# The Elements of Euclid for the Use of Schools and Colleges/Book II

*BOOK II.*

DEFINITIONS.

1. Every right-angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which contain one of the right angles.

2. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon.

Thus the parallelogram *HG*, together with the complements *AF*,*FC*, is the gnomon, which is more briefly expressed by the letters *AGK*, or *EHC*, which are at the opposite angles of the parallelograms which make the gnomon.

*THEOREM*.

*If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undecided line, and the several parts of the divided line.*

Let *A* and *BC* be two straight lines; and let *BC* be divided into any number of parts at the points *D*, *E*: the rectangle contained by the straight lines *A*, *BC*, shall be equal to the rectangle contained by *A*, *BD*, together with that contained by *A*, *DE*, and that contained by *A*, *EC*.

From the point *B* draw *BF* at right angles to *BC*; [I. 11.

and make *BG* equal to *A*; [I. 3.

through *G* draw *GH* parallel to *BC*;, and through *D*, *E*, *C* draw *DK*, *EL*, *CH*, parallel to *BG*. [I. 31.

Then the rectangle *BH* is equal to the rectangles *BK*, *DL*, *EH*.

But *BH* is contained by *A*, *BC*, for it is contained by *GB*, *BC*, and *GB* is equal to *A*. [*Construction*.

And *BK* is contained by *A*, *BD*, for it is contained by *GB*, *BD*, and *GB* is equal to *A*;

and *DL* is contained by *A*, *DE*, because *DK* is equal to *BG*, which is equal to *A*; [I. 34.

and in like manner *EH* is contained by *A*, *EC*.

*A*,

*BC*is, equal to the rectangles contained by

*A*,

*BD*, and by

*A*,

*DE*, and by

*A*,

*EC*. Wherefore,

*if there be two straight lines*&c. q.e.d.

*THEOREM*.

*If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square on the whole line*.

Let the straight line *AB* be divided into any two parts at the point *C*: the rectangle contained by *AB*, *BC*, together with the rectangle *AB*, *AC*, shall be equal to the square on *AB*.

[*Note*. To avoid repeating the word contained too frequently, the rectangle contained by two straight lines *AB*, *AC* is sometimes simply called the rectangle *AB*, *AC*.]

On *AB* describe the square *ADEB*; [I.46.

and through *C* draw *CF* parallel to *AD* or *BE*. [I.31.

Then *AE* is equal to the rectangles *AF*, *CE*.

But *AE* is the square on *AB*.

And *AF* is the rectangle contained by *BA*, *AC*, for it is contained by *DA*, *AC*, of which *DA* is equal to *BA*;

and *CE* is contained by *AB*, *BC*, for *BE* is equal to *AB*.

Therefore the rectangle *AB*, *AC*, together with the rectangle *AB*, *BC*, is equal to the square on *AB*.

Wherefore, *if a straight line* &c. q.e.d.

*THEOREM*.

*If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square on the aforesaid part*.

Let the straight line *AB* be divided into any two parts at the point *C*: the rectangle *AB*, *BC* shall be equal to the rectangle *AC*, *CB*, together with the square on *BC*.

On *BC* describe the square *CDEB*; [I. 46.

produce *ED* to *F*, and through *A* draw *AF* parallel to *CD* or *BE*. [I. 31.

Then the rectangle *AE* is equal to the rectangles *AD*, *CE*.

But *AE* is the rectangle contained by *AB*, *BC*, for it is contained by *AB*, *BE*, of which *BE* is equal to *BC*;

and *AD* is contained by *AC*, *CB*, for *CD* is equal to *CB*;

and *CE* is the square on *BC*.

Therefore the rectangle *AB*, *BC* is equal to the rectangle *AC*, *CB*, together with the square on *BC*.

Wherefore, *if a straight line* &c. q.e.d.

*THEOREM*.

*If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the two parts*.

Let the straight line *AB* be divided into any two parts at the point *C*: the square on *AB* shall be equal to the squares on *AC*, *CB*, together with twice the rectangle contained by *AC*, *CB*.

On *AB* describe the square *ADEB*; [I. 46.

join *BD*; through *C* draw *CGF* parallel to *AD* or *BE*, and through *G* draw *HK* parallel to *AB* or *DE*. [I. 31.

