5. But the square on EB is equal to the squares on AE and AB, because the angle EAB is a right angle; (I. 47.) 6. Therefore the rectangle CF. FA, together with the square on AE, is equal to the squares on AE and AB. 7. Take away the square on AE, which is common to both; 8. Therefore the remaining rectangle CF.FA, is equal to the square on AB. (ax. 3.) 9. But the figure FK is the rectangle contained by CF and FA, for FA is equal to FG; (def. 30.) 10. And AD is the square on AB; 12. Take away the common part AK, and the remainder FH is equal to the remainder HD. (ax. 3.) 13. But HD is the rectangle contained by AB and BH, for AB is equal to BD; 14. And FH is the square on AH; 15. Therefore the rectangle AB. BH is equal to the square on AH. Conclusion.—Therefore the straight line AB is divided in H, so that the rectangle AB BH, is equal to the square on AH. Q. E. F. PROPOSITION 12.—THEOREM. In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side on which, when produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle. (References—Prop. I. 47; II. 4.) Hypothesis. — Let ABC be an obtuse-angled triangle, having the obtuse angle ACB; and from the point A let AD be drawn perpendicular to BC produced. Sequence.—The square on AB shall be greater than the squares on AC and CB, by twice the rectangle BC. CD. Demonstration. — 1. Because the straight line BD is divided into two parts in the point C. 2. The square on BD is equal to the squares on BC and CD, and twice the B rectangle BC. CD. (II. 4.) 3. To each of these equals add the square on DA. 4. Therefore the squares on BD and DA are equal to the squares on BC, CD, DA, and twice the rectangle BC. CD. 5. But the square on BA is equal to the squares on BD and DA, because the angle at D is a right angle; (I. 47.) 6. And the square on CA is equal to the squares on CD and DA; (I. 47.) 7. Therefore the square on BA is equal to the squares on BC and CA, and twice the rectangle BC. CD; that is, the square on BA is greater than the squares on BC and CA by twice the rectangle BC. CD. Conclusion.—Therefore, in obtuse-angled triangles, &c. Q. E. D. 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. (References—Prop. I. 12, 16, 47; II. 3, 7, 12.) Hypothesis.--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. (I. 12.) Sequence. The square on AC, opposite to the angle B, shall be less than the squares on CB and BA, by twice the rectangle CB. BD. Case I.–First, let AD fall within the triangle ABC. Demonstration. - 1. Because the straight line CB is divided into two parts in the point D, 2. The squares on CB and BD are equal to twice the rectangle contained by CB, BD, and the square on DC. (II.7.) 3. To each of these equals add the square on DA. 4. Therefore the squares on CB, BD, DA, are equal to twice the rectangle CB. BD, and the squares on AD and DC. 5. But the square on AB is equal to the squares on BD and DA, because the angle BDA is a right angle; (I. 47.) 6. And the square on AC is equal to the squares on AD and DC; (I. 47.) 7. Therefore the squares on CB and 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 and BA, by twice the rectangle CB. BD. Case II.-Secondly, let AD fall without the triangle ABC. Demonstration.-l. Because the angle at D is a right angle (const.), the angle ACB is greater than a right angle; (I. 16.) 2. Therefore the square on AB is equal to the squares on AC and CB, and twice the rectangle BC.CD. (II. 12.) 3. To each of these equals add the square on BC. 4. Therefore the squares on AB and B BC are equal to the square on AC, and twice the square on BC, and twice the rectangle BC.CD. (ax. 2.) 5. But because BD is divided into two parts at C, 6. The rectangle DB.BC is equal to the rectangle BC.CD, and the square on BC; (II. 3.) 7. 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; 8. Therefore the squares on AB and BC are equal to the square on AC, and twice the rectangle DB. BC; that is, the square on AC alone is less than the squares on AB and BC, by twice the rectangle DB BC. Case III.-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 on AB and BC are equal to the square on AC, and twice the square B on BC. (I. 47, and ax. 2.) Conclusion.—Therefore, in every triangle, &c. Q. E. D. PROPOSITION 14.–PROBLEM. To describe a square that shall be equal to a given rectilineal figure. (References-Prop. I. 3, 10, 45, 47; II. 5. Given.-Let A be the given rectilineal figure. Sought.—It is required to describe a square that shall be equal to A. Construction.-1. Describe the rectangular parallelogram BCDE equal to the rectilineal figure A. (1. 45.) A If then the sides of it, BE, ED, are equal to one another, it is a square, and what was required is now done. 2. But if they are not equal, produce one of them, BE, to F, and make EF equal to ED. (I. 3.) 3. Bisect BF in G (I. 10), and from the centre G, at the distance GB, or GF, describe the semicircle BHF; 4. Produce DE to H, and join GH; Then the square described upon EH shall be equal to the rectilineal figure A. Proof.-1. Because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in the point E, 2. The rectangle BE. EF, together with the square on GE, is equal to the square on GF. (II. 5.) 3. But GF is equal to GH; 4. Therefore the rectangle BE. EF, together with the square on GE, is equal to the square on GH. 5. But the square on GH is equal to the squares on GE and EH; (I. 47.) 6. Therefore the rectangle BE. EF, together with the square on GE, is equal to the squares on GE and EH. 7. Take away the square on GE, which is common to both; 8. Therefore the rectangle BE- EF is equal to the square on EH. (ax. 3.) 9. But the rectangle contained by BE and EF is the parallelogram BD, because EF is equal to ED; (const.) 10. Therefore BD is equal to the square on EH. 11. But BD is equal to the rectilineal figure A; (const.) 12. Therefore the square on EH is equal to the rectilineal figure A. Conclusion.—Therefore, a square has been made equal to the given rectilineal figure A, viz. the square described on EH. Q. E. F. EXERCISES ON BOOK II. Prop. 1 to 8. 1. Divide a given straight line into two parts, so that the rectangle contained by them may be the greatest possible. 2. Divide a given straight line into two parts, so that the rectangle contained by them may be equal to a given square. 3. Produce a given straight line, so that the rectangle contained by the whole line so produced, and the part produced, may be equal to a given square. 4. Prove that the difference of two squares is equal to the rectangle contained by the sum and difference of their sides. 5. In any triangle the rectangle contained by the sum and uifference of the sides, is equal to the rectangle contained by the sum and difference of the segments into which the base is divided by a perpendicular from the vertical angle. 6. Construct a rectangle equal to the difference of two given squares, and having one of its sides equal to a given straight line. 7. Divide a given straight line into two parts, so that the difference of their squares may be equal to a given square. Prono to Prop. 9 to 14. 8. Divide a given straight line into two parts, so that the sum of their squares may be equal to a given square. 9. Produce a given straight line so that the sum of the squares of the whole line so produced, and of the part produced, may be equal to a given square. 10. In any triangle, if perpendiculars be drawn from the extremities of the base to the sides, the rectangles contained by each side and its segment between the perpendicular and the base, are equal. 11. Given the base of a triangle 26, and the sides 15 and 37 respectively, find whether it will be acute or obtuse-angled. Find also the segments of the base, the perpendicular, and the area. 12. Prove that half the sum of any two straight lines added to half their difference is equal to the greater; and half the difference taken from half the sum is equal to the less. 13. Divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to a given square. 14. The squares of the two sides of a triangle are together double of the squares of half the base, and of the line joining its middle point with the vertex. |