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* 6.2

thereto, the Rectangle under A D and D C, toge her with the Square of E C; tha11 * he equal to the square of E D. But E C is equal to E B; wherefore the Rectangle under A D anu DC, together with the Square

of E B, is equal to the Square of E D. Bur the square † 47. 1. of E D is fequal to the Squares of E B and-B D, tor

the Angle E B D is a Right Angle: Therefore the Rectangle wider A D and DC, cogerker with the Square of E B, is equal to the Squares of E B and B D; and if the common Square of È B be taken away, the Rectárrgle under A D and D C remaining, will be equal to the Square of the Tangent Line B D.

Now, let D CA not pass through the Centre of the | 1 of ebis. Circle A BC; and find I the Centre E thereof, and

draw E F perpendicular to AC, and join E B, E C, ED. Therefore EPD is a Right Anyle. And because a Right Line E F, drawn thro' the Centre, cuts

a Right Line A C, not drawn thro' the Centre, at Right 3 of tbis. Angles, it will * bifect the fame; and fo Ah is equal

to F C. Again, fince the Right Line A C is biseéted in F, and C D is added thereto, the Rectangle under A D and D C, together with the Square of F C, will be * equal to the Square of F D. And if the common Square of EF be added, then the Rectangle under A D and DC, together with the Squares of FC and F E, is equal to the Squares of D F and F E. But the Square of D, E is equal to the Squares of D F and

FE; for the Angle E F D is a Right one ; and the † 47.1. Square of C E is + equal to the Squares of CF and

FE. Therefore the Rectangle under A D and DC, together with the Square of C E, is equal to the Square of ED ; but GE is equal to E B. Wherefore the Rectangle urder A D and DC, together with the Square of E B, is equal to the Square of E D. But the Squares or E B and B D are + equal to the Square of ED; since the Angle EBD is a Right one Wherefore the Rectangle under AD and D C, together with the Square of E B, is equal to the Squares of E B and B D. And if the common Square of E B bè taken away, the Rectangle under AD and D C, remaining, will be equal to the Square of D B. There. fore, if any Point be taken without a Circle, and from that Puini two Right Lines fall to the Circle, one of which cuts the Circle, and the other touches it ; the Rectangle contained

• 6.2.

under

under the whole Sacant Line, and its Part between the Convexity of the Grile and the offumed Part, will be equal to rhe Square of the Tangent Line ; which was to be demonstrated

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PROPOSITION XXXVII;' :

THEOREM.
If some point be taken without a Circle, and iwo

Right Lines be drawn from it to the Circle, so
that one cuts it, and the other falls upon it ;
and if the ReEtangle under the whole Secant
Line, and the Part thereof, without the Circle,

be equal to the Square of the Line falling upon lidbe Circles then this laft Line will touch the

Circle. Ta
LET fome Port D be affumed without the Circle

ABC, and from it draw two Right Lines DCA, D B, to che Circle, in such a manner, that DCA cuts the Circle, and D B falls upon it: And let the Rectangle under A D and D C be equal to the Square of DB. I say, the Right Line D B touches the Circle.

For, let the Right Line D E be drawn * touching * 17 of this. the Circle ABC, and find F the Centre I of the Cir- | 1 of tbis. cle; and join EF, FB, FD.

Then leke Angle F E D is. tа Right Angle. And † 18 of this. because D E touches the Circle AB C, and DCA cuts it, the Rectangle under A D and D C will be equal to the Square of D E. But the Rectangle under AD and DC is I equal to the Square of D B.

I By Hyp. Wherefore the Square of D E shall be equal to the Square of D B. And so the Line D E will be equal to the Line D B. But E F is equal to FB: Therefore the two Sides DE, EF, are equal to the twn Sides DB, BF; and the Base F D is common. Where fore the Angle D E F is equal * to the Angle DBF;.8. 1. But D E F is a Right Angle; wherefore D B F is allo a Right Angle, and F B 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 B D neceffarily touches the Circle this in the same manner, if the Centre of the Circle be

We prove

in the Right Line CA. If, therefore, any Point be alfumed without a Circle, and two Right Lines be drawn from it to the Circle, so 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 demonstrated.

: Caroll. Hence, if from any Point, without a Circle, ; feveral Righe Lines A B, A C, are drawn, cutung

the Circle, the Rectangles comprehended under the whole Lines AB, AC, and their external Parts A E, A F, are equal between themselves. For, if the Tangent AD be drawn, the Rectangle under B A and A D is equal to the Square of AD; and the Rectangle under C A and A F is equal to the same Square of AD: Therefore the Rectangles fhall be equal.

Tbe End of the THIRD Book,

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