Then, because *CF* is parallel to *AD*, and *BD* falls on them, the exterior angle *CGB* is equal to the interior and opposite angle *ADB*; [I. 29.

but the angle *ADB* is equal to the angle *ABD*, [I. 5.

because *BA* is equal to *AD*, being sides of a square; therefore the angle *CGB* is equal to the angle *CBG*; [*Ax*. 1.

and therefore the side *CG* is equal to the side *CB*. [I. 6.

But *CB* is also equal to *GK*, and *CG* to *BK*; [I. 34.

therefore the figure *CGKB* is equilateral. It is likewise rectangular. For since *CG* is parallel to
*BK*, and *CB* meets them, the angles *KBC*, *GCB* are together equal to two right angles. [I. 29.

But *KBC* is a right angle. [I. *Definition* 30.

Therefore *GCB* is a right angle. [*Axiom* 3.

And therefore also the angles *CGK*, *GKB* opposite to
these are right angles. [I. 34. and *Axiom* 1.

Therefore *CGKB* is rectangular ;
and it has been shewn to be equi-
lateral ; therefore it is a square, and
it is on the side *CB*.

For the same reason *HF* is also a
square, and it is on the side *HG*,
which is equal to *AC*. [I. 34.

Therefore *HF*, *CK* are the squares
on *AC*,*CB*.

And because the complement *AG* is equal to the complement *GE*; [I.43.

and that *AG* is the rectangle contained by *AC*, *CB*, for
*CG* is equal to *CB*;

therefore *GE* is also equal to the rectangle *AC*, *CB*. [*Ax*. 1.

Therefore *AG*, *GE* are equal to twice the rectangle *AC*, *CB*.

And *HF', *CK* are the squares on *AC*, *CB*. *
Therefore the four figures

*HF*,

*CK*,

*AG*,

*GE*are equal to the squares on

*AC*,

*CB*, together with twice the rectangle

*AC*,

*CB*.

But

*HF*,

*CK*,

*AG*,

*GE*make up the whole figure

*ADEB*, which is the square on

*AB*.

Therefore the square on

*AB*is equal to the squares on

*AC*,

*CB*, together with twice the rectangle

*AC*,

*CB*.

Wherefore, *if a straight line* &c. q.e.d.

Corollary. From the demonstration it is manifest, that parallellograms about the diameter of a square are likewise squares.

*THEOREM*.

*If a straight line be divided into two equal parts and also into two unequal parts, the rectangle contained by the * *unequal parts, together with the square on the line between the points of section, is equal to the square on half the line*.

Let the straight line *AB* be divided into two equal
parts at the point *C*, and into two unequal parts at the
point *D* : the rectangle *AD*,*DB*, together with the square
on *CD*, shall be equal to the square on *CB*.

On *CB* describe the
square *CEFB* ; [l- 46.

join *BE*; through *D* draw
*DHG* parallel to *CE* or *BF*;

through *H* draw *KLM* paral-
lel to *CB* or *EF* ; and through
*A* draw *AK* parallel to *CL*
or *BM*. [I. 31.

Then the complement *CH* is equal to the complement *HF*, [I. 43.

to each of these add *DM* therefore the whole *CM* is equal
to the whole *DF*. [Axiom 2.

But *CM* is equal to *AL*, [I. 36.

because *AC* is equal to *CB*. [*Hypothesis*.

Therefore also *AL* is equal to *DF*, [*Axiom* 1.

To each of these add *CH*; therefore the whole *AH* is equal
to *DF* and *CH*. [*Axiom* 2.

But *AH* is the rectangle contained by *AD*, *DB*, for *DH* is equal to *DB* ; [II. 4, *Corollary*.

and *DF* together with *CH* is the gnomon *CMG* ;

therefore the gnomon *CMG* is equal to the rectangle *AD*,*DB*

To each of these add *LG*, which is equal to the square on
*CD*. [II. 4, *Corollary*, and I. 34.

Therefore the gnomon *CMG*, together with *LG*, is equal to
the rectangle *AD*,*DB*, together with the square on *CD*. [*Ax*.2.

But the gnomon *CMG* and *LG* make up the whole figure
*CEFB*, which is the square on *CB*.

Therefore the rectangle *AD*,*DB*, together with the square
on *CD*, is equal to the square on *CB*.

Wherefore, *if a straight line* &c. q.e.d.

Prom this proposition it is manifest that the difference of
the squares on two unequal straight lines *AC*, *CD*, is equal

*THEOREM*.

*If a straight line he bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced*.

