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4 AB BC

+AC2=

Four times the rectangle AB, BC, together with the square

(AB+BC), on AC, shall be equal to the square on the straight line

[blocks in formation]

made up of AB and BC together.

CONSTRUCTION.-Produce AB to D, so that BD may be equal to CB (Post. 2, and I. 3).

Upon AD describe the square AEFD ♣ (I. 46),

And construct two figures such as in the preceding propositions.

PROOF. Because CB is equal to BD (Const.), CB to GK, and BD to KN (Ax. 1),

G

M

K

P

R

Σ

For the same reason PR is equal E to RO.

And because CB is equal to BD, and GK to KN, Therefore the rectangle CK is equal to BN, and GR to RN (I. 36).

But CK is equal to RN, because they are the complements of the parallelogram CO (I. 43);

Therefore also BN is equal to GR (Ax. 1).

Therefore the four rectangles BN, CK, GR, RN are equal to one another, and so the four are quadruple of one of them, CK.

Again, because CB is equal to BD (Const.);

And that BD is equal to BK, that is CG (II. 4, Cor., and I. 34);

And that CB is equal to GK, that is GP (I. 34, and II. 4, cor.);

Therefore CG is equal to GP (Ax. 1).

And because CG is equal to GP, and PR to RO,

The rectangle AG is equal to MP, and PL to RF (I. 36). But MP is equal to PL, because they are complements of the parallelogram ML (I. 43), and AG is equal to RF (Ax. 1); Therefore the four rectangles AG, MP, PL, RF are equal to one another, and so the four are quadruple of one of them, AG.

And it was demonstrated that the four CK, BN, GR, and RN are quadruple of CK;

Therefore the eight rectangles which make up the gnomon AOH are quadruple of AK.

And because AK is the rectangle contained by AB and

BC, for BK is equal to BC;

Therefore four times the rectangle AB, BC is quadruple of

AK.

But the gnomon AOH was demonstrated to be quadruple of AK;

Therefore four times the rectangle AB, BC is equal to the gnomon AOH (Ax. 1).

[blocks in formation]

To each of these add XH, which is equal to the square Hence on AC (II. 4, cor., and I. 34);

Therefore four times the rectangle AB, BC, together with the square on AC, is equal to the gnomon AOH and_the square XH.

But the gnomon AOH and the square XH make up the figure AEFD, which is the square on AD;

adding
XH or AC2,

+ AC2

Therefore four times the rectangle AB, BC, together with 4 AB BC the square on AC, is equal to the square on AD, that is, on AF = the line made up of AB and BC together.

Therefore, if a straight line, &c. Q. E. D.

Proposition 9.-Theorem.

If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of section.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D;

(AB+BC)

The squares on AD and DB shall be together double of AD2+DB2 the squares on AC and CD.

CONSTRUCTION.-From the point C draw CE at right

angles to AB (I. 11), and make it equal to AC or CB (I. 3), and join EA, ÉB. Through D draw DF parallel to CE (I. 31).

Through F draw FG parallel to BA (I. 31), and join AF.

PROOF.-Because AC is equal to CE

E

C

(Const.), the angle EAC is equal to the angle AEC (I. 5).

=2 (AC2+ CD2).

For

AEC=} a
right Z
= CEB.

... AEB is

a right 4.

EG=GF.

And

DFDB.

But AE22AC3.

So also
EF2-2GF

=2 CD2.

AE2+ EF2 = 2(AC +CD2)

And because the angle ACE is a right angle (Const.), the angles AEC and EAC together make one right angle (I. 32), and they are equal to one another;

Therefore each of the angles AEC and EAC is half a right angle.

For the same reason, each of the angles CEB and EBC is half a right angle;

Therefore the whole angle AEB is a right angle.

And because the angle GEF is half a right angle, and the angle EGF a right angle, for it is equal to the interior and opposite angle ECB (I. 29),

Therefore the remaining angle EFG is half a right angle; Therefore the angle GEF is equal to the angle EFG, and the side EG is equal to the side GF (I. 6).

Again, because the angle at B is half a right angle, and the angle FDB a right angle, for it is equal to the interior and opposite angle ECB (I. 29),

Therefore the remaining angle BFD is half a right angle; Therefore the angle at B is equal to the angle BFD, and the side DF is equal to the side DB (I. 6).

