*

the Point of Contact B; the Angle FBC will be equal* 32 of this. to that in the alternate Segment of the Circle; but the Angle F B C is equal to the Angle D. Therefore the Angle in the Segment BAC, will be equal to the Angle D. Therefore the Segment BAC is cut off from the given Circle ABC, containing an Angle equal to the given Right-lined Angle D; which was to be done.

PROPOSITION XXXV.

THEOREM

If two Right Lines in a Circle mutually cut each other, the Rectangle contained under the Segments of the one, is equal to the Rectangle under the Segments of the other.

IN

IN the Circle ABCD, let two Right Lines mutually cut each other in the Point E. I fay the Rectangle contained under AE, and EC, is equal to the Rectangle contained under DE, EB.

If AC and DB pass thro' the Center, fo that E be the Center of the Circle ABCD; it is manifeft, fince AE, EC, DE, EB, are equal; that the Rectangle, under AE, EC, is equal to the Rectangle under DE, EB.

But if AC, DB, do not pafs thro' the Center, affume the Center of the Circle F; from which draw FG, FH, perpendicular to the Right Lines AC, DB, and join FB, FC, FE.

Then because the Right Line GF, drawn through the Center, cuts the Right Line AC, not drawn thro' the Center at Right Angles, it will alfo bifect the* 4 of this. fame. Wherefore AG is equal to GC: And because the Right Line AC is cut into two equal Parts in the Point G, and into two unequal Parts in E, the Rectangle under AE, EC, together with the Square of EG, ist equal to the Square of G C. And ift 5.2. the common Square of GF be added, then the Rectangle under AE, EC, together with the Squares of EG, GF, is equal to the Squares of CG, GF. the Square of FE is † equal to the Squares of EG, GF, and the Square of FC equal to the Squares ‡ 47. 1.

But

of

7

of CG,GF. Therefore the Rectangle under AE, EC, together with the Square of F E, is equal to the Square of F C, but CF is equal to F B. Therefore the Rectangle under AE, EC, together with the Square of EF, is equal to the Square of FB. For the fame Reason, the Rectangle under DE, EB, together with the Square of F E, is equal to the Square of FB. But it has been proved, that the Rectangle under AE, EC, together with the Square of F E, is alfo equal to the Square of FB. Therefore the Rectangle under AE, EC, together with the Square of FE, is equal to the Rectangle under DE, EB, together with the Square of FE. And if the common Square of FE be taken away, then there will remain the Rectangle under AE, E C, equal to the Rectangle under DE, EB. Wherefore, if two Right Lines in a Circle mutually cut each other, the Rectangle contained under the Segments of the one, is equal to the Rectangle under the Segments of the other; which was to be demonstrated.

PROPOSITION XXXVI.

THEOREM.

If fome Point be taken without a Circle, and from that Point two Right Lines fall to the Circle, one of which cuts the Circle, and the other touches it ; the Rectangle contained under the whole Secant Line, and its Part between the Convexity of the Circle and the affumed Point, will be equal to the Square of the Tangent Line.

LET any

Point D be affumed without the Circle ABC, and let two Right Lines DCA, DB, fall from the faid Point to the Circle; whereof DCA cuts the Circle, and DB touches it. I say the Rectangle under AD, DC, is equal to the Square of DB.

Now DCA either paffes thro' the Center, or not. In the first place, let it pass thro' the Center of the Circle ABC, which let be E, and join EB. Then * 18 of this. the Angle EBD is a Right Angle. And fo fince the Right Line AC is bifected in E, and CD is added thereto, the Rectangle under AD, DC, together

*

with the Square of EC, fhall be equal the Square* 6. 2. of ED. But EC is equal to EB; wherefore the Rectangle under AD, DC, together with the Square of E B, is equal to the Square of ED. But the Square of ED is equal to the Square of EB, and BD. For † 47. 1. the Angle EBD, is a Right Angle: Therefore the Rectangle under AD, DC, together with the Square of EB, is equal to the Squares of EB and BD; and if the common Square of EB be taken away, the Rectangle under AD, DC, remaining, will be equal to the Square of the Tangent Line BD.

