(QX,XE,DQE)
Collinear(XQDT)
Cocircular(D,AQET)
Conditions (CDL):
Equal(LengthOfLine(DE),7)
Equal(LengthOfLine(EX),24)
IsCentreOfCircle(D,D)
PerpendicularBetweenLine(DA,XA)
PerpendicularBetweenLine(XE,DE)
Goal: Value(LengthOfLine(QX))
Answer: 18
Theorem sequence:
radius_of_circle_property_length_equal(1,DE,D)
radius_of_circle_property_length_equal(1,DQ,D)
right_triangle_judgment_angle(1,XED)
right_triangle_property_pythagorean(1,XED)
line_addition(1,XQ,QD)
Problem: As shown in the diagram, AB=41, AC=30, NB=19, NC=27, AC⊥BC. Find the perimeter of △BAN.
问题：如图所示，AB=41，AC=30，NB=19，NC=27，AC垂直于BC。求△BAN的周长。
Construction (CDL):
Shape(BA,AN,NB)
Shape(NA,AC,CN)
Collinear(BNC)
Conditions (CDL):
Equal(LengthOfLine(AB),41)
Equal(LengthOfLine(AC),30)
Equal(LengthOfLine(NB),19)
Equal(LengthOfLine(NC),27)
PerpendicularBetweenLine(AC,BC)
Goal: Value(PerimeterOfTriangle(BAN))
Answer: 3*sqrt(181)+60
Theorem sequence:
right_triangle_judgment_angle(1,ACN)
right_triangle_property_pythagorean(1,ACN)
triangle_perimeter_formula(1,BAN)
Problem: As shown in the diagram, AD=12, DC=10, ∠DCB=60°, CD and BA are opposite sides of the ▱ DCBA, AE is perpendicular to BE. Find the area of quadrilateral DCBA.
问题：如图所示，AD=12，DC=10，∠DCB=60°，DCBA是▱，AE垂直于BE。求DCBA的面积。
Construction (CDL):
Shape(DC,CB,BE,ED)
Shape(EB,BA,AE)
Collinear(DEA)
Conditions (CDL):
Equal(LengthOfLine(AD),12)
Equal(LengthOfLine(DC),10)
Equal(MeasureOfAngle(DCB),60)
Parallelogram(DCBA)
PerpendicularBetweenLine(AE,BE)
Goal: Value(AreaOfQuadrilateral(DCBA))
Answer: 60*sqrt(3)
Theorem sequence:
altitude_of_quadrilateral_judgment_left_vertex(1,BE,BADC)
parallelogram_property_opposite_angle_equal(1,CBAD)
parallelogram_property_opposite_line_equal(1,DCBA)
sine_theorem(1,BAE)
parallelogram_area_formula_common(1,BADC)
Problem: As shown in the diagram, AB=5, BC=5, CB⊥AB. Find the length of line CA.
问题：如图所示，AB=5，BC=5，CB⊥AB。求直线CA的长度。
Construction (CDL):
Shape(CB,BA,AC)
Conditions (CDL):
Equal(LengthOfLine(AB),5)
Equal(LengthOfLine(BC),5)
PerpendicularBetweenLine(CB,AB)
Goal: Value(LengthOfLine(CA))
Answer: 5*sqrt(2)
Theorem sequence:
right_triangle_judgment_angle(1,CBA)
right_triangle_property_pythagorean(1,CBA)
Problem: As shown in the diagram, QR=10, QX=QZ, ZQ=8, XY∥QR. Find the length of line XY.
问题：如图所示，QR=10，QX=QZ，ZQ=8，XY∥QR。求直线XY的长度。
Construction (CDL):
Shape(XQ,QR,RY,YX)
Shape(QZ,ZR,RQ)
Collinear(ZQX)
Collinear(ZRY)
Conditions (CDL):
Equal(LengthOfLine(QR),10)
Equal(LengthOfLine(QX),LengthOfLine(QZ))
Equal(LengthOfLine(ZQ),8)
ParallelBetweenLine(XY,QR)
Goal: Value(LengthOfLine(XY))
Answer: 20
Theorem sequence:
line_addition(1,ZQ,QX)
parallel_property_corresponding_angle(1,RQ,YX,Z)
similar_triangle_judgment_aa(1,QZR,XZY)
similar_triangle_property_line_ratio(1,ZRQ,ZYX)
similar_triangle_property_line_ratio(1,RQZ,YXZ)
Problem: As shown in the diagram, the area of △JKL is 30, JK=12, PQ=15, △JKL is similar to △PQR.. Find the area of triangle PQR.
