238 
Appendix. Let us suppose that the distance AB is to be divided 
Fig. 183. 
into three equal parts. Take the distance AC equal to 
three times the distance AB, (by Prop. 1.), and in ge- 
neral, whatever be the number of equal into 
which AB is to. be divided, take AC equal to the 
same number of times AB; in other respects, the con- 
struction is the same in all cases. On C as a centre, 
with CA as a radius, describe an are PAp; andon A 
as a centre, with AB as a radius, describe another are 
PBp, meeting the former in P and p. In the circle 
pBP, beginning from the point p, place three chords 
pm, mn, n Q, each equal to the radius AB. On Pasa 
centre, with a radius equal to AB or AP, describe the 
arc AV, and on A as a centre, with a radius equal to 
the chord of PQ, describe another arc, meeting the 
former in V; then V shall be in a straight line joming 
A and B; and the distance AV shall be one third of 
the distance AB. z 
Join AP, CP, AQ, PQ, AV, PV. | The triangles 
CAP, PAQ are manifestly isosceles ; and. because the 
arcs pm, mn,nQ are each one-sixth of the circumfe- 
(2. 5.) the are p BQ is half the circumference ; hence it 
is the measure of the three angles of the triangle CAP, 
(31, 4. and 24. 1.) that is of the angle C, and twice the 
angle CAP; but the arc PB p is the measure of twice 
the angle PAB, because arc BP = are Bp ; therefore 
the remaining arc PQ is the measure.of the angle C. 
Now the same arc is also the measure of the angle 
PAQ; therefore the angle C. is equal to. the asigle 
PAQ; and since PC:CA:: PA: AQ, the -triangles 
PCA, PAQ are similar (20. 4.) ; hence the angle APQ 
is equal to’ the angle PAC ; but the angle APQ=PAV, 
because by construction AP is common to the triangles 
APQ, PAV, and PQ = AV, and AQ = PV; there- 
fore the angle PAC is equal to the angle PAV, and 
consequently V is in the straight line AC. And be- 
cause CA: AP:: AP: AV, that is, CA: AB: :AB: AV, 
therefore whatever part ABis of AC, the very same 
part will AV be of AB. 
The remaining points of division X, &c. may be 
found by making AX = 2 AV, &c. as taught in Prop. 1. 
Scuotium. The point V might also have been 
found by determining the points P and p as in the 
above construction, and then describing arcs on P and 
as centres to pass through the point A, these would. 
five intersected each other again in the point V. This 
construction, however, is not so good as the other, as 
a practical method, because the arcs cut each other 
obliquely. 
Prop, III. 
Having given two points in a straight line, to deter- 
mine the direction of a perpendicular to it, which shall 
pass through one of the points. 
Pros. 
Let the given points be A, B. On these points, with 
any radius gor than half AB, describe arcs to inter- 
sect each other in C. OnC, as a centre, describe a 
circle to pass through A and B, and determine the se- 
micircle ABP, as in the former problems, by cutting 
off successively three arcs A m, mn, n P, with a radius 
in the compasses equal to that of the circle ; then P will 
be a point in the perpendicular PB. For the angle 
APB, which is in a semicircle, is a right angle, 
Prop. IV. Pros. 
Having given two points in a straight line, to deter- 
GEOMETRY. 
- AC*+-CD?.4.2 EC x CD (13. 4.) =AC*--2 CD®, there- 
_CD*; hence DCV is a right-angled triangle (11. 4.) 
mine the direction of a perpendicular drawn from a 
point without it, and also the point in which the per- | 
pendicular meets the line. ' te eet 
Let A,B be the given points in the line AB, and P 
the point without it. On A and B, as centres, describe 
ares to ugh P, and meet each other at p, a 
point on the other side of AB. Because each of the 
peas A and B is equally distant from P and p, the 
ine AB is perpendicular to the line which 
through P and p, (17.1). It also bisects Pp at C ; there- 
fore C, the intersection of the line AB, and the perpen- 
dicular, may be found by Prop. 2. i ‘aug, 42 
- 
Prop. V. Pros. 
- Having given two points A and B in a straight line, p; 
and a in P without it, to determine the wok Ta a 
a line that passes through P, and is.parallel to AB... 
On P as a centre, with a radius equal to AB, de- 
scribe an arc of a circle ;/and on B:as a centre, with a 
radius equal to PA, describe another arc, cutting the 
former in Q; a line passin P and Q will be 
parallel to AB. For if AP, BP, BQ, be joined, 
the triangles PAB, BQP, will be, in all respects, equal; - 
therefore the angles, QPB, PBA, are equal, and PQis - 
parallel to AB, 
Prop. VI. Pros. 
To find the side of a square, that shall be equal to the 
difference of two given squares. 
Let AB and AC be the sides. of the given squares,’ 
In the line AB produced, take Ba= BA, ( Prop, Je 
and on A and a, as centres, with a radius equal to AC, 
describe ares cutting each other in D, The distance- 
from B to D will be the side of the square required. —_ 
Because ADa is an isosceles triangle, a straight line’ 
drawn from D,-the vertex of the triangle, to B, the 
middle of the base, will be perpendicular to the hase, 
(12.1) ; therefore AD? = AB? + BD’, and BD? =. 
AD? — AB: sais 
nell 
Prop. VII. Pros. 
To bisect a given are of a circle. 
Let AB be the given arc, and C the centre of the Fiz, 1 
circle. On B as a centre, with a radius equal to CA, 
describe .an are of a circle; and on C as a centre, with 
a radius equal to BA, describe another arc, cutting the’ - 
former in D. Find, m the side of a square, that shall. 
be equal to the difference of the squares of the lines: 
DA, DC, (Prop. 6.) On D, as a centre, with a ra- © 
dius equal to m, describe an arc, to meet the arc AB. 
in V; then V shall be the middle of thearc AB. 
Draw AE perpendicular to CD, and Pee ye 
dicular to AB, Spe by construction AB=C€D, and ° 
AC=BD, the figure ABDC is a pecele ners ; there- 
fore also eee is a rectangle, and CE= AF, but» 
AF=3 AB (6.2.); therefore CE = 3 AB=4CD, and — 
2CE=CD. In the triangle ACD, we have AD*=— 
fore AD*—CD* = AC? + CD*; but by construction» _ 
AD? — CD?=DV?; therefore DV?=AC?4CD*; and, 
if.a straight line be drawn from V toC, DV?=VC? 4. - 
a ee 
