Thesauros Edebiyat mathematics Στοιχεῖα

Στοιχεῖα

Στοιχεῖα Euclid

Kitap 3

36 Let the square on FE be added to each; therefore the rectangle AD, DC together with the squares on CF, FE is equal to the squares on FD, FE.
36 But the square on EC is equal to the squares on CF, FE, for the angle EFC is right; [I. 47] and the square on ED is equal to the squares on DF, FE; therefore the rectangle AD, DC together with the square on EC is equal to the square on ED.
36 And EC is equal to EB; therefore the rectangle AD, DC together with the square on EB is equal to the square on ED.
36 But the squares on EB, BD are equal to the square on ED, for the angle EBD is right; [I. 47] therefore the rectangle AD, DC together with the square on EB is equal to the squares on EB, BD.
36 Let the square on EB be subtracted from each; therefore the rectangle AD, DC which remains is equal to the square on DB.
36 Therefore etc. Q. E. D.

PROPOSITION 37.

37 If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle.
37 For let a point D be taken outside the circle ABC; from D let the two straight lines DCA, DB fall on the circle ACB; let DCA cut the circle and DB fall on it; and let the rectangle AD, DC be equal to the square on DB.
37 I say that DB touches the circle ABC.
37 For let DE be drawn touching ABC; let the centre of the circle ABC be taken, and let it be F; let FE, FB, FD be joined.
37 Thus the angle FED is right. [III. 18]
37 Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE. [III. 36]
37 But the rectangle AD, DC was also equal to the square on DB; therefore the square on DE is equal to the square on DB; therefore DE is equal to DB.

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