COROLLARY 1. THEOREM. If in an isosceles triangle (CBA) a perpendicular be drawn from either angle of the base to the opposite side, double the rectangle under that side (CB) and the segment (DB) between the perpendicular and the base is equal in area to the square on the base (AB). For the sum of the squares on AB and CB is equal in area to the square on AC together with double the rectangle under CB and DB (a); but AC is equal to CB (b), and therefore the squares on them are equal (c); and taking away these equals, we have the square on AB equal in area to double the rectangle under CB and DB (d). PROPOSITION XIV. (a) II. 13. PROBLEM.-To construct a square that shall be equal in area to a given rectilineal figure (A). SOLUTION. Construct a rectangle BCDE equal in area to the given rectilineal figure (a); if the adjacent sides be equal, the problem is solved. If not, produce either side BE, and make EF equal to the other side ED (b); bisect BF in G (c), and from the center G, at the distance GF, describe the semicircle BHF; produce DE to H, and join GH; then a square constructed on EH shall be equal in area to the given rectilineal figure. DEMONSTRATION. Because the straight line BF is bisected in G, and also cut into two unequal parts in E, the rectangle under BE and EF, together with the square on GE, is equal in area to the square on GF (d), or of GH which is equal to GF (e). But the square on GH is equal to the square on GE together with the square on EH (ƒ); and taking the square on GE away from both, the rectangle under BE and EF is equal in area to the square on EH; but EF is equal to DE, therefore the rectangle BD is equal in area to the square on EH; and therefore the square on EĤ is equal in area to the rectilineal figure A. SCHOLIUM. From this proposition we derive the following theorem:If a perpendicular be drawn from any point in the circumference of a semicircular to the diameter, the square on the perpendicular is equal in area to the rectangle under the segments into which it divides the diameter. THE ELEMENTS OF EUCLID. BOOK III. DEFINITIONS. 1. EQUAL circles are those of which the diameters are equal, or those from the centers of which straight lines drawn to the circumferences are equal. SCHOLIUM. 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 centers coincide, the circles must likewise coincide, since the straight lines from the centers are equal. 2. A TANGENT to a circle is a straight line which meets the circumference, but being produced, does not cut the circle. 3. Circles are said to touch one another when their circumferences meet, but do not cut one another. SCHOLIUM. They are said to touch externally when each circle is entirely without the other, as A and B; and the circumference of one circle is said to touch that of the other internally when one circle is entirely within the other, as B and C. 4. An ARC of a circle is any portion of the circumference. B C 5. A CHORD is a straight line drawn within a circle, whose extremities touch the circumference. arc SCHOLIUM. The portion of the circumference of a circle cut off by any chord is said to be the subtended by that chord;" thus the arc ACB is subtended by the chord AB. 6. Straight lines (or chords) are said to be equally distant from the center of a circle when the perpendiculars drawn to them from the center are equal. SCHOLIUM. And chords are said to be "farther from" or "nearer to" the center, according as the perpendiculars from them to the center are greater or less. 7. A SEGMENT of a circle is that portion of a circle contained by a chord and its arc. 8. An angle in a segment is the angle contained by the straight lines drawn from any point in the circumference of the segment to the extremities of its chord. SCHOLIUM. An angle is said to stand upon the arc intercepted between the straight lines which contain the angle; thus the angle ABC is said to be the angle at the circumference standing upon the arc AC, and the angle ADC is said to be the angle at the center standing upon the same arc. B 9. A SECTOR of a circle is that portion of a circle contained by two radii and the arc between them. D е SCHOLIUM. When the radii are at right angles, the sector is termed a quadrant, being then equal to the fourth part of the whole circle. 10. A segment of a circle is said to be similar to another segment, when an angle contained in one is equal to an angle contained in the other. SCHOLIUM. The word "similar" is used in Geometry in a more strict sense than in ordinary conversation; in the latter it signifies mere resemblance, whereas in the former it is only used to express perfect identity and sameness of form. This definition, as given by Euclid, is an anticipation of the twenty-first proposition. We have above restricted the criterion of similarity to consist in one angle contained in each segment being equal. 