Now let DCA not pafs through the Center of the Circle ABC; and find the Center E thereof, and ‡ 1 of this draw EF perpendicular to AC, and join EB, EC, ED. Therefore EFD is a Right Angle. And becaufe a Right Line EF, drawn through the Center, cuts a Right Line AC at Right Angles, not drawn through the Center, it will bifect the fame at Right* 3 of this Angles; and fo AF is equal to FC. Again, fince the Right Line AC is bifected in F, and CD is added thereto, the Rectangle under AD, DC, together with the Square of F C, will be * equal to the Square of FD. And if the common Square of EF be added, then the Rectangle under AD, DC, together with the Squares of FC and FE, is equal to the Squares of DF and FE. But the Square of DE, is equal to the Squares of DF and FE; for the Angle EFD is a Right one: And the Square of CE is equal to the Square of CF and FE. Therefore the Rectangle under AD, DC, together with the Square of CE, is equal to the Square of ED; but CE is equal to EB. Wherefore the Rectangle under AD, DC, together with the Square of E B, is equal to the Square of ED. But the Squares of EB and BD are t equal to the Square of ED; fince the Angle EBD is a Right one. Wherefore the Rectangle under AD and DC, together with the Square of EB, is equal to the Squares of EB and BD. And if the common Square of E B be taken away, the Rectangle under AD and DC, remaining, will be equal to the Square of DB. Therefore, if any Point be taken without a Circle, and from that Point two Right Lines fall to the Circle, one of which cuts the Circle, and the other touchet it; the Rectangle contained under the whole Secant Line, and

its

* 17 of this.

† 18 of this.

its Part between the Convexity of the Circle and th affum'd Point, will be equal to the Square of the Tar gent Line; which was to be demonftrated.

PROPOSITION XXXVII.

THEOREM.

If fome Point be taken without a Circle, and tw Right Lines be drawn from it to the Circle, that one cuts it, and the other falls upon it and if the Retlangle under the whole Secant Line and the Part thereof, without the Circle, b equal to the Square of the Line falling upon the Circle, then this laft Line will touch the Circle.

ET fome Point D be affumed without the Circle ABC, and from it draw two Right Lines DCA, DB, to the Circle in fuch Manner that DCA cuts the Circle, and DB falls upon it: And let the Rectangle under AD, DC, be equal to the Square of DB. I fay, the Right Line DB touches the Circle.

For let the Right Line DE be drawn touching the Circle ABC, and find F the Center of the Circle, and join EF, FB, FD.

Then the Angle FED is † a Right Angle. And because DE touches the Circle ABC, and DCA cuts it, the Rectangle under AD, and DC, will be equal to the Square of D E. But the Rectangle unBy Hyp. der AD and DC, is equal to the Square of DB. Wherefore the Square of DE fhall be equal to the Square of DB. And fo the Line DE will be equal to the Line DB. But EF is equal to FB: Therefore the two Sides DE, EF, are equal to the two Sides DB, BF and the Base FD is common. Wherefore the Angle DEF is equal to the Angle DBF; but DEF is a Right Angle; wherefore DBF is also a Right Angle, and FB produced is a Diameter. But a Right Line drawn at Right Angles, on the End of the Diameter of a Circle, touches the Circle; therefore BD neceffarily touches the Circle. We prove this in the fame Manner, if the Center of the Circle be in the Right Line CA. If therefore any Point be affumed without a Circle, and two Right

Lines be drawn from it to the Circle, fo that one cuts it, and the other falls upon it; and if the Rectangle under the whole Secant Line, and the Part thereof, without the Circle, be equal to the Square of the Line falling upon the Circle; then this laft Line will touch the Circle; which was to be demonftrated.

Coroll. Hence, if from any Point without a Circle, feveral Right Lines AB, AC, are drawn cutting the Circle; the Rectangles comprehended under the whole Lines AB, AC, and their external Parts AE, AF, are equal between themselves. For if the Tangent AD be drawn, the Rectangle under BA and AE, is equal to the Square of AD; and the Rectangle under CA and AF, is equal to the fame Square of AD: Therefore the Rectangles fhall be equal.

The END of the THIRD BOOK,

H

EUCLID'S