Let the straight line *AB* be bisected at the point *C*,
and produced to the point *D* : the rectangle *AD*, *DB*,
together with the square on *CB*, shall be equal to the
square on *CD*.

On *CD* describe the
square *CEFD* ; [I. 46.

join *DE* through *B* draw
*BHG* parallel to *CE* or
*DF*; through *H* draw
*KLM* parallel to *AD* or
*EF* ; and through *A* draw
*AK* parallel to *CL* or *DM*. [I. 31.

Then, because *AC* equal to *CB*, [*Hypothesis*.

the rectangle *AL* is equal to the rectangle *CH*; [I. 36.

but *CH* is equal to *HF*; [I. 43.

therefore also *AL* is equal to *HF*. [*Axiom* 1.

To each of these add CM ;

therefore the whole *AM* is equal to the gnomon *CMG*. [*Ax*. 2.

But *AM* is, the rectangle contained by *AD*, *DB*,
for *DM* is equal to *DB*. [II. 4, *Corollary*.

Therefore the rectangle *AD*, *DB* is equal to the gnomon
*CMG*. [*Axiom* 1.

To each of these add *LG*, which is equal to the square on
*CB*. [II. 4, *Corollary*, and I. 34.

Therefore the rectangle *AD*, *DB*, together with the square
on *CB*, is equal to the gnomon *CMG* and the figure *LG*.

But the gnomon *CMG* and *LG* make up the whole figure
*CEFD*, which is the square on *CD*.

Therefore the rectangle *AD*,*DB*, together with the square
on *CB*, is equal to the square on *CD*.

*if a straight line*&c. q.e.d.

*THEOREM*.

*If a straight line he divided into any two parts, the squares on the whole line, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part.*

Let the straight line *AB* be divided into any two
parts at the point *C*: the squares on *AB*, *BC* shall be
equal to twice the rectangle *AB*, *BC* together with the
square on *AC*.

On *AB* describe the square
*ADEB*, and construct the figure
as in the preceding propositions.

Then *AG* is equal to *GE* ; [1. 43.

to each of these add CK;

therefore the whole AK is equal to
the whole CE ;

therefore *AK*, *CE* are double of
*AK*.

But *AK*, *CE* are the gnomon *AKF*, together with the square *CK*;

therefore the gnomon *AKF*, together with the square *CK*,
is double of *AK*.

But twice the rectangle *AB*, *BC* is double of *AK*,
for *BK* is equal to *BC*. [II. 4, *Corollary*.

Therefore the gnomon *AKF*, together with the square *CK*,
is equal to twice the rectangle *AB*, *BC*.

To each of these equals add *HF*, which is equal to the
square on *AG*. [II. 4, *Corollary*, and I. 34.

Therefore the gnomon *AKF*, together with the squares
*GK*, *HF*, is equal to twice the rectangle *AB*, *BC*, together
with the square on *AC*.

But the gnomon *AKF* together with the squares *GK*, *HF*,
make up the whole figure *ADEB* and *GK*, which are the squares on *AB* and *BC*.

Therefore the squares on *AB*, *BC*, are equal to twice the
rectangle *AB*, *BC*, together with the square on *AC*.

*if a straight line*&c. q.e.d.

PROPOSITION 8. *THEOREM*.

If a straight line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts, together with the square on the other part, is equal to the square on the straight line which is made up of the whole and that part.

Let the straight line *AB* be divided into any two parts at the point *C*: four times the rectangle *AB*, *BC*, together with the square on *AC* shall be equal to the square on the straight line made up of *AB* and *BC* together.

Produce *AB* to *D*, so

*BD*may be equal to

*CB*; [

*Post*. 2. and I. 3.

on

*AD*describe the square

*AEFD*;

and construct two figures such as in the preceding propositions.

Then, because *CB* is equal to *BD*,[*Construction*.

and that *CB* is equal to *GK*, and *BD* to *KN*,[I. 34.

therefore *GK* is equal to *KN*.[*Axiom* 1.

For the same reason *PR* is equal to *RO*.

And because *CB* is equal to *BD*, and *GK* to *KN*, the rectangle *CK* is equal to the rectangle *BN*, and the rectangle *GR* to the rectangle *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*.[*Axiom* 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*, [*Construction*.

and that *BD* is equal to *BK*, [II. 4, *Corollary*.

that is to *CG*, [I. 34.

and that CB is equal to GK, [I. 34.

that is to *GP*; [II. 4, *Corollary*.

therefore *CG* is equal to *GP*. [*Axiom* 1.