问题：如图所示，三角形JKL的面积为30，JK=12，PQ=15，三角形JKL与三角形PQR是相似三角形。求△PQR的面积。
Construction (CDL):
Shape(JK,KL,LJ)
Shape(PQ,QR,RP)
Conditions (CDL):
Equal(AreaOfTriangle(JKL),30)
Equal(LengthOfLine(JK),12)
Equal(LengthOfLine(PQ),15)
SimilarBetweenTriangle(JKL,PQR)
Goal: Value(AreaOfTriangle(PQR))
Answer: 375/8
Theorem sequence:
similar_triangle_property_line_ratio(1,LJK,RPQ)
similar_triangle_property_area_square_ratio(1,LJK,RPQ)
Problem: As shown in the diagram, ∠DFH=81°. Find the measure of arc AHD.
问题：如图所示，∠DFH=81°。求⌒AHD的角度。
Construction (CDL):
Shape(DF,FH,HD)
Shape(FD,ADF)
Shape(HF,AFH)
Shape(DH,AHD)
Cocircular(A,DFH)
Conditions (CDL):
Equal(MeasureOfAngle(DFH),81)
Goal: Value(MeasureOfArc(AHD))
Answer: 162
Theorem sequence:
arc_property_circumference_angle_external(1,AHD,F)
Problem: As shown in the diagram, BC=4, BO=OT, CO=OA, ∠ABO=58°, CO⊥BO. Find the perimeter of BCTA.
问题：如图所示，BC=4，BO=OT，CO=OA，∠ABO=58°，CO⊥BO。求BCTA的周长。
Construction (CDL):
Shape(BC,CO,OB)
Shape(OC,CT,TO)
Shape(OT,TA,AO)
Shape(BO,OA,AB)
Collinear(COA)
Collinear(BOT)
Conditions (CDL):
Equal(LengthOfLine(BC),4)
Equal(LengthOfLine(BO),LengthOfLine(OT))
Equal(LengthOfLine(CO),LengthOfLine(OA))
Equal(MeasureOfAngle(ABO),58)
PerpendicularBetweenLine(CO,BO)
Goal: Value(PerimeterOfQuadrilateral(BCTA))
Answer: 16
Theorem sequence:
midpoint_of_line_judgment(1,O,CA)
midpoint_of_line_judgment(1,O,BT)
parallelogram_judgment_diagonal_bisection(1,BCTA,O)
vertical_angle(1,COB,AOT)
perpendicular_bisector_judgment_per_and_mid(1,BO,CA)
perpendicular_bisector_judgment_per_and_mid(1,TO,AC)
perpendicular_bisector_property_distance_equal(1,BO,CA)
perpendicular_bisector_property_distance_equal(1,TO,AC)
kite_judgment_equal_and_equal(1,BCTA)
rhombus_judgment_parallelogram_and_kite(1,BCTA)
quadrilateral_perimeter_formula(1,BCTA)
Problem: As shown in the diagram, MN=60, RV=6*x+11, ST=4*x-1, the midsegment of quadrilateral RSTV is MN. Find the value of x.
问题：如图所示，MN=60，RV=6*x+11，ST=4*x-1，MN是RSTV的中位线。求x的值。
Construction (CDL):
Shape(RM,MN,NV,VR)
Shape(MS,ST,TN,NM)
Collinear(RMS)
Collinear(VNT)
Conditions (CDL):
Equal(LengthOfLine(MN),60)
Equal(LengthOfLine(RV),6*x+11)
Equal(LengthOfLine(ST),4*x-1)
IsMidsegmentOfQuadrilateral(MN,RSTV)
Goal: Value(x)
Answer: 11
Theorem sequence:
midsegment_of_quadrilateral_property_length(1,MN,RSTV)
Problem: As shown in the diagram, AD=8, BD=12, BC is perpendicular to AC, CD is perpendicular to BD. Find the length of line CD.
