A) 
GEOMETRY. 
and CV is ita wt to CD, and consequently is — 
oe emt to the chord AB; therefore C\ bisects 
care AVB in V. ; ' 
Shint Prop. VIII. | Pros. 
To find the sum or difference of two given distances 
AB, CD. ~ ” 
On B, one ews of either of the given distances 
as a centre, with a radius equal to the other given dis- 
tance CD, describe a circle. On A, the other extremi- 
, with any radius, describe an arc to cut the circum- 
erence in mand. Bisect the arcs of the circle be- 
tween m and n in E and F, (last Prop.) then ABE is the 
sum of the distances AB, CD, and AF their difference. 
For if Am, Bm, Em, An, Bn, En be joined, the 
triangles on each side of AB will be equal; hence, as 
the prolongation of AB, and the line drawn from B to 
E will bisect the angle m Bn, the points A, B, E will 
be in a straight line, In like manner it appears that 
the points A, B, F are in a straight line; consequently 
AE=AB+ BE=AB + CD, and AF = AB— BE= 
AB—CD. 
Scnotium. By this problem, a line may be produ- 
eed to any given distance. Also from the greater of 
two lines a part may be cut off equal to the less. 
Prop. IX. Pros. 
To find the centre of a given circle. 
» Let ADB be the circle. Take any point B in the 
circumference, and on B as a centre, with any radius 
less than the diameter of the given circle, and greater 
than the fourth of that diameter, describe a circle ADC, 
cutting the other circle in D. Determine C, the oppo- 
site extremity of the diameter AC, as in the former 
ome On BC construct the isosceles triangle BEC, 
ving its sides BE, CE each equal to CD. On Easa 
centre, with EB or EC as a radius, describe an are, cut« 
ing the circle ADC in F; then the distance from A 
to F shall be the radius of the circle ADB; and arcs 
described on any two points in its circumference as 
centres, with AF as a radius, will evidently intersect 
each other in its centre. 
Su O the centre of the circle. Draw OA, OB, 
and the other lines as in the re. Because the chord 
AB = chord BD, the are AB is equal to the are BD, 
and the angle BAD = angle BDA; now the angle 
BAD, or CAD, is half the angle CBD, (16. 2), and the 
ns BDA is half the angle BOA ; therefore the angles. 
CBD, BOA are equal, and since CB = BD and BO= 
OA, the triangles CBD, BOA are similar; and CD: CB:: 
BA: BO; that is, CE: CB: : AB: BO. 
Again, because the isosceles triangles EBC, EBF are 
manifestly in all equal, the angle CBF is double 
the angle CBE ; but in the isosceles triangle ABF, of 
which a side AB is produced, the exterior angle CBF 
is equal to the two angles BAF, BFA, that is, to 2 BAF, 
thereforethe angle CBE = angle BAF, and BCE=BFA. 
Hence the isosceles triangles EBC, BFA are similar ; 
and CE : CB :: AB: AF. But it was shewn that 
CE: CB:: AB: BO; therefore AF = BO, that is, AF 
is the radius of the given cirele ABD. 
Prop. X. Pros, 
To determine the intersections of a line which 
Nr two given points A,B, and. a circle given by 
position. 
Case I. 
C the centre of the circle. From C, set off CF and Cf, 
in opposite directions, each equal to the radius of the 
circle, so that F and f may bé in the line AB (by Prop. 
8.) ; and the points F, f will manifestly be the intersec- 
tion of the straight line and circle. | 
Case 2. (Fig. 191.) When the line AB does not 
pass through the centre C. Draw CD perpendicular 
to AB (Prop. 4.), and produce it, so that DE = DC, 
(Prop. 1.).. On C and E as centres, with the radius 
of the circle in the compasses, describe arcs to intersect 
each other in F and /; and these points will be the in- 
tersection of the straight line and the circle. For the 
points F and f are in the line which bisects CE at right 
angles (17. 1.); therefore they are in the line AB; 
and the same points F, fare manifestly in the circum. 
ference of the circle; therefore they are the intersec« 
tions of the straight line and circle. 
mye Prop. XI. Pros. 
To find a third proportional to two given lines P, Q: 
On any point C as a centre, with a radius equal to P, 
the first of the three proportionals, describe an are ADB. 
In this are place the chord. AD equal to: Q, the second 
term. On Dras a centre, with DA-as a radius, describe 
a circle ABE, and find E, the opposite end of the dia- 
meter passing through A, as in the former problems. 
The distance from B to E shall be the third proportional 
sought. 
or, the isosceles triangles CAD, CBD being equal, 
the angle ADB is double the angle ADC; but in the 
isosceles triangle DBE, the outward angle ADB is the 
sum of the angles DEB, DBE, and therefore is. double 
the angle DEB. Hence the angles ADC, DBE are 
equal, and consequently the other angles of the trian~ 
les CAD, DBE are equal, and the triangles are similar. 
herefore CA: AD:: DB: BE, that is P: Q:: Q: BE. 
This construction can only apply, when the first term 
is greater than half the second ; when it is not, it may. 
be doubled or quadrupled, &c. by Prop. }. untila mul- 
tiple of it be found that exceeds the half of Q, and ther 
a like multiple of a third proportional to this multiple 
of P, and the line Q will evidently be a third propor- 
tional to P and Q. 5 
, Prope. XII.. Pros, 
To find a fourth proportional to three given lines P, 
R. 
On any point C as a centre, with radii equal to P and 
R, the first and third terms of the proportionals, describe 
concentric circles AB, DE. In the first of these, place 
the chord AB equal.to:the second term Q. Take any 
pe D in the circumference of the other circle, and 
rom B place between the two circumferences a line BE 
equal to AD. Then the distance between D and E 
shall be the fourth proportional sought. 
For, by construction, the three sides of the triangle 
ACD are equal to the three sides of the triangle BCE, 
each to each. Hence the angle ACD) is equal to the 
angle BCE, and: adding the common angle BCD, 
the angle ACB is equal to the angle DCE ; therefore 
the isosceles triangles ACB, DCE are similar, and CA: 
AB:: CD: DE; that is, P: Q:: R: DE. 
If the third term is more than double of the first, this 
construction will not immediately apply ; but it may be 
modified, as in last proposition, by taking a multiple of 
the first, and.then, the line required will be a like muly 
239 
(Fig. 190.) When the line passes through Appendix. 
Fig. 190, 
Fig. 191. 
Fig. 19%: 
Fig. 193: 
