K B H 2. Through E draw EH parallel to AB or DC, and through F draw FK parallel to AD or BC, cutting EH in G. (I. 31.) 3. From the centre G, at the distance GE, GF, or GK, describe the circle EFHK. Then EFHK shall be the circle required. Proof.-1. Each of the figures AK, KB, AH, HD, AG, GC, BG, GD, is a parallelogram (const.); and their opposite sides are equal. (I. 34.) 2. Then, because AD is equal to AB (I. def. 30), and that AE is half of AD, and AF half of AB, (const.) 3. Therefore AE is equal to AF. (ax. 7.) 4. Therefore the sides opposite to these are equal, namely, FG equal to GE. (I. 34.) 5. In the same manner it may be shown that the straight lines GH, GK, are each of them equal to FG or GE. 6. Therefore the four straight lines GE, GF, GH, GK, are equal to one another, and the circle described from the centre G, at the distance of any one of them, will pass through the extremities of the other three. 7. And because the angles at the points E, F, H, K, are right angles (I. 29); and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle; (III. 16, cor.) 8. Therefore each of the straight lines AB, BC, CD, DA, touches the circle EFHK. Conclusion.-- Therefore, the circle EFHK is inscribed in the square ABCD. Q. E. F. PROPOSITION 9.-PROBLEM. (References—Prop. I. 6, 8.) Given.—Let ABCD be the given square. Sought.-It is required to describe a circle about ABCD. Construction.—1. Join AC, BD, cutting one another in E. 2. From the centre E, at the distance EA, EB, EC, or ED, describe the circle ABCD. Then ABCD shall be the circle required. Proof.-1. Because AB is equal to AD, and AC common to the two triangles BAC, DAC; 2. The two sides BA, AC, are equal to the two sides DA, AC, each to each; B 3. And the base BC is equal to the base DC; 4. Therefore the angle BAC is equal to the angle DAC (I. 8), and the angle BAD is bisected by the straight line AC. 5. In the same manner it may be shown that the angles ABC, BCD, CDA, are severally bisected by the straight lines BD, AC. 6. Then, because the angle DAB is equal to the angle ABC, and that the angle EAB is half the angle DAB, and the angle EBA the half of ABC; 7. Therefore the angle EAB is equal to the angle EBA; (ax. 7.) 8. And therefore the side EA is equal to the side EB. (I. 6.) 9. In the same manner it may be shown that the straight lines EC, ED, are each of them equal to EA or EB. 10. Therefore the four straight lines EA, EB, EC, ED, are equal to one another, and the circle described from the centre E, at the distance of one of them, will pass through the extremities of the other three. Conclusion.—Therefore, the circle ABCD is described about the square ABCD. Q. E. T. PROPOSITION 10,-PROBLEM. To describe an isosceles triangle, having each of the angles at the base double of the third angle. (References—Prop. I. 5, 6, 32; II. 11; III. 32, 37; IV. 1, 5.) Construction.-1. Take any straight line AB, and divide it in the point C, so that the rectangle AB BC may be equal to the square on AC. (II. 11.) 2. From the centre A, at the distance AB, describe the circle BDE, in which place the straight line BD equal to AC, which is not greater than the diameter of the circle BDE (IV. 1); and join DA. Then the triangle ABD shall be such as is required; that is, each of the angles ABD, ADB, shall be double of the third angle BAD. 3. Join DC; and about the triangle ACD describe the circle ACD. (IV.5.) Proof.-1. Because the rectangle AB BC is equal to the square on AC, and that AC is equal to BD, (const.) B 2. Therefore the rectangle AB · BC is equal to the square on BD. 3. And because from the point B, without the circle ACD, two straight lines BCA, BD, are drawn to the circumference, one of which cuts the circle, and the other meets it, 4. And that the rectangle AB BC, contained by the whole of the cutting line, and the part of it without the circle, is equal to the square on BD, which meets it; 5. Therefore the straight line BD touches the circle ACD. (III. 37.) 6. And because BD touches the circle ACD, and DC is drawn from the point of contact D, 7. Therefore the angle BDC is equal to the angle DAC in the alternate seginent of the circle. (III. 32.) 8. To each of these equals add the angle CDA; 9. Therefore the whole angle BDA is equal to the two angles CDA, DAC. (ax. 2.) 