问题：如图所示，AD=8，BD=12，BC⊥AC，CD⊥BD。求直线CD的长度。
Construction (CDL):
Shape(CA,AD,DC)
Shape(CD,DB,BC)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),8)
Equal(LengthOfLine(BD),12)
PerpendicularBetweenLine(BC,AC)
PerpendicularBetweenLine(CD,BD)
Goal: Value(LengthOfLine(CD))
Answer: 4*sqrt(6)
Theorem sequence:
adjacent_complementary_angle(1,ADC,CDB)
mirror_similar_triangle_judgment_aa(1,CAD,BCA)
mirror_similar_triangle_judgment_aa(1,CDB,ABC)
mirror_similar_triangle_property_line_ratio(1,CAD,BCA)
mirror_similar_triangle_property_line_ratio(1,ADC,ABC)
mirror_similar_triangle_property_line_ratio(1,DCA,CAB)
mirror_similar_triangle_property_line_ratio(1,CDB,ABC)
mirror_similar_triangle_property_line_ratio(1,DBC,CAB)
mirror_similar_triangle_property_line_ratio(1,BCD,BCA)
Problem: As shown in the diagram, AC=7*x-4, AD=6*x+1, ∠ABC=30°, ∠ADB=90°, ∠BCA=90°, ∠DBA=30°. Find the value of x.
问题：如图所示，AC=7*x-4，AD=6*x+1，∠ABC=30°，∠ADB=90°，∠BCA=90°，∠DBA=30°。求x的值。
Construction (CDL):
Shape(BC,CA,AB)
Shape(BA,AD,DB)
Conditions (CDL):
Equal(LengthOfLine(AC),7*x-4)
Equal(LengthOfLine(AD),6*x+1)
Equal(MeasureOfAngle(ABC),30)
Equal(MeasureOfAngle(ADB),90)
Equal(MeasureOfAngle(BCA),90)
Equal(MeasureOfAngle(DBA),30)
Goal: Value(x)
Answer: 5
Theorem sequence:
mirror_congruent_triangle_judgment_aas(3,ADB,ABC)
mirror_congruent_triangle_property_line_equal(1,BAD,BCA)
Problem: As shown in the diagram, the diameter of circle O is equal to the length of line QS, ∠SVT=75°, ∠TVP=72°, the center of ⊙V is V, RV is perpendicular to SV. Find the measure of ⌒VRP.
问题：如图所示，圆O的直径与直线QS的长度相等，∠SVT=75°，∠TVP=72°，圆V的圆心为V，RV垂直于SV。求弧VRP的角度。
Construction (CDL):
Shape(VQ,VQP,PV)
Shape(VP,VPT,TV)
Shape(VT,VTS,SV)
Shape(VS,VSR,RV)
Shape(VR,VRQ,QV)
Collinear(QVS)
Cocircular(V,RQPTS)
Conditions (CDL):
Equal(DiameterOfCircle(O),LengthOfLine(QS))
Equal(MeasureOfAngle(SVT),75)
Equal(MeasureOfAngle(TVP),72)
IsCentreOfCircle(V,V)
PerpendicularBetweenLine(RV,SV)
Goal: Value(MeasureOfArc(VRP))
Answer: 123
Theorem sequence:
adjacent_complementary_angle(1,QVR,RVS)
adjacent_complementary_angle(1,SVP,PVQ)
angle_addition(1,SVT,TVP)
angle_addition(1,PVQ,QVR)
arc_property_center_angle(1,VRP,V)
Problem: As shown in the diagram, AD=3, CD=x, DB=12, AC is perpendicular to BC, CD is perpendicular to AD. Find the value of x.
问题：如图所示，AD=3，CD=x，DB=12，AC垂直于BC，CD⊥AD。求x的值。
Construction (CDL):
Shape(AC,CD,DA)
Shape(DC,CB,BD)
Collinear(ADB)
Conditions (CDL):
Equal(LengthOfLine(AD),3)
Equal(LengthOfLine(CD),x)
Equal(LengthOfLine(DB),12)
PerpendicularBetweenLine(AC,BC)
PerpendicularBetweenLine(CD,AD)
Goal: Value(x)
Answer: 6
Theorem sequence:
adjacent_complementary_angle(1,BDC,CDA)
mirror_similar_triangle_judgment_aa(1,CDA,BAC)
mirror_similar_triangle_judgment_aa(1,CBD,ACB)
mirror_similar_triangle_property_line_ratio(1,CDA,BAC)
mirror_similar_triangle_property_line_ratio(1,DAC,CBA)
mirror_similar_triangle_property_line_ratio(1,ACD,ACB)
mirror_similar_triangle_property_line_ratio(1,CBD,ACB)
mirror_similar_triangle_property_line_ratio(1,BDC,BAC)
mirror_similar_triangle_property_line_ratio(1,DCB,CBA)
Problem: As shown in the diagram, AB=20, AD=18, CE=16, BADC is a parallelogram, CE is perpendicular to DE. Find the area of BADC.
