angles: therefore the angles F E B, E F D will be less than E F two right angles. But right lines infinitely produced from A angles less than two right angles [by ax. 11.] will meet one another: Therefore E B, FD produced towards BD, G will meet. Let them be pro duced, and meet in the point G: join a G. Now becaufe A c is equal to c E, the angle A E C [by 5. 1.] will be equal to the angle EAC. But the angle at c is a right angle: wherefore E A C, A E C are [by 32. 1.] each half a right angle. By the fame reafon, CEB, EBC are each one half a right angle: therefore A E B is a right angle. And because E B C is one half a right angle, the angle DBG by 15. 1.] will be one half a right angle. But [by 29. 1.] BDG is a right angle too; for it is equal to the alternate angle DCE: wherefore the remaining angle DG B is one half a right angle; and fo the angle DGB is equal to the angle DB G: wherefore the fide B D will be [by 6. 1.] equal to the fide G D. Again, becaufe E G F is one half a right angle, and the angle at F is a right angle, for [by 34. 1.] it is equal to the oppofite angle at c; the remaining angle F E G will be one half a right angle: wherefore the angle EGF is equal to the angle FEG; and fo the fide GF [by 6. 1.] is equal to the fide E F. And fince E c is equal to CA, and the fquare of E c equal to the fquare of A C; the fquares of E C, C A are double to the fquare of CA. But the fquare of E A [by 47. 1.] is equal to the fquares of E C, CA: therefore the fquare of E A is double to the fquare of A c. Again, becaufe GF is equal to EF, and the fquare of G F is equal to the fquare of EF; the fquares of GF, EF are double to the fquare of E F. But [by 47. 1.] the fquare of E G is equal to the fquares of GF, EF; therefore the fquare of EG will be double to the fquare of EF: but E F is equal to CD: wherefore the fquare of EG will be double to the fquare of c D. But it has been demonftrated, that the fquare of E A is double to the fquare of AC: therefore the fquares of A E, GE are double to the fquares of A C, C D. But the fquare of A G [by 47. 1.] is equal to the fquares of A E, EG; therefore the fquare of AG is double to the squares of A C, CD. But [by [by 47. 1.] the fquare of AG is equal to the fquares of AD, DG: therefore the fquares of A D, DG are double to the fquares of A C, CD. But DG is equal to D B. Wherefore the squares of A D, D B are double to the fquares of A c, C D. If therefore a right line be divided into two equal parts, and any right line be joined to it in the fame direction, the fquare of the whole and joined line, together with the square of the joined line, are equal to twice the fquare of one half the line, and the fquare of the line made of the half line, and the joined line taken as one line. Which was to be demonftrated. h This theorem may be demonftrated otherwise thus: Let the right line A B be bifected in c, and let any right line B D be added to it in the fame direction: I fay, the fquares of A D, BD together, are double to the fquares of AC, CD together; for having added the right line AE to AB on the contrary fide thereof [viz. towards a] equal to BD EA then the whole ED will be C B D cut equally in c and unequally in B. Because EA, A C are equal to D B, B C, the right line E B will also be equal to the right line AD; fince E A is equal to D B, and A в common to both. Therefore [by 9. 2.] the fquares of EB, BD, that is, of a d, b d, will be double to the fquares of EC, CB, that is, of CD, AC. Which was to be demonftrated. To divide a right line into two fuch parts, that the rectangle contained under the whole line and one of the parts, fhall be equal to the Square of the other part. Let A B be a given right line: It is required to divide it into two fuch parts, that the rectangle contained under the whole line and one of thofe parts, fhall be equal to the fquare of the other part. For [by 46. 1.] defcribe the fquare A B D C of AB, and [by 10. 1.] divide AC into two equal parts at £; join BE; produce c A to F ; and [by 3. 1.] make E F equal to BE; alfo defcribe F H the fquare of A F, and produce G H to K: I fay, AB is fo divided in H, that the rectangle under A B, BH is equal to the, fquare of A H. For F A E G HR Book II. For because the right line AC is divided into two equal Parts at E, and AF is joined to it in the fame direction; the rectangle under C F, FA together with the fquare of AE [by 6. 2.] will be equal to the fquare of E F. But E F is equal to EB: wherefore the rectangle under CF, FA, together with the fquare of A E, is equal to the square of EB. But [by 47. 