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The line FG intersects the parallels E G and AD, therefore the < ADF is equal to the < EGF. Consequently ADF is a right 2.

3. To prove that G E is equal to G F.
Since E F intersects the parallels EC and GF,

the alternate Z8 CEF and EFG are equal. But CEF is half a right 2, therefore E F G is half a right Z. And since the three <s of the A EFG are together equal to two right <s, and EGF is a right Z, and EFG is half a right Z, it follows that GEF is half a right Z, and is equal to the <GFE.

Consequently (1. 6) G E is equal to G F. 4. To prove that D F is equal to D B.

In the ABDF, the < BDF is a right Z, DFB is half a right 4, therefore D BF is half a right Z, and is equal to D F B.

Therefore D F is equal to D B.

5. To prove that the a on AF is equal to the sum of the Os on A D and D B. ADF is a right Z, therefore in the A ADF the

on AF is equal to the sum of the Os on A D and DF.

But D F is equal to D B; therefore the O on DF is equal to the on D B.

Therefore the on AF is equal to the sum of the Os on A D and D B.

6. To prove that the on AF is also equal to twice the O on A C, together with twice the D on CD.

In the A AEF, the < AEF is a right Z; therefore the on AF is equal to the sum of the is on AE and EF.

In the A ACE the < ACE is a right Z; therefore the on A E is equal to the sum of the Os on AC and C E, and is therefore equal to twice the on A C, since AC and C E are equal.

Again, in the A EGF, the EGF is a right Zi

therefore the on E F is equal to the sum of the Os on EG and GF,

and

therefore equal to twice the o on EG, since EG is equal to G F.

But E G and C D are equal, because they are opposite sides of the parallelogram C G. Therefore twice the Q on EG is equal to twice the on CD.

It follows therefore that the I on AF is equal to twice the O on A O together with twice the OD.

7. It has thus been proved that the D on AF is equal to the sum of the Os on A D and D B, and also to twice the Ion A C, together with twice the o

on

on CD.

It follows therefore that the sum of the Os on AD and D B is equal to twice the I on A C, together with twice the O on CD.

1

A

D B

It is an excellent mental exercise to master the elaborate proof given by Euclid of the ninth and tenth propositions ; but the result might have been arrived at in a much shorter way. The following is one of the proofs that have been given : Let A B be bisected in C, and divided unequally in D.

It has been proved in the fourth proposition that O on AD

- O on AC + on CD + 2° rect. A C, CD.

But AC=BC.
Substitute B C for A C in this equation, and we get,-

on AD= Oon B C + D on CD + 2ce rect. BC, CD.

Add to both sides the on D B, then on AD + 0 on D B= on B C + D on CD + 2ce rect. BC, CD + O on D B.

But 2ce rect. BC, CD + O on DB=0 on BC + O on CD. (II. 7.)

A

Substitute the second side of this equation for the first in the previous equation, and we get,

on AD + O on DB=Oon B C + D on CD + on BC + on CD; i.e., on AD + Oon DB= 20e un B C + 2ce on CD. For the 10th proposition, let A B be bisected in C and

produced to D.

o, on AD= on AC + on BD

CD + 2ce rect. A C, C D. But CBS AC.

Substitute C B for AC in the previous equation, 'and we get,

on AD= on CB + O on CD + 2ce rect. CB, C D. Add the on D B to both sides, then on AD + O on DB= on CB+ on CD + 2ce rect. CB, CD + O on D B.

But 2ce rect. C B, C D + o oo DB= on CD + O on C B. (II. 7.)

Substitute the second side of this equation for the first in the previous equation, and we get,-

on AD + on D B = Don CB + oon CD + D on CB + on CD; i.e., on AD + O on DB=2ce on CB + 2ce

on CD.

If in Prop. IX. A D and D B be considered as two separate lines, it is clear that A B is their sum, and AC is half their sum. CD is what is left when D B is taken from CB, therefore it is equal to what would be left if D B were taken from AC. Therefore, if A E be cut off from

AC equal to D B, ED will be D B

bisected in C, But ED is the

difference between AD and AE or D B, therefore CD is half the difference between A D and D B. It follows, therefore, that the 9th proposition may be thus expressed :-“ The sum of the squares on two tines is equal to twice the square on half their sum, together with twice the

square on half their difference.

A

E

C

1 B

The very same relation is proved in the 10th proposition.

Let AD and D B be regarded as separate lines.

E A

Ć Let B A be produced, and AE cut off equal to BD. It is then evident that E D is the sum of AD and D B, and consequently C D is half their sum.

AB is the difference between AD and DB, and therefore A C is half their difference.

PROPOSITION XI.

To divide a given finite right line so that the rectangle contained by the whole line and one segment shall be equal to the square on the other segment.

For the construction required we must be able --
1. To produce a straight line to any length.
2. From the greater of two given straight lines to

cut off a part equal to the less. (I. 3.)
3. To draw a straght line parallel to a given

straight line. (I. 31.) 4. On a given straight line to describe a square.

(I. 46.) For the proof we must know, 1. That if a line be bisected and produced, the

rectangle contained by the whole line and the produced part, together with the on half the line, is equal to the on the line made up of

the half and the produced part. (II. 6.) 2. In a right-angled A, the 0 on the hypotenuse

is equal to the sum of the Os on the other two sides,

M M

K

E

A

B

G

F

H

on

Let A B be the given line. On A B describe the o

ABCD. Bisect DA in E, and join C E and B.

Produce E A, and cut off ET equal to EB; and from A B cut off AG, equal to AF.

G is the point of section required.

Now complete the square A H by drawing G H parallel to A F, and

FH parallel to A G, and produce HG to meet D O in K.

DA is bisected in E, and produced to F; consequently (II. 6), the rect. DF, FA (that is, the rectangle D H), together with the Don EA, is equal to the O on EF.

But E F is equal to E B, therefore the O on EF is equal to the I on E B.

Therefore the rect. DH, together with the E A, is equal to the on EB.

But EAB is a right-angled A, therefore the on E B is equal to the sum of the Os on EA and AB; and consequently the rect. D H, together with the o on E A, is equal to the sum of the son E A and A B. Take

away the O on E A from both these equals, and we have left the rect. D H equal to the square on A B, that is, the square D B.

From these take away the rect. DG, which is common to both, and we have left the OAH, equal to the rect. KB.

But A H is the on A G, and K B is the rect. AB, BG, for it is contained by CB and BG, and C B is equal to A B.

Therefore the line AB has been divided so that the rect. A B, B G is equal to the O on A G.

A line thus divided is said to be divided in extreme and mean ratio.

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