ELEMENTS

OF

GEOMETRY.

EUCLID. BOOK II.

DEFINITIONS.

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.

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E

D

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, at the opposite angles of the parallelograms which make the gnomon.'

PROP. I. THEOR.

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G

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

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A

by A, BD, together with that contained by A, DE, and that contained by A, EC. From the point в draw (11. 1.) BF at right angles to BC, and make BG equal (3.1.) to A; also through G draw (31. 1.) & GH parallel to BC; and through D, E, C, draw (31. 1.) DK, EL, CH parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH: but 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, 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. fore, if there be two straight lines, &c. Q. E. D.

PROP. II. THEOR.

Where

Ir 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 parts in the point c; the rectangle contained by AB, BC, together with the rectangle* A B, AC, shall be equal to the square of AB.

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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.

PROP. III. THEOR.

Ir a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts

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

is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

Let the straight line AB be divided into two parts in the point c: the rectangle AB, BC is equal to the rectangle ac, CB, together with the square of BC.

A C

B

Upon BC describe (46.1.) the square CDEB; produce ED to F, and through à draw (31. 1.) AF parallel to CD or BE; then the rectangle A E 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 equal to BC; and AD is contained by AC, CB, for CD is equal to 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.

PROP. IV. THEOR.

F D

Ir a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in C: the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB.

Upon AB describe (46. 1.) the square ADEB, and join B, D; through c draw (31.1.) CGF parallel to AD or BE, and through & draw HK parallel to AB or

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Then because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (29.1.) to the interior and opposite angle ADB; but ADB is equal (5.1.) to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle GBC; and therefore the side BC is equal (6.1.) to the side CG: but CB is equal (34.1.) also to GK, and CG to BK; wherefore the figure CGKB is equilateral: it is likewise rectangular; for CG is parallel to BK, and CB meets them; the angles KBC, GCB are therefore equal to two right angles; and KBC is a right angle; wherefore GCB is a right angle; and therefore also the angles (34.1.) CGK, GKB, opposite to these, are right angles, and CGKB

is rectangular; but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB. For the same reason HF also is a square, and it is upon the side HG, which is equal to AC: therefore HF, CK are the squares of AC, CB; and because the complement AG is equal (43.1.) to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB: and HF, CK are the squares of AC, CB; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB. But HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore, if a straight line, &c. Q. E. D.

COR. From the demonstration it is manifest that the parallelograms about the diameter of a square are likewise squares.

PROP. V. THEOR.

If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equivalent to the square of half the line.

Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D; the rectangle AD, DB, together with the square of CD, is equivalent to the square of CB.

Upon CB describe (46. 1.) the square CEFB; join B, E; and through D draw (31. 1.) DHG parallel to CE, or BF; and through H draw KLM parallel to CB, or EF; and also through a draw AK parallel to CL, or B M. Then because

the complement CH is equal (43. 1.) to the complement HF, to each of these add DM; therefore the whole

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D B

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CM is equal to the whole DF; but CM is equal (36.1.) to AL, because CB is equal to AC; therefore also AL is equal to DF: to each of these add CH, and the whole AH is equal to DF and CH: but AH is the rectangle contained by AD, DB, for DH is equal (Cor. 4.2.) to DB; and DF together

with CH is the gnomon CMG; therefore the gnomon CMG is equivalent to the rectangle AD, DB: to each of these add LG, which is equal (Cor. 4. 2.) to the square of CD; therefore the gnomon CMG, together with LG, is equivalent to the rectangle AD, DB, together with the square of CD: but the gnomon CMG and LG make up the whole figure CEF B, which is the square of CB: therefore the rectangle AD, DB, together with the square of CD, is equivalent to the square of CB. Wherefore, if a straight line, &c. Q. E. D.

From this proposition it is manifest, that the difference of the squares of two unequal lines AC, CD is equivalent to the rectangle contained by their sum and difference.

PROP. VI. THEOR.

Ir a straight line be bisected, and produced to any point: the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equivalent to the square of the straight line which is made up of the half and the part produced.

Let the straight line AB be bisected in c, and produced to the point D: the rectangle AD, DB, together with the square of CB, is equivalent to the square of CD.

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Upon CD describe (46. 1.) the square CEFD; join D, E; and through B draw (31. 1.) BHG parallel to CE, or DF, and through H draw KLM parallel to AD, or EF, and also through a draw AK parallel to CL, or DM. Then because AC is equal to CB, the rectangle AL is equal (36. 1.) to CH; but CH is equal (43. 1.) to HF; therefore also AL is equal to HF: to each of these add CM; therefore the whole AM is equivalent to the gnomon CMG: and AM is the rectangle contained by AD, DB, for DM is equal (Cor. 4. 2.) to DB: therefore the gnomon CMG is equivalent to the rectangle AD, DB; add to each of these LG, which is equal to the square of CB, therefore the rectangle AD, DB, together with the square of CB, is equivalent to the gnomon CMG, and the figure LG: but the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equivalent to the square of CD. Wherefore, if a straight line, &c. Q. E. D.