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to the straight line GH apply (44. 1.) the parallelog am GM equal to the triangle DBC, having the angle GHM equal to the angle E ; and because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM : add to each of these the angle KHG; therefore the angles FKH KHG are equal to the angles KÉG GHM; but FKH, KHG equal (29. 1.) to two right angles ; Therefore also KHG GHM are equal to two right angles; and because at the point H in the straight line GH, the two straight lines KH,HM, upon the opposite sides of it make the adjacent angles equal to two right angles, KH, is in the same straight line (14. 1.) with HM ; and because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal; (29. 1.) Add to each of these the angle HGL : Therefore the angles MHG, HGL, are equal to the angles HGF, HGL: But the angles MHG, HGL, are equal (29. 1.) to two right angles; wherefore also the angles HGF, HGL are equal to two right angles; and FG is therefore in the same straight line with GL ; and because KF is parallel to HG, and HG to ML; KP is parallel (30. 1.) to ML; and KM, FL are parallels ; wherefore KFLM is a parallelogram ; and because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM; the whole rectilineal figure ABCD, is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

Cur. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying (44. 1.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle.

Proposition. XLVI. Problem. To describe a square upon a given straight line.

Let AB be the given straight line; It is required to describe a square upon AB.

From the point A draw (11. 1.) AC at right angles to AB; and make, (3. 1.) AD equal to AB, and through the point D draw DE parallel (31. 1.) to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram : whence AB is equal (34. 1.) to. DE, and AD to BE : But BA is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral.

Likewise all its angles are right angles, because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE are equal (29. 1.) to two right angles : But BAD is a right angle ;

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B

but the opposite angles of parallelograms are equal ; (34. 1.) therefore each of the opposite angles ABE, BED is a right angle ; wherefore the figure ADEB is rectangular.

And it has been demonstrated that it is equilateral; it is therefore a aquare, and it is described upon the given straight line AB : which was to be done.

Cor. Hence every parallelogram that has one right angle has all its angles right angles.

Proposition XLVII. Theorem. In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Let ABC be a right-angled triangle having the right angle, BAC the square described upon the side BC is equal to the squares described upon BA, AC.

On BC describe (46. 1.) the square BDEC, and on BA; AC the squares GB, HC; and through A draw (31. 1.) AL parallel to BD, or CE, and join AD, FC ; then, because each of the angles BAC, BAG is a right angle, (30. Def.) the · two straight lines AC, AG, upon the opposite sides of AB, make with it

H at the point A the adjacent angles equal to two right angles; therefore CA is in the same straight line

K (14. 1.) with AG; for the same reason, AB and AH are in the same straight line ;

And because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each the angle ABC, and the whole angle DBA is equal (2. Ax.) to the whole FBC;

And because the two sides AB, BD are equal to the two FB, BC each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal (4. 1.) to the base FC, and the triangle ABD to the triangle FBC:

Now the parallelogram BL is double (41. 1.) of the triangle ABD, because they are upon the same base BD, and between the same pa• rallels, BD, AL; and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But ihe doubles of equals are equal (6. Ax.) to one another : Therefore the parallelogram BL is equal to the square GB :

And, in the same manner by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square HC: Therefore the whole square BDEC is equal to the two squares GB, HC; and the square EDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC:

Wherefore the square upon the side BC is equal to the squares upon

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the sides BA, AC. Therefore, in any right-angled triangle, &c. Q. E. D.

Proposition XLVIII. Theorem. If the square described upon one of the sides of a triangle, be equal to the squares

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upon the other two sides of it, the angle contained by these two sides is a right angle.

If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle.

From the point A draw (11. 1.) AD at right angles to AC, and make AD, equal to BA, and join DC: Then, because DA is equal to AB, the square of DA is equal to the

To each of these add the square of AC; therefore the squares of DA, AC are equal to the squares of BA, AC: But the square of DC is equal (47. 1.) to the squares of DA, AC; because DAC is a right angle; and the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore the square of DC is equal to the square of BC; and therefore also the side DC is equal to the side BC.

And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal (8. 1.) to the angle BAC; but DAC is a right angle, therefore also BAC is a right angle. Therefore, if the square, &c. Q. E. D.

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THE

ELEMENTS OF EUCLID.

BOOK II.

Definitions.

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II

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I. Every right-angled parallelogram is said to be contained by any two of the straight lines which contain one of the right angles.

II. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon. •Thus the parallelogram HG, together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon.'

Proposition. I. Theorem. If there be two straight lines, one of which is divided into any number of parts ; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Let A and BC be two straight lines ; and let BC be divided into any parts in the points D, E; the rectangle contained by the straight lines A, BC is equal to the rectangle contained by A, BD, together with that contained by A, в. DEC DE, and that contained by A, EC.

From the point B draw (11. 1.) BF at right angles to BC, and make BG equal

KLH (3 1.) to A, and through G draw (31. 1) GH parallel to BC; and through D, E, C,

F F draw (31. 1.) DK, EL, CH parallels to BG; then the rectangle BH is equal to the rectangles BK, DL, EH ; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A; and BK is contained by A, BD, for it is contained by GB,

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BD, of which GB is equal to A; and DL is contained by A, DE, because DK, that is (34. 1.) BG is equal to A ; and in like manner the rectangle EH is contained by A, EC: therefore the rectangle contained by A, BC, is equal to the several rectangles contained by A, BD, and by A, DE: and also by A, EC. Wherefore, if there be two straight lines, &c. Q. E. D.

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Proposition n. Theorem. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line. Let the straight line AB be divided into any two

CB parts in the point C; the rectangle contained by AB, BC, together with the rectangle* AB, AC, shall be equal to the square of AB.

Upon A B describe (46. 1.) the square ADEB, and through C draw (31. 1.) CF parallel to AD or BE. Then AE is equal to the rectangles AF, CE, and AE is the square of AB; and AF is the rectangle contained by BA, AC ; for it is contained by DA, AC, of which AD is equal to AB ; and CE is contained by AB, BC, for BE is equal to AB ; therefore the rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. If therefore a straight line, &c. Q. E. D.

Proposition III. Theorem. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle conSained by the two parts together with the square of the aforesaid part.

Let the straight line AB be divided into any two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC.

Upon BC describe (46. 1.) the square CDEB, and produce ED to F, and through A draw AC

B (31. 1.) AF, parallel to CD or BE; then the rectangle AE is equal to the rectangles AD, CE; and AE, is the rectangle contained by AB, BC, for it is contained by AB, BE of which BE is

FD equal to BC ; and AD is contained by AC,CB, for CD is equal 10 CB; and DB is the square of BC; therefore the rectangle AB, BC, is equal to the rectangle AC, CB, together with the square of BC. If therefore a straight line, &c. Q. E. D.

Proposition IV. Theorem. | If a straight line be divided into any two parts, the square of the

* N. B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC.

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