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If two circles (DAM, GAN,) touch each other externally ; the straight

PROPOSITION XII. THEOREM XI.
F , )
line (BC), which joins their centers, shall pass thro' the point of contact (A).
Hypothesis.

Thesis.
The straight line BC joins the centers

This straight line BC pases tbro' of ihe iwo O DAM, GAN, which

the point of contact of the touch each other externally in A.

two O. DEMONSTRATION.

If not,

Pos. 1.

This straight line, which joins the centers, will pass otherwise,
as BDGC.

Preparction.
Draw from the centers B & C to the point of contact A, the

rays BA, CA.
Because

ECAUSE BA is = to BD, & CA = to CG (D. 15. B. i.).
1. The straight lines BA + CA are = to the straight lines BD + CG. Ax.2. B. 1.

And if the part DC be added to the straight lines BD + CG.
2. BD + DG +CG, or the base BC of the ABAC is ; the two sides
BA+ CA, which is impossible.

P. 20. B. .
3. Therefore, the straight line BC, which joins the centers, will pass
thro’ the point of contact A.

Which was to be demonstrated.

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Two

PROPOSITION XIII. THEOREM XII. W O circles (ABCD, AGDF or ABCD, BECH,) which touch each other; whether internally; or externally: cannot touch in more points than one. Hypothefis.

Thesis. 1. O ABCD touches AGDF internally. The O ABCD, AGDF, or ABCD, II. O ABCD touches O BECH externally. BECH, touch only in one point, · If not,

DEMONSTRATION.
1. Either the O ABCD, AGDF, touch each other internally

in more points than one, as in A& in D.
2. Or the O ABCD, BECH, touch each other externally in
more points than one, as in B & in C.

1. Preparation.
1. Find the centers M & N of the O ABCD, AGDF.

P. 1. B. z. 2. Thro’ the centers, draw the line MN, & produce it to the O. Pof. 1. & 2.

BECAUSE MN joins the centers M & N of the two o ABCD,

AGDF, (Prep. 2.) which touch on the inside (Sup. 1.). 1. This straight line will pass thro' the points of contact A & D. P. 11, B.3.

But AM is = to MD (1. Prep. 2. & D. 15. B. 1.). 2. Therefore, the straight line AM is > ND, & AN is much > ND. Ax. 8. B. I.

But since AN is = to ND (1. Prep. 2. & D. 15. B. 1.).
3. The line AN will be > ND & = to ND; which is imposible.
4. Consequently, two O ABCD, AGDF, which touch each other in-
ternally, cannot touch each other in more points than one.

II. Preparation.
Thro' the points of contact B & C of the O ABCD, BECH,
draw the straight line BC.

Pof. 1.
ECAUSE the line BC joins the two points B & C in the O of the
O ABCD, BECH, (II. Prep.).
1. This straight line will fall within the two o ABCD, BECH. P. 2. B. 3.

But the O BECH touching externally the O ABCD (Sup. 2.). 2. BC, drawn in the O BECH, will fall without the O ABCD. D. 3. B. 33. Consequently, BC will, at the same time, fall within the O ABCD

( Arg. 1,), & without the same o (Arg. 2.); which is impoflibe. 4. Wherefore, two O ABCD, BCEH, which touch each other externally, cannot touch each other in more points than one.

Which was to be demonstrated.

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I

Cor. 3.

B.1.

PROPOSITION XIV. THEO RE M XIII.
N a circle (ABED) the equal chords (AB, DF,) are equally diftant from
the center (C); & the chords (AB, DE,) equally distant from the center
(C), are equal to one another.
Hypothesis.
CASE I.

Thesis.
The cbords AB, DE, are equal.

They are equally distant from the center C.

Preparation. 1. Find the center C of the O ABED.

P. 1. B.

3. 2. Let fall upon the chords AB, DE, the I CF, CG.

P. 12. B. 1. 3. From the center C to the points E & B, draw the rays CE, CB. Pos. 1.

DEMONSTRATION.
HE chords AB, DE, being = to one another (Hyp.) & bisected
in F&G (Prep. 2. & P. 3. B. 3.).

Ax 7. B.1. 1. Their halves FB, GE, are also equal.

P. 46, B. 1. 2. Consequently, the l] 01 FB is = to the of GE.

But because of CB = of CE (Prep. 3. & P. 46. Cor. 3.).
3. It follows, that of FB + of FC is = to of Ge+of ca: 52;. 47.
Therefore the equal of FB & of GE ( Arg. 2.) being taken away.

