PROPOSITIONS.

PROP. 1.-PROB.

Into a given circle to fit exactly a right line equal to a given right line, which is not greater than the diameter of the circle.

SOL. 1, III. To find the centre of a given circle.

3, I. From the greater of two given lines to cut off a part equal to the less.

Pst. 3 and 1. A circle may be described from any centre at any distance from that centre. A st. line may be drawn from any one point to any other point.

DEM. 15, I. A circle is a plane figure contained by one line, which is called the circumference, and is such that all st. lines drawn from a certain point within the figure to the circumference, are equal to one another.

Ax. 1. Things which are equal to the same thing, are equal to one another.

Def. 7, IV. A st. line is said to be fitted exactly into a circle, or to be applied in it, when the extremities of it are on the circumference of the circle.

Let ABC be the given;
and D the st. line CB
diam. of ABC;
in ABC to place a
st. line = D.
Find E the cen. of ABC,

and draw any diam., BC,
through it;

if BCD, the required thing is done;
but if not, and BC is > D;

from CB cut off CF = D;

and from C, with CF, desc. O GFA, and join CA;

then CA is the line required.

. C is cen. of O GFA .. CA = CF; but CFD, .. D = CA.

in ABC, a st. line has been placed, CA = the given st. line D.

Q. E. F.

USE AND APP.-I. Within a given O, ABC, to place a line of a given length, D, not greater than the diam. of the given, which line shall pass through A, a given point in the Oce.

II. To draw that diam. of a which shall pass at a given distance, N, from a given point A

C. 1 Pst. 3. & 1, IV.| With OA = N, desc. a ○ ABC, and in it place AK= N ; 2 Sol. then KO produced to L is the diam. required.

KL is a diam., and AK = A0=N;
.. KL, a diam., passes at the given distance from A.

PROP. 2.-PROB.

In a given circle to inscribe a triangle equiangular to a given triangle.

CON. 17, III. To draw a st. line from a given point, either without or in the circumference, which shall touch a given circle.

23, I. At a given point in a given line to make a rectil. ▲ equal to a given rectil. . Pst. 1.

DEM. 32, III. If a st. line touches a O, and from the point of contact a st. line be drawn cutting the circle, the s which this line makes with the line touching the Ŏ, shall be equal to the s which are in the altr. segs. of the O. Ax. 1.

Cor. 3, 32, I.

If two As have two

two s of the other, then the third
third of the other.

s of the one respectively equal to of the one shall be equal to the

Def. 3, IV. A rectil. figure is said to be inscribed in a circle when each angular point of the inscribed figure touches the Oce of the circle.

HAC = < DEF;

3 23, I.

4 Pst. 1. 5 Sol.

join BC;

and at A, in AH, make / GAB = ▲ DFE;

then A ABC is the ▲ required.

HAG is a tang., and AC from A cuts the ○; Z HACZ ABC in the alternate seg.; HAC = ▲ DEF; .. / ABC= / DEF; In like manner ACB = / DFE;

3 C. 2. Ax. 1. but

4 Sim.

5 Cor. 3.32,I... rem. ▲ BAC rem. / EDF;
6 Def. 3, IV. Hence ▲ ABC is eq. ang. with ▲ EDF,

is inscribed in the O.

Q. E. F.

and

SCH.-The Analysis of a problem is a very useful exercise, and, that the learner may become accustomed to the method, some examples will be given. Thus, of Prop. 2, setting out with the admission that the A ABC has its angles respectively equal to the s D, E, F, the Analysis will be

Through the point A draw A GH, a tangent to the

;

then, . CAH ▲ ABC= E, .. the line AC is given in position; and, being cut by the Oce, the point C is given.

In the same way it will appear that the point B also is given;

and the three points, A, B, C, are given, .. their junction forms A ABC, inscribed in the circle.

USE AND APP.-An eq. lat. A ABC, being inscribed in a circle, and through the angular points A, B, C, tangents, DE, EF, FD, being drawn, these tangents will also form an eq. lat. ▲, DEF, the area of which is four times that of the inscribed eq. lat. ▲.