And because *CG* is equal to *GP*, and *PR* to *RO*, the rectangle *AG* is equal to the rectangle *MP*, and the rectangle *PL* to the rectangle *RF*, [I. 36.

But *MP* is equal to *PL*, because they are the complements of the parallelogram *ML*; [I. 43.

therefore also *AG* is equal to *RF*. [*Axiom* 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 shewn 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*, *BC*; for *BK* is equal to *BC*;

therefore four times the rectangle *AB*, *BC* is quadruple of *AK*.

But the gnomon *AOH* was shewn to be quadruple of *AK*,

Therefore four times the rectangle *AB*, *BC* is equal to the gnomon *AOH*. [*Axiom* 1.

To each of these add *XH*, which is equal to the square on *AC*. [II. 4, *Corollary*, 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*.

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

Wherefore, *if a straight line* &c. q.e.d.

*THEOREM*.

*If a straight line he divided into two equals 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 at the point *C*, and into two unequal parts at the
point *D* : the squares on *AD*, *DB* shall be together double
of the squares on *AC*, *CD*.

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*, *EB* ; through
*D* draw *DF* parallel to *CE*, and
through *F* draw *FG* parallel
to *BA* ; [I. 31.

and join *AF*.

Then, because *AC* is equal to *CE*, [*Construction*.

the angle *EAC* is equal to the angle *AEC*. [I. 5.

And because the angle *ACE* is a right angle, [*Construction*.
the two other angles *AEC*, *EAC* are together equal to one
right angle ; [I. 32.

and they are equal to one another ;

therefore each of them is half a right angle.

For the same reason each of the angles *CEB*, *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.

*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* [*Construction*.

the square on *AC* is equal to the square on *CE*;

therefore the squares on *AC*, *CE* are double of the square on *AC*.

But the square on *AE* is equal to the squares on *AC*, *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*, [*Construction*.

the square on *EG* is equal to the square on *GF*;

therefore the squares on *EG*, *GF* are double of the square on *GF*.

But the square on *EF* is equal to the squares on *EG*, *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 shewn that the square on *AE* is also double of the square on *AC*.

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

But the square on *AF* is equal to the squares on *AE*, *EF*, because the angle *AEF* is a right angle, [I. 47.

Therefore the square on *AF* is double of the squares on *AC*, *CD*.

But the squares on *AD*, *DF* are equal to the square on *AF*, because the angle *ADF* is a right angle. [I. 47.

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

And *DF* is equal to *DB*;

therefore the squares on *AD*, *DB* are double of the squares on *AC*, *CD*.

Wherefore, *if a straight line* &c. q.e.d.

*THEOREM*.

*If a straight line he 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*.

Let the straight line *AB* be bisected at *C*, and pro-
duced to *D* : the squares on *AD*, *DB* shall be together
double of the squares on *AC*, *CD*.

From the point *C* draw *CE* at right angles to *AB*, [I, 11.

and make it equal to *AC*
or *CB*; [1.3.

and join *AE*,*EB* ; through
*E* draw *EF* parallel to
*AB*, and through *D* draw
*DF* parallel to *CE*. [1.31

Then because the straight
line *EF* meets the parallels
*EC*, *FD*, the angles *CEF*, *CFD* are together equal to two
right angles ; [I. 29.

and therefore the angles *BEF*, *EFD* are together less
than two right angles.

Therefore the straight lines *EB*, *FD* will meet, if produced,
towards *B*, *D*, [*Axiom* 12.

Let them meet at *G*, and join *AG*.

Then because *AC* is equal to *CE*, [*Construction*.

the angle *CEA* is equal to the angle *EAC* ; [I. 5.

and the angle *ACE* is a right angle ; [*Construction*.
therefore each of the angles *CEA*, *EAC* is half a right angle. [I. 32.

For the same reason each of the angles *CEB*, *EBC* is half
a right angle.

Therefore the angle *AEB* is a right angle.

And because the angle *EBC* is half a right angle,
the angle *DBG* is also half a right angle, for they are verti-
cally opposite ; [I. 15.

but the angle *BDG* is a right angle, because it is equal to
the alternate angle *DCE* ; [I. 29.

*DGB*is half aright angle, [1. 32.

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 is 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*,*CA* are double of the square on *CA*.

*AE*is equal to the squares on

*EC*,

*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*, *FE* are double of the square on *FE*.

