contained by the whole and one of the parts shall be equal to the square of the other part. Upon AB describe the square ABDC; (1. 46) in E, so he rectangle AB, BH, is equal to the square of AH. Produce GH to K. DEMONSTRATION Then, because the straight line AC is bisected in E, and produced to the point F, the rectangle CF, FA, together with the square of AE, is equal to the square of EF; (11. 6) but EF is equal to EB; therefore the rectangle CF, FA, together with the square of AE, is equal to the square of EB; and the squares of BA, AE, are equal to the square of EB, (1. 47) because the angle EAB is a right angle ; therefore the rectangle CF, FA, together with the square of AE, is equal to the squares of BA, AE; take away the square of AE, which is common to both; therefore the remaining rectangle CF, FA, is equal to the square of AB. But the figure FK is the rectangle contained by CF, FA, since AF is equal to FG; (def. 30) and AD is the square of AB; therefore the figure FK is equal to AD; take away the common part AK, and the remainder FH is equal to the remainder HD but HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the square of AH; therefore the rectangle AB, BH, is equal to the square of AH. Wherefore the straight line AB is divided in H, so that the rectangle AB, BH, is equal to the square of AH. Q. E. F. PROP. XII. - THEOREM. In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced ; then 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. (References — Prop. 1. 47; II. 4.) Let ABC be an obtuse-angle triangle, having the obtuse angle ACB; and from the point A let AD be drawn perpendicular to BC produced. (1. 12.) Then the square of AB shall be 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 point C, the square of BD is equal to the squares of BC, CD, and twice the rect angle BC, CD; (11. 4) to each of these equals add the square of DA; then the squares of BD, DA, are equal to the squares of BC, CD, DA, and twice the rectangles BC, CD; but the square of BA is equal to the squares of BD, DA, (1. 47) because the angle at D is a right angle ; and the square of CA is equal to the squares of CD, DA; (1. 47) 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. Wherefore, in obtuse-angled triangles, &c. Q. E. D. PROP. XIII.- THEOREM. In every triangle, the square of the side subtending either of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either of those sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite anyle, and the acute angle. (References — Prop. 1. 16, 47; 11. 3, 7, 12.) 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. (1. 12.) Then the square of AC opposite to the angle B, shall be less than the squares of CB, BA, by twice the rectangle CB, BD. First, Let AD fall within the triangle ABC. point D, CB, BD, and the square of DC; (11. 7) 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 to the squares of BD, DA, (I. 47) 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. Secondly, Let AD fall without the triangle ABC. Then, because the angle at D is a right angle, the angle ACB is greater than a right angle; (1. 16) and therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle BC, CD; (II. 12) to each of these equals add the square of BC, then the squares of AB, BC, are equal to the square of AC, and twice 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; (11. 3) 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 of BC; therefore the squares of AB, BC, are equal to the square of AC, and twice the rectangle DB, BC; wherefore 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. DEMONSTRATION Then BC is 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. PROP. XIV.- PROBLEM. To describe a square that shall be equal to a given rectilineal figure. (References - Prop. I. 3, 10, 45, 47; 11. 5.) Describe the rectangular parallelogram BCDE equal to the rectilineal figure A. (1. 45.) If, then, the sides of it, BE, ED, are equal to one another, it is a square, and what was required is now done. But if they are not equal, produce one of them, BE to F, and make EF equal to ED; bisect BF in G, (1. 10) and from the centre G, at the distance GB, or GF, describe the semicircle BHF, produce DE to meet the circumference in H. Then the square described upon EH shall be equal to the given rectilineal figure A. Join GH. DEMONSTRATION Because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in E, |