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PROPOSITION IV. PROBLEM IV.

To infcribe a circle (EFG) in a given triangle (ABD).

1. Bifect the V BAD, BDA by the ftraight lines AC, DC pro-
duced until they meet one another in C.

2. From the point C let fall upon AD the LCE.

3. And from the center C at the diftance CE, defcribe the O
EFG.

Preparation.

From the point Clet fall upon AB & DB the LCF, CG.

BECAUSE

DEMONSTRATION.

ECAUSE in the AAFC, ACE, the V FAC is to the V

CAE (Ref. 1), V CFA

AC common to the two A.

to V CEA (Prep. Ref. 2 & Ax. 10. B. 1); &

to CE.

2. The ftraight line CG is

3. Confequently, the straight lines CF, CE, CG are to one another;
and the defcribed from the center C at the diftance CE will
alfo pafs thro' the points F & G.

And fince the fides AD, AB, DB are at the extremities E, F,
G, of the rays CE, CF, CG (Ref. 2 & Prep.).

4. These fides will touch the O in the points E, F, G.
5. Therefore the EFG is infcribed in the ▲ ABD.

P. 9. B. 1.
P. 12. B. 1.

Pof. 3.

P. 12. B. I.

P. 26. B. 1.

Ax. 1. B. 1.
D. 15. B. 1.

P. 16. B. 3.
Cor.

D. 5. B. 4.

Which was to be done.

To

PROPOSITION V. PROBLEM V

O describe a circle (ABDH), about a given triangle (ABD).

Given.

The A ABD.

Refolution.

Sought.

The ABDH defcribed about
O
the AABD.

1. Bifect the fides AB, DB in the points E & F.

2. At the points E & F in AB, DB, erect the L EC, FC,
produced until they meet in C.

3. And whether the point C falls within (fig. 1.) or without
(fig. 3.) or in one of the fides (fig. 2). of the ▲ ABD,
from the center C at the diftance CA defcribe the
O ABDH.

OECAUSE in the ▲ AEC, BEC, the fide AE is (Ref. 1), EC common to the two A, & VAEC = to Ax. 10. B. 1).

1. The ftraight line CB is to CA.

It may be demonftrated after the fame manner, that

2. The ftraight line CB is to CD.

P. 10. B. I.

P. 11. B. I.

Pof. 3.

Pof. 1.

to the fide EB VBEC (Ref. 2

P. 4. B. 1.

3. Confequently, the ftraight lines CA, CB, CD are to one another; and the ABDH described from the center C at the distance ( Ax. 1. B. 1. CA, will pass alfo thro' the points B & D.

D. 15. B. 1.

D. 6. B. 4.

Which was to be done.

4. Therefore this O ABDH is described about the ▲ ABD.

I

COROLLARY

F the triangle ABD be acute angled, the point C falls within it (fig. 1); but if this triangle be obtufe angled, the point C falls without it (fig. 3); in fine if it be a right angled triangle, the point C is in one of the fides (fig. 2).

T

PROPOSITION VI. PROBLEM VI

O inscribe a Square (ABDE), in a given Circle (ABDE).

Given

The O ABDE.

Sought.

The ABDE infcribed in this O.

To

Refolution.

1. Draw the Diameters AD, BE, fo as to cut each other at L.
2. Join their Extremities by the straight Lines AB, BD, DE, EA.

DEMONSTRATION.

BECAUSE in the A ABC, DBC the fide AC is to the fide

CD (Ref. 1. & D. 15. B. 1.), BC common to the two A, & the
VBCA to V BCD (Ref. 1. & Ax. 10. B. 1).

1. The ftraight Line AB is to BD.

=

P. 11. B. 1.
Pof. 1.

P. 4. B. 1.

to AB. to one

Ax. 1. B. 1,

It may be demonftrated after the fame manner, that 2. The straight line BD is to DE, DE to EA & EA 3. Confequently, the straight lines AB, BD, DE, EA are another, or the quadrilateral figure ABDE is equilateral. And because each of the V ABD, BDE, DEA, EAB is placed in a femi.O.

4. Thefe V are L, & the equilateral qadrilateral, figure ABDE is rectangular.

5. Wherefore this quadrilateral figure i sa fquare infcribed in the S O ABDE.

P. 31. B. 3. D. 30. B. 1. D. 3.B.4

Which was to be done.

To

PROPOSITION VII. PROBLEM VII.

O describe a Square (ABCD) about a given Circle (HEFI).

4.

5.

But EID of the Pgme. AI being a L (Ref. 2).

7. The VA, which is diagonally oppofite to it, is alfo a L.

It may be proved after the fame manner, that

8. The VB, C, D are L.

9. Confequently, there has been defcribed about the quadrilateral figure ABCD equilateral (Arg. 6.) & (Arg. 7.8); or a fquare.

6. And fince El is to HF (D. 15. B. 1.), the ftraight lines AD, BC, AB, DC are equal.

1. Draw the diameters EI, HF fo as to cut each other at L
2. At the Extremities H, E, F, I of those diameters erect the
LAD, AB, BC, CD.

DEMONSTRATION.

1. THE lines DA, AB, BC, CD, are tangents of the HEFI.

2. And the straight line AD, is Plle. to EI, as alfo the straight line
BC; because the HGE+ GHA, & V FGE + GFB are
2 L (Ref. 1. & 2).

3. Confequently, AD is Plle. to BC, likewife AB, HF, DC are Plles.
Wherefore the quadrilateral figures AI, EC, AF, HC, AC are Pgmes.
From whence it follows, that the straight lines AD, EI, BC, also AB,
HF, DC, are to one another.

P.11. B.I.

P.11. B.1.

SP.16. B.3.
Cor.

to

P.28. B.1.

P 30. B.1.
D.35. B.1.

P.34. B.1.

Ax.1. B.1.

P.34. B.1.

HEFI a rectangular

SD. 4. B.1.
D.30. B.1.

Which was to be done.

To

O infcribe a circle (ABDE) in a given fquare (FGHI).

2. Thro' the points of section A & B, draw AD Plle. to FG or
IH & BE Plle. to FI or GH.
3. From the center C, where AD, BE interfect each other, at
the distance CA defcribe the

BECAUSE

ABDE.

DEMONSTRATION.

ECAUSE the figures FE, BH, FD, AH, FC, AE, BD, CH are

But the whole lines FI, FG being equal (D. 30. B. 1.) and FA, FB being the halves of thofe ftraight lines (Ref. 1).

P.10. B.1.

P.31. B.1.
Pof.3.

Pgies. (Ref. 2. & D. 35. B. 1).

1. The straight line FA is

to BC & FB to AC.

P.34, B.I.

2. The ftraight line FA is

to FB.

Ax.7. B.1.

3. Confequently, BC is alfo

to AC; and likewife AC is to CE &

BC to CD.

Ax.1. B.I.

4. From whence it follows, that the ftraight lines AC, BC, CE, CD
are to one another, and the defcribed from the center
Cat the diftance CA; paffes alfo thro' the points B, D, E.
But the DAF, EBG, ADH, BEI being L (P. 34. B. 1.) as being
interior oppofite to the LGFA, HGB, ÎHD, FIE (D. 30. B. 1).
The ftraight lines FI, FG, GH, HI are tangents of the
ABDE.

6. Wherefore this is infcribed in the square FGHI.

Which was to be done.