Fig. 67. 
Fig. 68, 
Fig. 69. 
208 
rhont “AEB be greater than a semicircle, (Fig. 65). 
Draw CA, CB to the ends of ‘the base of the segment, 
then each of the angles ADB, AEB will be half of the 
angle ACB, (16,); therefore the angles ADB, AEB 
are equal. re 
Next, let the segment AEB be less than a semicircle ; 
draw the diameter DCF, and join EF ; and because the 
segment ADEF is greater than a semicirele by the first 
case, the angle ADF=AEF. In like manner, because 
the segment BEDF is greater than a semicircle, the an- 
gle BDF is equal to the angle BEF; therefore the 
whole angle ADB is equal to the whole angle AEB. 
Pror. XVIII. 
The opposite angles of any quadrilateral ABCD in- 
scribed-in a circle, are together equal to two right an- 
gles. 
THEOR. 
‘Draw the diagonals AC, BD. In the segment ABCD, 
the angle ABD=ACD;; and in the: segment’ CBAD, 
the angle CBD=CAD, (17.) ; therefore the whole an- 
gle ABC is equal to the sum ACD4CAD; and add- 
ing’ ADC, we get the sum ABC+ADC equal to 
the sum ACD 4+ CAD 4 ADC; now these three 
angles are the angles of the triangle ADC, and 
therefore equal to two right angles, (24. 1.): There- 
fore the sum of the angles ABC, ADC is equal to two 
right angles. In like manner it may be demonstrated, 
that the sum of the angles BAD, BCD is equal to two 
right angles. 
Prov. XIX. Tneor. 
An angle ABD in a semicircle is a right angle; an 
angle BAD in a‘segment greater than a semicircle, is 
less than a right angle; and an angle BED in a seg- 
ment less than a semicircle, is greater than a right an- 
gle. 
Produce AB to F, and draw BC to the centre; and 
because CA=CB, the angle CBA=CAB, (12. 1.), in 
like manner, because CD=CB, the angle CBD=CDB, 
therefore the sum CBA4+CBD=CAB+4CDB;; that is, 
ABD=CAB-+-CDB ; but this last sum is equal to the 
angle DBF, (24. 1.) therefore the angle ABD=DBF. 
Hence-each isa right angle, (9. def. 1.) ; therefore the 
angle ABD in a semicircle is a right angle. 
And because in the triangle ABD the angle ABD is 
a right.angle; therefore BAD, which is manifestly in 
a segment-less than a semicircle, is less than a right 
angle, (2. Cor. 24. .1.). Again, because ABED is a qua- 
drilateral in a-circle, we have A+ E= two right an- 
gles, (18.) ; but A is less ‘than a right angle ; therefore 
&, which is in a segment greater than .a semicircle, is 
greater than a right angle. 
Prop. XX. Tueor. 
The angle BAE, contained by a tangent AE toa cir= 
cle, and a chord AB drawn from the point of contact, 
is equal to the angle AGB in the alternate segment. 
Let the diameter ACF be drawn, and GF be joined ; 
and because FGA and FAE are right angles, (19. and 
9.), and of these, FGB a part of the one, is equal FAB 
a part ofthe other, (17.). The remainders BAE, BGA 
are equal. 
GEOMETRY, 
Problems relating to the two First Sections. © 
Geometrical problems, like geometrical theorems, — 
may be multiplied without end. They are divided into 
classes, according to the nature of the lines employed 
in their solution. The most simple, called Plane Pron 
blems, require only straight lines, and circles, and they _ 
may be all ultimately reduced to three. prin 
1. To draw a straight line from one given point to 
another given point, = F.C WOE GD 
2. To prolong’a straight line. | ") walieoils 
8. To describe a circle on any point as a centre, with 
any given radius. a i oecti oie: ati 
hese are resolved by the mechanical contrivances of 
a ruler and compasses, which are commonly known ; — 
and they belong rather to mechanics than to geometry, 
which does not teach how to resolve them, but takes 
for granted that the manner of solving them is known. — 
This assumption is formally made in the postulates, (be- 
ginning of Sect. I.) Nags kse db MR) fy 
The elements of geometry treat only of the more 
simple plane problems, to which the complex may be 
cedvinwile Some of these are now tobe considered. 
Prostem I. 
From the greater CD of two unequal lines AB, CD, Fi 
to cut off a part equal to the less. 
From C asia centre, with “a radius ‘equal to AB, let 
the circumference of a circle be described cutting CD 
in E, (3d Postulate), and thething isdone, | 
Pros. II. 
At a given point A to draw a line equal to a given i 
line CD. . vi 
Draw the indefinite line AF (Post. 
1. and 2.); from 
which cut off AB=CD, (Prob. 1.) ‘ 
Pros. III. at 
To bisect a given straight line AB,| that is!'todivid 
it into two equal parts. aad 
i 
On A and B as centres, with any radius greater than 
the half of AB, deseribe two-ares of circles to meet in 
D, (Post. 3. and Prop. 12.), this point will be equally — 
distant from A and B. In like manner, find another 
int E, either on the same or on the other side of the — 
ine, which may be equally distant from A and B. 
Through D and E draw a straight line to meet AB in © 
C; the point C will be the middle of AB. a4 
For the points D and E are ina straight line, per- 
ndicular to the middle of AB (17. 1.) ; therefore the — 
a = is that perpendicular, and C is the middle of — 
e line, ars 
Prop.IV. Sa f , 
To draw a perpendicular to ‘a given line BC from a F 
given point A in that line. } 10 eae 
Take the points B and C at equal distances from A ; 
and on B and.C as centres, with a radius greater than 
BA, describe ares to meet in D. Draw AD, whieh 
will be the perpendicular required. Pac 
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