PROP. C. THEOR. If a straight line be drawn from any point in the base of an isosceles triangle, or the base produced, to the opposite angle, the rectangle contained by the segments between the point and the extremities of the base is equal to the difference between the square of the line drawn to the opposite angle, and the square of one of the equal sides. Fig. 1. A Fig. 2. A Let ABC be an isosceles triangle, and let a straight line be drawn from any point D in the base (fig. 1), or in the base produced (fig. 2), to the opposite angle A; the rectangle BD.DC is equal to the difference between the squares of AD and AB. Bisect the base BC in E (I. 10), and join EA; then, because BE is equal to CE, and EA common to the two trianthere are B gles BEA, CEA, two sides in the one equal to two sides in the other; also the base BA is equal to the base CA; therefore the angle BEA is equal to the angle CEA (I. 8), and each is a right angle (I. Def. 7). E D C B E C And, first, Let D be between E, the middle of the base, and one of its extremities; then BD.DC+DE2=(II. 5) BE2; and, adding_ AE2 to these equals, BD.DC+DE2+EA2=BE2+EA2: but DE2+EA2=DA2 (I. 47), and BE2+EA2=BA2; therefore BD.DC+DA2=BA2, and hence the rectangle BD.DC is equal to the excess of BA2 above DA.2 Secondly, Let D be in BC produced; then BD.DC+BE2=DE2 (II. 6); and, adding AE2 to these equals, BD.DC+BE2+EA2= DE2+EA2: but BE2+EA2-BA2 (I. 47) and DE2+EA2-DA2; therefore BD.DC+BA2=DA2, and the rectangle BD.DC is the excess of DA2 above BA2. Thirdly, When the point D is in the middle of the base, the truth of the proposition is manifest (I. 47.) 'BOOK THIRD. DEFINITIONS. A. The radius of a circle is the straight line drawn from the centre to the circumference. I. A straight line is said to touch a circle, when it meets the circle, and being produced, does not cut it. II. Circles are said to touch one another, which meet, but do not cut one another. III. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. IV. And the straight line, on which the greater perpendicular falls, is said to be farther from the centre. B. An arc of a circle is any part of the circumference. V. A segment of a circle is the figure contained by a straight line, and the arc which it cuts off. VI. An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment. VII. And an angle is said to insist or stand upon the arc intercepted between the straight lines which contain the angle. VIII. The sector of a circle is the figure contained by two straight lines drawn from the centre, and the arc of the circumference between them. IX. Similar segments of a circle are those in which the angles are equal, or which contain equal angles. PROP. I. PROB. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisect (I. 10) it in D; from the point D draw (I. 11) DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the circle ABC. F C For, if it be not, let, if possible, G be the centre, and join GA, GD, GB; then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; but the base GA is also equal to the base GB, because they are radii of the same circle; therefore the angle ADG is equal (I. 8) to the angle GDB. But when a straight line, standing upon another straight line, makes the adjacent angles equal to one another, each of the angles is a right angle (I. Def. 7); therefore the angle GDB is a right angle: but FDB is likewise a right angle; wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impossible; therefore G is not the centre of the circle ABC. In the same manner it can be shown that no other point than F is the centre; that is, F is the centre of the circle ABC. Which was to be found. E D B COR. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other. If any two points be taken in the circumference of a circle, the straight line which joins them will fall within the circle. Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from A to B will fall within the circle. the Take any point in AB as E; find D, centre of the circle ABC; join AD, DB, and DE, and let DE meet the circumference in F. Then because DA is equal to DB, the angle DAB is equal (I. 5) to the angle DBA; and because AE, a side of the triangle DAE, is produced to B, the angle of DEB is greater (I. 16) than the angle DAE: but DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE. Now to the greater angle the greater side A B F is opposite (I. 19); DB is therefore greater than DE: but DB is equal to DF; wherefore DF is greater than DE, and the point E is therefore within the circle. The same may be demonstrated of any other point between A and B; therefore AB is within the circle. Wherefore, if any two points, &c. Q. E. D. PROP. III. THEOR. If a straight line drawn through the centre of a circle bisect a straight line in the circle, which does not pass through the centre, it will cut that line at right angles; and, if it cut it at right angles, it will bisect it. Let ABC be a circle, and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre in the point F: it also cuts it at right angles. Take (III. 1) E, the centre of the circle, and join EA, EB; then because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in C E the one equal to two sides in the other: but the base EA is equal to the base EB; therefore the angle AFE is equal (I. 8) to the angle BFE. And when a straight line standing upon another makes the adjacent angles equal to one another, each of them is a right angle (I. Def.7); therefore each of the angles AFE, BFE is a right angle: wherefore the straight line CD, drawn through the centre, bisecting AB, which does not pass through the centre, cuts AB at right angles. D F B Again, let CD cut AB at right angles; CD also bisects AB; that is, AF is equal to FB. The same construction being made, because the radii EA, EB are equal to one another, the angle EAF is equal (I. 5) to the angle EBF; and the right angle AFE is equal to the right angle BFE. Therefore, in the two triangles EAF, EBF there are two angles in the one equal to two angles in the other; and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal (I. 26). AF therefore is equal to FB. Wherefore, if a straight line, &c. Q. E. D. PROP. IV. THEOR. If, in a circle, two straight lines cut one another, in a point which is not the centre, they cannot bisect each other. Let ABCD be a circle, and AC, BD two straight lines in it, which cut one another in a point E, which is not the centre; AC, BD do not bisect one another. For, if possible, let AE be equal to EC, and BE to ED; if one of the lines pass through the centre, it is plain that it cannot be bisected by the other, which does not pass through the centre; but if neither of them pass through the centre, take (III. 1) F, the centre of the circle, and join EF; and because FE, a straight line through the centre, bisects another, AC, which does not pass through the centre, it must cut it at right (III. 3) angles; wherefore FEA is a right angle. Again, because the straight line FE bisects the straight line BJ, which does not pass through the centre, it must cut it at right (III. 3) angles; where- A fore FEB is a right angle: and FEA was shown to be a right angle; therefore FEA is equal to the angle FEB, the less to the B greater, which is impossible; therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q. E. D. D E If two circles cut one another, they cannot have the same centre. Let the two circles ABC, CDG cut one another in the points B, C; they have not the same centre. G F E For, if it be possible, let E be their centre; join EC, and draw any straight line EFG meeting the circles in F and G; and because E is the centre of the circle ABC, CE is equal to EF. Again, because E is the centre of the circle CDG, CE is equal to EG; but CE A was shown to be equal to EF; therefore EF is equal to EG, the less to the greater, which is impossible; therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. PROP. VI. THEOR. B If two circles touch one another internally, they cannot have the same centre. Let the two circles ABC, CDE touch one another internally in the point C; they have not the same centre. For, if they have, let it be F; join FC, and draw any straight line FEB, meeting the circles in E and B; and because F is the centre of the circle ABC, CF is equal to FB. Also, because F is the centre of the circle CDE, CF is equal to FE; but CF was shown A to be equal to FB; therefore FE is equal to FB, the less to the greater, which is impossi C D E B ble; wherefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D. |