Book II. See Note, a 12. 1. b 7. 12. PROP. XIII. THEOR. IN every triangle, the square of the side subtending 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 opposite 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 sides containing it, let fall the perpendicular a 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. A A First, Let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D, the squares of CB, BD are equalb to twice the rectangle contained by CB, BD, and the square of DC: to each of these equals add the square of AD; therefore the squares of CB, BD, DA are equal to twice the rectangle CB, BD, and the squares of AD, DC: but the square of AB is equal 47. 1. 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, BA 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. d 16. 1. e 12. 2. Secondly, Let AD fall with- B D A squares of AB, BC are equal to the square of AC, and twice Book 11. 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 equalf 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 less than the squares of AB, BC by twice the rectangle DB, BC. Lastly, Let the side AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifest that the squares of AB, BC are equal * to the square of AC and twice the square of BC. Therefore, in every triangle, &c. Q. E. D. A 847.1. B PROP. XIV. PROB. TO describe a square that shall be equal to a gi- See Note. ven rectilineal figure. Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A. Describe 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 what was requir ed is now done: but if they are not equal,produceone of them BE to F, and make EF equal to ED, and bisect 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, 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, c 47. 1. 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, 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 described upon EH. Which was to be done. EQUAL circles are those of which the diameters are equal, or Book III. from the centres of which the straight lines to the circumferences are equal. 'This 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 must likewise coincide, since 'the straight lines from the centres are equal.' "The angle of a segment is that which is contained by the straight line and the circumference." VIII. a 10. 1. 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; b11. 1. from the point D drawb DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the circle ABC. |