问题：如图所示，AB=20，AD=18，CE=16，BC和AD是▱BADC的一组对边，CE垂直于DE。求四边形BADC的面积。
Construction (CDL):
Shape(BA,AE,EC,CB)
Shape(CE,ED,DC)
Collinear(AED)
Conditions (CDL):
Equal(LengthOfLine(AB),20)
Equal(LengthOfLine(AD),18)
Equal(LengthOfLine(CE),16)
Parallelogram(BADC)
PerpendicularBetweenLine(CE,DE)
Goal: Value(AreaOfQuadrilateral(BADC))
Answer: 288
Theorem sequence:
adjacent_complementary_angle(1,AEC,CED)
altitude_of_quadrilateral_judgment_right_vertex(1,CE,BADC)
parallelogram_area_formula_common(1,BADC)
Problem: As shown in the diagram, BC=17*sqrt(2), ∠BAC=45°, ∠CBA=45°, AC⊥BC. Find the length of line AB.
问题：如图所示，BC=17*sqrt(2)，∠BAC=45°，∠CBA=45°，AC垂直于BC。求直线AB的长度。
Construction (CDL):
Shape(BA,AC,CB)
Conditions (CDL):
Equal(LengthOfLine(BC),17*sqrt(2))
Equal(MeasureOfAngle(BAC),45)
Equal(MeasureOfAngle(CBA),45)
PerpendicularBetweenLine(AC,BC)
Goal: Value(LengthOfLine(AB))
Answer: 34
Theorem sequence:
sine_theorem(1,BAC)
Problem: As shown in the diagram, BC=11, CA=23, OA=25, OB=35, OA⊥CA. Find the perimeter of △OCB.
问题：如图所示，BC=11，CA=23，OA=25，OB=35，OA垂直于CA。求△OCB的周长。
Construction (CDL):
Shape(OA,AC,CO)
Shape(OC,CB,BO)
Collinear(ACB)
Conditions (CDL):
Equal(LengthOfLine(BC),11)
Equal(LengthOfLine(CA),23)
Equal(LengthOfLine(OA),25)
Equal(LengthOfLine(OB),35)
PerpendicularBetweenLine(OA,CA)
Goal: Value(PerimeterOfTriangle(OCB))
Answer: sqrt(1154)+46
Theorem sequence:
right_triangle_judgment_angle(1,OAC)
right_triangle_property_pythagorean(1,OAC)
triangle_perimeter_formula(1,OCB)
Problem: As shown in the diagram, ∠UTS=47°, RU is perpendicular to SU, TS is perpendicular to RS, TV is perpendicular to UV. Find the measure of ∠SRU.
问题：如图所示，∠UTS=47°，RU垂直于SU，TS垂直于RS，TV⊥UV。求∠SRU的大小。
Construction (CDL):
Shape(SR,RU,US)
Shape(SU,UV,VS)
Shape(VU,UT,TV)
Collinear(RUT)
Collinear(SVT)
Conditions (CDL):
Equal(MeasureOfAngle(UTS),47)
PerpendicularBetweenLine(RU,SU)
PerpendicularBetweenLine(TS,RS)
PerpendicularBetweenLine(TV,UV)
Goal: Value(MeasureOfAngle(SRU))
Answer: 43
Theorem sequence:
triangle_property_angle_sum(1,TSR)
Problem: As shown in the diagram, ∠ACB=5*x+21°, ∠CDG=12*x+6°, PA∥EG. Find the value of x.
问题：如图所示，∠ACB=5*x+21°，∠CDG=12*x+6°，PA∥EG。求x的值。
Construction (CDL):
Shape(FC,CA)
Shape(PC,CF)
Shape(AC,CD)
Shape(DC,CP)
Shape(CD,DG)
Shape(ED,DC)
Shape(GD,DB)
Shape(BD,DE)
Collinear(ECDB)
Collinear(ACP)
Collinear(GDE)
Conditions (CDL):
Equal(MeasureOfAngle(ACB),5*x+21)
Equal(MeasureOfAngle(CDG),12*x+6)
ParallelBetweenLine(PA,EG)
Goal: Value(x)
Answer: 9
Theorem sequence:
parallel_property_collinear_extend(3,PA,EG,C)
parallel_property_collinear_extend(3,GE,AC,D)
parallel_property_ipsilateral_internal_angle(1,CA,DG)