Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[ocr errors]

g

In bd take any point f, and from a e the greater, cut off ag equal (i. 3) to a f, the less, and join fc, gb.

Because a f is equal to ag, and ab to a c, the two sides fa, a c are equal to the two ga, a b, each to each ; and they contain the angle fag common to the two triangles afc, a gb; there

а. fore the base fc is equal (i. 4) to the base g b, and the triangle afc to the triangle a gb; and the remaining angles of the one are equal (i. 4) to the remaining angles of the other, each to each, to which the equal sides are opposite ; viz. the angle acf to the angle a bg, and the angle a fc

b to the angle a gb: and because the whole a f is equal to the whole ag, of which the parts a b, a c, are equal ; the remainder bf shall be equal (3 ax.)

f to the remainder cg; and fc was proved to be equal to gb; therefore the two sides bf, fc are

/d equal to the two cg, gb, each to each ; and the angle bfc is equal to the angle cgb, and the base bc is common to the two triangles bfc, cgb; wherefore the triangles are equal (i. 4), and their remaining angles, each to each, to which the equal sides are opposite ; therefore the angle fb c is equal to the angle g cb, and the angle b cf to the angle cbg: and since it has been demonstrated, that the whole angle abg is equal to the whole a cf, the parts of which, the angles cbg, bcf are also equal; the remaining angle abc is therefore equal to the remaining angle a cb, which are the angles at the base of the triangle a bc: and it has also been proved that the angle f bc is equal to the angle gcb, which are the angles upon the other side of the base. Therefore the angles at the base, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is also equiangular.

PROPOSITION VI.-THEOREM. If two angles of a triangle be equal to one another, the sides which subtend,

or are opposite to, the equal angles, shall also be equal to one another. LET a bc be a triangle having the angle a b c equal to the angle acb; the side a b is also equal to the side a c. For, if a b be not equal to a c, one them is

a greater than the other : let ab be the greater; and from it cut (i. 3) off d b equal to a c, the less, and join dc; therefore, because in the triangles d b c,

d ac b, d b is equal to a c, and b c common to both, the two sides, db, b c are equal to the two ac, cb, each to each ; and the angle dbc is equal to the angle a cb; therefore the base dc is equal to the base a b, and the triangle dbc is equal to the triangle (i. 4) a cb, the less to the greater ; which is absurd.

b , Therefore a b is not unequal to a C, that is, it is equal to it. Wherefore, if two angles, &c. Q. E. D.

COR. Hence every equiangular triangle is also equilateral.

PROPOSITION VII.-THEOREM.

Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which cre terminated in the other

extremity. If it be possible, let there be two triangles a cb, a d b, upon the same base

a b, and upon the same side of it, which have d

their sides ca, da terminated in the extremity a of the base equal to one another, and likewise their sides cb, db, that are terminated

in b.

Join cd; then, in the case in which the vertex of each of the triangles is without the other triangle, because a c is equal to a d, the

angle a cd is equal (i. 5) to the angle a dc: but а.

b the angle a cd is greater than the angle bed;

therefore the angle adc is greater also than bcd; much greater then is the angle bdc than the angle bed. Again, because c b is equal to db, the angle b dc is equal (i. 5) to the angle bed; but it has been demonstrated to be greater than it ; which is impossible.

But if one of the vertices, as d, be within the other triangle a cb; produce a c, ad to e, f; therefore, because a c is equal to ad in the

triangle acd, the angles ecd, fdc upon the

other side of the base cd are equal (i. 5) to one f

another : but the angle ecd is greater than the angle bed: wherefore the angle fdc is likewise greater than b c d ; much greater then is the angle bd c than the angle bed. Again, because cb is equal to db, the angle bdc is equal (i. 5) to the angle bcd; but bdc has been proved

to be greater than the same bcd; which is ima

To possible.

The case in which the vertex of one triangle is upon a side of the other, needs no demonstration.

Therefore, upon the same base, and on the same side of it, &c. Q. E. D.

PROPOSITION VIII.—THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each

to each, and have likeurise their bases equal; the angle which is contained by the two sides of the one shall equal to the angle contained by the two

sides equal to them, of the other. LET a b c, def be two triangles, having the two sides a b, a c, equal to

the two sides de, df, each to each, viz. a b to d e, and a c to df; and also the base b c equal to the base e f. The angle bac is a

d 8 equal to the angle ed f.

For, if the triangle a b c be applied to def, so that the pcint b be on e, and the straight line bc upon ef; the point c shall also coincide with the point f, be

b
с

f cause bc is equal to e f. Therefore b c coinciding with ef; ba and a c shall coincide with ed and df; for, if the base b c coincides with the base e f, but the sides ba, ca do not coincide with the sides ed, fd, but have a different situation, as eg, fg, then upon the same base e f, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity ; but this is impossible (i. 7); therefore, if the base b c coincides with the base ef, the sides ba, a c, cannot but coincide with the sides ed, df; wherefore likewise the angle bac coincides with the angle ed f, and is equal (8ax.) to it. Therefore if two triangles, &c. Q.E.D.

