I

PROP. IX. THEO R.

Fa ftraight line be divided into two equal, and alfo into two unequal parts; the squares of the two unequal parts, are together double of the fquare of half the line, and of the square of the line between the points of fection.

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

51

Book II.

From the point C draw CE at right angles to AB, and make it a. 11. 1. equal to AC or CB, and join EA, EB; thro' D drawb 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 becaufe the angle ACE is a right angle, the two others AEC, EAC together make one right angle; and they are equal d. 33, 1, to one another; each of them therefore is half of a right angle. for the fame reafon each of the angles CEB, EBC is half a right angle; 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

fite angle ECB, the remaining an

A

Ε

e. 29. 1,

C D

B

gle EFG is half a right angle; therefore the angle GEF is equal

to the angle EFG, and the fide EG equal to the fide GF. again, f. 6. r, because the angle at B is half a right angle, and FDB a right an

e

gle, for it is equal to the interior and oppofite angle ECB, the
remaining angle BFD is half a right angle; therefore the angle at
B is equal to the angle BFD, and the fide DF to f the fide DB. and
because AC is equal to CE, the fquare of AC is equal to the fquare
of CE; therefore the fquares of AC, CE are double of the square
of AC. but the fquare of EA is equal to the fquares of AC, CE, 8. 47. 1,
because ACE is a right angle; therefore the square of EA is double
of the fquare of AC. again, because EG is equal to GF, the fquare
of EG is equal to the fquare of GF; therefore the fquares of EG,
GF are double of the fquare of GF; but the fquare of EF is

Book II. equal to the fquares of EG, GF; therefore the fquare of EF is

h. 34. 1.

8. 47. I.

2. II. I.

b. 31. 1.

h

E

F

CD

B

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

PROP. X. THEOR.

IF a straight line be bifected, and produced to any point,

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

Let the ftraight line AB be bifected in C, and produced to the point D; the fquares of AD, DB are double of the fquares of AC, CD.

From the point C draw * CE at right angles to AB, and make it equal to AC or CB, and join AE, EB; thro' E draw 6 EF parallel to AB, and thro' 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. but straight lines which with another straight line make the interior angles upon the fame d. 12. Ax. fide lefs than two right angles, do meet & if produced far enough.

C. 29. I.

C. 5. I.

therefore EB, FD fhall 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

e

right angle. for the fame reafon, each of the angles CEB, EBC Book II. is half a right angle; therefore AEB is a right angle, and becaufe EBC is half a right angle, DBG is alfo f half a right angle, for 32. 1. they are vertically oppofite; but BDG is a right angle, because it f. 15. 1. is equal to the alternate angle DCE; therefore the remaining angle 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. 1. again, because EGF is half a

C. 29. I.

the fide FE. And because EC is equal to CA, the square 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 i i. 47. 3, to the squares of EC, CA; therefore the fquare of EA is double of the fquare of AC. again, because GF is equal to FE, the fquare of GF is equal to the fquare of FE; and therefore the squares of GF, FE are double of the square of EF. but the fquare of EG is equal to the fquares of GF, FE; therefore the square 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 square of EA is double of the fquare of AC; therefore the squares of AE, EG are double of the fquares of AC, CD. and the fquare of AG is equal i to the fquares of AE, EG ; i. 47. ke therefore the square 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 squares of AD, DB are double of the fquares of AC, CD. Wherefore if a straight line, &c. Q. E. D.

D 3

Book II.

2. 46. 1. b. 10. 1. C. 3. I.

4. 6. 2.

Tod

PROP. XI. PRO B.

O divide a given ftraight line 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.

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.

a

C

Upon AB defcribe the fquare ABDC, biêct ↳ AC in E, and join BE; produce CA to F, and make EF equal to EB; and upon AF describe 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.

Because the straight 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 fquare of EF. but EF is equal to EB; there fore the rectangle CF, FA, together with F the fquare of AE is equal to the fquare of 47. 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, E
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 C
fquare of AB; therefore FK is equal to

G

HB

K D

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 fquare 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.

PROP. XII. THEOR.

IN obtufe angled triangles, if a perpendicular be

drawn from any of the acute angles to the oppofite fide produced, the fquare of the fide fubtending the obtufe 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.

Book II.

Let ABC be an obtufe angled triangle, having the obtufe angle ACB, and from the point A let AD be drawn perpendicular to a. 12. I. BC produced. the fquare 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 fquare of BD is equal

to the fquares of BC, CD, and twice the rectangle BC, CD. to each of thefe equals add the fquare 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

to the fquares of BD, DA, be-B cause the angle at D is a right an

c

A b. 4. 20

D

C. 47. I.

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 squares of BC, CA, by twice the rectangle BC, CD Therefore in obtufe angled triangles, &c. Q. E. D.

D 4