10. But the exterior angle BCD is equal to the angles CDA, DAC; (I. 32.) 11. Therefore also the angle BDA is equal to the angle BCD. (ax. 1.) 12. But the angle BDA is equal to the angle DBA (I. 5), because AD is equal to AB. 13. Therefore each of the angles BDA, DBA, is equal to the angle BCD. (ax. 1.) 14. And because the angle DBC is equal to the angle BCD, the side DB is equal to the side DC; (I. 6.) 15. But DB was made equal to CA; therefore CA is equal to CD; (ax. 6.) 16. And therefore the angle CAD is equal to the angle CDA. (I. 5.) 17. Therefore the angles CAD, CDA, are together double of the angle CAD. 18. But the angle BCD is equal to the angles CAD, CDA. (I. 32.) 19. Therefore the angle BCD is double of the angle CAD. 20. And the angle BCD has been shown to be equal to each of the angles BDA, DBA; 21. Therefore each of the angles BDA, DBA, is double of the angle BAD. Conclusion.—Therefore an isosceles triangle ABD has been described, having each of the angles at the base double of the third angle. Q. E. F. PROPOSITION 11.-PROBLEM. E To inscribe a regular pentagon, that is, an equilateral and equi angular pentagon, in a given circle. (References Prop. I. 9; III. 26, 27, 29; IV. 2, 10.) Given. Let ABCDE be the given circle. Sought. It is required to inscribe in it a regular pentagon. Construction.-1. Describe an isosceles triangle FGH, having each of the angles at G, H, double of the angle at F. (IV. 10.) 2. In the circle ABCDE, inscribe the triangle ACD, equiangular to the triangle FGH, so that the angle CAD may be equal to the angle at F, and each of the angles ACD, ADC, equal to the angle at G or H; (IV. 2.) Therefore each of the angles ACD, ADC, is double of the angle CAD. 3. Bisect the angles ACD, ADC, by the straight lines CE, DB. (I. 9.) 4. Join AB, BC, AE, ED. Proof.—1. Because each of the angles ACD, ADC, is double of the angle CAD, and that they are bisected by the straight lines CE, DB, 2. Therefore the five angles ADB, BDC, CAD, DCE, ECA, are equal to one another. 3. But equal angles stand upon equal arcs; _ (III. 26.) 4. Therefore the five arcs AB, BC, CD, DE, EA, are equal to one another. 5. And equal arcs are subtended by equal chords; (III. 29.) 6. Therefore the five straight lines AB, BC, CD, DE, EA, are equal to one another; 7. Therefore the pentagon ABCDE is equilateral. 11. Therefore the whole arc ABCD is equal to the whole arc BCDE. (ax. 2.) 12. And the angle AED stands on the arc ABCD, and the angle BAE on the arc BCDE. 13. Therefore the angle AED is equal to the angle BAE. (III. 27.) 14. For the same reason each of the angles ABC, BCD, CDE, is equal to the angle AED or BAE; 15. Therefore the pentagon ABCDE is equiangular. 16. And it has been shown to be equilateral. Conclusion. Therefore, in the given circle a regular pentagon has been inscribed. Q. E. F. H M B PROPOSITION 12.-PROBLEM. To describe a regular pentagon about a given circle. (References—Prop. I. 4, 8, 26, 47; III. 17, 18, 27; IV. 11.) Given.—Let ABCDE be the given circle. Sought.-It is required to describe a regular pentagon about it. Construction.-1. Let the angles of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that the arcs AB, BC, CD, DE, EA are equal. (IV. 11.) 2. Through the points A, B, C, D, E, draw GH, HK, KL, LM, MG, touching the circle. (III. 17.) Then the figure GHÅLM shall be the pentagon required. 3. Take the centre F, and join FB, FK, FC, FL, FD. Proof.-1. Because the straight line KL touches the circle ABCDE, in the point C, to which FC is drawn from the centre F, FC is perpendicular to KL; (III. 18.) 2. Therefore each of the angles at C is a right angle. 3. For the same reason the angles at the points B, D, are right angles. 4. And because the angle FCK is a right angle, the square on FK is equal to the squares on FC, CK. (I. 47.) 5. For the same reason the square on FK is equal to the squares on FB, BK. 6. Therefore the squares on FC, CK, are equal to the squares on FB, BK (ax. 1); of which the square on FC is equal to the square on FB; 7. Therefore the remaining square on CK is equal to the remaining square on BK; (ax. 3.) 8. Therefore the straight line CK is equal to the straight line BK. 9. And because FB is equal to FC, and FK common to the two triangles BFK, CFK; |