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of contact B, draw BA at right angles to EF; take any roint Book III. C in the circumference, and join AD, DC, CB.

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Now, because BA is drawn from B, at right angles, to EF, the a 11. 1. center of the circle is in AB'; and, because ADCB is a femicircle, b 19. the angle ADB is a right angle ; therefore ADB is equal to the c 31. two angles DBA, DAB 4; but ABF is likewife a right angle; der. 32. 1. therefore the angle ABF is equal to the angles DBA, DAB; but the angle ABF is likewife equal to the angles DBF, DBA; therefore the angles DBA, DBF, are equal to the angles DAB, DBA. Take the common angle DBA from both, there re- e ax. 1. 1, mains the angle DBF equal to DAB, the angle in the alternate fegment.

Likewife the angle DCB is equal to the angle DBE; for, DCB, DAB, are equal to two right angles f, and DBF, DB£ §, f 22. equal to two right angles; but DAB is proved equal to DBF;8 13. 1, therefore the remainder, DCB, is equal to DBE. Wherefore,

&c.

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angle.

PROP. XXXIII. PROB. ·

PON a given right line to deferibe a fegment of a circle,
that will contain an angle equal to agiven right lined

It is required, upon AB, to defcribe a fegment of a circle, that will contain an angle equal to a given angle, C.

b II. I.

C IO. I.

At the point A, with the right line AB, make the angle BAD equal to C2; draw AE at right angles to AD; bisect a 23. 1. AB in F, and draw FG, at right angles, to AB, cutting AE in the point G; join GB; with the center G, and distance GA, defcribe the circle ABE, which will pass through the point B; for, because AB is bifected in F, and GF drawn, at right angles, to AB, the right lines AF, FG., are. equal to BF, FG; and the angle AFG equal to BFG; therefore AG is equal to GB d. Now, because AD is a tangent to the d 4. 1. circle, the angle BAD is equal to the angle in the alternate e 16. fegment BEA; but the angle DAB is equal to the angle Cs therefore the angle AEB is equal to the angle C. Wherefore,

&c.

T

PRO P. XXXIV. PRO B.

O cut off a fegment from a given circle that fall contain an
angle equal to a given right lined angle.

f

32.

g Conft.

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b 3.

€ 47. J. d 5. 2.

It is required to cut off a fegment from the given circle ABC, that fhall contain an angle equal to the given angle D.

Draw the line EF, touching the circle in B"; from which draw BC, making an angle FBC equal to the angle D'; then the angle FBC will be equal to the angle in the alternate fegment, viz. BAC; but FBC is equal to the angle D; therefore BAC is cqual to the angle D. Wherefore, &c.

PROP. XXXV. THE OR.

IF two right lines in a circle mutually cut each other, the rectangle contained under the fegment of the one, is equal to the rectangle contained under the fegarents of the other.

Let the two right lines AC, DB, in the circle ABCD, mutually cut each other in E; then the rectangle under AE, EC, is equal to the rectangle under DE, EB; if AC, DB, pafs cach through the center, then the rectangle under AE, EC, is cqual to the rectangle under DE, EB, for the lines are equala.

C

d

2dly, Ií AC, paffing through the center, cut BD, not paffing through the center, at right angles, in the point E, find the center F, and join FD; for, becaufe BE is equal to ED, and the angle DEF is a right one, the fquares of DE, EF, are equal to the fquare of FD ; but the rectangle under AE, EC, together with the fquare of EF, is equal to the fquare of FC 4, or TD. Take the fquare of FE, which is common, from both, there remains the rectangle under AE, EC, equal to the fquare of ED, that is, the rectangle under BE, ED. If the right linc, AC, paffing through the center, cut BD, not paffing through the center, and not at right angles, draw FG at right angles to BD, and join FD; then BG is equal to GD; the rectangle under BE, LD, together with the fquare of GE, equal to the fquare of GDa. Add the fquare of GF to both, then the setangle under BE, ED, with the fquares of EG, GF, or the fquale of EF, are cqual to the fquare of FD; but the rectangle under AE, EC, together with the fquare of EF, are likewife equal to the fquare of FD. Take the fquare of EF from both, then the rectangle under AE, EC, is equal to the rectangle under DE, EL.

