PROP. XIV. PROB L.

To describe or make a square equal to a given rightlined figure.

Let the given right-lined figure be A: It is required to defcribe or make a fquare equal to that right-lined figure. Make [by 45. 1.] the right-angled parallelogram BD equal to the right-lined fi

gure A.

therefore BE

be equal to

H

If

ED, what was

required will

be

already

done. For the fquare B D is made equal to the right-lined figure A. If not, either B E, or ED, muft be greater. Loc BE be the greater, which continue out at F. So that E F [by 3. 1.] be equal to E D: then divide F B into two equal parts in G, and with the centre G, and diftance equal to either G B, or GF [by poft. 3.] defcribe a femicircle B H F, produce D E to H, and join G H.

Now because the right-line BF is divided into two equal parts at G, and into unequal ones at E: the rectangle under BE, E F, together with the fquare of EG [by 5. 2.] will be equal to the fquare of G F. But FG is equal to GH: Wherefore the rectangle under B E, EF, together with the fquare of GE, is equal to the fquare of G H. But [by 47. 1.] the fquares of H E, GE are equal to the fquare of GH: therefore the rectangle under B E, E F, together with the fquare of GE, is equal to the fquares of H E, G E, take away the fquare of GE from both, then the remaining rectangle under B E, EF, is equal to the fquare of E H. But the rectangle under B E, E F, is the parallelogram B D, becaufe EF is equal to ED: therefore the parallelogram BD is equal to the fquare of H E. But the parallelogram BD [by conftr.] is equal to the right-lined figure A: therefore the right-lined figure A will be equal to the fquare of EH.

Wherefore the fquare defcribed upon the right line E H, will be equal to the given right-lined figure A. Which was to be done. Additions

Additions to the Second Book.

If

PROP. I. THEOR.

a perpendicular be drawn from the oppofite angle of any triangle upon the bafe, making two fegments thereupon, the fquare of cne of the fides, together with the Square of that Segment next to the other fide, will be equal to the square of the other fide, together with the fquare of that fegment of the bafe next to the first mentioned fide. And the difference of the fquares of the fides will be equal to the difference of the fquares of the faid fegments of the bafe.

Let there be any triangle AB C, and let the perpendicular CD be drawn from the angle c oppofite to the base upon the bafe, making the two fegments A D, DB of the bafe: I fay, the fquare of the fide AC, together with the fquare of the fegment D B of the bafe, next to the fide C B, will be equal to the fquare of the other fide CB, together with the fquare of the fegment A D of the bafe. And the difference of the fquares of the fides A C, CB, will be equal to the difference of the fquares of the fegments of the bafe A D, DB.

For let the fide AC be greater than the fide Bc. Then because the angle ADC is a right angle, the fquare of the fide AC [by 47. 1.] will be equal to the fquares of the fegment A D of the bafe, and of the perpendicular CD; alfo

A

C

C

because the angle

CDB is a right angle, the fquare of Dthe fide CB [by 47. 1.] will be equal to

the fquares of the fegment D B of the bafe, and of the perpendicular c D. But fince, if equal things be added to equal ones, the fums will be equal. Therefore the fquares of AC,

RD,

BD, CD will be equal to the fquares of C B, A D, CD, and taking away the common fquare of CD from both, there will remain the fquares of A C, B D, equal to the fquares of CB, AD, which was one thing to be demonftrated. And because, if from equal things be taken away equal ones, the remainders are equal. Therefore the difference between the fquares of AC and C B, will be equal to the difference between the fquares of A D, D B. Which was the other thing to be demonftrated.

Therefore, &c. Which was to be demonftrated.

PROP. II. THEO R.

If the angle oppofite to the base of a triangle be two third parts of two right angles, the fquare of the bafe of that triangle will be equal to the fquares of the two fides, together with the rectangle under the two fides; and if that angle be one third part of two right angles, the Square of the base will be equal to the difference of the fquares of the two fides, and the rectangle under them.

Let there be a triangle A B C, having the angle ABC oppofite to the bafe, two third parts of two right angles: I fay, the fquares of the fides AB, BC, together with the rectangle under the fides A B, B C, will be equal to the fquare of the bafe AC,

Continue one of the fides A B to E, and upon A E let the perpendicular CD be drawn [by 12. 1.] from the angle A C B make DE equal to B D, and join the right line c E.