11. A rectilineal figure is said to be contained by a circle, when its circumference passes through all its angular points. SCHOLIUM. And the circle is said to be described about the figure. LEMMA. THEOREM. If a straight line (CE) bisects the chord of a circle (AB) perpendicularly, it contains the center of that circle. DEMONSTRATION. For if not, let any point G without the line CE be the center of the circle. Draw GA, GD, and GB. Because in the triangles GAD and GBD, the side GA is equal to GB (a), DA equal to DB (a), and GD common to both, the angles GDA and GDB are equal (b), and therefore are right angles (c); but the angle CDB is also a right angle (a) therefore the angle GDB is equal to CDB (d), a part equal to the whole, which is absurd; therefore the point G is not the center of the circle ACB, and in like manner it may be shown that no other point without the line CE is the center, therefore the line CE contains the center of the circle. D E (a) Hypoth. SCHOLIUM. The above theorem has been introduced by way of lemma to the first proposition, to meet the objection which has been urged against Euclid's demonstration, because he therein assumes the position of the center previous to the perfect solution of the problem. Euclid's demonstration is further defective in two other respects; first, that it does not apply when the point G is assumed in the line CE itself, in which it may be without coinciding with the point F; and secondly, his demonstration is deficient, inasmuch as he only proves as much as is proved in the foregoing proposition, namely that the center must be contained in the line CE but he omits to prove in what part of that line it is situated. PROPOSITION I. PROBLEM.-To find the center of a given circle (ACB). SOLUTION. Draw within the circle any straight line AB, bisect it in D (a), and from D draw DC perpendicular to AB (b), and produce it to E; bisect CE in F (a), and F is the center of the circle. DEMONSTRATION. Because the line CE bisects the chord AB of the circle ACB perpendicularly (c), it contains the center of the circle; and because it is bisected in F (d), FE is equal to FC, therefore F is the center of the circle (e). THEOREM. If three points (A, B, and C) are not in the same straight line, a circle may be described whose circumference shall pass through them. CONSTRUCTION. Join AB and BC, bisect AB by the perpendicular DE (a), and BC by the perpendicular FE (a). Э (a) I. 10. 1 E B (c) Preceding lemma. DEMONSTRATION. The lines DE and FE will meet; for join DF, then the angles FDE and DFE are together less than the angles BDE and BFE, or than two right angles (b), and therefore they will meet in some point E (c); then every circle which passes through the points A and B has its center in the perpendicular DE (d), and every circle which passes through the points B and C has its center in the perpendicular FE (d); therefore the circumference of the circle whose center is E passes through the three points A, B, and C. PROPOSITION II. THEOREM.-If any two points (A and B) be taken in the circumference of a circle, the straight line which joins them falls within the circle. A E C DEMONSTRATION. For, if it be possible, let the straight line AEB have a point E without the circle ACB. Find the center D (a), and join DA, DB, and DE, cutting the circumference in F. Because in the triangle ADB the side AD is equal to BD (b), the angle DAB is equal to the angle DBA (c); but the external angle DEA is greater than the internal angle DBA (d), and therefore greater than its equal DAB. Then in the triangle DAE, because the angle DEA is greater than the angle DAE, the opposite side DA is greater than DF (e), and therefore greater than DF; but DA and DF, being both radii of the same circle, are equal (6), which is absurd; therefore the point E is not without the circle, and in a similar manner it may be shown that no point in the line AB is without the circle. SCHOLIUM. The foregoing proof being by the method "reductio ad absurdum," is necessarily indirect. By the admission of the axiom "That the extremity of a straight line less than the radius is within the circle," the following direct proof may be given. Take any point E, join DE, and produce it, if necessary, to meet the circumference in F; then it may be proved, as above, that DA is greater than DE, but DA and DF are equal (a), therefore DF is greater than DE, and E is by the above axiom within the circle. (a) III. 1. (b) I. Def. 15. (c) I. 5. (d) I. 16. (e) I. 19. A F C B (a) I. Def. 15. |