3y, If

3dly, If neither país through the center, draw GH, paffing Book III, through the center F, and cutting AC, BD, in E; then the rectangle under AE, EC, is equal to the rectangle under BE, ED; for cach is equal to the rectangle under GE, EH. Wherefore, &c.

PRO P. XXXVI. THEOR.

IF F fome point be taken without a circle, and from that point two right lines be drawn, one of which touches the circle, and the other cuts it, the rectangle under the whole fecant line, and the part between the point and convexity of the circle, is equal to the Square of the tangent line.

Let ABC be the circle, D the given point, and DCA, DB, the two given right lines, of which DB touches the circle, and DCA cuts it; the reangle under AD, DC, is equal to the fquare of DB.

Now, DCA either paffes thro' the center, or not. First, let it pass thro' the center E, and join BE; then, because AC is bifected in E, and DC added, the rectangle under AD, DC, together with the fquare of CE, are equal to the fquare of DE 2; a 6. 2. but the fquare of DE is equal to the fquares of DB, BE; for b 47. the angle DBE is a right angle; therefore the rectangle under 18. AD, DC, together with the fquare of CE, are equal to the fquares of DB, BE. Take the equal fquares of BE, CE, from both, there remains the rectangle AD, DC, equal to the fquare of DB.

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€ 12. IA

2dly, Lct DA not pafs through the center of the circle ABC; find the center Ed, and join ED, EC, EB; draw EF, at right angles, to AC, cutting it in F; then AF is equal to FC; f. therefore the rectangle under AD, DC, together with the fquare. of CF, are equal to the fquare of FD. Add the fquare of FE to both; then the rectangle under AD, DC, with the fquares of CF, FE, are equal to the fquares of DF, FE; but the fquare of CE is equal to the fquares of CF, FEb, and the fquare of DE equal to the fquares of DF, FE; therefore the rectangle under AD, DC, with the fquare of CE, are equal to the fquare of DE; but the fquare of DE is equal to the fquares of DE, BE; therefore the rectangle under AD, DC, with the fquare of CE, are equal to the fquares of DB, BE. Take the cqual fquares of BE, CE, from both, and the rectangle under AD, DC, is equal to the fquare of DB. Wherefore, &c.

PROP.

BOOK III.

a 17.

b I.

C 18.

d 36.
e Hyp.

£8. I.

$ 16.

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[F, from a point without a circle two right lines be drawn, one of which cuts the circle, and the other falls upon it; and, if the rectangle under the whole fecant line, and part betwixt the point and circle, be equal to the fquare of the other line, this laft line fhall be a tangent to the circle.

Let fome point D, be affumed without the circle ABC, and from it draw the lines DCA, DB, fo that DCA cut the circle, and DB fall upon it; and, if the rectangle under AD, DC, be equal to the fquare of DB, then DB will touch the circle in the point B.

For, let DE be drawn a tangent to the circle in the point E 2; find the center Fb, and join BF, FL, and DF.

Then the angle DEF is a right angle ; therefore the rectangle under AĎ, DC, is equal to the square of D£ a; but the rectangle under AD, DC, is equal to the fquare of DB ; therefore the fquare of DB is equal to the fquare of DE, and DB equal to DE; therefore the right lines DE, EF, are equal to DB, BF; and FD common; therefore the angle DBF is equal to the angle DEF f; but DEF is a right angle; therefore DBF is likewife a right angle: Therefore DB is a tangent to the circles. Wherefore, &c.

COR.I. Hence, if any number of right lines, as DA, DG, be drawn from the point Á, cutting the circle in C and H, the rectangles under AD, DC, and GD, DH, are equal to one another; for each of them is equal to the fquare of BD.

II. If, from any two points in the circumference of a circle, two tangents be drawn, fo that, being produced, they will meet one another; then these tangents will be equal to one another; for each of their fquares, viz. of BD, DE, is equal to the rectangle contained under AD, DC.

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