54, I.

6 26, I. 73, III.

8 Sim.

= DC, and

LOCD;

OBD

B

A

.. A OBD = A OCD, ▲ BOD = COD,
and BDO / CDO;

i. e. Zs BOC and BDC are each bisected by DO;
also DO bisects the line BC;

..DO bis. BC at rt. s, and passes through the vertex A.
So, EO bis. AC, and passes to the vertex B;

and FO bis. AB, and passes to the vertex C.

9 Cor. 5, 1. 32, I. Now Z OBC = of art. ;.. 4 DBC = } of a rt. Z.;

10 Sim.

11 Cor. 6, I.

12 Sim.

13 Conc.

14 D. 11, 12.

15

16

17 Conc.

So s DCB, BDC, each

.. A BDC is eq. lat. and

of a rt. L,

ДАВС;

So AS ACE, ABF are each eq. lat. and ▲ ABC; .. A DEF 4 A ABC.

Also DE 2 DC 2 BC;

=

EF 2 AE2 AC= 2 BC;

and FD 2 FB 2 AB - 2 BC;
.. A DEF is equilateral.

Q. E. F.

PROP. 3. PROB.

About a given circle to circumscribe a triangle equiangular to a given triangle.

SOL. Pst. 2. 1, III. 23, I. 17, III.

DEM. 18, III. If a st. line touches a O, the st. line drawn from the
centre to the point of contact, shall be perpendicular to the line touching
the circle.
Cor. 1. 32, I. All the interior s of any rectil. fig., together with four
rt. s, are equal to twice as many rt. s as the figure has sides.
Ax. 3, I. If equals be taken from equals the remainders are equal.
13, I. The s which one st. line makes with another upon one side of
it are either rt. s, or together equal to two rt. s. Ax. 1.

Def. 4, IV. A rectil. fig. is said to be described about a ☹, when each
side of the circumscribed fig. touches the Oce of the O.

H

3 23, I.

to desc. a

eq. ang. to the

A DEF.

Produce EF both

ways to G, H; M

B

21,III.Pst.1 find K the cen. of ○ ABC, and draw KB; at K in BK make ▲ BKA = / DEG, and BKC = / DFH; and through A, B, C, draw LM, MN, NL, tangs. to ABC;

4 17, III.

5 Sol.

D.1 C. 4.

2 C. 2, 3.

318, III.
4 Cor.1,32,I.

5 D 2.
6 Ax. 3.
713, I.
8 Ax. 1.

9 C.3. Ax.3.

10 Sim.

then ALMN shall be the A required.

LM, MN, NL, are tangs to

ABC;

and KA, KB, KC, lines from the cen. to A,B,C;
.. thes at A, B, and C are all rt. ≤s.
And the 4s of the qu. lat. AMBK

= 4rt. Zs;

= 2 rt. Zs;

and two of the four, KAM, KBM, are rt. Zs;
..the other two Zs,
AKB, AMB
but s DEG, DEF

= 2 rt. /s;

.. Zs A KB, AM B, = s DEG, DEF;
and ▲ AKB = ▲ DEG, .. / AMB= / DEF.
In like manner ▲ LNM = / DFE;

11 32,I.Ax.3. .. rem. MLN = rem. EDF;

12 Def. 4. IV... ▲ LMN is eq. ang. with ▲ DEF; and is described about the ABC.

Q. E. F.

SCH.-Analysis: We suppose the problem to have been solved, the A LMN being described about the given ABC, so that / L = L D, ME, and N =

F.

Join K the cen. of the to the tang.
In the qu. lat. BKCN, the four
and .8 KBN, KCN, are 2 rt.

s

s; ../s BKC, BNC = 2 rt. s.

But N being given, its supplement / BKC is also given;
consequently, KB and KC, the two radii, are given in position.
Thus, it may be shown that the Z AKB is given, and the line KA
given in position.

The inters. of KA, KB, KC, with the Oce, or the points A, B, C, are given; .. the tangs. MN, NL, LM, are given in position.

Thus the ALMN is given.