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* 32. I.

right angle *. for the same reason, each of the angles CEB, EBC Book II. is half a right angle; therefore AEB is a right angle, and because EBC is half a right angle, DBG is also f half a right angle, for they are vertically opposite; but BDG is a right angle, because it

C

f. 15. I.

is equal to the alternate angle DCE; therefore the remaining an- c. 29. I. gle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore alfo the fide BD is equal to the fide DG. g. 6. I. again, because EGF is half a

right angle, and that the angle at F is a right angle, because it is equal to the oppofite angle ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF; wherefore alfo the fide GF is equal

to

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i i. 47. I.

the fide FE. And becaufe EC is equal to CA, the fquare of EC is equal to the square of CA; therefore the fquares of EC, CA are double of the fquare of CA. but the fquare of EA is equal to the fquares of EC, CA; therefore 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 fquare of FE; and therefore the squares of GF, FE are double of the fquare of EF. but the fquare of EG is equal to the squares of GF, FE; therefore the fquare of EG is double of the fquare of EF. and EF is equal to CD, wherefore the fquare of EG is double of the fquare of CD. but it was demonstrated that the fquare of EA is double of the square of AC; therefore the squares of AE, EG are double of the fquares of AC, CD. and the fquare of AG is equal to the fquares of AE, EG; . 47. L. therefore the fquare of AG is double of the fquares of AC, CD. but the fquares of AD, DG are equal to the fquare of AG; therefore the squares of AD, DG are double of the fquares of AC, CD. but DG is equal to DB; therefore the fquares of AD, DB are double of the fquares of AC, CD. Wherefore if a straight line, &c. Q. E, D.

D 3

i

i.

Book IT.

a. 46. I.

b. 10. I. C. 3. I.

d. 6. z.

e. 47. I.

T

PROP. XI. PROB.

O divide a given ftraight line into two parts, fo that the rectangle contained by the whole, and one of the parts, shall be equal to the fquare of the other part.

Let AB be the given ftraight line; it is required to divide it into two parts, fo that the rectangle contained by the whole, and one of the parts, fhall be equal to the fquare of the other part.

Upon AB defcribe a the fquare ABDC, bifect AC in E, and join BE; produce CA to F, and make EF equal to EB; and upon AF defcribe the fquare FGHA, and produce GH to K. AB is divided in H fo, that the rectangle AB, BH is equal to the fquare of AH.

with F

e

G

HB

Because the ftraight line AC is bifected in E, and produced to the point F, the rectangle CF, FA, together with the square of AE, is equal to the square of EF. but EF is equal to EB; there fore the rectangle CF, FA, together with the fquare of AE is equal to the fquare of EB. and the fquares of BA, AE are equal to the fquare of EB, because the angle EAB is a right angle; therefore the rectangle CF, A FA, together with the fquare of AE is equal to the fquares of BA, AE. take away the fquare of AE, which is common to both, therefore the remaining rectangle CF, FA is equal to the fquare of AB. and the figure FK is the rectangle contained by CF, FA, for AF is equal to FG; and AD is the fquare of AB; therefore FK is equal to AD. take away the common 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 fquare of AH. wherefore the ftraight line AB is divided in H, fo that the rectangle AB, BH is equal to the fquare of AH. Which was to be done.

K D

PROP. XII. THEOR.

N obtufe angled triangles, if a perpendicular be

IN

drawn from any of the acute angles to the oppofite fide produced, the fquare of the fide fubtending the obtuse angle, is greater than the fquares of the fides containing the obtufe angle, by twice the rectangle contained by the fide upon which when produced the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtufe angle.

Let ABC be an obtufe angled triangle, having the obtufe angle ACB, and from the point A let AD be drawn a perpendicular to BC produced. the square of AB is greater than the squares of AC, CB by twice the rectangle BC, CD.

Book II.

2. 12. I

A b. 4. 20

Because the straight line BD is divided into two parts in the point C, the fquare of BD is equal b to the fquares of BC, CD, and twice the rectangle BC, CD. to each of thefe equals add the square of DA; and the fquares of BD, DA are equal to the fquares of BC, CD, DA, and twice the rectangle BC, CD. but the fquare of BA is equal

C

C.

D c. 47. I.

* to the squares of BD, DA, be-B
cause the angle at D is a right an-
gle; and the fquare of CA is equal to the fquares of CD, DA.
therefore the fquare of BA is equal to the fquares of BC, CA,
and twice the rectangle BC, CD; that is, the fquare of BA is
greater than the fquares of BC, CA, by twice the rectangle BC,
CD. Therefore in obtufe angled triangles, &c. Q. E. D.

D 4

Book II.

See N.

IN

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every triangle the fquare of the fide fubtending any of the acute angles, is lefs than the fquares of the fides containing that angle, by twice the rectangle contained by either of these fides, and the straight line intercepted between the perpendicular let fall upon it from the oppofite 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 fides containing it let fall the . 12. 1. perpendicular a AD from the oppofite angle. the fquare of AC oppofite to the angle B, is less than the squares of CB, BA by twice the rectangle CB, BD.

b. 7. 2

A

First, Let AD fall within the triangle ABC; and because the ftraight line CB is divided into two parts in the point D, the fquares of CB, BD are equal to twice the rectangle contained by CB, BD, and the fquare of DC. to each of thefe equals add the fquare of AD, ` therefore the fquares of CB, BD, DA are equal to twice the rectangle CB, BD, and the fquares of AD,

c. 47. 1. DC. but the fquare of AB is equal B

C

D

to the fquares of BD, DA, because the angle BDA is a right angle; and the fquare of AC is equal to the fquares of AD, DC, therefore the fquares of CB, BA are equal to the fquare of AC, and twice the rectangle CB, BD; that is, the fquare of AC alone is less than the fquares of CB, BA by twice the rectangle CB, BD.

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are equal to the fquare of AC, and twice the fquare of BC, and Book II. 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, f. 3. 2. CD and the square of BC. and the doubles of these are equal. therefore the fquares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC. therefore the fquare of AC alone, is less than the fquares of AB, BC, by twice the rectangle DB, BC.

Laftly, let the fide AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B. and it

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A

C. 47. I.

To

PROP. XIV. PROB.

O defcribe a fquare that fhall be equal to a given See N. rectilineal figure.

Let A be the given rectilineal figure; it is required to describe

a fquare that shall be equal to A.

Defcribe the rectangular parallelogram BCDE equal to the rec- a. 45. Ia tilineal figure A. If then the fides of it BE, ED are equal to one another, it is a fquare,

re

and what was quired is now done. but if they are not equal, produce one of them BE to F, and make EF equal

to ED, and bifect

H

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BF in G; and from the center G, at the distance GB or GF defcribe the femicircle 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 fquare of EG, is equal to the fquare of GF. be 5. 2.1 but GF is equal to GH; therefore the rectangle BE, EF, together

b

with the fquare of EG, is equal to the fquare of GH. but the fquares

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