PROPOSITION IX.-PROBLEM.

To bisect a given rectilineal angle, that is, to divide it into two equal angles. LET bac be the given rectilineal angle, it is required to bisect it.

Take any point d in a b, and from a c cut (i. 3) off a e equal to ad; join de, and upon it describe (i. 1) an equilateral triangle def; then join a f; the straight line af

a bisects the angle bac.

Because a d is equal to a e, and a f is common to the two triangles da f, eaf; the two sides da,

d af, are equal to the two sides ea, a f, each to each ; and the base df is equal to the base ef; therefore the angle daf is equal (i. 8) to the angle eaf; wherefore the given rectilineal angle bac is bisected b

f by the straight line a f. Which was to be done.

PROPOSITION X.-PROBLEM.

To bisect a given finite straight line, that is, to divide it into two equal

parts. LET a b be the given straight line, it is required to divide it into two equal parts.

Describe (i. 1) upon a b an equilateral triangle a b c, and bisect (i. 9) the angle a c b by the straight line cd. ab is cut into two equal parts in the point d.

Because a c is equal to cb, and cd common to the two triangles a cd, bcd; the two sides a C, cd are equal to bc, cd, each to each ; and the angle a c d is equal to the angle bed; therefore

the base a d is equal to the base (i. 4) d b, and the b straight line a b is divided into two equal parts in the point d.

Which was to be done.

[blocks in formation]

PROPOSITION XI.- PROBLEM.

a

с

e

To draw a straight line at right angles to a given straight line, from a

given point in the same. LET a b be a given straight line, and c a point given in it; it is required to draw a straight line from the point c at right angles to a b. Take any point d in a c, and make (i. 3) ce equal to cd, and upon de

describe (i. 1) the equilateral triangle dfe, and join fc, the straight line fc drawn from the given point c is at right angles to the given straight line a b.

Because dc is equal to ce, and fc common to the two triangles dcf, ecf; the two sides dc, cf, are equal to the

two ec, cf, each to each ; and the base d

b d f is equal to the base e f; therefore the

angle d cf is equal (i. 8) to the angle ecf; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right angle (10 def.); therefore each of the angles dcf, ecf, is a right angle. Wherefore, from the given point c, in the given straight line a b, fc has been drawn at right angles to a b. Which was to be done.

COR. By help of this problem, it may be demonstrated, that two straight lines cannot have a common segment. If it be possible, let the two straight lines a b c, abd have the seg

ment a b common to both of them. је

From the point b draw be at right angles to a b; and because a b c is a straight line, the angle cbe is equal (10 def.) to the angle e ba; in the same

manner, because a bd is a straight line, d

the angle d be is equal to the angle

eba ; wherefore the angle d be is a b

с

equal to the angle cbe, the less to the

greater, which is impossible ; therefore two straight lines cannot have a common segment.

ز

PROPOSITION XII.—PROBLEM. To draw a straight line perpendicular to a given straight line of an

unlimited length, from a given point without it. LET a b be the given straight line, which may be produced to any length both ways, and let c be a point without it. It is required to draw a straight line perpendicular to a b from the point c.

Take any point d upon the other side of a b, and from the centre c, at the distance cd, describe (3 post.) the circle egf meeting a b in f, g; and bisect (i. 10) fg in h, and join

h cf, ch, cg; the straight a f

g

b line ch, drawn from the

d given point c, is perpendicular to the given straight line a b.

Because fh is equal to hg, and hc common to the two triangles fh c, ghc, the two sides fh, hc are equal to the two gh, hc, each to each; and the base cf is equal (15 def.) to the base cg; therefore the angle chf is equal (i. 8) to the angle chg; and they are adjacent angles; but when (10 def.) a straight line standing on another straight line makes the adjacent angles equal to one another, each of them is a right angle; and the straight line which stands upon the other is called a perpendicular to it ; therefore from the given point ca perpendicular ch has been drawn to the given straight line a b. Which was to be done.

PROPOSITION XIII.-THEOREM. The angles which one straight line makes with another upon one side of it

are either two right angles, or are together equal to two right angles. LET the straight line a b make with cd, upon one side of it, the angles cba, abd : these are either two right angles, or are together equal to two right angles.

For if the angle c ba be equal to a b d, each of them is a right angle

[ocr errors]

d b

d

ъ (10 def.); but if not, from the point b draw be at right angles (i. 11) to cd; therefore the angles c be, ebd are two right angles (10 def.); and because cbe is equal to the two angles cba, a be together, add the angle e bd

« ΠροηγούμενηΣυνέχεια »