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The line F G intersects the parallels E G and A D, therefore the ADF is equal to the Consequently ADF is a right Z.

3. To prove that GE is equal to G F.

EGF.

Since E F intersects the parallels E C and G F, the alternate s CEF and EFG are equal. But CEF is half a right, therefore EFG is half a right . And since the threes of the ▲ EFG are together equal to two rights, and EGF is a right ▲, and EFG is half a right, it follows that G E F is half a right, and is equal to the GFE.

Consequently (I. 6) G E is equal to G F.

4. To prove that D F is equal to D B. In the ▲ BDF, the

BDF is a right, DFB is half a right, therefore DBF is half a right ▲, and is equal to D F B.

Therefore D F is equal to D B.

5. To prove that the □ on AF is equal to the sum of thes on AD and D B.

ADF is a right, therefore in the ▲ ADF the on AF is equal to the sum of the s on AD and DF.

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

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

the

6. To prove that the☐ on A F is also equal to twice on AC, together with twice the on CD.

In the fore the

AEF, the AEF is a right; thereon AF is equal to the sum of the ☐s on

A E and E F.
In the
fore the

ACE the ACE is a right; thereon AE is equal to the sum of the s on A C and C E, and is therefore equal to twice the ☐ on A C, since AC and CE are equal.

Again, in the AEGF, the EGF is a right,

therefore the

on EF is equal to the sum of the s on E G and GF, and is therefore equal to twice the ☐ on E G, since E G is equal to G F.

But E G and C D are equal, because they are opposite sides of the parallelogram C G. twice the

on E G is equal to twice the

Therefore

on CD. It follows therefore that the on A F is equal to twice the on AC together with twice the C D.

on

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

on CD.

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

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.

A

A

C D B

It has been proved in the fourth proposition that on AD = on AC

+ on CD + 2c rect. A C, C D. But ACB C.

Substitute B C for A C in this equation, and we get,on A Don BC + ☐ on CD + 2ce rect. B C, CD. Add to both sides the on BC +

on DB

on D B.

on D B, then

on AD + O on CD + 2ce rect. B C, CD +

=

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

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

on AD + on D B = ☐ on B C +

on BC

on CD; i.e., □ on AD +

un BC + 2ce on CD.

on CD + on DB = 2ce ☐

For the 10th proposition, let A B be bisected in C and produced to D.

A

C

BD

on AD☐ on AC + on CD2ce rect. A C, C D.

But CBA C.

Substitute CB for AC in the previous equation, and

we get,

on AD=☐ on CB+

Add the

on DB

on D B.

on CD + 2ce rect. CB, C D.

on DB to both sides, then on A D + O on CB+ on CD + 2ce rect. CB, CD +

But 2ce rect. CB, CD + □ on DB on CD + 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 =

on CB+

on CB+2ce

on CB+

on CD + on CD; i.e., ☐ on A D + ☐ on D B = 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 A C is half their sum. CD is what is left when DB is taken from CB, therefore it is equal to what would be left if DB were taken from A C. Therefore, if A E be cut off from AC equal to DB, ED will be bisected in C. But ED is the difference between AD and AE or D B, therefore CD is half the difference between AD and DB. It follows, therefore, that the 9th proposition may be thus expressed :-" The sum of the squares on two lines is equal to twice the square on half their sum, together with twice the square on half their difference."

A E

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E A

1

B D

The very same relation is proved in the 10th proposition. Let AD and D B be regarded as separate lines. Let B A be produced, and A E cut off equal to BD. It is then evident that ED is the sum of AD and D B, and consequently CD 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 ▲, the
is equal to the sum of the
sides,

on the hypotenuse

s on the other two

M

Let A B be the given line. On A B describe the
ABCD. Bisect DA in E, and join
E and B.

D

K

c

[blocks in formation]

Produce E A, and cut off EF equal to EB; and from A B cut off AG, equal to A F.

G is the point of section required.

Now complete the square AH by drawing GH parallel to A F, and FH parallel to AG, and produce

HG to meet D C in K.

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

the on EF.

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

Therefore the rect. DH, together with the on EA, is equal to the

on E B.

But EA B is a right-angled ▲, therefore the ☐ on EB is equal to the sum of the Os on EA and A B; and consequently the rect. DH, together with the ☐ on E A, is equal to the sum of the □s on E A and A B.

Take away theon EA 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. D G, which is common to both, and we have left the AH, equal

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

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

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

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