but DG is equal to DB; therefore 20. AC, CD. The squares of AD, DB, are double of the squares of Wherefore, if a straight line, &c. Q.E.D. PROP. XI.-PROBLEM. To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part. Let AB be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part. Upon AB describe (I. 46.) the square ABDC; bisect (I. 10.) AC in E, and join BE; produce CA to F, and make (I. 3.) EF equal to EB; and upon AF describe (I. 46.) the square FGHA; AB is divided in H, so that the rectangle AB, BH, is equal to the square of AH. Produce GH to K. Then because the straight line AC is bisected in E, and produced to the point F, (II. 6.) 1. The rectangle CF, FA, together with the square of AE, is equal to the square of EF; but EF is equal to EB; therefore 2. 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 (I. 47.) to the square of EB, because the angle EAB is a right angle; therefore take 3. The rectangle CF, FA, together with the square of AE, away the square of AE, which is common to both; therefore of AB; and the figure FK is the rectangle contained by CF, FA, for AF is equal (Def. 30.) to FG; and AD is the square of AB; therefore 5. The figure FK is equal to AD. Take away 6. the common part AK, and The remainder FH is equal to the remainder HD; and HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the square of AH; therefore 7. 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. Which was to be done. PROP. XII.-THEOREM. In obtuse-angled triangles, if a perpendicular be drawn from either 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 (I. 12.) perpendicular 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 point C, (II. 4.) 1. The square of BD is equal to the squares of BC, CD, and twice the rectangle BC, CD; to each of these equals add the square of DA; and 2. The squares of DB, DA, are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD; but the square of BA is equal (I. 47.) to the squares of BD, DA, because the angle at D is a right angle; and the square of CA is equal (I. 47.) to the squares of CD, DA; therefore 3. 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.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 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 (I. 12.) 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. B D First, let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D, (II. 7.) 1. The squares of CB, BD, are equal to twice the rectangle contained by CB, BD, and the square of DC; to each of these equals add the square of AD; therefore 2. 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 (I. 47.) 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 3. 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. A B Then, because the angle at D is a right angle, (I. 16.) and therefore (II. 12.) 2. The square of AB is equal to the squares of AC, CB, and twice the rectangle BC, CD; to each of these equals add the square of BC, and 3. 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, (II. 3.) 4. The rectangle DB, BC, is equal to the rectangle BC, CD, and the square of BC; and the doubles of these are equal; therefore 5. 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 BC is the straight line between the perpendicular and the acute angle at B; and it is manifest that (I. 47.) 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. Let A be the given rectilineal figure; it is required to describe a square that shall be equal to 4. Describe (I. 45.) the rectangular parallelogram BCDE equal to the rectilineal figure 4. 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, and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H: the square described upon EH shall be equal to the given rectilineal figure 4. Join GH; and because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, (II. 5.) 1. The rectangle BE, EF, together with the square of EG, is equal to the square of GF; but GF is equal (Def. 15.) to GH; therefore 2. The rectangle BE, EF, together with the square of EG, is equal to the square of GH; but the squares of HE, EG, are equal (I. 47.) to the square of GH; therefore take 3. The rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EG; away the square of EG, which is common to both, and 4. 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 5. BD is equal to the square of EH; but BD is equal to the rectilineal figure 4; therefore 6. 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. |