But the square on *EG* is equal to the squares on *GF*, *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 shewn that the square on *AE* is double of the square on *AC*.

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

But the square on *AG* is equal to the squares on *AE*,*EG*. [I. 47.

Therefore the square on *AG* is double of the squares on *AC*, *CD*.

But the squares on *AD*, *DG* are equal to the square on *AG*. [I. 47.

Therefore the squares on *AD*, *DG* are double of the squares on *AC*, *CD*.

And *DG* equal to *DB*;

therefore the squares on *AD*, *DB* are double of the squares on *AC*,*CD*.

Wherefore, *if a straight line* &c. Q.E.D.

*PROBLEM*.

*To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part*.

Let *AB* be the given straight line: it is required to
divide it into two parts, so that the rectangle contained by
the whole and one of the parts may be equal to the scpiare
on the other part.

On *AB* describe the square
*ABDC*; [I. 46.

bisect *AC* at *E*; [I. 10.

join *BE*; produce *CA* to *F*, and
make *EF* equal to *EB*; [I. 3.

and on *AF* describe the square
*AFGH*. [1. 46.*AB* shall be divided at *H * so
that the rectangle *AB*, *BH* is
equal to the square on *AH*.

Produce *GH* to *K*.
Then, because the straight line
*AC* is bisected at *E*, and pro-
duced to *F*, the rectangle *CF*, *FA*, together with the
square on *AE*, is equal to the square on *EF*. [II. 6.

But *EF* is equal to *EB*. [*Construction*.

Therefore the rectangle *CF*, *FA*, together with the square
on *AE*, is equal to the square on *EB*.

But the square on *EB* is equal to the squares on *AE*,*AB*,
because the angle *EAB* is a right angle. [I. 47.

Therefore the rectangle *CF*, *FA*, together with the square
on *AE*, is equal to the squares on *AE*, *AB*.

Take away the square on *AE*, which is common to both;

therefore the remainder, the rectangle *CF*, *FA*, is equal to
the square on *AB*. [*Axiom* 3.

But the figure *FK* is the rectangle contained by *CF*, *FA*,
for *FG* is equal to *FA*;

and *AD* is the square on *AB*;

therefore *FK* is equal to *AD*.

Take away the common part *AK*, and the remainder *FH*
is equal to the remainder *HD*. [*Axiom* 3.

But *HD* is the rectangle contained by *AB*, *BH*, for *AB* is
equal to *BD* ;

and *FH* is the square on *AH*;

therefore the rectangle *AB*,*BH* is equal to the square on *AH*.

Wherefore *the straight line AB is divided at H, so that *
the rectangle AB, BH is equal to the square on AH*. q.e.f. *

*THEOREM*.

*In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side on which, then produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle*.

Let *ABC* be an obtuse-angled triangle, having the
obtuse angle *ACB*, and from the point *A* let *AD*
be drawn
perpendicular to *BC* produced ; the square on *AB* shall be
greater than the squares on *AC*, *CB*, by twice the rectangle *BC*, *CD*.

Because the straight line
*BD* is divided into two parts
it the point *C*, the square on
*BD* is equal to the squares on
*BC*, *CD*, and twice the rectangle
*BC*, *CD*. [II. 4.

To each of these equals add the
quare on DA.

Therefore the squares on *BD*, *DA* are equal to the squares on
*BC*, *CD*, *DA*, and twice the rectangle *BC*, *CD*. [*Axiom* 2.

But the square on *BA* is equal to the squares on *BD*, *DA*,
because the angle at *D* is a right angle ; [I. 47.

and the square on *CA* is equal to the squares on *CD*, *DA*. [1. 47.

therefore the square on *BA* is equal to the squares on
*BC*, *CA*, and twice the rectangle *BC*, *CD* ;

that is, the square on *BA* is greater than the squares on
*BC*, *CA* by twice the rectangle *BC*, *CD*.

*in obtuse-angled triangles*&c. q.e.d.

*THEOREM*.

*In every triangle, the square on the side subtending an acute angle, is less than the squares on the sides containing that angle, by twice the rectangle contained by eithier of these sides, and the straight line intercepted *
between the perpendicular let fall on it from the opposite angle, and the acute angle*. *

Let *ABC* be any triangle, and the angle at *B* an acute
angle ; and on *BC* one of the sides containing it, let fall
the perpendicular *AD* from the opposite angle: the square
on *AC*, opposite to the angle *B*, shall be less than the
squares on *CB*, *BA*, by twice the rectangle *CB*, *BD*.

