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of the remaining parts of the first greater than the corresponding part of the other it is required to compare (as to greater and less) all the other parts of one triangle with the corresponding parts of the other.

26. From the right angle of a triangle draw a line to meet the hypothenuse, making an angle with one side equal to the adjacent acute angle of the triangle: then

(a) This line will make with the other side an angle equal to the other acute angle of the triangle;

(b) It will make with the perpendicular from the right angle upon the hypothenuse, an angle equal to the difference of the acute angles of the triangle.

(c) It will itself be equal to each of the segments into which it divides the hypothenuse;

27. Only one perpendicular can be drawn from a given point to a given straight line, whether the point be in the line or without it.

28 The perpendicular is the shortest line that can be drawn to a line from a point without it.

29. Of straight lines drawn to a given straight line from a given point without it, that which is nearer to the perpendicular is less than the more remote; and those which are equally distant are equal.

30. Equal straight lines drawn from a given point to a given straight line, meet it at equal distances from the perpendicular; they also make equal angles with the perpendicular; and equal angles with the given straight line.

31. Given two sides and one of the angles of a triangle to construct the triangle.

32. Two right-angled triangles are wholly equal, when :---

(a) The two sides including the right angles are equal each to each ;

(b) The hypothenuse and one side are equal each to each; (c) The hypothenuse and either acute angle are equal each to each;

(d) An acute angle and the side opposite are equal, each to each; (e) An acute angle and the side adjacent are equal, each to each. 33. If two lines be perpendicular to two others which meet one another, these will intersect and contain an angle equal to that of the first two lines.

BOOK II.

DEFINITIONS.

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

2. 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 H opposite angles of the parallelograms which make the gnomon."

K

PROPOSITIONS.

PROPOSITION I.

THEOR. 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, DE, and that contained by A, EC.

B

E с

From the point B draw (11. 1.) BF at right angles to BC, and make BG equal (3. 1.) to A; and 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;

And BH is contained by A, BC, for it is contained by GB, BC, and GB is equal (Constr.)

to A ;

F

A

K

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:

VOL. II.

D

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, etc.

PROPOSITION II.

Q. E. D.

THEOR. 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 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 AB 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;

D

C B

F E

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, etc. Q. E. D.

PROPOSITION III.

THEOR. 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 contained 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 (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 equal to BC;

A

And AD is contained by AC, CB, for CD is F equal to CB;

And DB is the square of BC;

c

E

Therefore the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. If therefore a straight line, etc.

Q. E. D.

*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.

PROPOSITION IV.

THEOR. If 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 BD, and, through C draw (31. 1.) CGF parallel to AD or BE, and through G draw HK parallel to AB or DE:

H

Λ

c

B

K

And 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 (30 Def.) to AD, being sides of a square; wherefore the angle CGB is equal (1 Ax.) 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:

D

F

Ε

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 (34. 1.) 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, etc. Q. E. D.

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

PROPOSITION 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 equal 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 equal to the square of CB.

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Upon CB describe (46. 1.) the square CEFB, join BE, 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 BM:

C

D

B

And because the complement CH is equal (43. 1.) to the complement HF, to each of these add DM; therefore the whole CM is equal (2 Ax.) to the whole DF; but CM is equal (36. 1.) to AL, because AC is equal to CB; therefore also AL is equal to DF.

To each of these add CH, and the whole AH is equal to DF and CH:

K

H

M

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But AH is the rectangle contained by AD, DB, for DH is equal (4. Cor. 11.) to DB; and DF together with CH is the gnomon CMG; Therefore the gnomon CMG is equal to the rectangle AD, DB:

To each of these add LG, which is equal (4. Cor. 11.) to the square of CD:

Therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square of CD ;

But the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB: therefore the rectangle AD, DB, together with the square of CD is equal to the square of CB. Wherefore, if a straight line, etc. Q. E. D.

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

PROPOSITION VI.

THEOR. If 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 equal 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 equal to the square of CD.

Upon CD describe (46. 1.) the square CEFD, join DE, 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;

And 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:

C

B D

L

H

M

G F

To each of these add CM; therefore the whole AM is equal to the gnomon CMG :

And AM is the rectangle contained by AD, DB for DM is equal (4. Cor. 11.) to DB: therefore the gnomon CMG is equal to the rectangle AD, DB:

Add to each of these LG, which is equal to the square of CB;

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