IF

F in a circle (DAEB) two chords (AB, DE,) cut one another; the rectangle contained by the fegments (AF, FB,) of one of them, is equal to the rectangle contained by the fegments (DF, FE,) of the other.

Hypothefis.

I. AB, DE, are two chords of the fame O DAEB.
II. And thefe chords cut one another in a point F.

Thefis.

The Rgle AF. FB is = to the Rgle DF. FE.

CASE I. If the two chords pass thro' the center F of the O. Fig. 1.

THEN,

DEMONSTRATION.

HEN, the ftraight lines AF, FB, DF, FE, are to one

another.

2. Confequently, the Rgle AF. FB is to the Rgle DF. FE.
CASE II. If one of the chords AE, paffes thro' the center &
cuts the other DE which does not pass thro' the
center at (Fig. 2.).
Preparation.

D. 15. B. 1
Ax. 2 B. 2

Draw the ray CE.

DEMONSTRATION.

BECAUSE

ECAUSE the ftraight line AB is cut equally in C & unequally

Pof. I.

(P. 47. B. 1.).

to the of FE+ the of CF.

2. From whence it follows, that the Rgle AF. FB +the□ of CF is

3. Confequently, the Rgle AF. FB is to the of FE.

And fince DF is to FE (P. 3. B. 3.), or DF. FE to the

=

Ax. 1. B. 1.

Ax. 3. B. 1.

Ax, 1. B. 1.

135

BECAUS

DEMONSTRATION.

ECAUSE DH is to HE (Prep. 1. & P. 3. B. 3.).

1. The Rgle DF. FE+ the

2. Wherefore, the Rgle DF. FE+

theof DH+ of CH.

P. 12. B. 1. Pof. 1.

of FH is

to the

of DH.

P. 5. B. 2.

Ax. 2. B. 1.

Moreover, the Rgle AF. FB +□of CF being to the fame

of CB (P. 5. B. 2.).

4. The Rgle DF. FE +□ of CF is also to the Rgle AF. FB + of CF.

5. Or taking away the common of CF, the Rgle DF. FE is to the Rgle AF. FB.

CASE IV.

If neither of the chords AB, DE, paffes thro'
the center (Fig. 4.).

Preparation.

Thro' the point F, draw the diameter GH.

ECAUSE each of the Rgles AF. FB & DF. FE is =

Rgle GF. FH (Cafe III.).

Ax. 1. B. 1.

Ax. 3. B. 1.

Pof. 1.

to the

Ax. 1. B. 1.

Which was to be demonstrated,

1. These Rgles AF. FB & DF. FE are alfo to one another.

I

F from any point (E) without a circle (ABD) two ftraight lines be drawn, one of which (DE) touches the circle, & the other (EA) cuts it; the rectangle contained by the whole fecant (AE), & the part of it (EB) without the circle, fhall be equal to the fquare of the tangent (ED).

Hypothefis.

1. The point E is taken without the O ABD. II. From this point E, a tangent ED & a fecant EA, have been drawn.

Thefis.

The Rgle AE. EB is = to the

☐ of ED.

CASE I. If the fecant AE paffes thro' the center (Fig. 1.).

Preparation.

From the point of contact D, Draw the ray CD.

Pof. 1.

DEMONSTRATION.

THE

HE ray CD is then to the tangent ED.

And because the straight line AB is bifected in C, & produced to the point E.

the of DE+the□ of CD

2. The Rgle AE. EB + the of CB is to the of CE.
Moreover, the of CE being alfo to
(P.47.B.1.), or to the of DE+ the

3. The Rgle AE. EB + the of CB is
of CB.

of CB (P. 46. Cor. 3. B. 1.).
to the of DE+the

The of CB being taken away from both fides. 4. The Rgle AE. EB will be to the of DE.

CASE II. If the fecant AE does not pafs thro' the center.

Preparation.

1. Let fall from the center C upon AE, the L CF.
2. Draw the rays CB, CD, & the straight line CE.

P. 18. B. 3.

P. 6. B. 2.

Ax. 1. B. 1.

Ax. 3. B. 1.
Fig. 2.

P. 12. B. L
Pof, 1.

BECAUSE

ECAUSE the ftraight line AB is bifected in F (Prep. 1. & P. 3.

B. 3.) and produced to the point E.

=

P. 6. B. 2.

of FB+

of FC is = of CE.

SAx. 2. B. 1.
P. 47. B. 1.

1. The Rgle AE.EB+ of FB is to the of FE.
2. Confequently, the Rgle AE. EB+
to the of FE+ of FC, or is to the
But fince the of DE+ of CD is to the of CE, and
the of FB of FC is to the of CB (P. 47. B. 1.), or is
to the of CD (D 15. & P. 46. Cor. 3. R. 1.)
3. The Rgle AE.EB+ of CD is to the of DE+ of CD.
to the of DE.

4. Confequently, the Rgle AE.EB is

Which was to be demonstrated.

COROLLARY I.

Ax. 3. B. 1.

IF (fig. 3.) from any point (Ė) without a circle (ADBF), there be drawn several

ftraight lines (AE, EG, &c). cutting it in (B & F, &c): the rectangles contained by the whole fecants (AE, GE), and the parts of them (EB, EF) without the circle, are equal to one another; for drawing from the point (E) the tangent (ED), these rectangles will be equal to the fquare of the fame tangent (ED).

IF

COROLLARY II.

F from any point (E), without a circle (ADBF), there be drawn to this circle two tangents (ED, EC), they will be equal to one another, because the fquare of each is equal to the fame rectangle (AE.EB).

PROPOSITION XXXVII. THEOREM XXXI.

IF from a point (E), without a circle (ADH), there be drawn two straight

lines, one of which (AE) cuts the circle, and the other (ED) meets it; if the rectangle contained by the whole fecant (AE) and the part of it without the circle (EB), be equal to the fquare of the line (ED) which meets it: the line which meets fhail touch the circle in D.

DEMONSTRATION,

BECAUSE the Rgle of AE.EB is to the

of ED (Hyp. 3.) and

P. 17. B. 3. Pof. 1.

{Cor. 3.

the Rgle AE.EB is alfo to the of EH (Prep. 1 & P. 36. B. 3 ) 1. The of ED is: to the of EH (Ax. 1. B. 1.) or ED = EH.P.46. B. 1. And moreover, fince in the A CDE, CHE, the fide CD is to the fide CH (D. 15. B. 1), and CE is common to the two A. 2. The CDE is to the V CHE.

3. Wherefore, V CHE being a L (Prep. 1. & P. 18. B. 3), \ CDE is alfo a L.

4. And the ftraight line ED touches the ADH in the point D.

P. 8. B. 1.

Ax. 1. B.1. P. 16. B. 3. Cor. 3.