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For draw* CE from the Point C at Right Angles * 31. 1. to AB, which make equal to AC, or CB, and draw AE, EB; likewife thro' E let EF be drawn parallelt 31. 1. to AD, and thro D, DF + parallel to CE.

Then becaufe the Right Line EF falls upon the Parallels EC, FD, the Angles CEF, EFD, are f equal † 29. 1. to two Right Angles. Therefore the Angles FEB, EFD, are together less than two Right Angles. But Right Lines making, with a third Line, Angles together less than two Right Angles, being infinitely produced, will meet * Wherefore EB, FD, pro- * Ax. 12. duced, will meet towards BD, Now let them be produced, and meet each other in the Point G, and let AG be drawn.

And then because AC is equal to CE, the Angle AEC will be equal to the Angle EACT: But the † 5. to Angle at C is a Right Angle. Therefore the Angle CĂE, or AEC, is half a Right one. By the same way of Reasoning, the Angle CEB, or EBC, is half a Right one. Therefore AEB is a Right Angle, And since EBC is half a Right Angle, DBG will falfo | 15. . be half a Right Angle, since it is vertical to CBE, But BDG is a Right Angle also; for it is * equal to * 29. 1. the alternate Angle DCE. Therefore the remaining Angle D GB is half a Right Angle, and so equal to DB G. Wherefore the Side BD is + equal to the † 6. I. Side DG. Again, because E G F is half a Right Angle, and the Angle at F is a Right Angle, for it is equal to the opposite Angle at C; the remaining Angle FEG will be also half a Right one, and is equal to the Angle EGF; and so the Side GF is f equal to the Side EF. And since EC is equal to CA, and the Square of EC equal to the Square of CA; there fore the Squares of EC, CA, together, are double to the Square of CA. But the Square of EA is I equal $ 47. to to the Squares of EC, CA. Wherefore the Square of EA is double to the Square of AC. Again, becaufe GF is equal to FE, the Square of GF also is equal to the Square of FE. Wherefore the Squares of GF, FE, are double to the Square of FE. But the Square of EG ist equal to the Squares of GF, FE. Therefore the Square of EG is double to the Square of EF: But. EF is equal to CD. Wherefore the Square of EG fhall be double to the Square of CD.

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But the Square of E A has been proved to be double to the Square of A C. Therefore the Squares of AE, EG, are double the Squares of AC, CD. But the Square of AG ist equal to the Squares of AE, EG; and consequently the Square of AG is double to the Squares of AC, CD. But the Squares of AD, DG, are equal to the Square AG. Therefore the Squares of AD, DG, are double the Squares of AC, CD. But DG is equal to DB. Wherefore the Squares of AD, DB, are double to the Squares of AC, CD. Therefore if a Right Line be cut into two equal Parts, and to it be direčtly added another; the Square made on (the Line compounded of] the whole Line, and the added one, together with the Square of the added Line, hall be double to the Square of the half Line, and the Square of. [ that Line which is compounded of ] the half, and the added Line.

PROPOSITION XI.

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PROBLEM.
To cut a given Right Line so, that the Rectangle

contained under the whole Line, and one Segment,

be equal to the Square of the other Segment. L

ET AB be a given Right Line. It is required

to cut the fame so, that the Rectangle contained under the whole, and one Segment thereof, be equal

to the Square of the other Segment. 46. I.

Defcribe * ABCD the Square of AB, bifect AC in E, and draw BE: Also, produce CA to F, so that EF be equal to EB. Describe FGHA the Square of AF, and produce GH to K. I say, AB is cut in H so, that the Rectangle under AB, BH, is equal to the Square of A H.

For since the Right Line AC is bifected in E, and AF is directly added thereto, the Rectangle under

CF, FA, together with the Square of AĚ, will be + 6 of this. f equal to the Square of EF. But EF is equal to

EB. Therefore the Rectangle under CF, FA, to

gether with the Square of AE, is equal to the Square 47. I.

of EB. But the Squares of BA, AE, are f equal to the Square of EB; for the Angle at A is a Right An

gle.

gle. Therefore the Rectangle under CF, FA, to-
gether with the Square of AE, is equal to the Squares
of BA, AE. And taking away the Square of AE,
which is common, the remaining Rectangle under
CF, FA, is equal to the Square of A B. But FK
is the Rectangle under CF, FA; since AF is equal
to FG; and the Square of AB is AD. Wherefore
the Rectangle FK is equal to the Square AD. And
if AK, which is common, be taken from both, then
the remaining Square FH is equal to the remaining
Rectangle HD. But HD is the Rectangle under
AB, BH, fince AB is equal to BD, and FH is
the Square of AH. Therefore the Rectangle under
AB, BH shall be equal to the Square of AH. And
so the given Right Line AB is cut in H, so that the
Rectangle under A B, BH, is equal to the Square of
AH; which was to be done.

