Book II: PRO P. XIII. THEOR. See N. IN every triangle, the square of the fide fubtending any of the acute angles, is less than the squares of the fides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the oppofite angle, and the acute angle. a 12. I. b 7. 2. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular. AD from the opposite angle: The square of AC, opposite to the angle B, is less than the squares of CB, BA by twice the rectangle CB, BD. First, Let AD fall within the triangle ABC ; and because А C A out the triangle ABC: Then, becaule the angle at D is a right angle, the angle ACB is greater d than a right angle; and there. fore the square of AB is equal to the squares of AC, CB, and twice the rectangle BC, CD: To thefe equals add the square of BC, and the B D squares squares of AB, BC are equal to the square of AC, and twice Book II. the square of BC, and twice the rectangle BC, CD: But because BD is divided into two parts in C, the rectangle DB, BC is equal to the rectangle BC, CD and the square of BC: And ? 3. 2. the doubles of these are equal : Therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: Therefore the square of AC alone is less than the squares of AB,BC bytwice the rectangle DB, BC. А Laftly, Let the lide AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifeft that the squares of AB, BC are e. qual to the square of AC, and twice the C 47. I. {quare of BC : Therefore, in every triangle, &c. Q. E, D. B с O describe a square that thall be equal to a given sce N. Trectilineal figure. Let A be the given rectilineal figure ; it is required to de• {cribe a square that shall be equal to A. Describe a the rectangular parallelogram BCDE equal to the a 45. 1. rectilineal figure A. If then the sides of it BE, ED are equal to one another, it is a square, and H what was required is now done : But if they are not e. A qual, produce one of them BE to F, B G E and make E Fequal to ED, and bifect C D BF in G, and from the centre G, at the distance GB, or GF, describe the semicircle 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 ac E, the rectangle BE, EF, together with the square of EG, is equal to the square of b 5. 2. GF: But GF is equal to GH; therefore the rectangle BE, EF, E to. ? Book II. together with the square of EG, is equal to the square of GH: mo But the squares of HE, EG are equal to the square of GH: 0 47. 1. Therefore the rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EG: Take away the square of EG, which is common to both ; and the remaining rectangle BE, EF is equal to the square 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 square of EH; but BD is equal to the rectilineal figure A ; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square deferibed upon EH. Which was to be done. ES I. or from the centres of which the straight lines to the circumferences are equal. • This is not a definition but a theorem, the truth of which II. a circle, when it meets the UI. another, which meet, but IV. ftant from the centre of a circle, V. greater perpendicular falls, is said to VI. C E 2 Book III. VI. tained by a straight line and the cir- VII. VIII. tained by two straight lines drawn IX. upon the circumference intercepted X. ed by two straight lines drawn from the XI. are those in which the an- PRO P. I. :P R, 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 straight line AB, and bisect it in D; from the point D drawb DC at right angles to AB, and pro: duce it to E, and bisca CE in F: The point F is the centre of the circle ABC. a 10. 1. 11. 1. Pors |