difference (II. 17.), the rectangle contained by the sum and difference of AC, CD is equivalent to the rectangle contained by the sum and difference of AI, ID. But since the radius CG is equal to CH, the sum of AC and CD is AH, and their difference is AG; and because the perpendicular CI bisects the chord BD (III. 4.), the sum of AI and ID is AD, and their difference AB. Wherefore the rectangle AH, AG is equivalent to the rectangle AB, AD. In the same way it is proved, that the rectangle AH, AG is equivalent to the rectangle AE, AF; and consequently the rectangle AE, AF is equivalent to the rectangle AB, AD. . Cor. 1. If the vertex A of the straight lines lie within the circle and the point I coincide with it, BD, being then at right angles to CA, is bisected at A (III. 4.), and the rect- es EI angle AB, AD is the same as the square \ of AB. Consequently the square of a perpendicular AB limited by the circumference is equivalent to the rectangle under the segments AG, AH of the diameter. Cor. 2. If the vertex A lie without the circle and the point I coincide with B or D, the angle ABC being then a right angle, the incident line AB must be a tangent A (III. 20.), and consequently the two points of section B and D must coa- IB lesce in a single point of contact. Wherefore the rectangle under the distances AB, AD becomes the same as the square of AB; and consequently H H. the rectangle contained by the segments AG, AH of the diameter, is equivalent to the square of the tangent AB. PROP. xxvii. PROB. To construct a square equivalent to a given rectilineal figure. Let the rectilineal figure be reduced by Proposition 6. Book II. to an equivalent rectangle, of which A and B are the two contain- R ing sides; draw an indefinite straight line CE, in which take the part CD equal to A and DE to B, on C de scribe a semicircle, and erect the per- C D E pendicular DF from the diameter to meet the circumference: DF is the side of the square equivalent to the given rectilineal figure. For, by Cor. 1. to the last Proposition, the square of the perpendicular DF is equivalent to the rectangle under the segments CD, DE of the diameter, and is consequently equivalent to the rectangle contained by the sides A and B of a rectangle that was made equivalent to the rectilineal figure. - PROP. xxviii. THEOR. A quadrilateral figure may have a circle described about it, if the rectangles under the segments made by the intersection of its diagonals be equi valent, or if those rectangles are equivalent which are contained by the external segments formed by producing its opposite sides. Let ABCD be a quadrilateral figure, of which AC and BD are the diagonals, and such that the rectangle AE, EC is equivalent to the rectangle BE, ED; a circle may be made to pass through the four points A, B, C, and D. For describe a circle through the three points A, B, C (III. 9. cor.), and let it cut BD in G. Because AC and BG intersect each other within a circle, the rectangle AE, EC is equivalent to the rectangle BE, EG (III. 26.); but the rectangle AE, EC is by hypothesis equivalent to the rectangle BE, ED. Wherefore BE, EG is equivalent to BE, ED; and these rectangles have a common base BE, consequently (II. 3. cor.) their altitudes EG and ED are equal, and hence the point G is the same as D, or the circle passes through all the four points A, B, C, and D. Again, if the opposite sides CB and DA be produced to meet at F, and the rectangle CF, FB be equal to DF, FA, a circle may be described about the figure. For, as before, let a circle pass through the three points A, B, C, but cut AD in H. And from the property of the circle, the rectangle CF, FB is equivalent to HF, FA; but the rectangle CF, FB is also equivalent to DF, FA ; whence the rectangle HF, FA is equivalent to DF, FA, and the base HF equal to DF, or the point H is the same as D. |