Of the 
Circle. 
Fig, 52, 
Fig. 52. 
Fig. 53. 
206 
Prop. IV. Tueor. 
In the same circle, or in equal circles, equal ares are 
subtended by equal chords; and conversely, equal 
chords, subtend equal ares. 
If the radii AC, EO are equal, and the arc AMD is 
equal to the are ENG; the chord AD shall be equal to 
the chord EG.. For the semicircle AMDB may evident- 
ly be applied exactly upon the semicircle ENGF ; and 
then the curve line AMDB will coincide entirely with 
the curve ENGF; but we suppose the are AMD = 
arc ENG;; therefore the point D will fall on G, and 
the chord AD on the chord EG. 
Conversely, if the chord AD be equal to the chord 
EG, then the are AMD= are ENG. For, drawing 
the radii CD, OG; the triangles ACD, EOG have 
AC = EO, CD = OG, AD = EG;; therefore they are 
equal, (11. 1.); and the angle ACD = EOG; conse- 
quently, if the semicircle ADB be placed on the equal 
semicircle EGF ; so that AC may coincide with EO ; 
the point D will manifestly fallon G, and the’ are AMD 
will be equal to ENG. ~ , 
Prop. V. Tueor. 
In the same, or in equal circles, a greater arc is sub- 
tended by a greater chord, and conversely ; provided 
that the ares are each less than a semicircumference. 
_ Let the are AH be greater'than the are AD, and let 
the chords AD, AH, and the radii CD, CH be drawn. 
The two sides AC, CH of the triangle ACH are equal 
to the two sides AC, CD of the triangle ACD ; and the 
angle ACH is greater than the angle ACD; therefore 
AH => AD (10. 1.) Thus it appears, that the chord 
that subtends the greater arc is the greater. 
Conversely, if the chord AH is supposed greater than 
AD, it may be inferred’ from‘the ‘same triangles, that 
the angle ACH is greater than the angle ACD, and 
therefore that the are AH is greater than AD. 
Scuo.ium. If the arcs, instead of being less, were 
greater than a semicircumference, the opposite property 
ae rite true ; thatis, the greater the arc, the smaller 
the chord. 
Prop, VI. 
The radius CG; which is perpendicular to a’ chord 
AB, bisects the chord, and the arc it subtends ; that is, 
divides each into two’ équal parts. 
THEoR, 
Draw the radii CA, CB; these, “with respect to the 
perpendicular CD, are two equal oblique lines; there- 
fore they mett AB at equal: distances from the perpen- 
dicular (16.'1'); “that' is, ‘AD = ‘DB. 
Next, if AD = DB; then, ‘because CG is ’a perpen- 
dicular to AB, every point in CG is equally distant 
from A and B (17. 1:);! therefore, if AG and BG be 
drawn, the chord AG = chord BG; hence the are 
AG'=are'BG' (4.) . 
Scuottum. From this proposition, it appears that'the 
centre C, the middle D of the chord, and the. middle 
G of'the are, "are three points situated in’a straight line 
perpendicular to the chord. » Now, ‘two points are suf- 
ficient to fix the ‘position ofa-straight line. Therefore, 
any straight ‘line’ which ‘passes’ through two ‘of’ these 
GEOMETRY. 
ints will necessarily pass through the third, and be — 
F rpentiealil tothe chord. == ; 
Hence it also follows, that the ticular to the 
middle of a chord passes through the centre and the mid- 
dle of the arc subtended by the chord. = 
Prop. VII. Tuxor. 
‘One circle, and Her 8 one, can be described through 
three given points A, B, C, which are not in a strai 
line. - 
: ya . 
Join AB, BC, and ue Ngee pe each into two equal 
parts by the pe iculars DE, FG. Now, if ABC; 
or DBF, be fi eight wala then, if the points D, F be 
joined, ‘DBF will be a no angled triangle, and there- 
fore each of the angles BED, BDF will be less than a 
right angle (2. Cor. 24. 1.) ; and consequently, each of 
the angles FDE, DFG will be less than a right angle, 
and their sum will be less than two right angles : ‘her 
DE, FG will meet, if produced in a point O (Schol. 
21.1.) Butif DBF is not a right angle, the twollines 
GF, AB will make with BI’ two angles on one side, 
which will be less than Saat ero therefore those 
lines will meet in a point K; and as BFK is a right 
angle, BK ‘will be less than’a right angle (2. Cor: 24. 
1.); therefore EDK and GKD will ‘be together less 
than two right angles, and consequently will meet at a 
point O, as in the other case. Now this point O, con- 
sidered as in the perpendicular’ DE, will’ be equally dis- 
tant from A and'B  (17.-1.) ; and ‘considered ‘as in’ the 
sae ndicular FG, it will be equally distant’from C ‘and 
‘y therefore it will be'equally distant from A, B, and 
C, ‘and these three points will be in the circumference 
of'a circle, of which O-is the centre. at apace 
Again, every circle that passes through A and B must 
have its centre in DE; and every circle that wt 
through C and. B must have its centre in FG (Schol. 6.) ; 
but these two lines can only have one common point ; 
therefore only one circle can pass through the points.A, 
Bren Mise GW ted ai"; FSA Hee Sete tue 
Cor. Two ciréles cannot cut each other in more than 
two points; for, if they could have three common 
points, they would have the same centre, and would 
coincide. Pei: 
J 
Prop, VIII, Tyson, = 
Two equal chords are equally distant froin:the een. 
tre; and of two uriequal chords, the’shortest is farthest 
from the centre. 
1. Let the chord AB=DE ; suppose them divided 
into two equal parts by the perpendiculars CF, CG from 
the centre ; and draw*the radii) CA, CD. The right 
angled triangles CAF, CDG have equal hypothenuses 
CA, CD; also AF, the half of AB, equal to DG, the 
half of DE; therefore the triangles are equal (18. 1.), 
and CF=CG ;' therefore AB,\ DE are equally distant — 
from the’ centre. y ¥l to diving 
2, Next let the chord AH be greater than the chord _ 
DE, so that the are AKH is.greater than there DME. 
In AKH take AKB equal: to DME; draw the chord 
AB, and CF a perpendicular from the'centre' upon AB, 
meeting AH in O ;:and’Clia dicular upon AH; 
It is evident that CF is greater than CO, and CO great« 
er'than CI (16. 1!) } much  iore’thén iscCF- > CI: 
But CF=CG, ‘since the chords“AB, ‘DE are'equaly — 
