to the triangles ECD, EGC; that is, the trapezium AFGE will be equal to the trapezium EGC D. And fince the triangles FBG, B G C are equal, the trapezium ABGE will be equal to the figure EGBCD: confequently, the trapezium A B G E will be one half the given trapezium A B C D. Again, because [by conftr.] G H is parallel to E B, the triangles EGB, EBH will [by 37.1.] be equal to one another. Therefore the triangles ABE, EBH, will be equal to the triangles A B E, E G B. But the trapezium ABGE confifts of the triangles A B E, E B G, and the trapezium A B H E of the triangles A B E, E BH: wherefore the trapeziums ABHE, ABGE will be equal to one another. But the trapezium ABGE has been proved to be equal to one half of the given trapezium ABC D. Therefore the trapezium A B HE will also be equal to one half of the given trapezium A B C D.

Therefore a given trapezium is divided into two equal parts, by a right line drawn from a given point in the middle of its fide. Which was to be done.

The problem is almoft as eafily refolved, when the given point is not in the middle of one of the fides; for it is only drawing a right line from fuch a point joining the point н, found as above, and through the middle point E drawing a right line parallel to this line. Then a right line drawn from the given point to that wherein this parallel interfects the oppofite fide B C will divide the trapezium into two equal parts.

EUCLI D'

EUCLID's

ELEMENTS.

BOOK II.

1.

E

DEFINITIONS.

VERY right-angled parallelogram is faid to be contained under the two right lines, comprehending a right angle a.

2. In every parallelogram, any one of the parallelograms that are about a diameter, with the two complements, is called a gnomon.

a Here Euclid, for brevity's fake, fays, Every right-angled parallelogram is contained under two right lines forming a right angle; though, indeed, every parallelogram is comprehended or contained under four right lines. But as the oppofite fides of every parallelogram are equal to one another, any of the two fides which contain a right angle, may not unaptly be faid to contain the whole parallelogram.

PROPOSITION I. THEOREM. If there be two right lines, and one of them is cut or divided into any parts whatsoever, the rectangle contained under the two right lines is equal to all the rectangles contained under the undivided line, and the feveral fegments or parts of the other line.

L

ET there be two right lines, A and B C, and let B C be any how cut or divided in the points D, E: I fay, the rectangle contained under the right lines A, BC, is equal

to

B

D

E C

to the rectangle which is contained under the right lines A, BD, and to the rectangle contained under A, D E, and to that which is contained under A, E C.

K

L H

A

For [by prop. 11. b. I.] from the point B draw the right line G BF at right angles to в C, and make B G equal to A. And thro' G [by 31. 1.] draw GH parallel to BC. And through the points D, E, c draw the right lines D K, EL, C H parallel to в G.

F

Then the rectangle B H is equal to the rectangles BK, DL, EH. And B H is the rectangle under A, BC; for it is contained under G B, BC, and BG is equal to A: but the rectangle BK is contained under A, B D; for it is contained under G B, BD, and GB is equal to A: and the rectangle D L is contained under the right lines A, DE; because DK, that is, BC, is equal to A: And in like manner, the rectangle E H, is contained under the right lines A, EC: Therefore the rectangle under the right lines A, BC, is equal to all the rectangles under A, B D, and under A and D E, and alfo under A and E C.

Therefore if a right line be cut or divided into any parts foever, the rectangle contained under the two right lines is equal to all the rectangles contained under the undivided line, and the several fegments or parts of the divided line. Which was to be demonstrated.

If a right line be any how divided into two parts, the rectangles contained under the whole line and each of the parts, are equal to the * Square of the whole line b.

For let the right line A B be any how divided into two parts at c I fay, the rectangle contained under the right lines A, B C, together with the rectangle contained under BA, A C, are equal to the fquare of the right line A B.

For brevity's fake, we call the fquare defcribed upon a line, the fquare of a line,

For

A

C B

For [by 46. 1.] defcribe the fquare A DEB of AB; and through c [by 31. 1.] draw CF parallel to A D or B E. Then is A E equal to both the rectangles A F, CE: But AE is the fquare of the right line AB, and the rectangle A F is contained under the right lines B A, A C; for it is contained under the right lines DA, AC, whereof A D is equal to A B; and the rectangle CE is contained under the right lines A B, BC; for the right line T BE is equal to the right line AB: Therefore the reSangle contained under the right lines A B, AC, together with the rectangle contained under A B, BC, is equal to the fqua e of a B.

D

F

If the efore a right line be any how divided into two parts, the rectangles contained under the whole line and each o. the parts will be equal to the square of the whole line. Which was to be demonftrated.

Although Euclid propofes this fecond theorem of a right line divided any how only into two parts; the fame thing may be demonftrated after the fame manner, if a right line be divided into any number of parts.

If a right line be divided into any two parts, the rectangle contained under the whole line, and one of the parts, is equal to the rectangle contained under the parts, and the fquare of the part aforefaid.

For let the right line A B be divided in the point c into any two parts: I fay, the rectangle under ▲ B, B C is equal B to the rectangle under A C, CB, together with the square of the right line BC.

C

D

For [by 46. 1.] defcribe the fquare CDE B of в C, produce ED to F, and through A draw [by 31. 1.] A F parallel to CD

or B E.

Them

Then the rectangle A E will be equal to the rectangles AD, CE, and AE is the rectangle contained under the right lines A B, BC; for it is contained under the right lines A B, B E, whereof B E is equal to BC: But the rectangle A D is contained under A C, CB, fince D C is equal to C B, and D B is the fquare of BC: Therefore the rectangle under A B, B C is equal to the rectangle under ▲ C, CB, together with the fquare of B C.

If therefore a right line be divided into any two parts, the rectangle contained under the whole line and one of the parts, is equal to the rectangle contained under the parts, and the fquare of the part aforefaid. Which was to be demonstrated.

PROP. IV. THEOR.

If a right line be divided into any two parts, the Square of the whole line is equal to both the fquares of the parts, together with twice the rectangle contained under the parts.

For let the right line A B be divided by the point c into any two parts: I fay, the fquare of A B is equal to both the fquares of A C, CB, together with twice the rectangle contained under the parts A C, C B.

For [by 46. i.] defcribe the fquare A DEB of A B. Join B D, and through c draw [by A 31. 1.] the right line C G F parallel to A D or BE; but through G, the right line H K parallel to A B, or D E.

H

Then because C F is parallel to A D, and the right line B D falls upon them, the outward angle BGC will be [by 29. 1.] equal to the inward oppofite D

C B

G

K

angle A D B. But [by 5. 1.] the angle A D B is equal to the angle ABD, becaufe the fide BA is equal to the fide AD; and fo the angle C G B is equal to the angle GBC: therefore the fide B C [by 6. 1.] is equal to the fide CG. But likewife C B is [by 34. 1.] equal to the fide GK, and CG to BK. Therefore the fide GK is equal to K B, and the parallelogram C G K B is equilateral: I fay, it is rightangled too. For because CG is parallel to B K, and the