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

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

PROP. E. PROB.

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 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 draw EF at rt. 4s to CD. (Prop. XI.) F is the point required. Join CF, DF.

ED,

A

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

[blocks in formation]

base DF. (Prop. IV.)

Fig. 1.

C

E

D

B

F

Fig. 2.

[blocks in formation]

Wherefore a pt. F has been found in the gn. line AB, equidistant from the gn. pts. c and D.-Q. E. F.

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 describing an isosceles ▲ 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 AB be the gn. st. line; then it is required to bisect it. CONST.-Describe upon AB an

equilateral

ABC (Prop. I.),

and bisect the ACB by the st. line CD (Prop. IX.); then AB is cut

into two =

A

B

D

parts in the pt. D. DEMONST.-Because AC = CB, and CD is common to the ▲ ACD, BCD, .. the two sides AC, CD = BC, CD, each to each, and the ACD = BCD (Const.); .. the base AD = base DB (Prop. IV.): i. e. the st. bisected in the point D.-Q. E. F.

PROP. XI. PROB.

line AB is

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

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. Zs to AB.

Δ

CONST.-In AC take any pt. D, a and make CE = CD. (Prop. III.)

D

Upon DE describe an equilateral ▲ DEF. (Prop. I.) Join cr (Post. 1). The st. line CF, drawn through the gn. pt. c, is at rt. s to AB.

DEMONST.-1. Because DC = CE, and CF is common to the two As DCF, ECF; .. the two sides DC, CF = EC, CF, each to each; also base DF = base EF; ... ≤ DCF = ≤ ECF (Prop. VIII.), and they are adjacent

s.

2. But when the adjacent

s which one st. line makes with another are equal, each of them (Def. 10); ... each of the s

is called a rt.

DCF, ECF is a rt. .

3. Wherefore from the gn. pt. c, in the gn. st. line AB, the line CF has been drawn at rt. ▲s 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. в draw BE at rt. ≤s to AB (Prop. XI.); then, because ABC is a st. line, the CBE = ZEBA; and if ABD is a st.

PART. ENUN.-Let the sides of any ▲ ABC be bisected in

the pts. DEF; through D and E let DG, EG
be drawn at rt. 2 to AB, BC, and meeting
in G (Prop. XI.); then the st. line drawn
through rat rt. 4s to AC shall also
pass through G, and this pt. G is equi-
distant from each of the 4s of the AB

ABC.

G

A

F

C

E

CONST.-Join GF, GA, GB, GC (Post. 1). DEMONST.-1. Prove that AG = BG. Because AD = DB, and DG is common to the As ADG, BDG; .. the two sides AD, dg = BD, DG, each to each, and the ADG = Z BDG (Def. 10); .'. base AG = BG. (Prop. IV.)

base

2. In the same manner it may be shewn that CG = BG. 3... also AG = CG.

4. Prove that ▲ AFG = L CFG = rt. 4.

Again, because AF = 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 ▲ CFG (Prop. VIII.), and they are adjacant them is a rt. 2.

AFG = L

;

... each of

5. Hence the line FG, drawn at rt. to AC through F, passes through G, and it has been shown that AG = BG

CG.

=

Wherefore, if the three sides of a ▲ be 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.

C

CONST. Take any pt. D upon the other side of AB. From cr. c, with rad. CD, describe the

EGF,

meeting AB in the pts.
F and G (Post. 3). Bisect A
FG in H (Prop. X.), and
join CH. The line CH, drawn
is to the gn. st. line AB.

[blocks in formation]

from the pt. c,

Join CF, CG.

DEMONST.-Because FH = HG, and HC

is common to the AS FHC, GHC; .. the two sides FH, HC = GH, HC, each to each; and by the property of O, the base Fc = the base CG (Def. 15); .. the CHF = ≤ CHG (Prop. VIII), and they are adjacent s: but when a st. line standing on another st. line makes the adjacent to one another, each of them is art. 4, and the st. line which stands upon the other is called a to it; .. from the gn. pt. c a st. line сH has been drawn 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 could not be described cutting the line in two pts., but the 1 would fall beyond the extremity of the line. In order to drop a upon a gn. finite line at or near its extremity, the 31st Prop. of Book III. must first be proved.

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