IF

PROPOSITION XII. THEOREM XI.

F two circles (DAM, GAN,) touch each other externally; the straight line (BC), which joins their centers, fhall pafs thro' the point of contact (A).

This ftraight line, which joins the centers, will pafs otherwife,
as BDGC.

Preparation.

Draw from the centers B & C to the point of contact A, the
rays BA, CA.

BECAUSE

AUSE BA is — to BD, & CA — to CG (D. 15. B. 1.).

Pof. I.

1. The ftraight lines BA + CA are to the ftraight lines BD + CG. Ax. 2. B. 1. And if the part DC be added to the ftraight lines BD + CG.

2. BD + DG+CG, or the base BC of the A BAC is > the two fides BACA, which is impoffible.

3. Therefore, the ftraight line BC, which joins the centers, will pafs thro' the point of contact A.

P. 20. B. 1.

Which was to be demonstrated.

B

MN

A

F

Tw

PROPOSITION XIII THEOREM XII.

WO circles (ABCD, AGDF or ABCD, BECH,) which touch each other; whether internally; or externally cannot touch in more points than one.

Hypothefis.

1. O ABCD touches O AGDF internally. II. O ABCD touches O BECH externally.

If not,

1. Either the

Thefis.

The ABCD, AGDF, or ABCD,

BECH, touch only in one point.

DEMONSTRATION.

ABCD, AGDF, touch each other internally

in more points than one, as in A & in D.

2. Or the ABCD, BECH, touch each other externally in

more points than one, as in B & in C.

I. Preparation.

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

P. 1. B. 3.

2. Thro' the centers, draw the line MN, & produce it to the O. Pof. 1. & 2.

BECAU

ECAUSE MN joins the centers M & N of the two © ABCD,

AGDF, (Prep. 2.) which touch on the inside (Sup. 1.).

1. This ftraight line will pafs thro' the points of contact A & D. But AM is to MD (I. Frep. 2. & D. 15. B. 1.).

2. Therefore, the ftraight line AM is > ND, & AN is much > ND.
But fince AN is to ND (I. Prep. 2. & D. 15. B. 1.).

3. The line AN will be > ND & to ND; which is impoffible.
4. Confequently, 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 ABCD, BECH,
draw the straight line BC.

BECAUSE the line BC joins the two points B & C in the ○ of the

O ABCD, BECH, (II. Prep.).

ABCD

1. This ftraight line will fall within the two O ABCD, BECH.
But the BECH touching externally the ABCD (Sup. 2.).
2. BC, drawn in the O BECH, will fall without the O ABCD.
3. Confequently, BC will, at the fame time, fall within the
(Arg. 1,), & without the fame (Arg. 2.); which is impoffibe.
4. Wherefore, two O ABCD, BCEH, which touch each other exter-
nally, cannot touch each other in more points than one.

Which was to be demonstrated.

IN

BE

PROPOSITION XIV. THEOREM XIII.

N a circle (ABED) the equal chords (AB, DE,) are equally distant from the center (C); & the chords (AB, DE,) equally diftant from the center (C), are equal to one another.

Hypothefis.

The chords AB, DE, are equal.

TH

1. Find the center C of the O ABED.

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

P. 1. B. 3.

P. 12. B. I.

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

DEMONSTRATION.

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

1. Their halves FB, GE, are alfo equal. 2. Confequently, the of FB is

to the of GE.

But becaufe of CB of CE (Prep. 3. & P. 46. Cor. 3.). 3. It follows, that of FB of FC is to of GE + of CG, Therefore the equal of FB & of GE (Arg. 2.) being taken away. the of GC (Ax. 3. B. 1.), or FC = GC. DE, are equally distant from the cenWhich was to be demonstrated.

4. The of FC will be

5. Confequently, the chords AB, ter C of the O ABED.

BECAU

ECAUSE FC

DEMONSTRATION.

Ax 7. B. 1.

P. 46, B. 1.
Cor. 3.

SP. 47. B. 1.
Ax. 1. B. 1.

GC (Hyp. & D. 4. B. 3.), & CB = CE

S P. 46. B. 1.

S

(Prep. 3. & D. 15, B. 1.).
1. The of FC the of CG, & the of CB the of CE.
2. Confequently, of FC+ of FB, of CG + of GE.
Therefore, the equal of FC & of CG (Arg. 1.) being taken away.
3. The of FB will be the of GE (Ax. 3. B. 1.) or FBGE. (§
4. Confequently, FE, GE, being the femichords (Prep. 2. P. 3. B. 3.),
the whole chords AB, DE, are alfo to one another.

Which was to be demonstrated.

Cor. 3.
P. 47. B. 1.
Ax, 1. B. 1.
P. 46. B. 1.
Cor. 3.
Ax. 6. B. 1.

THE

PROPOSITION XV. THEOREM XIV.

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

P. 12. B. I.

CM

1. From the center C let fall upon HI & FK the LCG, CN.
2. From CN, the greatest of those L, take away a part

to CG.

3. At the point M in CN, erect the DM & produce it to E.
4. Draw the rays CD, CF, CE, CK.

DEMONSTRATION.

BECAUSE the ftraight lines CD, CE, CA, CB, are — to one

another (Prep. 4. & D. 15. B. 1.).

1. It follows, that CD + CE is to CA + CB or AB. But CD+CE is > DE (P. 20. B. 1.).

2. Wherefore, AB is alfo > DE or > HI, becaufe HI

(Prep. 2.).

3. It may be proved after the fame manner, that AB is also > FK.

P. 3. B. 1.
P. 11. B. 1.
Pof. 1.

Ax. 2. B. 1.

Which was to be demonftrated. I.

Moreover, the ▲ CDE, CFK, having two fides CD, CE, = to the two fides CF, CK, each to each (Prep. 4. & D. 15. B. 1.), & the \ DCE > \ FCK (Ax. 8. B. 1.).

4. The bafe DE will be > the base FK. 5. And because HI is to DE (Prep:

HI is alfo > FK.

Which was to be demonftrated. II,

THE

PROPOSITION XVI. THEOREM XV.

H E ftraight line (AB) perpendicular to the diameter of a circle (AHD) at the extremity of it (A), falls without the circle; & no ftraight line can be drawn between this perpendicular (AB) & the circumference from the extremity, fo as not to cut the circle; alfo the angle (HAD) formed by a part of the circumference (HEA) & the diameter (AD), is greater than any acute re&tilineal angle; & the angle (HAB) formed by the perpendicular (AB) & the fame part of the circumference (HEA), is lefs than any acute rectilineal angle. Hypothefis.

I. AB is drawn perpendicular to the extremity A of the diametr.

II. And makes with the arch HEA the mixtilineal HAB.

III. The diameter AD makes with the fame arch HEA the mixtilineal Y HAD.

I. If not,

Thefis.

I. The LAB falls without the O AHD. II. No ftraight line can be drawn between the AB & the arch HEA. III. The mixtilineal \ HAD is > any acute rectilineal .

IV. The mixtilineal V HAB is < any acute redilineal V.

DEMONSTRATION.

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

Preparation.

From the center C to the point of section E, draw the ray CE. Pof. 1.

BECAUSE CA is to CE (D. 15. B. 1.).

1. The V CAE will be to the V CEA.

P. 5. B. 1.

2. And because the VCAE is a L (Sup.); V CEA is also a L. 3. Wherefore, the two V CAE+CEA, of the A AEC will not be <2; which is impoffible.

Ax. 1. B. 1.

P. 17. B. 1.

4. Therefore, the L AB falls without the circle.

Which was to be demonftrated. I.

P