B F K H D E I NOTE P. Demonstration of ART. 34. Suppose EH greater than AD, and EI taken equal to AD. Also, take EK equal to EI, and draw IL parallel to EK, and KL parallel to El. Then EKLI is a square, for it is a parallelogram whose adjacent sides are equal, and whose angle KEI is a right angle. (DEF. XIII.) Consequently, on the above supposition, EKLI would be equal to ABCD, by preceding ART.; and therefore to EFGH, inasmuch as ABCD and EFGH are given equal. But as a part cannot be equal to the whole, this is impossible, and the above supposition must be false. In the same way it can be proved that AD is not greater than EH. Hence AD and EH must be equal. This, &c. NOTE Q. In order to prevent splitting this demonstration into cases, we form the equal angle ABH with that side of ABC which is "not the greater;" for if we formed it with the greater side, then the other side might fall either above, upon, or below the base Ac, rendering a proof for each case necessary. Geometers add to their demonstration of this theorem a Note, proving that when the side not the greater is chosen, the point G will always fall below the base ac; they generally divide it into three cases, and use indirect proofs. But the substance of this Note is embodied in our demonstration, and a direct proof is given in few words. Б E Simson's Note in his edition of Euclid is erroneous, and totally inconclusive. He says, that, "because c is in the circumference of a circle described from the centre в with the distance BG, [See figure, ART. 44,] it must be in that part of it which falls above AG, the angle ABC being greater than ABG." But the very same argument would prove that G would fall below ac, whichever side we had formed the angle with. Thus, if вс be the greater side, and that the angle CBG equal to FED be formed with it, as the extremity of the lesser side A is in the circumference of the circle described with the centre в and the distance BG, and as the angle CBA is greater than AN D F CBG, it would follow from Simson's reasoning that & must fall below AC, which it evidently does not. figure. The learned Editor was deceived by his PART II. Of the Circle. LESSON VI. Observation. One extremity of a definite right line remaining fixed, if the line revolve about this point, it is evident that the other extremity will trace out a line which is every where equally distant from the fixed point; and that if the line revolve progressively to its first direction, the line which it traces out will return into itself, so as totally to include a surface. Thus if AB be the right line, A its fixed extremity, the line BCDB will be traced out by the progressive revolution of AB about the point A, through the points B C, D, to its first direction AB. Also every point of this line will be equally distant from A. DEF. XIV. A Circle is a plane figure bounded by one line, such, that all right lines drawn from it to one and the same point are equal to each other. DEF. XV. In a circle the bounding line is called the Circumference, and the point to which the equal lines are drawn from the circumference is called the Centre. [See NOTE R.] ART. 48. The centre of a circle falls within the circumference. This is evident. [See NOTE S.] ART. 49. A circle cannot have more than one centre. This is evident. [See NOTE T.] DEF. XIV. Any right line terminated both ways in a circle is called a Chord of the circle, or of the arch it cuts off. Thus in the circle (ART. 50) ABC, AB is a chord of the arch ACB, or AFB. ART. 50. Aright line perpendicular to a chord through its middle point, will pass, if produced, through the centre of the circle. Let DC be perpendicular to AB at its, middle point D. Then the centre of the circle ABC must be in the direction of DC. DEM. If the chord AB pass through the centre, its middle point must be the centre (DEF. XIV.) as is plain; therefore the line DC, in this case, would also F B pass through the centre. If the chord AB do not pass through the centre, the line DC passes through it. For suppose the centre to lie any where out of the direction nc, as at E; and draw EA, ED, EB. Then in the triangles EDB, EDA, EB would be equal to EA, since E is the centre; DB is given equal to DA; and ED is a common side. Therefore, by ART. 6, the angle EDB would be equal to EDA, and consequently each would be a right angle by DEF. VI. But CDB is a right angle, because CD is a perpendicular. Consequently, on the above supposition, the angle EDB would be equal to the angle CDB, by ART. 8,-the whole to a part, which is impossible. Hence, the above supposition is false; that is, the centre does not lie out of the perpendicular DC, but in it. This, &c. PROB. XI. To find the centre af a given circle. Let ABC be a circle. It is required to find the centre of it. CONSTRUCTION. Draw any chord Ac in the circle, and divide it at the point D into two equal parts, by PROB. V. Through D draw the chord BE perpendicular to ac, by PROB. VII., and divide BE into two equal parts at F. Then F is the centre of the circle. B E DEMONSTRATION. By ART. 50, the chord BE passes through the centre of the circle.-Hence, by DEF. XIV., its middle point is the centre of the circle. This was what was required by the present Article. ART. 51. In a circle, a right line from the centre perpendicular to a chord divides it into two equal parts. Let ABC be a circle, D its centre, and DE a perpendicular to the chord AB. DEM. Draw DA, DB. In the triangles DEA, DEB, the sides DA and DB are equal, by DEF. XIV.; and therefore also the angles at B and A, by ART. 4. Moreover the angles DEA, DEB are equal by DEF. VI. Hence, by ART. 46, the sides EA and EB are equal. This, &c. ART. 52. In a circle, a right line through the centre dividing a chord which does not pass through the centre into two equal parts, is perpendicular to it. Let ABC be a circle, D its centre, and DE a right line dividing the chord AB into two equal parts at E. Then DE is perpendicular to AB. DEM. Draw DA, DB. In the triangles DEA, DEB, the sides DA and DB are equal by DEF. XIV.; the sides EA and EB are given equal; and the side DE is com E B mon.-Hence, by ART. 6, the angles DEA and DEB are equal; that is, DE is perpendicular to AB, by DEF. VI. This, &c. ART. 53. A right line cannot meet the circumference of a circle in more than two points. Let ABC be a circle. A right line cannot meet its circumference in more than two points. DEM. For suppose AC to be a right line and to meet the circumference in three points, A, B, C; and draw DA, DB, DC, from the centre D. Then as DA and DC are equal, the angles DAC, A DCA, are equal, by ART. 4; but the angle DBC is greater than Con But DAC by ART. 36, and therefore greater than DCA. sequently DC would be greater than DB, by ART. 42. DC is also equal to DB by DEF. XIV., which is impossible. Hence the above supposition is false; AC does not meet the circle in three points; that is, it does not meet it in more than two. This, &c. ART. 54. If a right line meet a circle in two points, that part of it between the points lies wholly within, and those parts of it not between the points lie wholly without the circle. By preceding ART.; for the line cannot meet the circumference again. DEF. XVII. A right line which meets a circle in two points but is not terminated in both, is called a Secant. Thus, in the annexed figure, [See D DEF. XVIII. A right line drawn from the centre of a circle and terminated in the circumference, is called a Radius. DEF. XIX. A right line drawn through the centre of a circle, and terminated both ways in the circumference, is called a Diameter. ART. 55. In a circle, a perpendicular to a diameter, at its extremity, meets the circle in but one point. Let BCD be a circle, DB a diameter of it, and BE a perpendicular to it at its extremity B. Then BE meets the circle but at the one point B. DEM. For suppose BE to meet the circle again at the point F; then if a be the centre, as AB and AF would be equal, the angles ABF, AFB, would be equal, and consequently, by ART. 38, both would he acute. But ABF is given a right angle. D BF E |