text: ## 11 Apr 2014
  
  ### Archimedes’ “On the Equilibrium of Planes”, Prop 15
  
  by [Search Blog Here. Index-tags are found on the bottom of the left column.]
  
  [Mathematics, Calculus, Geometry, Entry Directory]
  
  [Archimedes Entry Directory]
  
  [“On the Equilibrium of Planes”, Entry Directory]
  
  Archimedes
  
  On the Equilibrium of Planes
  
  or
  
  The Centres of Gravity of Planes,
  
  Book I
  
  Proposition 15
  
  [The following is quotation]
  
  Proposition 15
  
  P15. If AD, BC he the two parallel sides of a trapezium ABCD, AD being the smaller, and if AD, BC be bisected at E, F respectively, then the centre of gravity of the trapezium is at a point G on EF such that
  
  Produce BA, CD to meet at O. Then FE produced will also pass through 0, since AE = ED, and BF = FC.
  
  Now the centre of gravity of the triangle OAD will lie on OE, and that of the triangle OBC will lie on OF. [Prop. 13]
  
  It follows that the centre of gravity of the remainder, the trapezium ABCD, will also lie on OF. [Prop. 8]
  
  Join BD, and divide it at L, M into three equal parts. Through L, M draw PQ, RS parallel to BC meeting BA in P, R, FE in W, V, and CD in Q, S respectively.
  
  Join DF, BE meeting PQ in H and RS in K respectively.
  
  Now, since
  
  BL = 1/3BD,
  
  FH=1/3FD.
  
  (Heath 201)
  
  Therefore H is the centre of gravity of the triangle DBC*. [* This easy deduction from Prop. 14 is assumed by Archimedes without proof.]
  
  Similarly, since EK = 1/3BE, it follows that K is the centre of gravity of the triangle ADB.
  
  Therefore the centre of gravity of the triangles DBC, ADB together, i.e. of the trapezium, lies on the line HK.
  
  But it also lies on OF.
  
  Therefore, if OF, HK meet in G, G is the centre of gravity of the trapezium.
  
  Hence [Props. 6, 7]
  
  ΔDBC : ΔABD = KG : GH
  
  = VG: GW.
  
  But ΔDBC : ΔABD = BC: AD.
  
  Therefore BC : AD = VG : GW.
  
  It follows that (2BC + AD) : (2AD + BC) = (2VG + GW) : (2GW + VG) = EG : GF.
  
  Q. E. D.
  
  (Heath 202)
  
  From:
  
  Archimedes. “On the Equilibrium of Planes or The Centres of Gravity of Planes, Book I”. In The Works of Archimedes. Ed. T.L. Heath. Cambridge UP, 1897. Obtained at
  
  https://archive.org/details/worksofarchimede00arch
explanation: The document is an excerpt from Archimedes' "On the Equilibrium of Planes, Book I," specifically Proposition 15. It provides a detailed geometric proof to determine the center of gravity of a trapezium. The text does not contain a question but rather a complete mathematical proof.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## 11 Apr 2014
  
  ### Archimedes’ “On the Equilibrium of Planes”, Prop 15
  
  by [Search Blog Here. Index-tags are found on the bottom of the left column.]
  
  [Mathematics, Calculus, Geometry, Entry Directory]
  
  [Archimedes Entry Directory]
  
  [“On the Equilibrium of Planes”, Entry Directory]
  
  Archimedes
  
  On the Equilibrium of Planes
  
  or
  
  The Centres of Gravity of Planes,
  
  Book I
  
  Proposition 15
  
  [The following is quotation]
  
  Proposition 15
  
  P15. If AD, BC he the two parallel sides of a trapezium ABCD, AD being the smaller, and if AD, BC be bisected at E, F respectively, then the centre of gravity of the trapezium is at a point G on EF such that
  
  Produce BA, CD to meet at O. Then FE produced will also pass through 0, since AE = ED, and BF = FC.
  
  Now the centre of gravity of the triangle OAD will lie on OE, and that of the triangle OBC will lie on OF. [Prop. 13]
  
  It follows that the centre of gravity of the remainder, the trapezium ABCD, will also lie on OF. [Prop. 8]
  
  Join BD, and divide it at L, M into three equal parts. Through L, M draw PQ, RS parallel to BC meeting BA in P, R, FE in W, V, and CD in Q, S respectively.
  
  Join DF, BE meeting PQ in H and RS in K respectively.
  
  Now, since
  
  BL = 1/3BD,
  
  FH=1/3FD.
  
  (Heath 201)
  
  Therefore H is the centre of gravity of the triangle DBC*. [* This easy deduction from Prop. 14 is assumed by Archimedes without proof.]
  
  Similarly, since EK = 1/3BE, it follows that K is the centre of gravity of the triangle ADB.
  
  Therefore the centre of gravity of the triangles DBC, ADB together, i.e. of the trapezium, lies on the line HK.
  
  But it also lies on OF.
  
  Therefore, if OF, HK meet in G, G is the centre of gravity of the trapezium.
  
  Hence [Props. 6, 7]
  
  ΔDBC : ΔABD = KG : GH
  
  = VG: GW.
  
  But ΔDBC : ΔABD = BC: AD.
  
  Therefore BC : AD = VG : GW.
  
  It follows that (2BC + AD) : (2AD + BC) = (2VG + GW) : (2GW + VG) = EG : GF.
  
  Q. E. D.
  
  (Heath 202)
  
  From:
  
  Archimedes. “On the Equilibrium of Planes or The Centres of Gravity of Planes, Book I”. In The Works of Archimedes. Ed. T.L. Heath. Cambridge UP, 1897. Obtained at
  
  https://archive.org/details/worksofarchimede00arch
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is an excerpt from Archimedes' "On the Equilibrium of Planes, Book I," specifically Proposition 15. It provides a detailed geometric proof to determine the center of gravity of a trapezium. The text does not contain a question but rather a complete mathematical proof.
  
  Label: no-question