Ax. 1. B. 1.
4. The of FC will be = the of GC (Ax. 3. B. 1.), or FC=GC.S. 46. B.1.
5. Consequently, the chords AB, DE, are equally distant from the cen-
ter C of the O ABED.
Which was to be demonstrated. D. 4.

B. 3.
Hypothesis.
CASE II.

Thesis.
The chords AB, DE, are equally distant

These chords are equal. from the center c of the O ABED.

.
ECAUSE FC = GC (Hyp. & D. 4. B. 3.), & CB = CE
(Prep. 3. & D. 15, B. 1.).

SP. 46. B. 1.
1. The of FC =the of CG, & the of CB = the of CE. I Cor. 3.
2. Consequently, of FC + of FB, = of CG +of GE. SP.

47.

B. 1. There!ore, the equal of FC & of CG (Arg. 1.) being taken away. | Ax. 1. B. 1. 3. The o of FB will be the of GE (AX. 3. B. 1.) or FB=-GE. S P. 46. B. 1. 4. Consequently, FB, GE, being the femichords ( Prep. 2. P. 3. B. 3.), I Cor. 3. the whole chords AB, DE, are also = to one another.

Ax. 6. B... Which was to be demonstrated.

Cor. 3.

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B. 1. P.11.

PROPOSITION XV. THEOREM XIV. HE diameter (AB) is the greatest straight line in a circle (AIK); all others that (HI), which is nearer the diameter, is always greater than one (FK) more remote. Hypothesis.

Thesis. 1. AB is the diameter of the O AIK.

1. The diameter AB is each of II. The chord HI is nearer the diame

the chords HI, FK. ter than the chord FK.

II. The chord HI is the chord FK.

Preparation.
1. From the center C let fall upon HI & FK the ICG, CN.

P. 12. B.1. 2. From CN, the greatest of those 1, take

away a part CM = to CG.

P.

3. 3. At the point Min CN, erect the I DM& produce it to E.

B.1. 4. Draw the rays CD, CF, CE, CK.

Pol. I.
DeMONSTRATION.
ECA

,
another (Prep. 4. & D. 15. B. 1.).
1. It follows, that CD + CE is = to CA + CB or AB.

Ax.2. B.1. But CD + CE is > DE (P. 20. B. 1.). 2. Wherefore, AB is also > DE or > HI, because HI = DES D. 4.

B. 3. ( Prep. 2.).

P. 14. B. 3: 3. It may be proved after the same manner, that AB is also > FK.

Which was to be demonstrated. I. Moreover, the ACDE, CFK, having two sides CD, CE, = to the two fides CF, CK, each to each (Prep. 4. & D. 15. B. 1.), & the

V DCE > V FCK (Ax. 8. B. 1.). 4. The base DE will be > the base FK.

P. 24. B.1. 5. And because HI is = to DE (Prep. HI is also > FK. S0.4.B. 3:

P. 14. B. 3: Which was to be demonstrated. II.

Because the fraight lines CD, CE, CA, CB, are = 10 one

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PROPOSITION XVI. THEOREM XV. H E straight line (AB) perpendicular to the diameter of a circle (AHD) at the extremity of it (A), falls without the circle ; & no straight line can be drawn between this perpendicular (AB) & the circumference from the extremity, so as not to cut the circle; also the angle (HAD) formed by a part of the circumference (HEA) & the diameter (AD), is greater than any acute rectilineal angle; & the angle (HAB) formed by the perpendicular (AB) & the fame part of the circumference (HEA), is less than any acute rectilineal angle. Hypothesis.

Thelis. I. AB is drawn perpendicular to the 1. The I AB falls without the O AHD.

extremity A of the diametr. II. No straight line can be drawn beII. And makes with the arch HEA tween the L AB Es the arch HEA. the mixtilineal Y HAB.

III. The mixtilineal Y HAD is > any III. The diameter AD makes with acute rectilineal V.

the same arch HEA the mixtili IV. The mixtilineal V HAB is < any neal Y HAD.

acute redilineal V.

DEMONSTRATION.

1. If not,

BECAUS

The I AB will fall within the O AHD, & will cut it some-
where in E, as AE.

Preparation.
From the center C to the point of section E, draw the ray CE. Pos. 1.
ECAUSE CA is = to CE (D. 15. B. 1.).
1. The V CAE will be = to the V CEA.

P. 5. B. I. 2. And because the V CAE is a L (Sup.); V CEA is also a L. Ax. 1. 3. Wherefore, the two V CAE + CEA, of the A AEC will not be <2 L i which is impossible.

P. 17. B. 1. 4. Therefore, the I AB falls without the circle.

Which was to be demonstrated. I.

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