Then because [by 13. 1.] the angles ABC, EBC are equal to two right angles; and ABC [by fuppofition] is equal to two third parts of two right angles; the angle EBC will be one third part of two right angles; and because the angles

BD C, EDC are

right angles, for CD is perpendicular to AE; and BD is made equal to DE,

E

D

and

and the fide CD is common. Therefore [by 4. 1.] the fides BC, CE of the right-angled triangles B D C, DEC will be equal to one another; and accordingly [by 6. 1.] the angle BEC will be equal to the angle EBC. But E B C has been proved to be equal to one third part of two right angles. Therefore the angle B E C is alfo one third part of two right angles. And fo because the three angles of every triangle are [by 32. 1.] equal to two right angles, the remaining angle BCE will be one third part of two right angles. Confequently, the triangle B EC will be equilateral; and fo the fides B E, B C will be equal; and twice B D will be equal to the fide B C. Again, because in the obtufe-angled triangle A B C, the perpendicular C D falls upon the continuation of the fide A B; the fquare of AC [by 12. 2.] is equal to the fquares of A B, BC, together with twice the rectangle under AB, DB; and because twice the rectangle under A B, BD is equal to the rectangle under AB, and twice B D ; that is, BE, and BE has been proved to be equal to the fide B C. Therefore twice the rectangle under A B, B D, will be equal to the rectangle under the fides A B, BC of the triangle. Confequently, the fquares of the fides A B, B C, together with twice the rectangle under A P, B D, will be equal to the fquares of A B, BC, together with the rectangle under A B, BC; and therefore the fquare of the bafe A c, will be equal to the fquares of the fides A B, BC, together with the rectangle under the fides A B, B C. Which was the first thing to be demonftrated.

Secondly, Let the angle ABC of the triangle ABC oppofite to the bafe, be one third part of two right angles :`I fay, the fquare of the bafe A C will be equal to the difference between the fquares of the fides A B, B C, and the rectangle

under them.

For the confiruction being the fame as before, only C D being here perpendicular to the fide A B, and not its continuation, we demonftrate as before, that the rectangle under the fides A B, BC, is equal to twice the rectangle under the fide A B, and the fegment DB. And therefore because [by the note to 13. 2.] the difference between the fquares of the fides A E, BC, and twice the rectangle under the fide A B, and the fegment DB, is equal to the fquare of the bafe AC; the fquare of the bafe AC will be equal to the difference of the fquares of the fides AB, BC, and the rect

angle

angle under them. Which was the second thing to be demonftrated.

Therefore if, &c. Which was to be demonftrated.

SCHOLI U M.

If the angle oppofite to the base be one fixth part of two right angles; it is easily demonftrated by the forty-feventh propofition of the firft book, that the fquare of the bafe is equal to the difference between the fum of the fquares of the fides and the rectangle under them, multiplied by the fquare root of the number 3.

PROP. III. THEOR.

In any triangle, if the bafe be bifected, and a right line be drawn from the oppofite angle to the point of bifection: I fay, the fquares of the two fides of the triangle will be equal to twice the Square of the bifetting line, together with twice the Square of one half the bafe.

Let there be a triangle A B C, whose base AC is bifected in the point E by the right line B E drawn from the angle в oppofite to the bafe: I fay, the fquares of the fides AB, B C are equal to twice the fquare of the bifecting line в E, together with twice the fquare of the half base A E, or

E C.

For [by 12. 1.] from the angle в, draw the perpendi cular B D

upon the

base A C.

B

B

Then [by

47. I.]

the squares A

of AB, B C are equal to the fquares of A D, D C, together with twice the fquare of B D. But fince AC is divided

equally in E, and unequally in D; the fquares of A D, DC will be [by 9. 2.] equal to twice the fquares of A E, E C; therefore the fquares of A B, B C will be equal to twice the fquares of B D, A E, E D. But fince twice the fquare of B is equal to twice the fquares of ED, DB. For [by 47. 1.]