First, let *AD* fall within the
triangle *ABC*.

Then, because the straight line
*CB* is divided into two parts
at the point *D*, the squares on
*CB*, *BD* are equal to twice the
rectangle contained by *CB*, *BD*
and the square on *CD*. [II. 7.

To each of these equals add the
square on *DA*.

Therefore the squares on *CB*, *BD*, *DA* are equal to twice
the rectangle *CB*, *BD* and the squares on *CD*, *DA*. [*Ax*. 2.

But the square on *AB* is equal to the squares on *BD*, *DA*,
because the angle *BDA* is a right angle ; [I. 47.

and the square on *AC* is equal to the squares on *CD*, *DA*. [I. 47.
Therefore the squares on *CB*, *BA* are equal to the square
on *AC* and twice the rectangle *CB*, *BD* ;

that is, the square on *AC* alone is less than the squares on
*CB*, *BA* by twice the rectangle *CB*, *BD*.

Secondly, let *AD* fall without
the triangle *ABC*.
Then because the angle at *D* is
a right angle, [*Construction*.

the angle *ACB* is greater than
a right angle ; [I. 1.

and therefore the square on *AB* is equal to the squares on *AC*, *CB*, and twice the rectangle *BC*, *CD*. [II. 12.

To each of these equals add the square on *BC*.

Therefore the squares on *AB*, *BC* are equal to the square on *AC*, and twice the square on *BC*, and twice the rectangle *BC*, *CD*. [*Axiom* 2.

But because *BD* is divided into two parts at *C*, the rectangle *DB*, *BC* is equal to the rectangle *BC*, *CD* and the square on *BC*; [II. 3.

and the doubles of these are equal,

that is, twice the rectangle *DB*, *BC* is equal to twice the rectangle *BC*, *CD* and twice the square on *BC*.

Therefore the squares on *AB*, *BC* are equal to the square on *AC*, and twice the rectangle *DB*, *BC*;

that is, the square on *AC* alone is less than the squares on *AB*, *BC* by twice the rectangle *DB*, *BC*.

Lastly, let the side *AC* be perpendicular to *BC*.

Then *BC* is the straight line between the perpendicular and the acute angle at *B*;

and it is manifest, that the squares on *AB*, *BC* are equal to the square on *AC*, and twice the square on *BC*. [I. 47 and *Ax*. 2.

Wherefore, *in every triangle* &c. q.e.d.

*PROBLEM*.

*To describe a square that shall he equal to a given rectilineal figure*.

Let *A* be the given rectilineal figure: it is required to describe a square that shall be equal to *A*.
Describe the rectangular parallelogram *BCDE* equal to the rectilineal figure *A*. [1. 45.

Then if the sides of it, *BE*, *ED* are equal to one another, it is a square, and what was required is now done.

But if they are not equal, produce one of them *BE* to *F* make *EF* equal to *ED*, [I. 3.

and bisect *BF* at *G*; [I. 10.

from the centre *G*, at the distance *GB*, or *GF*, describe the semicircle *BHF*, and produce *DE* to *H*.

The square described on *EH* shall be equal to the given rectilineal figure *A*.

Join *GH*. Then, because the straight line *BF* is divided into two equal parts at the point *G*, and into two unequal parts at the point *E*, the rectangle *BE*, *EF*, together with the square on *GE*, is equal to the square on *GF*. [II. 5.

But *GF* is equal to *GH*.

Therefore the rectangle *BE*, *EF*, together with the square on *GE*, is equal to the square on *GH*.

But the square on *GH* is equal to the squares on *GE*, *EH*;[I.47.

therefore the rectangle *BE*, *EF*, together with the square on *GE*, is equal to the squares on *GE*, *EH*.

Take away the square on *GE*, which is common to both;

therefore the rectangle *BE*, *EF* is equal to the square on *EH*. [*Axiom* 3.

But the rectangle contained by *BE*, *EF* is the parallelogram *BD*,

because *EF* is equal to *ED*. [*Construction*.

Therefore *BD* is equal to the square on *EH*.

But *BD* is equal to the rectilineal figure *A*. [*Construction*.

Therefore the square on *EH* is equal to the rectilineal figure *A*.

Wherefore *a square has been made equal to the given rectilineal figure *A*, namely, the square described on EH*.q.e.f