PROPOSITION XII.

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THIORE M.
In obtufe angled Triangles, the Square of the Side

fubtending the obtufe Angle, is greater than the
Squares of the Sides containing the obtufe Angle,
by twice the Rectangle under one of the Sides,
containing the obtufe Angle, viz. tbat on which,
produced, the Perpendicular falls, and the Line
taken without, between the perpendicular and
the obtufe Angle.
ET ABC be an obtuse angled Triangle, hav-

ing the obtufe Angle BAC; and * from the * 12. 20
Point B draw BD perpendicular to the Side CA
produced. I say the Square of BC is greater than
the Squares of B A and AC, by twice the Rectangle
contained under CA, and AD.

For because the Right Line CD is any how cut in the Point A, the Square of CD shall be f equal to the t 4 of {bis. Squares of CA, and AD, together with twice the Rectangle under CA, and AD. And if the Square of BD, which is common, be added, then the Squares of CD, DB, are equal to the Squares of CA, AD,

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47. 1.

and D B, and twice the Rectangle contained under CA and AD. But the Square of CB is * equal to the Squares of CD, DB; for the Angle at D is a Right one, since BD is perpendicular, and the Square of A B is * equal to the Squares of AD, and DB: Therefore the Square of CB is equal to the Squares of CA and AB, together with twice a Rectangle under CA and AD. Therefore in obtufe angled Trian. gles, the Square of the Side fubtending the obtufe Angle, is greater than the Squares of the Sides containing the obtufe Angle, by twice the Rectangle under one of the Sides containing the obtufe Angle, viz. that on which, produced, the Perpendicular falls, and the Line taken without, between the perpendicular and the obtufe Angle; which was to be demonstrated.

PROPOSITION XIII.

THE OREM II.
In acute angled Triangles, the Square of the Side fub.

tending the acute Angles, is less than the Squares
of the Sides containing the acute Angle, by twice
a Rectangle under one of the Sides about the acute
Angle, viz. on which the Perpendicular falls,
and the Line assumed within the Triangle, from
the Perpendicular to the acute Angle.

12. I.

ET ABC be an acute angled Triangle, having drawn AD perpendicular to BC. I say the Square of A C is less than the Squares of CB and BX by twice a Rectangle under CB and BD.

For because the Right Line CB is cut any how in + 7 of this. D, the Squares of CB and BD will be f equal to

twice a Rectangle under CB and BD, together with the Square of DC. And if the Square of AD bę added to both, then the Squares of CB, BD, and DA, are equal to twice the Rectangle contained un

der CB and BD, together with the Squares of AD 1 41

and DC. But the Square of AB is I equal to the Squares of B D and DA; for the Angle at D is a Right Angie. And the Square of AC is I equal to

the

the Squares of AD and DC. Therefore the Squares of CB and B A are equal to the Square of A C, together, with twice the Rectangle contained under CB and BD. Wherefore the Square of AC only, is less than the Squares of CB and BA, by twice the Rectangle under CB and BD. Therefore in acute angled Triangles, the Square of the Side subtending the acute Angles, is less than the Squares of the Sides containing the acute Angle, bay twice a Rectangle under one of the Sides about the acute Angley viz. on which the Perpendicular falls, and the Line assumed within the Triangle, from the perpendicular to the acute Angle; which was to be demonstrated,

PROPOSITION XIV.

PROBLEM.
To make a Square equal to a given Right-lined

Figure.
ET A be the given Right-lined Figure. It is

. required to make a Square equal thereto. Make * the Right-angled Parallelogram BCDE* 45. 3. equal to the Right-lined Figure A. Now if BE be equal to ED, what was proposed will be already done, since the Square BD is made equal to the Rightlined Figure A: But if it be not, let either BE or ED be the greater : Suppose BE, which let be produced to F; so that E F be equal to ED. This being done, let BE be + bisected in G, about which, as a Center, † 10. 1. with the Distance GB or GF, describe the Semicircle BHF; and let DE be produced to H, and draw GH, Now because the Right Line BF is divided into two equal Parts in G, and into two unequal ones in E, the Rectangle under BE and EF, together with the Square of GE, shall be f equal to the Square of GF.85 of tbis. But GF is equal to GH Therefore the Rectangle under BE, EF, together, with the Square of GE, is equal to the Square of GH. But the Squares of HE and EG are * equal to the Square of GH. * 47. 1. Wherefore the Rectangle under BE, EF, together with the Square of EG, is equal to the Squares of HE, EG, And if the Square of EG, which is com

mon

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