And because AC is equal to CE (Const.), the square on AC is equal to the square on CE;

Therefore the squares on AC and CE are double of the square on AC.

But the square on AE is equal to the squares on AC and CE, because the angle ACE is a right angle (I. 47);

Therefore the square on AE is double of the square on AC. Again, because EG is equal to GF (Const.), the square on EG is equal to the square on GF;

Therefore the squares on EG and GF are double of the square on GF.

But the square on EF is equal to the squares on EG and
GF, because the angle EGF is a right angle (I. 47);

Therefore the square on EF is double of the square on GF.
And GF is equal to CD (I. 34);

Therefore the square on EF is double of the square on CD. But it has been demonstrated that the square on AE is also double of the square on AC;

Therefore the squares on AE and EF are double of the squares on AC and CD.

AE2+EF2 =AF2

But the square on AF is equal to the squares on AE and But EF, because the angle AEF is a right angle (I. 47); Therefore the square on AF is double of the squares on AC AD2+ and CD.

But the squares on AD and DF are equal to the square on AF, because the angle ADF is a right angle (I. 47);

Therefore the squares on AD and DF are double of the squares on AC and CD.

DF2

=AD2+

DB2.

And DF is equal to DB; therefore the squares on AD and .. AD2+ DB are double of the squares on AC and CD. Therefore, if a straight line, &c. Q.E.D.

Proposition 10.—Theorem.

If a straight line be bisected and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced.

DB
=2(AC2+
CD2)

AD2+DB2 =2(AC2+

Let the straight line AB be bisected in C, and produced to D; The squares on AD and DB shall be together double of the CD). squares on AC and CD.

CONSTRUCTION.-From the point C draw CE at right angles to AB, and make it equal to AC or CB (I. 11, I. 3), and join AE, EB.

Through E draw EF parallel to AB, and through D draw DF parallel to CE (I. 31). Then because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right angles (I. 29); therefore the angles BEF, EFD are less than two right angles;

therefore EB, FD will meet, if produced towards B and D (Ax. 12).

Let them meet in G, and join A

B

D

AG.

PROOF.-Because AC is equal

to CE (Const.), the angle CEA is equal to the angle EAC (I. 5).

And the angle ACE is a right angle; therefore each of the angles CEA and EAC is half a right angle (I. 32).

As in Prop. 9.

AEB is a right 2.

And
BD=DG.

Also

GF=FE.

Again as in Prop. 9.

AE2= 2 AC2,

And
EG2 =
2 CD2.
...AE2+
E 2 =
2(AC2+

CD2).

But

AE2+EG2
AG2

For the same reason each of the angles CEB and EBC is half a right angle;

Therefore the whole angle AEB is a right angle.

And because the angle EBC is half a right angle, the angle DBG, which is vertically opposite, is also half a right angle (I. 15);

But the angle BDG is a right angle, because it is equal to the alternate angle DCE (I. 29);

Therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG;

Therefore also the side BD is equal to the side DG (I. 6). Again, because the angle EGF is half a right angle, and the angle at F a right angle, for it is equal to the opposite angle ECD (I. 34);

Therefore the remaining angle FEG is half a right angle (I. 32), and therefore equal to the angle EGF;

Therefore also the side GF is equal to the side FE (I. 6). And because EC is equal to CA, the square on EC is equal to the square on CA;

Therefore the squares on EC and CA are double of the square on CA.

But the square on AE is equal to the squares on EC and CA (I. 47);

Therefore the square on AE is double of the square on AC.

Again, because GF is equal to FE, the square on GF is equal to the square on FE;

Therefore the squares on GF and FE are double of the Square on FE.

But the square on EG is equal to the squares on GF and
FE (I. 47);

Therefore the square on EG is double of the square on FE.
And FE is equal to CD (I. 34),

Therefore the square on EG is double of the square on CD. But it has been demonstrated that the square on AE is double of the square on AC;

Therefore the squares on AE and EG are double of the squares on AC and CD.

But the square on AG is equal to the squares on AE and EG (I. 47);

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