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(i. 31) ae parallel to be, and join e c. The triangle a b c is equal (i. 37) to

d the triangle eb c, because it is upon the same base bc, and between the same parallels bc, a e. But the triangle a bc is equal to the triangle bdc, therefore also the triangle bde is equal to the triangle e bc the greater to the less, which is impossible. Therefore ae is not parallel to bc. In the same manner,

b it can be demonstrated, that no other line but ad is parallel to bc; ad is therefore parallel to it. Wherefore equal triangles, &c, Q. E. D.

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PROPOSITION XL.-THEOREM. Equal triangles upon equal bases, in the same straight line and towards the

same parts, are between the same parallels. LET the equal triangles abc, def be upon equal bases b c, ef, in the same straight line bf, and towards the

a

d same parts; they are between the same parallels.

Join a d; ad is parallel to bc. For, if it is not, through a draw (i. 31) ag parallel to bf, and join gf: the triangle a bc b is equal (i. 38) to the triangle gef, because they are upon equal bases b c, ef, and between the same parallels bf, ag: but the triangle a b c is equal to the triangle def; therefore also the triangle d ef is equal to the triangle gef, the greater

less, which is impossible: Therefore ag is not parallel to bf. And in the same manner it can be demonstrated that there is no other parallel to it but ad: ad is therefore parallel to bf. Wherefore equal triangles, &e. Q. E. D.

PROPOSITION XLI.-THEOREM.

If a parallelogram and triangle be upon the same base and between the same

parallels, the parallelogram shall be double of the triangle. LET the parallelogram a bed and the triangle ebc be upon the same

base bc, and between the same parallels b c, a e; a

d e

the parallelogram abcd is double of the triangle e bc.

Join ac ; then the triangle a bc is equal (i. 37) to the triangle ebc, because they are upon the same base bc, and between the same parallels b c, ae. But the parallelogram abcd is double (i. 34) of the triangle a bc, because the diameter a c

divides it into two equal parts; wherefore abcd b

is also double of the triangle e bc. Therefore, if a parallelogram, &c. Q. E. D.

PROPOSITION XLII.—PROBLEM.

09

To describe a parallelogram that shall be equal to a given triangle, and have

one of its angles equal to a given rectilineal angle. LET a b c be the given triangle, and d the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle a b c, and have one of its angles equal to d.

Bisect (i. 10) bc in e, join a e, and at the point e in the straight line ec make (i. 23) the angle cef equal to d; and through a draw (i. 31)

ag parallel to ec, and through c draw cg parallel to ef: therefore fecg is a parallelogram. And because be is equal to ec, the triangle a be is likewise equal (i. 38) to the

triangle a ec, since they d

are upon equal bases be, e c, and between the same parallels, bc, ag; therefore the triangle a bc is

double of the triangle b e с

a ec. And the parallelogram fecg is likewise double (i. 41) of the triangle a ec, because it is upon the same base, and between the same parallels : therefore the parallelogram fecg is equal to the triangle a bc, and it has one of its angles cef equal to the given angle d; wherefore there has been described a parallelogram fecg equal to a given triangle a bc, having one of its angles cef equal to the given angle d. Which was to be done.

PROPOSITION XLIII.—THEOREM. The complements of the parallelograms, which are about the diameter of any

parallelogram, are equal to one another.
LET abcd be a parallelogram, of which
the diameter is a c, and eh, fg,

the
pa-
h

d rallelograms about a c, that is, through which a c passes, and bk, kd, the other

Ik parallelograms which make up the whole figure a b c d, which are therefore called the complements. The complement b k is equal to the complement k d.

Because a b c d is a parallelogram, and a c its diameter, the triangle a b c is equal

с (i. 34) to the triangle adc: and, because ekha is a parallelogram, the diameter of which is a k, the triangle a ek is equal to the triangle ahk: for the same reason, the triangle kgc is equal to the triangle kfc. Then, because the triangle aek is equal to the triangle a hk, and the triangle kgc to kfc; the triangle a ek, together with the triangle kgc is equal to the triangle a hk together with the triangle kfc. But the whole triangle a b c is equal to the whole a dc; therefore the remaining complement bk is equal to the remaining complement kd. Wherefore the complements, &c. Q. E. D.

PROPOSITION XLIV.-PROBLEM. To a given straight line to apply a parallelogram, which shall be equal to

a given triangle, and have one of its angles equal to a given rectilineal

angle. LET a b be the given straight line, and c the given triangle, and d the given rectilineal angle. It is required to apply to the straight line ab a parallelogramı equal to the triangle c, and having an angle equal to d.

