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BOOK III. is bifected in E, and produced to the point D, the rectangle AD, DC, together with the fquare of EC, is equal to the fquare of ED, and CE is equal to EB: Therefore the rectangle AD, DC, together with the fquare of EB, is equal to the fquare of ED: But the fquare of ED € 47. 1. is equal to the fquares of EB, BD, becaufe EBD is a right angle: Therefore the rectangle AD, DC, together with the fquare of EB, is equal to the squares of EB, BD: Take the common away fquare of EB; therefore the remaining rectangle AD, DC is equal to the square of DB.

f 3.3.

But if DCA does not pafs through

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d1. 3. the centre of the circle ABC, take the centre E, and draw EF e 12. 1. perpendicular to AC, and join EB, EC, ED: And because EF, which paffes through the centre, cuts AC, which does not pafs through the centre, at right angles, it fhall bifect 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 fquare of FC, is equal to the fquare of FD: To each of thefe equals add the square of FE; therefore the rectangle AD, DC, together with the fquares of CF, FE, is equal to the fquares of DF, FE: But the fquare of ED is equal to the fquares of DF, FE, because EFD is a right angle; and the fquare of EC is equal to the squares of CF, FE; therefore the rectangle AD, DC, together with the fquare of EC, is equal to the fquare of ED: And CE is equal to EB; therefore the rectangle AD, DC, together with the fquare of EB, is equal to the fquare of ED: But € 47. 1. the fquares of EB, BD are equal to the fquare of ED, becaufe EBD is a right angle; therefore the rectangle AD, DC, together with the fquare of EB, is equal to the fquares of EB, BD: Take away the common fquare of EB; therefore the remaining rectangle AD, DC is equal to the fquare of DB. Wherefore, if from any point, &c. Q. E. D.

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COR.

COR. If from any point without a circle, there be drawn Book III. 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, AF to the rectangle CA, AE: For each of them is equal to D the fquare of the ftraight line AD which touches the circle.

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IF

PROP. XXXVII. THEOR.

F from a point without a circle there be drawn two ftraight 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; the line which meets shall touch the circle.

Let any point D be taken without the circle ABC, and from it let two ftraight 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 fquare of DB; DB touches the circle.

b b 18.3.

Draw the ftraight line DE touching the circle ABC, find a 17. 3. its centre F, and join FE, FB, FD; then FED is a right angle: And because DE touches the circle ABC, and DCA cuts

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it, the rectangle AD, DC is equal to the fquare of DE: But c 36. 3. the rectangle AD, DC is, by hypothefis, equal to the fquare

of DB: Therefore the fquare of DE

is equal to the fquare of DB; and the
ftraight line DE equal to the straight
line DB And FE is equal to FB,
wherefore DE, EF are equal to DB,
BF; and the bafe FD is common to the B
two triangles DEF, DBF; therefore
the angle DEF is equal to the angle
DBF; but DEF is a right angle, there-
fore allo DBF is a right angle: And
FB, if produced, is a diameter, and the
ftraight line which is drawn at right
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angles

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d 8. I.

BOOK III.

angles to a diameter, from the extremity of it, touches the circle: Therefore DB touches the circle ABC. Wherefore, if e 16. 3. from a point, &c. Q. E. D.

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COR. Hence ftraight lines DB, DE, drawn to touch a circle ABC from the fame point D without it, are equal to one. another.

IF

PROP. A. THEOR.

F the oppofite angles of a quadrilateral figure be together equal to two right angles, and the cir cumference of a circle pafs through three of these angles, it fhall alfo pafs through the fourth.

Let ABCD be a quadrilateral figure, of which the oppofite angles are equal to two right angles; and let the circumference of a circle pafs through the three points A, B, C; it fhall alfo pass through the point D.

