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CDE = rt. L. (by Const.) Wherefore through 'B, its extremity, a line be, has been drawn at rt. to AB.Q. E. F.

From the Eleventh Proposition, the two following are also deductions.

PROP. E. PROB.

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GEN. Enun.-In a straight line given in position, but indefinite in length, to find a point equally distant from two given points, either on the same or on opposite sides of the given line. PART. ENUN.-Let AB be the

Fig. 1. gn. line, and c, d the gn. pts. ; then it is required to find in the st. line AB a pt. equidistant from c and D.

Const.-Join cd, and bisect it in E (Prop. X.); through E Adraw EF at rt. %$ to cd. (Prop. XI.) F is the point required.

Fig. 2. Join CF, DF.

DEMONST.-Because ce = EP, and EF is common to A CEF, DEF; .. the two sides CE, EF = DE, EF, each to each, and the L

[ DEF (Def. 10); .. the base CF

base df. (Prop. IV.) Wherefore a pt. F has been found in the gn. line AB, equidistant from the gn. pts. c and D.-Q. E. F.

B

F

A

-B

E

CEF =

D

PROP. F. THEOR.

GEN. ENUN.-If the three sides of a triangle be bisected, and straight lines be drawn through the points of bisection at rt. angles to the sides, these lines will meet in a point equally distant from each of the angles.

Wherefore the < BAC is bisected by the st. line AF.—Q. E. F.

It is clear that the same result would follow from des. cribing an isosceles A upon DE (Prop. A or B); and that by repeating the operation, a rectilineal / may be divided in 4, 8, 16, 32, 64 equal parts; and so on, doubling continually. A similar remark applies to the following Proposition.

PROP. X. PROB.

Gen. Enun.-To bisect a given finite straight line; that is, to divide it into two equal parts.

Part. Enun.—Let Al be the gn. st. line; then it is required to bisect it.

Const.—Describe upon AB an equilateral a ABC (Prop. I.), and bisect the 2 ACB by the st. line cd (Prop. IX.); then AB is cut into two = parts in the pt. D. DEMONST. -Because AC

= CB, and cd is common to the AS ACD, BCD, .. the two sides AC, CD = BC, CD, each to each, and the < ACD = Z BCD (Const.); .. the base Ad = base dB (Prop. IV.): i.e. the st. line AB is bisected in the point D.-Q. E. F.

A

B

D

PROP. XI. PROB.

GEN. ENUN.-To draw a straight line at right angles to a given straight line from a given point in the same.

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Part. Enun.-Let AB be the gn. st. line, and c a gn. pt. in it; then it is required to draw from c a st. line at rt. Z to AB.

Const.-In ac take any pt. D, A. and make CE = ca. (Prop. III.) Upon de describe an equilateral A DEF. (Prop. I.) Join CF (Post. 1). The st. line cf, drawn through the gn. pt. c, is at rt. Zs to AB.

DEMONST.-1. Because Dc = CE, and cf is common to the two A$ DCF, ECF; .. the two sides DC, CF = EC, CF, each to each ; also base DF = base EF; .. ZDCF= 2 ECF (Prop. VIII.), and they are adjacent Z s.

2. But when the adjacent z $ which one st. line makes with another are equal, each of them is called a rt. < (Def. 10); .. each of the Zs DCF, ECF is a rt. 2.

3. Wherefore from the gn. pt. c, in the gn. st. line AB, the line of has been drawn at rt. Zs to AB.-Q. E. F.

CoR.-By help of this Proposition, it may be demonstrated that two straight lines cannot have a common segment.

Suppose, if possible, the two st. lines ABC, ABD to have the segment AB common to both of them.

From the pt. B draw be at rt. Z $ to AB (Prop. XI.); then, because ABC is a st. line, the < CBE Z EBA; and if ABD is a st.

Part. Enun.—Let the sides of any A ABC be bisected in the pts. DEF; through d and E let DG, EG

A be drawn at rt. Z $ to AB, BC, and meeting in G (Prop. XI.); then the st. line drawn through F at rt. s to AC shall also pass through g, and this pt. G is equidistant from each of the 48 of the AB

D

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E

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ABC.

Const.-Join GF, GA, GB, GC (Post. 1).
DEMONST.-1. Prove that AG = BG.

Because AD = DB, and do is common to the AS ADG, BDG; .. the two sides AD, DG = BD, DG, each to each, and the L ADG = BDG (Def. 10); .. base AG = base BG. (Prop. IV.)

2. In the same manner it may be shewn that CG = BG. 3. .. also AG = CG. 4. Prove that / AFG = L CFG = rt. L.

Again, because Ar = FC, and Fg is common to the As AFG, CFG; .. the two sides AF, FG = CF, FG, each to each, and the base AG = base cg; .. also / AFG = 2 CFG (Prop. VIII.), and they are adjacant 48 ; .. each of them is a rt. L.

5. Hence the line fg, drawn at rt. Z to Ac through F, passes through g, and it has been shown that ag = BG =

CG.

Wherefore, if the three sides of a Abe bisected, &c. Q. E. D.

PROP. XII. PROB.

Gen. Enun.To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.

PART. ENUN.-Let AB be the gn. st. line, which may be produced to any length both ways, and let c be a gn. pt. without it; then it is required to draw a st. line - to AB from the pt. c.

of AB.

E

H

A

-B

F

G

D

Join CF, cG.

HG, and

HC

Const.—Take any pt. D upon the other side

From cr. c, with rad.cd, describe the EGF, meeting AB in the pts. F and G (Post. 3). Bisect FG in - (Prop. X.), and join ch. The line ch, drawn from the pt. c, is I to the gn. st. line AB.

DEMONST.—Because FH is common to the AS FHC, GHC; .:. the two sides fa, hc = GH, Hc, each to each ; and by the property of o, the base Fc = the base cg (Def. 15); ::. the < chr = Z CHG (Prop. VIII:), and they are adjacent Zs: but when a st. line standing on another st. line makes the adjacent Z = to one another, each of them is a rt. L, and the st. line which stands upon the other is called a I to it; .. from the gn. pt. c a st. line ch has been drawn Ir to AB. -Q. E. F.

The condition that the st. line should be given in position only, and not in length, is manifestly essential ; for otherwise the gn. pt. might be so placed, not only that a O could not be described cutting the line in two pts., but the I would fall beyond the extremity of the line. In order to drop a I upon a gn.

finite line at or near its extremity, the 31st Prop. of Book III. must first be proved.

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