also EC is equal to the greater segment AC, and FC is equal to the less CB; therefore the rectangle under EC and EF is equal in area to the square on CF, and the line CE is cut in extreme and mean ratio. PROPOSITION XII. THEOREM.-If a perpendicular be drawn from any of the acute angles of an obtuse-angled triangle (ABC) to the opposite side (BC) produced, the square on the side (AB) subtending the obtuse angle is greater than the sum of the squares on the two sides (BC and CA), which contain the obtuse angle, by double the rectangle under the side (BC), which is produced, and the external segment (CD) between the obtuse angle and the perpendicular. CONSTRUCTION. Produce BC, and from the acute angle A draw AD perpendicular to BC produced (a). B (a) I. 12. (b) II. 4. (c) I. 47. DEMONSTRATION. Because the straight line BD is divided into two parts in C, the square on BD is equal in area to the sum of the squares on BC and CD, together with double the rectangle under BC and CD (b); to each of these equals add the square on AD, and the sum of the squares on BD and AD is equal in area to the sum of the squares on BC, CD, and AD, together with double the rectangle under BC and CD. But because D is a right angle, the square on AB is equal in area to the sum of the squares on BD and AD (c), and the square on CA is equal in area to the sum of the squares on CD and AD (c); therefore the square on AB is equal in area to the sum of the squares on BC and CA together with double the rectangle under BC and CD; that is, the square on AB is greater than the sum of the squares on BC and CA by double the rectangle under BC and CD. COROLLARY. If in any obtuse-angled triangle (ABC) the sides (AC and CB) which contain the obtuse angle be produced, and perpendiculars be drawn to the acute angles, the rectangle under one of those sides (BC) and the produced part (CD) between the obtuse angle and the perpendicular, is equal in area to the rectangle under the other side (AC) and its produced part (CE). For it may be proved by the foregoing proposition that double the rectangle under AC and CE is also equal in area to the excess of the square on AB above the sum of the squares on BC and CA; therefore the rectangle under AC and CE is equal in area to the rectangle under BC and CD. PROPOSITION XIII. THEOREM.-If in any triangle (ABC) a perpendicular be drawn to one of the sides (BC) which contains an acute angle, from the opposite angle, the square on the side (AC) subtending that acute angle is less than the sum of the squares on the sides (AB and BC) which contain that angle, by double the rectangle under the side (BC) to which the perpendicular is drawn, and the segment (BD) between the perpendicular and the acute angle. CONSTRUCTION. From A draw AD perpendicular to BC, produced if necessary (a). DEMONSTRATION. Because when a straight line is divided, the sum of the squares on the whole line and one of the segments is equal in B area to double the rectangle under the whole line and that segment together with the square on the other seg A D ment (6); therefore the sum of the squares on BD and CB is equal in area to double the rectangle under BD and CB, together with the square on CD; to each of these equals add the square on AD, and the sum of the squares on CB, BD, and AD is equal in area to the sum of the squares on CD and AD, together with double the rectangle under BD and CB. But, because the angles at D are right angles, the square on AB is equal in area to the sum of the squares on BD and AD (c), and the square on AC is equal in area to the sum of the squares on CD and AD (c); therefore the sum of the squares on AB and CB is equal in area to the square on AC together with double the rectangle under BD and CB; that is, the square on AC is less than the sum of the squares on AB and CB by double the rectangle under BD and CB. SCHOLIUM. Euclid has separated this proposition into three cases, depending upon whether the perpendicular falls within or without the triangle, and employs the twelfth proposition to prove the second case. This division is not, however, necessary, as all the cases may be demonstrated from the seventh proposition, as is done above. By comparing the demonstrations of this and the preceding propositions, it will be seen how nearly identical they are, and they may be combined in one general proposition in the following terms:-"The difference between the square on one side of a triangle and the sum of the squares on the other two sides, is equal in area to double the rectangle under either of these two sides and the segment between the perpendicular on it from the opposite angle and the angle included by the sides." 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. II. 13. (b) Hypoth. (c) I. 46, cor. 1. (d) Ax. 3. 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 BL, and make EF equal to the other side ED (6); 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. A о B 5. A CHORD is a straight line drawn within a circle, whose extremities touch the circumference. SCHOLIUM. The portion of the circumference of a circle cut off by any chord is said to be the "arc subtended by that chord;" thus the arc ACB is subtended by the chord AB. A 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. 66 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. 9. A SECTOR of a circle is that portion of a circle contained by two radii and the arc between them. 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. |