Book II. PROP. XII. THEOR. IN obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle. Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn perpen-a 12.5 dicular to BC produced : The square of AB is greater than the squares of AC, CB by twice the rectangle BC, CD. Because the straight line BD is divided into two parts in the A 6 4. Ze twice the rectangle BC, CD: To each of these equals add the square of DA; and the squares of BD, DA are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD: But the square of BA is equal to the squares of BD, DA, be € 47. I cause the angle at D is a right B С D angle; and the square of CA is equal to the squares of CD, DA: Therefore the square of BA is equal to the squares of BC, CA, and twice the rectangle BC, CD, that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BC, CD. Therefore, in obtuse angled triangles, &c. Q. Ě. D. PRO P. Book II. PRO P: XIII. THE O R. See N. IN every triangle, the square of the side fubtending any of the acute angles, is less than the squares of the sides 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 a. cute angles, and upon BC, one of the sides containing it, let fall the perpendicular a 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. Firit, Let AD fall within the triangle ABC; and because А C to the squares of BD, DA, because the angle BDA is a right angle; and the square of AC is equal to the squares of AD, DC: Therefore the squares of CB, 5A, are equal to the square of AC, and twice the rectangle CB, BD; that is, the square of AC alone is less than the squares of CB, BA by twice the rectangle CB, BD. Secondly, Let AD fall with A out the triangle ABC: Then, because the angle at D is a right angle, the angle ACB is greater d than a right angle ;. and therefore 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 С squares {quares 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 be. m caufe 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 f. 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 lefs than the íquares of AB, BC by twice the rectangle DB,BC. А Lastly, 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 equal to the square of AC, and twice the {quare of BC: Therefore, in every triangle, &c. R.E. D. B С C. '47. I. T describe a square that shall be equal to a given Sec No rectilineal figure. N Let A be the given rectilineal figure; it is required to defcribe 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 EF equal In ED, and bisect 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 at E, the rectangle BE, ILF, together with the square of EG, is equal to the square of 5 s.a. CF: But GF is equal to GH; therefore the rectange BE, EF, E C 47. I. Book II. together with the square of EG, is equal to the square of GH: But the squares of HE, EG are equal to the square of GH: Therefore the rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EĞ: 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 described upon EH. Which was to be done. THE E I. or from the centers of which the straight 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 centers coincide, the circles must likewise coincide, fince the straight lines from their centers are equal.' II. A straight line is said to touch a circle, when it meets the III. another, which meet, but IV. stant from the center of a circle, V. greater perpendicular falls, is said to. E 2 VI. A |