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47. 1.

the

Produce GH to K: and because the straight line AC is bi. sected in E, and produced to the point F, the rectangle CF, FA, together with the square of AE, is equal * to the square . 6. 2. of EF: but EF is equal t to EB ; therefore

+ Constr. the rectangle CF, FA, together with the square of AE, is equal to the square of EB:

HH B but the squares of BA, AE are equal to the

ΑΙ square of EB, because the angle EAB is a

E right angle; therefore the rectangle CF, FA, together with the square of AE, is equal to

K D the squares of BA, AE: take

away square of AE, which is common to both; therefore the remaining rectangle CF, FA is equal † to the square of AB: but the + 3 Ax. figure FK is the rectangle contained by CF, FA, for FA is equal + to FG; and AD is the square of AB ; therefore FK + 30 Def. is equal † to AD: take away the common part AK, and the † 1 Ax. remainder FH is equal † to the remainder HD: but hd is † 3 Ax. the rectangle contained by AB, BH, for AB is equal + to BD; + 30 Def. and FH is the square of AH; therefore the rectangle AB, BH is equal to the square of AH. Wherefore, the straight line AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. Which was to be done.

PROPOSITION XII.

THEOR.—In obtuse-angled triangles, if a perpendicular be

drawn from either of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle, is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side

upon nehich, 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 of AB shall be greater than the squares of AC, CB, by twice the rectangle BC, CD.

Because the straight line BD is divided into two parts in the point C, the square of BD is equal

to the

squares of BC, CD, and twice the rectangle BC, CD: to each of these equals add the

*

# 12. ).

B

* 4. 2.

+ 2 Ax.

47, 1.

*

to the squares

47. 1.

square of DA; therefore the squares of BD, DA are equal +
to the squares of BC, CD, DA, and twice the rectangle
BC, CD: but the square of BA is equal
of BD, DA, because the angle at D is a right angle; and the
square of CA is equal * to the squares of CD, DA ; therefore
the

square of BA is equal to the squares of BC, CA, and twice the rectangle BC, CD; that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BC, CD. Therefore, in obtuse-angled triangles, &c. Q. E. D.

PROPOSITION XIII.

See N.

Theor.— In every triangle, the square of the side subtending

either of the acute angles, is less than the squares of 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 upon it from the opposite angle, and the acute angle.

• 12. 1.

* 7. 2.

+ 2 Ax.

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

First, let AD fall within the triangle ABC: and because the straight line CB is divided into two parts in the point D, the squares of CB, BD are equal to twice the rectangle contained by CB, BD, and the square of DC: to each of these equals add the square

of AD; therefore the squares of CB, BD, DA are equal + to twice the rectangle CB, BD, and the squares of AD, DC: but the square

of AB is equal * to the squares of BD, DA, because the angle BDA is a right angle ; and the square of AC is equal to the squares of AD, DC; therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD; that is, the square of AC alone, is less than the squares of 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, the angle ACB is

* than a right angle; and therefore the square of

B.

D

# 47. 1.

• 16. 1.

greater

B

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AB is equal* to the squares of AC, CB, and twice the rectangle • 12. 2. BC, CD: to each of these equals add the square of BC; therefore the squares of AB, BC are equalt to the square of AC, † 2 Ax. and twice the square of BC, and twice the rectangle BC, CD: but because BD is divided into two parts in C, the rectangle,

* 3. 2. DB, BC is equal * to the rectangle BC, CD and the square of BC; and the doubles of these are equal; therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: therefore the square of AC alone, is less than the squares of 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 of AB, BC are equal * to the

• 47. 1. and

2 Ax. square of AC, and twice the square of BC. Therefore, in every triangle, &c.

À

B

Q. E. D.

PROPOSITION XIV.

PROB.—To describe a square that shall be equal to a given See N.

rectilineal figure.

• 45. 1.

D

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. Then if the sides

BH of it, BE, ED are equal to one an

G E

F other, it is t a square, and what

C

+ 30 Def. was required is now done : but if they are not equal, produce one of them BE to F, and make + EF equal to ED, and bisect + + 3. 1: BF in G; and from the centre G, at the distance GB,

† 10. 1. GF, describe, the semicircle BHF, and produce DE to H: the square described upon EH shall be equal to the given rectilineal figure A.

Join GH: and because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the square of EG, is equal * to the square of GF: but GF is equalt to GI; therefore the rectangle BE, EF, together with the square

of

+ 15 Def.

or

* 5. 2.

EG, is equal to the square of GH: but the squares of HE, EG are equal * to the square of GH; therefore the rectangle BE, EF, together with the square of EG, is equal + to the squares of HE, EG : take away the square of EG, which is common to both; and the remaining rectangle BE, EF is equal + to the square of EH: but the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH: but BD is equal + to the rectilineal figure A; therefore the rectilineal figure A is equal t to the square of EH. Wherefore, a square has been made equal to the given rectilineal figure A; viz. the square described upon

* 47.1.

† 1 Ax.

+ 3 Ax.

EH. Which was to be done.

+ Constr.

†1 Ax.

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1. EQUAL circles are those of which the diameters are equal,

or from the centres of which the straight lines to the circumferences are equal.

“ This is not a definition, but a theorem, the truth of which is evident ; for, if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centres are equal."

II. A straight line is said to touch a circle, when it meets

the circle, and being produced does not cut it.

III. Circles are said to touch one an

other, which meet but do not cut one another.

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IV. Straight lines are said to be equally distant from the

centre of a circle, when the perpendiculars drawn to them from the centre are equal.

V. And the straight line on which the greater

perpendicular falls, is said to be farther from the centre.

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