1.] the fquares of BA, A E are equal to the square of E B ; for the angle at A is a right angle: therefore the rectangle under CF, F A, together with the fquare of A E, is KD equal to the fquares of B A, A E. Take away from both the common fquare of A E; and then there will remain the rectangle under CF, FA equal to the fquare of A B. But the rectangle K F is the rectangle under C F, F A; for AF is equal to FG, and A D is the square, of AB therefore the rectangle F K is equal to the fquare A D. Take away the common rectangle AK; and there will remain F H equal to D H. But D H is the rectangle under A B, BH, fince A B is equal to B D, and H is the fquare of AH: therefore the rectangle under AB, BH is equal to the fquare of a H. Wherefore the given right line A B is fo divided in the point H, that the rectangle under A B, B H is equal to the fquare of A H. Which was to be demonstrated. i Numbers are applicable to all the other propofitions of this book but this, there being no number that can be divided into two others fuch, that the product of the whole by one of the parts, shall be equal to the fquare of the other part. PROP. XII. THEOR. In obtufe-angled triangles, the fquare of the fide oppofite to the obtufe angle, is greater than the fquares of the fides containing the obtufe angle, by twice the rectangle contained under one of the fides about the obtufe angle upon which a perpendicular falls, and the right line without between the obtuje angle and the point whereon the perpendicular falls. Let Let there be an obtufe-angled triangle B AC, having the obtuse angle B A C; and from the point B, let the perpendicular B D fall upon the continuation A D of the fide CA: I fay, the fquare of B C is greater than the fquares of B A, AC by twice a rectangle contained under the right lines CA, A D. For because the right line CD is divided any how at A, the fquare of CD [by 4. 2.] will be equal to the fquares of CA, A D, together with twice the Add the then the B rectangle under C A, A D. the fquares of CD, DB; for the angle at D is a right angle; and the fquare of A B is equal to the fquares of A D, DB: Therefore the fquare of CB is equal to the fquares of C A, A B, and to twice the rectangle under C A, A D: Wherefore the fquare of CB is greater than the fquares of CA, AB, by twice the rectangle contained under the right lines CA, A D. Wherefore in obtufe-angled triangles, the fquare of the fide oppofite to the obtufe ang'e is greater than the fquares of the fides containing the obtufe angle, by twice a reetangle contained under one of the fides containing the obtufe angle upon which a perpendicular falls, and the right line without between the obtufe angle, and the point upon which the perpendicular falls. Which was to be demonftrated. PROP. XIII. THEOR. In acute-angled triangles, the fquare of the fide oppofite to an acute angle is less than the Squares of the fides containing that acute angle, by twice a rectangle contained under that fide about the acute angle upon which a perpendicular falls, and the right line within, between the perpendicular and the acute angle. Let there be an acute-angled triangle ABC, having an acute angle at B; and from the point A draw the perpen dicular dicular A D upon BC: I fay, the fquare of A c is less than the fquares of C B, BA, by twice the rectangle contained under the right lines C B, B D. B D For because the right line C B is any how divided at D, the fquares of CB, B D [by 7. 2.] are equal to twice the rectangle contained under the right lines C B, BD, together with the fquare of DC. Add the fquare of A D to both then the fquares of c B, B D, DA are equal to twice the rectangle under C B, B D, together with the fquares of A D, DC. But [by 47. 1.] the fquare of AB is equal to the fquares of BD, DA, fince the angle at D is a right angle; and the fquare of AC is equal to the fquares of A D, DC: therefore the fquares of CB, B A are equal to the fquare of A C, together with twice the rectangle contained under the right lines C B, B D. Wherefore the fquare of A C alone is lefs than the fquares of CB, B A, by twice the rectangle contained under the right lines C B, в d. Therefore in acute-angled triangles, the fquare of the fide oppofite to any acute angle, is less than the fquares of the fides containing that acute angle, by twice the rectangle under that fide about the acute angle upon which the perpendicular falls, and the right line within between the acute angle and the perpendicular. Which was to be demonftrated. k Euclid has not mentioned exprefsly the following useful theorem: In any obtufe-angled triangle, if a perpendicular be let fall upon the bafe or fide oppofite to the obtufe angle, the Square of either of the fides containing the obtufe angle, together with twice the retangle under the bafe and that fegment of it (made by the falling of the perpendicular upon it) which lies under the remaining de containing the obtufe angle, will be equal to the Square of this remaining fide, and the Square of the base, taken together. Its demonftration is the very fame as that of this 13th propofition. The theorem likewife holds true in any right-angled triangle, when the perpendicular falls from the right angle upon the bafe, or fide oppofite to it. PROP. |