Join AC, BD and if the circumfe

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rence ABC do not pass through D, let it, if poffible, meet BD in the point E; and join AE, EC; therefore the B a 21. 1. angle AEC is unequal to the angle ADC: but because ABCE is a quadrilateral figure in a circle, the oppofite b 22. 3. angles ABC, AEC are equal b to two right angles and the two ABC, adc

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are alfo equal to two right angles; therefore the angles ABC, AEC are equal to the angles ABC, ADC: take away the common angle ABC, and the remaining angle AEC is equal to the remaining angle ADC: and they are alfo unequal, which is im poffible: Therefore the circumference ABC cannot but pafs through D; that is, it paffes through it. Wherefore, &c, Q. E. D.

COR. 1. And if two triangles ACD, ACE, on the same base AC, have their vertical angles at D, E equal to one another; the circumference which paffes through

the points A, C, D, fhall alfo pafs
through E.

In the circumference take any point
B on the other fide of AC, and join
AB, BC; therefore ABC, ADC are

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b 22, 3 equal to two right angles: but ADC B is equal to AEC; therefore ABC, AEC are equal to two right angles; wherefore the circumference ABC paffes through the point E.

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COR.

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COR. 2. If ADC be a right angle; the circumference of the Book III. circle, of which AC is the diameter, fhall pass through the point D. Take any point B in the circumference on the other

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fide of AC, and join AB, BC; therefore the angle ABC in the femicircle is a right angle ; and ABC, ADC are therefore equal c 31. 3. to two right angles. Wherefore the circumference ABC paffes through the point D.

PROP. B. THEOR.

[F two ftraight lines cut one another, and the rectangle contained by the fegments of one of them be equal to the rectangle contained by the fegments of the other; the circumference which paffes through three of their extremities, fhall alfo pass through the fourth.

Let the two ftraight lines AB, CD cut one another in E; and let the rectangle AE, EB be equal to the rectangle CE, ED; alfo let the circumference of a circle pafs through the three points A, C, B; it shall also pass through the point D. If not, let it, if poffible, cut CD in fome other point F: and because the ftraight lines AB, CF in the circle cut one another in E, the rectangle AE, EB is equal to the rectangle CE, EF: but the rectangle AE, EB is equal to the rectangle CE, ED; therefore the rectangle CE, EF is equal to the rectangle CE, ED; and EF is therefore equal to ED; the less to the

a 35. 3.

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greater, which is impoffible. Therefore the circumference ACB must pass through the point D.

Q. E. D.

Wherefore, &c.

COR. Likewife, if the ftraight line CD meet a circle in C,

and meet a ftraight line AB, which cuts the circle in the point E without it; and the rectangle CE, ED be equal to the rectangle AE, EB: it may be proved in the fame manner, that the point D is in the circumference; by ufing the Cor. to the 36th, inftead of the 35th.

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BOOK III.

IF

PROP. C. THEOR.

[F a ftraight line be drawn from the extremity of the diameter of a circle, and meet a perpendicular to that diameter; the rectangle contained by the segment of it within the circle, and the fegment between the extremity of the diameter and the perpendicular, is equal to the rectangle contained by the diameter, and the fegment of it between the fame extremity and the perpendicular.

Let ABC be a circle, of which AC is a diameter, and let DE be perpendicular to AC, and from the extremity A, let AB be drawn, meeting the circumference again in B, and the perpendicular in E; the rectangle BA, AE is equal to the rectangle CA, AD.

Join BC, CE; and because ABC is an angle in a femicircle, a 31.3. it is a right angle 2; therefore the circle of which CE is a diab 2. Cor. meter, fhall pass through B b : For the fame reason, it shall A. 3. pass through the point D: let this be the circle CBED: and

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because the ftraight lines BE, CD, either cut one another in the point A within the circle, or are drawn from the point A with€ 35. 3. or out it; the rectangle BA, AE is equal to the rectangle CA, Cor. 36. 3. AD. Wherefore, &c. Q. E. D.

IF

C

PROP. D. THEOR.

F from two angles of a triangle, perpendiculars be drawn to the oppofite fides, the ftraight line drawn from the third angle to the point where they meet, fhall be perpendicular to the third fide of the triangle.

Let

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