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therefore the rectangle AE, EC, together with the square of EG, is equal

to the square of AG; (II. 5) to each of these equals add the square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF,

is equal to the squares of AG, GF; (ax. 2) but the squares of EG, GF, are equal to the square of EF; (1. 47) and the squares of AG, GF, are equal to the square of AF; therefore the rectangle AE, EC, together with the square of EF, is equal

to the square of AF; that is, to the square of FB; but the square of FB is equal to the rectangle BE, ED, together with

the square of EF; (11. 5) therefore the rectangle AE, EC, together with the square of EF, is equal

to the rectangle BE, ED, together with the square of EF ; (ax. 1) take away the common square of EF, and therefore the remaining rectangle AE, EC, is equal to

the remaining rectangle BE, ED. (ax. 3.) Lastly, let neither of the straight lines AC, BD, pass through the centre.

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B

CONSTRUCTION

Take the centre F, (III. 1) and through E, the intersection of the

straight lines AC, DB, draw the diameter GEFH.

DEMONSTRATION

Because the rectangle AE, EC, is equal, as has been shown, to the

rectangle GE, EH; and for the same reason, the rectangle BE, ED, is equal to the same

rectangle GE, EH;

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therefore the rectangle AE, EC, is equal to the rectangle

BE, ED. (ax. 1.)
Wherefore, if two straight lines, &c.

Q. E. D.

PROP. XXXVI.-THEOREM.

If from any point without a circle, two straight lines be drawn, one of which cuts the circle, and the other touches it;

then the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

(References — Prop. 1. 12, 47; II. 6 ; III. 1, 3, 18.) Let D be any point without the circle ABC; and let DCA, DB, be two straight lines drawn from it, of which DCA cuts the circle, and DB touches the same. Then the rectangle AD, DC, shall be equal to the square

of BD. Either DCA passes through the centre, or it does not. First, let it pass through the centre E.

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From the point E to B, draw the straight line EB.

DEMONSTRATION

Then the angle EBD is a right angle; (III. 18) and because the straight line AC is bisected in E, and produced to the

point D, the rectangle AD, DC, together with the square of EC, is equal to the

square of ED; (1, 6) but CE is equal to EB ;

therefore the rectangle AD, DC, together with the square of EB, is equal

to the square of ED; but the square of ED is equal to the squares of EB, BD, because EBD

is a right angle; (1. 47) therefore the rectangle AD, DC, together with the square of EB, is equal

to the squares of EB, BD; (ax. 1) take away the common square of EB; therefore the remaining rectangle AD, DC, is equal to the

square of the tangent DB. (ax. 3.)

Secondly, let DCA not pass through the centre of the circle ABC.

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CONSTRUCTION Take the centre E, (I11. 1) and draw EF perpendicular to AC, (1. 12)

and join EB, EC, ED.

DEMONSTRATION

Because the straight line EF, which passes through the centre, cuts the

straight line AC, which does not pass through the centre, at right angles, it shall likewise bisect it ; (III. 3)

therefore AF is equal to FC; and because the straight line AC is bisected in F, and produced to D, the rectangle AD, DC, together with the square of FC, is equal to the

square of FD; (11. 6) to each of these equals add the square of FE; therefore the rectangle AD, DC, together with the squares of CF, FE,

is equal to the squares of DF, FE; (ax. 2) but the square of ED is equal to the squares of DF, FE, because EFD

is a right angle ; (1. 47) and for the same reason the square of EC is equal to the squares of

CF, FE; therefore the rectangle AD, DC, together with the square of EC, is

equal to the square of ED; (ax. 1)

but CE is equal to EB; therefore the rectangle AD, DC, together with the square of EB, is equal

to the square of ED, but the squares of EB, BD, are equal to the square of ED, because

EBD is a right angle ; (1. 47) therefore the rectangle AD, DC, together with the square of EB, is

equal to the squares of EB, BD; take away the common square of EB; therefore the remaining rectangle AD, DC, is equal to

the square of DB. (ax. 3.) Wherefore, if from any point, &c.

Q. E. D.

Cor. — If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another.

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Thus the rectangle BA, AE, is equal to the rectangle CA, AF, for each of them is equal to the square of the straight line AD, which touches the circle.

PROP. XXXVII.-THEOREM.

If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle be equal to the square of the line which meets it; then that line shall touch the circle.

(References Prop. I. 8; III. 1, 16 cor., 17, 18, 36.) Let any point D be taken without the circle ABC, and from it let two straight lines, DCA and DB, be drawn, of which DCA cuts the circle, and DB meets it.

If the rectangle AD, DC, be equal to the square of DB,
Then DB shall touch the circle.

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CONSTRUCTION Draw the straight line DE, touching the circle ABC in E; (III. 17) find F, the centre of the circle, (III. 1) and join FE, FB, FD. ·

DEMONSTRATION

Then FED is a right angle; (111. 18)
and because DE touches the circle ABC, and DCA cuts it;

the rectangle AD, DC, is equal to the square of DE; (III. 36) but the rectangle AD, DC, is, by hypothesis, equal to the square of

DB;

therefore the square of DE is equal to the square of DB; (ax. 1)

and the straight line DE equal to the straight line DB; and FE is equal to FB. (1. def. 15.) Hence in the two triangles DEF, DBF, because DE is equal to DB, and EF to FB, the two sides DE, EF, are equal to the two DB, BF, each to each; and the base FD is common to the two triangles ;

therefore the angle DEF is equal to the angle DBF; (1. 8) but DEF was shown to be a right angle,

therefore also DBF is a right angle ; (ax. 1) and FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from

the extremity of it, touches the circle;' (111. 16 cor.)

therefore DB touches the circle ABC. Wherefore, if from a point, &c.

Q. E. D.

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