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Therefore the square on AG is double of the squares on AG? AC and CD.

=AD2+

DB2.

But the squares on AD and DG are equal to the square on ..AD2+ AG (I. 47);

DB2

=2(AC+

Therefore the squares on AD and DG are double of the CD2), squares on AC and CD.

And DG is equal to DB; therefore the squares on AD and DB are double of the squares on AC

and CD.

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

Proposition 11.-Problem.

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

Let AB be the given straight line.

It is required to divide AB into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square on the other part. CONSTRUCTION.-Upon AB describe

the square ABDC (I. 46).

Bisect AC in E (I. 10), and join BE.
Produce CA to F, and make EF equal

to EB (I. 3).

F

G

A

Upon AF describe the

square AFGH

[blocks in formation]

square on AH.

[blocks in formation]

PROOF.-Because the straight line AC is bisected in E,

and produced to F,

The rectangle CF, FA, together with the square on AE, CF-FA+ is equal to the square on EF (II. 6). But EF is equal to EB (Const.);

Therefore the rectangle CF, FA, together with the square

on AE, is equal to the square on EB.

AE2
=EF2
EB2

But the square on EB is equal to the squares on AE and =AB2+ AB, because the angle EAB is a right angle (I. 47);

AE2.

CF FA =AB?,

...FK=AD

Take away
AK, then

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

Take away the square on AE, which is common to both; Therefore the remaining rectangle CF, FA is equal to the square on AB (Ax. 3).

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

And AD is the square on AB;

Therefore the figure FK is equal to AD.

Take away the common part AK, and the remainder FH FH=HD is equal to the remainder ĦD (Ax. 3).

or AB BH =AH2.

For
AB2=AC2
+CB2+

2 BC.CD.

BD2=BC2 +CD2+

2 BC CD.

But HD is the rectangle contained by AB and BII, 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.

Therefore the straight line AB is divided in H, so that the rectangle AB, BH is equal to the square on AH. Q.E.F.

Proposition 12.-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, when 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 and CB by twice the rectangle BC, CD.

PROOF.-Because the straight line BD
is divided into two parts in the point C,
The square on BD is equal to the
CD, and twice the rectangle BC, CD

squares on BC and
(II. 4).
To each of these equals add the square on DA;

or AB2

DA?)

Therefore the squares on BD and DA are equal to the ..BD+DA squares on BC, CD, DA, and twice the rectangle BC, CD. =BC2+ But the square on BA is equal to the squares on BD and (CD2+ DA, because the angle at D is a right angle (I. 47); And the square on CA is equal to the squares on CD and C+ DA (I. 47);

Therefore the square on BA is equal to the squares on BC and CA, and twice the rectangle BC, CD; that is, the square on BA is greater than the squares on BC and CA by twice the rectangle BC, CD.

Therefore, in obtuse-angled triangles, &c.

Proposition 13.-Theorem.

Q.E.D.

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 either 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 (I. 12).

+2BC CD

AC2
+2BC CD.

+AB2

The square on AC opposite to the angle B, shall be less AC2=CB2 than the squares on CB and BA, by twice the rectangle CB BD. CB, BD.

CASE I. First, let AD fall within the triangle ABC.

PROOF. Because the straight line CB is divided into two parts in the point D, The squares on CB and BD are equal to twice the rectangle contained by CB, BD, and the square on DC (II. 7).

A

To each of these equals add the square on DA.

For

CB2+BD2

=2CB BD

+ DC2.

Therefore the squares on CB, BD, DA are equal to twice..CB2+

the rectangle CB, BD, and the squares on AD and DC.

(BD2+ DA2)

+(AD2+ DC2),

But the square on AB is equal to the squares on BD and 2CB BD DA, because the angle BDA is a right angle (I. 47); And the square on AC is equal to the squares on AD and DC (I. 47);

Therefore the squares on CB and BA are equal to the

[ocr errors]

CB2+BA? =2CB.DB + AC2.

..AC =CB2

+ BA2 → 2CB BD.

AB2-AC2 +CB2+ BC CD.

...AB2+
BC2
=AC2+
2(BC2+
BC CD).

Now

DB BC=

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

Б

CASE II.-Secondly, let AD fall without the triangle ABC.
PROOF. Because the angle at D is a right angle (Const.),
A the angle ACB is greater than a right

C

angle (I. 16);

Therefore the square on AB is equal to the squares on AC and CB, and twice the rectangle BC, CD (II. 12).

To each of these equals add the square on BC.

Therefore the squares on AB and BC are equal to the square on AC, and twice the square on BC, and twice the rectangle BC, CD (Ax. 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 BC CD rectangle DB, BC is equal to twice the rectangle BC, CD and twice the square on BC;

BC2.

...AC3

=AB2+ BC22BC.DB.

Therefore the squares on AB and BC are equal to the square on AC, and twice the rectangle DB, BC; that is, the A square on AC alone is less than the squares on AB and BC by twice the rectangle DB, BC.

CASE III. Lastly, let the side AC be perpendicular to BC.

PROOF.-Then BC is the straight line between the perpendicular and the acute angle at B; and it is manifest that the squares on AB and BC are equal to the square on AC, and twice the square on BC (I. 47, and Ax. 2).

B

Therefore, in every triangle, &c. Q.E.D.

Proposition 14.-Problem.

To describe a square that shall be 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.

CONSTRUCTION.-Describe the rectangular parallelogram BCDE equal to the rectilineal figure A. (I. 45).

If then 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 cqual, produce one of them, BE, to F, and make EF equal to ED (I. 3).

[blocks in formation]

Bisect BF in G (I. 10), and from the centre G, at the distance GB, or GF, describe the semicircle BHF;

Produce DE to H, and join GH;

Then the square described upon EI shall be equal to the rectilineal figure A.

PROOF.-Because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in the point E,

The rectangle BE,EF, together with the square on GE, BE EF is equal to the square on GF (II. 5).

But GF is equal to GH;

+GE2

=GF2

=GH3 =GE2+

Therefore the rectangle BE,EF, together with the square EH

on GE, is equal to the square on GH.

But the square on GH is equal to the squares on GE and EH (I. 47);

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

Take away the square on GE, which is common to both; Therefore the rectangle BE,EF is equal to the square on EH (Ax. 3).

But the rectangle contained by BE and EF is the parallelogram BD, because EF is equal to ED (Const.); Therefore BD is equal to the square on EH.

or BD =EH2.

But BD is equal to the rectilineal figure A (Const.); Therefore the square on EH is equal to the rectilineal Hence figure A.

Therefore, a square has been made equal to the given rectilineal figure A, viz., the square described on EH. Q.E.F.

EH2=A.

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