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Book II.
PROP. IX.

IX. THE OR.
IF F a straight line be divided into two equal, and also

into two unequal parts; the squares of the two unequal parts, are together double of the square of half the line, and of the square of the line between the points of section.

Let the straight line AB be divided at the point C into two equal, and at D into two unequal parts. the squares of AD, DB are together double of the squares of AC, CD.

From the point C draw. CE at right angles to AB, and make it a. ř1. t. equal to AC or CB, and join EA, EB; thro' D draw b DF parallel b. 31. 1. to CE, and thro' F draw FG parallel to AB; and join AF. then because AC is equal to CE, the angle EAC is equal to the angle c. s: 1. AEC ; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angled; and they are equal d. 31. 1, to one another ; each of them therefore is half of a right angle. for the same reason each of the angles CEB, EBC is half a right an

E gle; and therefore the whole AEB is a right angle. and because the angle GEF is half a right angle, and EGF a right angle, for it is equal to the interior and oppo-A

. 29. 1,

C D B fite angle ECB, the remaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the side EG equal to the side GF. again, f. 6. 1, because the angle at B is half a right angle, and FDB a right angle, for it is equal e to the interior and opposite angle ECB, the remaining angle BFD is half a right angle; therefore the angle at B is equal to the angle BFD, and the side DF to f the side DB. and because AC is equal to CE, the square of AC is equal to the square of CE; thereforc the squares of AC, CE are double of the square of AC. but the square of EA is equal 8 to the squares of AC, CE, 8.47.86 because ACE is a right angle; therefore the square of EA is double of the square of AC. again, because EG is equal to GF, the square of EG is equal to the square of GF; therefore the squares of EG, GF are double of the square of GF; but the square of EF is

Book 11.

h. 34. 1.

equal to the squares of EG, GF; therefore the square of EF is
double of the square GF. and GF is equal h to CD; therefore the
square of EF is double of the
square of CD. but the square of E
AE is likewise double of the square
of AC; therefore the squares of
AE, EF are double of the squares
of AC, CD. and the square of
AF is equal 8 to the squares of

A C D B
AE, EF because AEF is a right
angle; therefore the square of AF is double of the squares of AC,
CD. but the squares of AD, DF are equal to the square of AF,
because the angle ADF is a right angle; therefore the squares of
AD, DF are double of the squares of AC, CD. and DF is equal
to DB ; therefore the squares of AD, DB are double of the squares
of AC, CD. If therefore a straight line, &c. Q. E. D.

8. 47. I.

2.1.1.

PROP. X. THEOR.
IF a straight line be bisected, and produced to any point,

the square of the whole line thus produced, and the square of the part of it produced are together double of the square of half the line bifected, and of the square of the line made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D; the squares of AD, DB are double of the squares of AC, CD.

From the point C draw . CE at right angles to AB, and make b. 31.1.

it equal to AC or CB, and join AE, EB; thro’E draw 6 EF parallel to AB, and thrơ D draw DF parallel to CE. and because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right angles; and therefore the angles BEF, EFD are less than two right angles. bat straight lines which

with another straight line make the interior angles upon the fame d. 12. Ax. fide less than two right angles, do meet & if produced far enough.

therefore EB, FD shall meet, if produced, toward BD. let them meet in G, and join AG. then because AC is equal to CE, the angle CEA is equal to the angle EAC; and the angle ACE is a right angle; therefore each of the angles CEA, EAC is half a

c. 29. 1.

6. S. 1.

A

BID

right angle for the same reafon, 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 * 32. 1. they are vertically opposite; but BDG is a right angle, because it f. 15. 1. is equal to the alternate angle DCE; therefore the remaining an-“ 29.1. gle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side BD is equal 8 to the side DG.g. 6.1. again, because EGF is half a right angle, and that the an

E gle at F is a right angle, because it is equal - to the op

h. 34. 8. posite angle ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF; wherefore also

G the side GF is equal 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 equal i i. 47. 1, to the squares 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 square of FE; and therefore the squares of GF, FE are double of the square of EF. but the square of EG is equal i 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 de monstrated that the square of EA is double of the square of AC; therefore the squares of AE, EG are double of the squares of AC, CD. and the square of AG is equal i to the squares of AE, EG ; i. 47. te therefore the square of AG is double of the squares of AC, CD. but the squares of AD, DG are equal i 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.

D 3

Book 11.

PROP. XI. PROB.

O divide a given straight line into two parts, fo that

the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.

C. 3. I.

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. 2. 46. I.

Upon AB describe : the square ABDC, bilet AC in E, and b. 10. 1. join BE ; produce CA to F, and make < EF equal to EB; and

upon AF describe : the square FGHA, and produce GH to K, AB is divided in H so, that the rectangle AB, BH is equal to the square of AH.

Because the straight line AC is biseEted in E, and produced to

the point F, the rectangle CF, FA, together with the square of d. 6. a.

AE, is equal d to the square of EF. but EF is equal to EB; there, fore the rectangie CF, FA, together with F G

the square of AE is equal to the square of 6:47.!. EB. and the squares of BA, AE are equal o

to the square of EB, because the angle EAB
is a right angle; therefore the rectangle CF, A HB
TA, together with the square of AE is equal
to the squares of BA, AE. take a way the
{quare of AE, which is common to both,

E
therefore the remaining rectangle CF, FA
is equal to the square of AB. and the figure
FK is the rectangle contained by CF, FA,
for AF is equal to FG ; and AD is the

K D
square 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 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.

PRO P. XII.

THEOR.

N obtuse angled 'triangles, if a perpendicular be

drawn from any of the acute angles to the oppofite fide produced, the square of the side subtending the obtuse angle, is greater than the squares of the fides 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 drawn · perpendicular to 2. 12. 1. 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 equal 5 to the squares of BC, CD, and

A 6.4.20 twice the rectangle BC, CD. to cach 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 equal ' to the squares of BD, DA, be-B cause the angle at D is a right angle; and the square of CA is equal o 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.

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