## 11 Apr 2014

### Archimedes’ “On the Equilibrium of Planes”, Prop 15

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[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
The text is a detailed geometric proof from Archimedes' "On the Equilibrium of Planes," specifically Proposition 15, which involves finding the center of gravity of a trapezium.
no_question

the passage does not contain a mathematical question but rather provides a detailed geometric proof. there is no specific question posed, and no answer to a question is provided.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## 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
user
The text is a detailed geometric proof from Archimedes' "On the Equilibrium of Planes," specifically Proposition 15, which involves finding the center of gravity of a trapezium.

TYPE: NO_QUESTION

The passage does not contain a mathematical question but rather provides a detailed geometric proof. There is no specific question posed, and no answer to a question is provided.