PROPOSITION 13. THEOREM. In every triangle, the square on the side subtending an acute angle, is less than the squares on 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 on it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B an acute angle; and on BC one of the sides containing it, let fall the perpendicular AD from the opposite angle: the square on AC, opposite to the angle B, shall be less than the squares on CB, BA, by twice the rectangle CB, BD. First, let AD fall within the triangle ABC. Then, because the straight line CB is divided into two parts at the point D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on CD. [II. 7. To each of these equals add the square on DA. Therefore the squares on CB, BD, DA are equal to twice the rectangle CB, BD and the squares on CD, DA. [Ax. 2. But the square on AB is equal to the squares on BD, DA, because the angle BDA is a right angle; [I. 47. and the square on AC is equal to the squares on CD,DA.[1.47. Therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD; that is, the square on AC alone is less than the squares on CB, BA by twice the rectangle CB, BD. Secondly, let AD fall without the triangle ABC. Then because the angle at D is a right angle, [Construction. D the angle ACB is greater than a right angle; [I. 16. [II. 12. and therefore the square on AB is equal to the squares on AC, CB, and twice the rectangle BC, CD. To each of these equals add the square on BC. Therefore the squares on AB, BC are equal to the square on AC, and twice the square on BC, and twice the rectangle BC, CD. [Axiom 2. But because BD is divided into two parts at C, angle DB, BC is equal to the rectangle BC, CD square on BC; and the doubles of these are equal, that is, twice the rectangle DB, BC is equal to twice the rectangle BC, CD and twice the square on BC. Therefore the squares on AB, BC are equal to the square on AC, and twice the rectangle DB, BC ; Then BC is the straight line between the perpendicular and the acute angle at B; that is, the square on AC alone is less than the squares on AB, BC by twice the rectangle DB, BC, Lastly, let the side AC be perpendicular to BC. and it is manifest, that the squares on AB, BC are equal to the square on AC, and twice the square on BC. [I. 47 and Ax, 2. Wherefore, in every triangle &c. Q.E.D. the rectand the [II. 3. Describe the rectangular parallelogram BCDE equal totherectilineal figure A. [1.45. Then if the sides of it, BE, ED are equal to one another, it is a square, and what was required is now done. PROPOSITION 14. PROBLEM. To describe a square that shall be equal to a given rectilineal figure. B B Let A be the given rectilineal figure: it is required to describe a square that shall be equal to A. G But if they are not equal, produce one of them BE to F, make EF equal to ED, [I. 3. and at G; bisect BF from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DEto H, H The square described on EH shall be equal to the given rectilineal figure A. Join GH. Then, because the straight line BF is divided into two equal parts at the point G, and into two unequal parts at the point E, the rectangle BE, EF, together with the square on GE, is equal to the square on GF. [II. 5. But GF is equal to GH. Therefore the rectangle BE, EF, together with the square on GE, is equal to the square on GH. But the square on GH is equal to the squares on GE, EH ;[1.47. therefore the rectangle BE, EF, together with the square on GE, is equal to the squares on GE, EH. Take away the square on GE, which is common to both; therefore the rectangle BE, EF is equal to the square on EH. [Axiom 3. But the rectangle contained by BE, EF is the parallelogram BD, [Construction. because EF is equal to ED. [Construction. Therefore BD is equal to the square on EH. Wherefore a square has been made equal to the given rectilineal figure A, namely, the square described on EH. Q.E.F. BOOK III. DEFINITIONS. 1. EQUAL circles are those of which the diameters are equal, 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 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. 2. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it. 3. Circles are said to touch one another, which meet but do not cut one another. 4. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. 5. And the straight line on which the greater perpendicular falls, is said to be farther from the centre. 6. A segment of a circle is the figure contained by a straight line and the circumference it cuts off. 7. The angle of a segment is that which is contained by the straight line and the circumference. 8. An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment to the extremities of the straight line which is the base of the segment. 9. And an angle is said to insist or stand on the circumference intercepted between the straight lines which contain the angle. 10. A sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them. 11. Similar segments of circles are those in which the angles are equal, or which contain equal angles. [Note. In the following propositions, whenever the expression "straight lines from the centre," or "drawn from the centre,' occurs, it is to be understood that the lines are drawn to the circumference. Any portion of the circumference is called an arc. PROPOSITION 1. PROBLEM. To find the centre of a given circle. Let ABC be the given circle: it is required to find its centre. |