Book II: See N. a 12. I. b 7.2. € 47. I. d 16. I. € 12. 2. PROP. XIII. THEOR. IN every triangle, the square of the fide fubtending any of the acute angles, is lefs than the fquares of the fides containing that angle, by twice the rectangle contained by either of these fides, and the straight line intercepted between the perpendicular let fall upon it from the oppofite angle, and the acute angle. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the fides containing it, let fall the perpendicular AD from the oppofite angle: The fquare of AC, oppofite to the angle B, is less than the fquares of CB, BA by twice the rectangle CB, BD. A Firft, Let AD fall within the triangle ABC; and because the ftraight line CB is divided into two parts in the point D, the fquares of CB, BD are equal to twice the rectangle contained by CB, BD, and the fquare of DC': To each of these equals add the fquare of AD; therefore the fquares of CB, BD, DA are equal to twice the rectangle CB, BD, and the fquares of AD, DC: B But the fquare of AB is equal D to the fquares of ED, DA, because the angle BDA is a right angle; and the square of AC is equal to the fquares of AD, DC: Therefore the fquares of CB, BA are equal to the fquare of AC, and twice the rectangle CB, BD; that is, the fquare of AC alone is lefs than the fquares of CB, BA by twice the rectangle CB, BD. Secondly, Let AD fall without the triangle ABC: Then, becaufe the angle at D is a right angle, the angle ACB is greater than a right angle; and therefore the fquare of AB is equal to the fquares of AC, CB, and twice the rectangle BC, CD: To these equals add the fquare of BC, and the B A D fquares fquares of AB, BC are equal to the fquare of AC, and twice Book II. the fquare of BC, and twice the rectangle BC, CD: But becaufe BD is divided into two parts in C, the rectangle DB, BC is equal to the rectangle BC, CD and the fquare of BC: And f 3. 2. the doubles of thefe are equal: Therefore the fquares of AB, BC are equal to the fquare of AC, and twice the rectangle DB, BC: Therefore the fquare of AC alone is less than the fquares of AB,BC by twice the rectangle DB, BC. Laftly, Let the fide AC be perpendicular to BC; then is BC the ftraight line between the perpendicular and the acute angle at B; and it is manifeft that the fquares of AB, BC are equal to the fquare of AC, and twice the fquare of BC: Therefore, in every triangle, &c. Q. E. D. A C 47. I. T PROP. XIV. PROB. O defcribe a fquare that thall be equal to a given see N. rectilineal figure. Let A be the given rectilineal figure; it is required to de• scribe a square that fhall be equal to A. Defcribe the rectangular parallelogram BCDE equal to the a 45. I. rectilineal figure A. If then the fides of it BE, ED are equal to one another, it is a fquare, and what was required is now done: But if they are not equal, produce one of them BE to F, and make EFequal to ED, and bifect BF in G; and from the centre G, at the diftance GB, or GF, defcribe the femicircle BHF, and produce DE to H, and join GH: Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the fquare of EG, is equal to the fquare of b 5. 2. GF: But GF is equal to GH; therefore the rectangle BE, EF, E to. C47. I. с Book II. together with the fquare of EG, is equal to the fquare of GH: But the fquares of HE, EG are equal to the fquare of GH: Therefore the rectangle BE, EF, together with the square of EG, is equal to the fquares of HE, EG: Take away the square of EG, which is common to both; and the remaining rectangle BE, EF is equal to the fquare of EH: But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the fquare of EH; but BD is equal to the rectilineal figure A; therefore the rectilineal fi gure A is equal to the fquare of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square defcribed upon EH. Which was to be done. THE E U C L I D. воок III. DEFINITIONS. I. QUAL circles are those of which the diameters are equal, or from the centres of which the ftraight lines to the circumferences are equal. This is not a definition but a theorem, the truth of which ' is evident; for, if the circles be applied to one another, so that ⚫ their centres coincide, the circles muft likewife coincide, fince the ftraight lines from the centres are equal. "The angle of a fegment is that which is contained by the "ftraight line and the circumference." VIII. a 10. 1. b 11. I. PROP. I. PR, O B. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. Draw within it any ftraight line AB, and bifect it in D; from the point D draw b DC at right angles to AB, and produce it to E, and bifect CE in F: The point F is the centre of the circle ABC. Fors |