66. 1.

h 34. 1.

For the same reason, each of the angles CEB, EBC is half Book II. a right angle; therefore AEB is a right angle: And because EBC is half a right angle, DBG is alsof half a right '15. 1. angle, for they are vertically opposite ; but BDG is a right angle, because it is equal to the alternate angle DCE; €29. 1. therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side BD is equals to the side DG. Again, because EGF is half a right angle, and that the angle at F is a right angle, because it is equalb

CI

B. to the opposite angle A

D ECD, the remaining angle FEG is half a

G right angle, and equal to the angle EGF; wherefore also the side GF is equals to the side FE. And because EC is equal to CA, the square of EC is equal to the square of CA; therefore the squares of EC, CA are double of the square of CA: But the square of EA is equali to the squares of EC, CA; there- i 47. 1. fore the square of EA is double of the square of AC: Again, because GF is equal to FE, the square of GF is equal to the square of FE; and therefore the squares of GF, FE are double of the square of EF: But the square of EG is equali to the squares of GF, FE; therefore the square of EG is double of the square of EF: And EF is equal to CD; wherefore the square of EG is double of the square of CD. But it was demonstrated, that the square of EA is double of the square of AC; therefore the squares of AE, EG, are double of the square of AC, CD: And the square of AG is equali to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD: But the squares of AD, GD are equal to the square of AG; therefore the squares of AD, DG are double of the squares of AC, CD: But DG is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD. Wherefore, if a straight line, &c. Q. E. D.

Book II.

PROP. XI. PROB.

a 46. 1.

b

10. J.

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 of 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, shall be equal to the square of the other part.

Upon AB describea the square ABDC; bisectb AC in E, and join BE; produce CA to F, and make EF equal 'to EB, and upon AF describea the square of FGHA; AB is divided in H, so that the rectangle AB, BH, is equal to the square of AH.

Produce GH to K; because the straight line AC is bi

sected in E, and produced to the point F, the rectangle d 6. 9. CF, FA, together with the square of AE, is equald to the

square of EF: But EF is equal to EB; therefore the rectangle CF, FA, together with the square of AE, is equal to the square of EB: And the squares of

F ? 47. 1. BA, AE, are equale to the square of

G
EB, because the angle EAB is a right
angle; therefore the rectangle CF,FA,
together with the square of AE, is
equal to the squares of BA, AE: Take A

HB
away the square of AE, which is com-
mon to both, therefore the remaining
rectangle CF, FA, is equal to the
square of AB; and the figure FK is E
the rectangle contained by CF, FA,
for AF is equal to FG; and AD is
the square of AB; therefore FK is
equal to AD: Take away the common

K D part AK, and the remainder FH is equal to the remainder HD: And HD is the rectangle contained by AB, BH, for AB is equal to BD; 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.

Book II.

PROP. XII. THEOR.

In obtuse angled triangles, if a perpendicular be drawn from any 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 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 drawna • 12. 1. perpendicular to BC produced: The square of AB is 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

А equal bto the squares of BC, CD,

4. 2. and twice the rectangle BC, CD: To each of these equals add the square

of DA; and 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 equale B to the squares 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.

• 47, 1.

Book II. .

PROP. XIII. THEOR.

See N. In every triangle, the square of the side subtending

any of the acute angles, is less than the squares of
the sides containing that angle, by twice the rect-
angle 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.

Let ABC be any triangle, and the angle at B one of its

acute angles, and upon BC, one of the sides containing it, * 12. 1. let fall the perpendiculara AD from the opposite angle:

The square of AC, opposite to the angle B, is 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

A
into two parts in the point D,
b7. 2. the squares of CB, BD are equalb
to twice the rectangle contained

ant
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,

B D
BD, and the squares of AD, DC:
47. 1. 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, BI), that is, the square of AC
alone is less than the squares of CB, BA by twice the rect-
angle CB, BD.

Secondly, Let AD fall without
the triangle ABC: Then, because

the angle at D is a right angle, . 16. 1. the angle ACB is greaterd than

a right angle; and therefore the * 12. 2. square of AB is equale to the

squares of AC, CB, and twice the
rectangleBC,CD:To these equals
add the square of BC, and the B

C

1

squares of AB, BC are equal to the square of AC, and twice Book II,
the square of BC, and twice the rectangle BC, CD: But
because BD is divided into two parts in C, the rectangle
DB, BC is equal to the rectangle BC, CD and the square f 3. 2.
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 rect-
angle DB, BC.
Lastly, let the side AC be perpendicular to

A
BC; then is BC the straight line between the
perpendicular and the acute angle at B; and
it is manifest, that the squares of AB, BC, are
equals to the square of AC and twice the
square of BC: Therefore, in every triangle, &c.
Q.E.D.

& 47. 1.

To describe a square that shall be equal to a given See N. rectilineal figure.

Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A.

Describea the rectangular parallelogram BCDE equal · 45. 1. to the rectilineal figure A. If then the sides of it, BE, ED, are equal to one another, it is a

H square,and what was required is now done : But A if they are not

BI equal, produce

G one of them BE

C to F, and make EF equal to ED, and bisect BF in G; and from the cevtre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H, and join GH : Therefore 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 1, 5.2. square of GF: But GF is equal to Gi: Therefore the rectangle BE, EF, together with the square of EG, is equal