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PROP. XXXV. THEOR.

If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E: the rectangle contained by AE, EC shall be equal to the rectangle contained by BE, ED.

If AC, BD pass, each of them, through the centre, so that E is the centre, it is evident that AE, EC, BE, ED, being all equal, the rectangle AE, EC is equal to the rectangle BE, ED.

But let one of them, BD, pass through

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the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E: Bisect BD in F, which must be the centre of the circle ABCD, and join AF: Then, because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, AC is bisected in E (3. 3) : And because the straight line BD is divided into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED, together with the square of EF, is equal to the square of FB (2. 5), that is, to the square of AF: But the squares of AE, EF are equal to the square of AF; therefore the rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF: Take away the common square of EF, and the remainder, the rectangle BE, ED, is equal to the remainder, the square of AE, that is, to the rectangle AE, EC.

Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles: bisect, as before, BD in F, the centre of the circle; join AF, and from F draw FG perpendicular to AC: Then AC is bisected in G; therefore the rectangle AE, EC, together

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with the square of EG, is equal to the square of AG: 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: But the squares of EG, GF are equal to the square of EF, 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; therefore the rectangle AE, EC, together with the square of EF, is equal to the rectangle BE, ED, together with the square of EF: Take away the common square of EF, and the remainder, the rectangle AE, EC, is equal to the remainder, the rectangle BE, ED. Lastly, let neither of the straight lines AC, BD pass through the centre: take the centre F, and through E, the intersection of the straight lines AC, BD, draw the diameter GEFH: Then, because, as

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has been shewn, the rectangle GE, EH is equal to the rectangle AE, EC, and also to the rectangle BE, ED, therefore the rectangle AE, EC is equal to the rectangle BE, ED.

Wherefore, If two straight lines &c.

Q. E. D.

PROP. XXXVI. THEOR.

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

Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts the circle, and DB touches it: the rectangle AD, DC shall be equal to the square of DB.

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First, let DCA pass through the centre E, and join EB: Then EBD is a right angle (3. 18): And because AC is bisected in E, and produced to D, the rectangle AD, DC, together with the square of EC, is equal to the square of ED (2.6): But EC 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; 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 remainder, the rectangle AD, DC, is equal to the square of DB.

Next, let DCA not pass through the centre of the circle ABC; take the centre E, and draw EF perpendicular to AC, and join EB, EC, ED: Then because EF, which passes through the centre, cuts AC, which does not pass through the centre, at right angles, it also bisects it; therefore AF is equal to FC: And because AC is bisected in F, and produced to D, the rectangle AD, DC, together with the square of

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FC, is equal to the square of FD: 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: But the square of CE is equal to the squares of CF, FE, because CFE is a right angle, and the square of DE is equal to the squares of DF, FE; therefore the rectangle AD, DC, together with the square of CE, is equal to the square of DE: And CE is equal to BE; therefore the rectangle AD, DC, together with the square of BE, is equal to the square of DE: But the squares of DB, BE are equal to the square of DE, because DBE is a right angle; therefore the rectangle AD, DC, together with the square of BE, is equal to the squares of DB, BE: Take away the common square of BE; therefore the remainder, the rectangle AD, DC, is equal to the square of DB. Wherefore, If from any point &c. Q.E. D.

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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, viz. the rectangle BA, AE, 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. THEOR.

If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it, then, 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, the line which meets shall touch the circle.

Let any point D be taken without the circle ABC, and from it let two straight lines, DCA, DB, be drawn, of which DCA cuts the circle, and DB meets it, and let the rectangle AD, DC be equal to the square of DB: DB shall touch the circle.

D

Draw the straight line DE, touching the circle ABC (3. 17); find its centre F, and join FB, FD, FE: Then FED is a right angle (3. 18): And because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square of DE (3. 36): But (hyp.) the rectangle AD, DC is equal to the square of DB; therefore the square of DE is equal to the square of DB, and the straight line DE to the straight line DB: And EF is equal to BF; therefore the two DE, EF are equal to the two DB, BF, each to each, and the base DF is common to the two triangles DEF, DBF-therefore the angle DEF is equal to the angle DBF: But DEF is a right angle; therefore also DBF is a right angle: And BF, 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 (3. 16); therefore DB touches the circle ABC.

Wherefore, If from a point &c.

Q. E. D.

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