therefore CG > C1 ; therefore of two un: _chords, 
the smaller is the farthest pyr “hag: pl ‘ 
1 , 2 
Prop. IX. THEOoR. 
line BD, drawn perpendicular to the ex- 
a radius CA, is a tangent to the circumfer- 
chan baiviir 
A strai 
- tremity o 
ence. 
F oblique line CE is longer than the perpen- 
dicular CA’ (16. 1); therefore the point E, must, be 
without the circle ; and as this holds true of every point 
in the line BD, except the point A, the line BD isa 
one tangent AD can be drawn from 
a point A in the circumference. For if it were possi- 
ble to:draw another tangent, as AG, then as CA would 
not be dicular to AG, another line CF would be 
endicular to AG ; and so, CF would. be less than 
A (16. ke therefore F would fall. within the cir- 
cle, and AP, if produced, would cut the. cireumferetice. 
Prop. X. TuHeor. 
Two parallel chords AB, DE in a circle intercept 
equal ares AD, BE. ' 
Draw the radius CH cular to AB; it. will 
also be perpendicular to DE (1. Cor. 21.1.) There- 
fore H. will be at the same time the middle of the arcs 
AHB.and DHE: Hence we have DH=HE and AH= 
HB; and therefore DA=BE... » 
Pror. XI... Tueor. 
If two circumferences. cut each other, the straight 
line which passes through their centres shall be per- 
pendicular to the chord which joins the points of inter- 
section, and shall divide it into two equal parts. 
For the line AB, which joins the. point of intersec- 
tion, being a common chord ,of the two circles; if, 
through the middle of this chord, a perpendicular be 
al wh, it will ss through C and D, the centres of 
’ 1 the circles. atv oaly one line can be drawn through 
_ two given points; therefore the straight line which 
| __ passes through the centres is a perpendicular.at the mid- 
it Me of the common chord. 
\Phor. XI . Tueror. 
_ Ifthe distance of the centres be less than the sum of 
their radii; and if, at the same time, the greater radius 
the sum of the lesser and the distance of the 
‘Regs only that CD 
Prop. XIE, Tueror 
‘Ifthe distance cD of the centres of two circles be 
ial to the sum of their radii CA, AD, the twoicir- 
es touch each other externally. x 
~ 
a. 
GEOMETRY. 
207 
It is evident that they will have a common point A, 
but they cannot have another common point ; for if they 
had two common points, (as in Fig. 58.) it would be 
necessary that the distance of their centres should be 
less than the sum of their radii. 
Of the 
Prop. XIV. Teor. 
If the distance of the centres. be: equal to the diffe- Fig. 61. 
rence of the radii CA, AD, the two circles will touch 
each other internally. 
In the first place, it is evident that they have a com- 
mon point A, and they cannot have another ; for, that 
this might be possible, it would be necessary that the 
eater radius AD should be less than the sum of the 
lesser and the, distance. of the centres C, D, which is 
inconsistent with the hypothesis. 
Cor. Therefore, if two circles touch each other, ei- 
ther externally or internally, their centres and the point 
of contact are in the same straight line. 
Prop. XV. Treor. 
In the same circle, or in equal circles, equal angles 
ACB, DCE at the centre; intercept equal arcs AB, DE 
on the circumference; and conversely, if the arcs AB, DE 
be equal, the angles ACB, DCE arevalso equal. 
Fig. 62. 
1._If the angle ACB=DCE, these two angles may 
be placed on each other, and-as their sides are equal, 
the point A will fall on D, and the point Bon E; but 
then the are AB must also fall on the are DE; for if 
the two ares did not coincide, there would be in the one 
or the other points unequally distant from the centre, 
which is impossible; therefore the arc: AB=DE. 
2. Next if the are AB=DE, the angle ACB shall be 
equal to DCE; for if they are not equal, let ACB be the 
eater. Take ACI=DCE, then by what has been 
emonstrated AI=DE; but by hypothesis, the are 
AB=DE;; therefore the arc) AImAB, which is im- 
possible ; wherefore the angle ACB=DCE. 
Prop. XVI. THEor,. 
An angle ACB at the centre of a circle, is double of Figs. 68 
the angle ADB at the circumference, upon the same 64. 
arc AB. : ; est st 
Draw DC, producing it to E. First, let the angle at Fig. 63. 
the centre be within the angle at the circumference, 
(Fig. 63,) then theangle ACE=CAD+4 CDA, (24. 1); 
but because CA=CD, the angle CAD=CDA, (12. 1,) 
therefore the angle ACE=2:'CDA. For:the same rea- 
son, the angle BCE=2 CDB : Therefore, the whole an- 
gle ACB is double the whole:angle‘:A DB. 
Next, let the angle at the centre be without the angle 
at the circumference, (Fig. 64). It may be demonstra- 
ted as in the first case, that the angle ECB=2 EDB; 
and that the angle ECA a part of the first, is equal to 
2EDA apart ofthe second ;' therefore the’ remainder 
ACB is doublethe remainder ADB. oh 
Fig. 64 
Prop. XVII. Tueor. 
The angles ADB, AEB, in the same segment AEB Figs. 65, 
of a circle, are equal to one another. 66, 
Let C be the centre of the circle, and first let the seg- 