Make (i. 42) the parallelogram befg equal to the triangle c, and having the angle e bg equal to the angle d, so that be be in the same straight line with a b, and produce

f e

k fg to h; and through a draw (i. 31) ah parallel to bg or ef, and join h b.

d Then, because the straight line hf falls upon

the

parallels a h, ef, the angles

с a hf, hfe, are together equal (i. 29) to two right angles; wherefore the an

h

1 gles bhf. hfe, are less than two right angles. But straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet (12 ax.) if produced far enough : therefore hb, fe shall meet if

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m

produced ; let them meet in k, and through k draw k1, parallel to e a or fh, and produce ha, gb to the points 1 m: then hikf is a parallelogram, of which the diameter is hk, and ag, me are the parallelograms about h k ; and lb, bf are the complements : therefore lb is equal (i. 43) to bf; but bf is equal to the triangle c; wherefore lb is equal to the triangle c; and because the angle gbe is equal (i. 15) to the angle a bm, and likewise to the angle d; the angle a bm is equal to the angle d. Therefore the parallelogram 1 b is applied to the straight line a b, is equal to the triangle c, and has the angle a bm equal to the angle d. Which was to be done.

PROPOSITION XLV.-PROBLEM. To describe a parallelogram equal to a given rectilineal figure, and having an

angle equal to a given rectilineal angle. LET à bcd be the given rectilineal figure, and e the given rectilineal angle. It is required to describe a parallelogram equal to abcd, and having an angle equal to ë.

Join db, and describe (i. 42) the parallelogram fh equal to the triangle adb, and having the angle hkf equal to the angle e; and to the straight line gh apply (i. 44) the parallelogram 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 fk h is equal to ghm: add to each

of these the angle khg; a d

1 therefore the angles fkh,

khg, are equal to the angles khg, ghm; but fkh, khgare equal (i. 29) to two right angles; therefore also khg, ghm, are equal to two right an

gles; and because at the b

h

point h in the straight

line gh, the two straight lines k h, hm upon the opposite sides of it make the adjacent angles equal to two right angles, k h is in the same straight line (i. 14) with hm ; and because the straight line hg meets the parallels km, fg, the alternate angles mhg, hgf are equal (i. 29): add to each of these the angle hgl: therefore the angles mhghg 1, are equal to the angles hgf hg 19. But the angles m hg, hgl, are equal (i. 29) to two right angles ; wherefore also the angles hgf, hg1 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;kfis parallel (i. 30) to ml; and km, fl are parallels; wherefore kfim is a parallelogram; and because the triangle a bd is equal to the parallelogram hf, and the triangle d bc to the parallelogram ĝm; the whole rectilineal figure a b c d is equal to the whole parallelogram kflm; therefore the parallelogram kflm has been described equal to the given rectilineal figure a b c d, having the angle fkm equal to the given angle e. Which was to be done.

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CÓR. 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 (i. 44) to the given straight line a parallelogram equal to the first triangle a bd, and having an angle equal to the given angle.

upon a b

PROPOSITION XLVI.- PROBLEM.

To describe a square upon a given straight line. Let a b be the given straight line; it is required to describe a square

From the point a draw (i. 11) ac at right angles to a b; and make (i. 3) ad equal to a b, and through the point d draw de parallel (i. 31) to a b, and through b draw be parallel to a d therefore a deb is a parallelogram: whence a b is equal (i. 34) to de, and a d to be: but ba is equal to ad; therefore the c 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; á because the straight line ad meeting the parallels ab, de, the angles b ad, ade are equal (i. 29) to two right angles : but bad is a right angle; therefore also a de is a right angle; but the opposite angles of parallelograms are equal (i. 34); therefore each of the opposite angles a be, bed is a right angle; wherefore the figure ad eb is

а.

b rectangular, and it has been demonstrated that it is equilateral ; it is therefore a square, and it is described upon the given straight line a be which was to be done.

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

SA

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 a b c 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 b c describe (i. 46) the square bd ec, and on ba, ac

the

squares gb, hc; and through a draw (i. 31) a 1 parallel to b d or ce, and join åd, fc. Then, because each of the angles bac, bag is a right angle (30 def.), the two straight lines a c, ag, upon the opposite sides of a b, make with it at the point a the adjacent angles equal to two right angles ; therefore cá is in the same straight line (i. 14) with a g; for the same reason, a b and a h are in the same straight line ; and because the angle dbc is equal to the angle f ba, each of them being a right angle, add to each the angle a bc, and the whole angle dba is equal (2 ax.) to the

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